#### NCERT Class 11 Chapter-11: Conic Sections

Exercise 11.1

Q.1: A circle is given with centre (0, 3) and radius 3. Obtain the equation of the given circle.

Answer: Based on formulae given in conic sections

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given in the question:

(h, k) = centre of a circle = (0, 3) and radius r = 3

The equation of a circle is:

(a – 0)2 + (b – 3)2 = 32

a2 + b2 – 6b + 9 = 9

a2 + b2 – 6b + 9 – 9 = 0

a2 + b2 – 6b = 0

Q.2: A circle is given with centre (-1, 2) and radius 4. Find the equation of the circle.

Answer: Based on formulae given in conic sections

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given in the question:

Centre of a circle = (h, k) = (-1, 2) and radius (r) = 4

The equation of a circle is:

(a – (- 1))2 + (b – 2)2 = 42

(a + 1)2 + (b – 2)2 = 42

a2 + 2a + 1 + b2 – 4b + 4 = 16

a2 + 2a + b2 – 4b + 5 = 16

a2 + 2a + b2 – 4b + 5 – 16 = 0

a2 + 2a + b2 – 4b – 11 = 0

Q.3: A circle is given with centre ($$\frac{1}{3}$$, $$\frac{1}{2}$$) and radius $$\frac{1}{14}$$. Find the equation of the circle.

Answer: Based on formulae given in conic sections

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given in the question:

Centre of a circle = (h, k) = ($$\frac{1}{3}$$, $$\frac{1}{2}$$) and the radius (r) = 4

The equation of a circle is:

(a – $$\frac{1}{3}$$)2 + (b – $$\frac{1}{2}$$)2 = $$\frac{1}{14}$$2

(a – $$\frac{1}{3}$$)2 + (b – $$\frac{1}{2}$$)2 = $$\frac{1}{14}$$2

$$a^{2} – 2.\frac{1}{3}.a + \frac{1}{9} + b^{2} – 2.\frac{1}{2}.b + \frac{1}{4} = (\frac{1}{14})^{2} \\ a^{2} – \frac{2a}{3} + \frac{1}{9} + b^{2} – b + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{1}{9} + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} – \frac{1}{196} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{637 – 9}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{628}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + 0.35 = 0 \\ 3a^{2} + 3b^{2} – 2a – 3b + 1.068 = 0 \\$$

Q.4: A circle is given with centre (2, 2) and radius $$\sqrt{5}$$. Obtain the equation of the given circle.

Answer: Based on formulae given in conic sections

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given in the question:

Centre of a circle = (h, k) = (2, 2) and radius (r) = $$\sqrt{5}$$

The equation of a circle is:

(a – 2)2 + (b – 2)2 = $$\sqrt{5}^{2}$$

(a – 2)2 + (b – 2)2 = $$\sqrt{5}^{2}$$

a2 – 4a + 4 + b2 – 4b + 4 = 5

a2 – 4a + b2 – 4b + 8 = 5

a2 – 4a + b2 – 4b + 8 – 5 = 0

a2 – 4a + b2 – 4b + 3 = 0

a2 + b2 – 4a – 4b + 3 = 0

Q.5: A circle is given with centre (- x, – y) and radius $$\sqrt{x^{2} – y^{2}}$$. Find the equation of the circle.

Answer: Based on formulae given in conic sections

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given in the question:

Centre of a circle = (h, k) = (- x, – y) and radius ‘r’ = $$\sqrt{x^{2} – y^{2}}$$

The equation of a circle is:

(a – (- x)2 + (b – (- y)2 = $$\sqrt{x^{2} – y^{2}}^{2}$$

(a + x)2 + (b + y)2 = $$\sqrt{x^{2} – y^{2}}^{2}$$

(a + x)2 + (b + y)2 = x2 – y2

a2 + 2ax + x2 + b2 + 2by + y2 = x2 – y2

a2 + 2ax + b2 + 2by + y2 = 0

Q.6: The equation of a given circle is given as (a + 5)2 + (b – 3)2 = 36. Find the centre and radius of the circle.

Answer: Based on formulae given in conic sections

Given in the question:

The equation of a given circle (a + 5)2 + (b – 3)2 = 36

(a + 5)2 + (b – 3)2 = 36

{a – (- 5)}2 + (b – 3) = 62

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = – 5, k = 3 and r = 6.

Hence, the centre of the circle is (-5, 3), and the radius of circle is 6.

Q.7: The equation of a given circle is given as a2 + b2 – 4a – 4b + 3 = 0

Find the radius and centre of the circle.

Answer: Based on formulae given in conic sections

Given in the question:

The equation of a given circle a2 + b2 – 4a – 4b + 3 = 0

a2 + b2 – 4a – 4b + 3 = 0

a2 – 4a + b2 – 4b + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22} – 22 – 22 + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22} – 8 + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22} = 5

(a – 2)2 + (b – 2)2 = $$\sqrt{5}^{2}$$

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 2 and r = $$\sqrt{5}$$.

Thus, the centre of the circle is (2, 2), and the radius of circle is $$\sqrt{5}$$.

Q.8: The equation of a given circle is given as a2 + b2 – 6a + 8b – 14 = 0

Find the centre and radius of the circle.

Answer: Based on formulae given in conic sections

Given in the question:

The equation of a given circle a2 + b2 – 6a + 8b – 14 = 0

a2 – 6a + b2 + 8b – 14 = 0

{a2 – 2.a.3 + 32} + {b2 + 2.b.4 + 42} – 32 – 42 – 14 = 0

(a – 3) 2 + (b + 4) 2 – 9 – 16 – 14 = 0

(a – 3) 2 + (b + 4) 2 – 39 = 0

(a – 3) 2 + (b + 4) 2 = 39

(a – 3) 2 + (b – (- 4)) 2 = 39

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 3, k = 4 and r = $$\sqrt{39}$$.

Thus, the centre of the circle is (3, 4), and the radius of circle is $$\sqrt{39}$$.

Q.9: The equation of a given circle is given as 2a2 + 2b2 – 8a = 0

Find the centre and radius of the circle.

Answer: Based on formulae given in conic sections

Given in the question:

The equation of a given circle 2a2 + 2b2 – 8a = 0

2a2 + 2b2 – 8a = 0

2a2 – 8a + 2b2 = 0

2[a2 – 4a + b2] = 0

{a2 – 2.a.2 + 22} – 22 + (b – 0)2 = 0

[a2 – 2.a.2 + 22] + [b – 0]2 – 22 = 0

(a – 2)2 + (b – 0)2 = 22 = 4

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 0 and r = $$\sqrt{4}$$ = 2.

Thus, the centre of the circle is (3, 4), and the radius of circle is 2.

Q.10: The circle passing through the points (6, 2) and (4, 3) and whose centre lies on the line 2a + y = 8. Find the equation of the circle.

Answer: Based on formulae given in conic sections

The equation of a required circle with centre (h, k) and radius r is given as

(a – h)2 + (b – k)2 = r2

As the circle passes through (6, 2) and (4, 3)

(6 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1)

(4 – h)2 + (3 – k)2 = r2 . . . . . . . . . . . . . . . . (2)

The centre (h, k) of the circle lies on the line 2a + y = 8,

2h + k = 8 . . . . . . . . . . . . . . . . . . . . . (3)

Now, On Comparing equations 1 and 2, we will get:

(6 – h)2 + (2 – k)2 = (4 – h)2 + (3 – k)2

36 – 12h + h2 + 4 – 4k + k2 = 16 – 8h + h2 + 9 – 6k + k2

36 – 12h + 4 – 4k = 16 – 8h + 9 – 6k

– 12h – 4k + 8h + 6k = 16 + 9 – 36 – 4

4h – 2k = 15 . . . . . . . . . . . . . . . . . . (4)

Now, on Answerving Equations (3) and (4), we will get:

h = $$\frac{31}{8}$$

k = $$\frac{1}{4}$$

Substituting the values of h and k in equation (1), we will get:

(6 – $$\frac{31}{8}$$)2 + (2 – $$\frac{1}{4}$$)2 = r2

$$\\(\frac{17}{8})^{2} + (\frac{7}{4})^{2} = r^{2}$$

$$\\\Rightarrow$$ $$\frac{289}{64} + \frac{49}{16} = r^{2}$$

$$\\\Rightarrow$$ $$\frac{289 + 196}{64} = r^{2}$$

$$\\\Rightarrow$$ $$\frac{485}{64} = r^{2}$$

$$\Rightarrow$$ r = 7.57

The required equation is:

(a – $$\frac{31}{8}$$)2 + (b – $$\frac{1}{4}$$)2 = r2

(a – $$\frac{31}{8}$$)2 + (b – $$\frac{1}{4}$$)2 = 7.572

$$(a – \frac{31}{8})^{2} + (b – \frac{1}{4})^{2} = 7.57^{2} \\$$

$$\\\Rightarrow$$ $$a^{2} – \frac{31}{4}a + (\frac{31}{8})^{2} + b^{2} – \frac{b}{2} + \frac{1}{16} = 57.30\\$$

$$\\\Rightarrow$$ $$a^{2} – \frac{31}{4}a + \frac{961}{64} + b^{2} – \frac{b}{2} + \frac{1}{16} =57.30$$

Therefore, the required equation is: 64a2– 496a + 64b2 -32b = 2702.51

Q.11: The circle passing through the points (3, 2) and (-2, 2) and whose centre lies on the line a – 3y – 13 = 0. Find the equation of the circle.

Answer: Based on formulae given in conic sections

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

As the circle passes through (3, 2) and (-2, 1)

(3 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1)

(-2 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . (2)

The centre (h, k) of the circle lies on the line a – 3y – 13 = 0,

h – 3k = 13 . . . . . . . . . . . . . . . (3)

Now, on comparing equations 1 and 2, we will get:

(3 – h)2 + (2 – k)2 = (-2 – h)2 + (2 – k)2

9 – 6h + h2 = 4 + 4h + h2

9 – 6h = 4 + 4h

h = 0.5 . . . . . . . . . . . . . . . . . . (4)

Now, on solving equation (3), we will get:

k = – $$\frac{25}{6}$$ = – 4.166

Now, on substituting the values of h and k in equation (1), we will get:

(3 – 0.5)2 + (2 – (- $$\frac{25}{6}$$))2 = r2

(2.5) 2 + (2 + $$\frac{25}{6}$$)2 = r2

(2.5) 2 + (2 + 4.16)2 = r2

(2.5) 2 + (6.16)2 = r2

r2 = 44.1956

r = 6.65

The required equation is:

(a – h)2 + (b – k)2 = r2

(a – 0.5)2 + (b – (- 4.16))2 = r2

(a – 0.5)2 + (b + 4.16)2 = r2

a2 – a + 6.25 + b2 + 8.32b + 17.30 = 44.19

a2 – a + b2 + 8.32b = 44.19 – 6.25 – 17.30

a2 + b2 – a + 8.32b – 20.64 = 0

Q.12: The radius of the circle is 6 whose centre lies on x – axis and passes through the point (3, 2). Find the equation.

Answer: Based on formulae given in conic sections

Given in the question:

The radius of the circle is 6 and passes through the point (3, 2)

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

As, the centre lies on x – axis, k = 0, the equation becomes:

(a – h) 2 + b 2 = 36

(3 – h) 2 + 22 = 36

(3 – h) 2 = 36 – 4

3 – h = $$\sqrt{32}$$

h = $$3 + 4\sqrt{2}$$ or h = $$3 – 4\sqrt{2}$$

Q.13: The circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Find the equation of the circle.

Answer: Based on formulae given in conic sections

Suppose, the equation of the required circle be (x – h) 2 + (y – k) 2 = r 2

The centre of the circle passes through (0, 0):

(0 – h) 2 + (0 – k) 2 = r 2

h 2 + k 2 = r 2

Therefore, the equation of the circle is:

(x – h) 2 + (y – k) 2 = h 2 + k 2 .

Given in the question, the circle makes intercepts a and b on the coordinate axes.

The circle passes through points (a, 0) and (0, b).

Therefore, (a – h) 2 + (0 – k) 2 = h 2 + k 2 . . . . . . . . . . . . . . . (1)

(0 – h) 2 + (b – k) 2 = h 2 + k 2 . . . . . . . . . . . . . . . . . . (2)

Now, from equation (1), we obtain:

a 2 – 2ah + h 2 + k 2 = h 2 + k 2

a 2 – 2ah = 0

a (a – 2h) = 0

a = 0 or (a – 2h) = 0

We know, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.

Now, from equation (2), we will get:

h 2 + b 2 – 2bk + k 2 = h 2 + k 2

b 2 – 2bk = 0

b (b – 2k) = 0

b = 0 or(b – 2k) = 0

However, b ≠ 0;

Therefore, (b – 2k) = 0 ⇒ k =b/2.

Thus, the equation of the required circle is:

$$(x – \frac{a}{2})^{2} + (y – \frac{b}{2})^{2} = (\frac{a}{2})^{2} + (\frac{b}{2})^{2} \\ (\frac{2x – a}{2})^{2} + (\frac{2y – b}{2})^{2} = \frac{a^{2} + b^{2}}{2} \\ 4x^{2} – 4ax + a^{2} + 4y^{2} – 4bx + b^{2} = a^{2} + b^{2} \\ 4x^{2} + 4y^{2} – 4ax – 4by = 0 \\ x^{2} + y^{2} – ax – by = 0$$

Q.14: The circle with centre (3, 3) and passes through the point (6, 5), find the equation of a circle?

Answer: Based on formulae given in conic sections

Given in the question:

The centre of the circle (h, k) = (3, 3).

As the circle passes through point (6, 5), the radius (r) of the circle is the distance between the points (2, 2) and (6, 5).

$$r = \sqrt{(3 – 6)^{2} + (3 – 5)^{2}} + \sqrt{(- 3)^{2} + (- 2)^{2}} = \sqrt{9 + 4} = \sqrt{13}\\$$

Therefore, the equation of the circle is:

$$\\(a – h)^{2} + (b – k)^{2} = r^{2} \\ (a – 3)^{2} + (b – 3)^{2} = \sqrt{13}^{2} \\ a^{2} – 6a + 9 + b^{2} – 6b + 9 = 13 \\ a^{2} + b^{2} – 6a – 6b + 18 – 13 = 0 \\ a^{2} + b^{2} – 6a – 6b + 5 = 0$$

Q.15: Check whether the point (–2.0, 3.0) lies inside, outside or on the circle x 2 + y 2 = 25?

Answer: Based on formulae given in conic sections

Given in the question:

The equation of the given circle is x 2 + y 2 = 25

x 2 + y 2 = 25

(x – 0)2 + (y – 0)2 = 52 , which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 0, k = 0, and r = 5.

Centre = (0, 0) and radius = 5

Distance between point (– 2.0, 3.0) and centre (0, 0)

$$\sqrt{(- 2 – 0)^{2} + (3 – 0)^{2}}\\$$

=$$\sqrt{4 + 9} = \sqrt{13} = 3.60 (approx.) < 5$$

As, the distance between point (– 2.0, 3.0) and centre (0, 0) of the circle is less than the radius of the circle, point (– 2.0, 3.0) lies inside the circle.

Exercise 11.2

Q.1: For the equation y2 = 16x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Answer: Based on formulae given in conic sections

Given in the question:

y2 = 16x

As the coefficient of x is positive, the given parabola has the opening on the right side.

By comparing this equation with y2 = 4ax, we get:

4a = 16 ⇒ a = 4

Therefore, focus coordinates = (a, 0) = (4, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 4

So, x + 4 = 0

Therefore, Length of the latus rectum = 4a = 4 × 4 = 16

Q.2: For the equation x2 = 8y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Answer: Based on formulae given in conic sections

Given in the question:

x2 = 8y

As the coefficient of y is positive, the given parabola has the opening upwards.

By comparing this equation with x2 = 4ay, we get

4a = 8 ⇒ a = 2

Therefore, focus coordinates = (0, a) = (0, 2)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y = – a = – 2

So, x + 2 = 0

Therefore, Length of the latus rectum = 4a = 4 × 2 = 8

Q.3: For the equation y2 = – 12x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Answer: Based on formulae given in conic sections

Given in the question:

y2 = – 12x

As the coefficient of x is negative, the given parabola has the opening on the left side.

By comparing this equation with y2 = – 4ax, we get

4a = 12 ⇒ a = 3

Therefore, focus coordinates = (- a, 0) = (- 3, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = a = 2

So, x + 4 = 0

Therefore, Length of the latus rectum = 4a = 4 × 3 = 12

Q.4: For the equation x2 = – 20y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Answer: Based on formulae given in conic sections

Given in the question:

x2 = – 20y

As the coefficient of y is negative, the given parabola has the opening downwards.

By comparing this equation with x2 = – 4ay, we get:

– 4a = – 20 ⇒ a = 5

Therefore, focus coordinates = (0, a) = (0, 5)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y = a = 5

Therefore, Length of the latus rectum = 4a = 4 × 5 = 20

Q.5: For the equation y2 = 24x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Answer: Based on formulae given in conic sections

Given in the question:

y2 = 24x

As the coefficient of x is positive, the given parabola has the opening on the right side.

By comparing this equation with y2 = 4ax, we get:

4a = 24 ⇒ a = 6

Therefore, focus coordinates = (a, 0) = (6, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 6

So, x + 6 = 0

Therefore, Length of the latus rectum = 4a = 4 × 6 = 24

Q.6: For the equation x2 = – 7y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Answer: Based on formulae given in conic sections

Given in the question:

x2 = – 7y

As the coefficient of y is negative, the given parabola has the opening downwards.

By comparing this equation with x2 = – 4ay, we get:

– 4a = – 7 ⇒ a = $$\frac{7}{4}$$

Therefore, focus coordinates = (0, a) = (0, $$\frac{7}{4}$$)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y = a = $$\frac{7}{4}$$

Therefore, Length of the latus rectum = 4a = 4 × $$\frac{7}{4}$$ = 7

Q7. Obtain the equation of the parabola which satisfies the given conditions below:

(i) Focus coordinates = (8, 0)

(ii) Directrix (x) = – 8

Answer: Based on formulae given in conic sections

Given in the question:

(i) Focus coordinates = (8, 0)

(ii) Directrix (x) = – 8

The axis of the given parabola is the x – axis as the focus is represented on the x – axis.

The required equation can be either y2 = 4ax and y2 = – 4ax

As we know, directrix (x) = – 8

As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (8, 0)

Therefore, the equation of the parabola = y2 = 4ax,

Here, a = 8

Thus, the equation of the parabola = y2 = 32 x.

Q.8: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Focus coordinates = (0, – 5)

(ii) Directrix (y) = 5

Answer: Based on formulae given in conic sections

Given in the question:

(i) Focus coordinates = (0, – 5)

(ii) Directrix (x) = 5

The axis of the given parabola is the y – axis as the focus is represented on the y – axis.

The required equation can be either x2 = 4ay and x2 = – 4ay

As we know, directrix (x) = 5

As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (0, -5)

Therefore, the equation of the parabola = x2 = – 4ay,

Here, a = 5

Thus, the equation of the parabola = x2 = – 20y

Q.9: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Focus coordinates = (4, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertex is at origin i.e., (0, 0)

(ii) Focus coordinates = (4, 0)

As the vertex of a parabola lies at origin and the focus is represented on the positive side of the x – axis, and hence, the axis of the parabola is x – axis.

The required equation is y2 = 4ax

As we know, focus is (4, 0)

Therefore, the equation of the parabola = y2 = 4ax

Here, a = 4

Thus, the equation of the parabola = y2 = 16 x

Q.10: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Focus coordinates = (- 3, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertex is at origin i.e., (0, 0)

(ii) Focus coordinates = (- 3, 0)

As the vertex of a parabola lies at origin and the focus is represented on the negative side of the x – axis, and hence, the axis of the parabola is x – axis.

The required equation is y2 = – 4ax

Therefore, the equation of the parabola = y2 = – 4ax

As we know, focus is (- 3, 0)

Here, a = 3

Thus, the equation of the parabola = y2 = – 12 x

Q.11: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Passing through coordinates = (3, 5)

The axis of the parabola is on x – axis.

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertex is at origin i.e., (0, 0)

(ii) Passing through coordinates = (3, 5)

As the vertex of a parabola lies at origin and the passing through coordinates = (3, 5).

The required equation is either y2 = 4ax or y2 = – 4ax

The coordinates (3, 5) is in first quadrant.

Therefore, the equation of the parabola = y2 = 4ax, whereas (3, 5) should satisfy the equation

So,

52 = 4a (3)

25 = 12 a

a = $$\frac{25}{12}$$

The parabola’s equation is:

y2 = 4 ($$\frac{25}{12}$$) x

3 y2 = 25 x

Q.12: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Passing through coordinates = (6, 4)

The equation is symmetric as considered with y – axis.

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertex is at origin i.e., (0, 0)

(ii) Passing through coordinates = (6, 4)

As the vertex of a parabola lies at origin and the passing through coordinates = (6, 4).

The required equation is either x2 = 4ay or x2 = – 4ay

The coordinates (6, 4) is in first quadrant.

Therefore, the equation of the parabola = x2 = 4ay, whereas (6, 4) should satisfy the equation

So,

62 = 4a (4)

36 = 16 a

a = $$\frac{36}{16}$$

a = $$\frac{9}{4}$$

The parabola’s equation is:

x2 = 4($$\frac{9}{4}$$)y

x2 = 9 y

Exercise 11.3

Q.1: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$$ . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{25}$$ > the denominator of $$\frac{y^{2}}{9}$$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 9} = \sqrt{16} = 4$$

We get:

m = 5, n = 3

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (4, 0) and (- 4, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 6

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.3^{2}}{5} = \frac{18}{5}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{4}{5}$$

Q.2: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$$ . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{9}$$ < the denominator of $$\frac{y^{2}}{16}$$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

$$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{16 – 9} = \sqrt{7}$$

We get,:

m = 4, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 4), (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = ($$\sqrt{7}$$, 0) and (-$$\sqrt{7}$$, 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 6

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.3^{2}}{4} = \frac{9}{2}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{7}}{4}$$

Q.3: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $$\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$$ . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{25}$$ > the denominator of $$\frac{y^{2}}{4}$$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 4} = \sqrt{21}$$

We get:

m = 5, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = ($$\sqrt{21}$$, 0) and (-$$\sqrt{21}$$, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 4

Length of the latus rectum = $$\frac{2n^{2}}{m}=\frac{2.2^{2}}{5} = \frac{8}{5}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{21}}{5}$$

Q.4: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $$\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:
$$\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1$$ . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{36}$$ < the denominator of $$\frac{y^{2}}{121}$$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

$$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{121 – 36} = \sqrt{85}$$

We get:

m = 11, n = 6

The vertices coordinates are (0, m) and (0, – m) = (0, 11), (0, – 11)

The foci’s coordinates are (0, c) and (0, – c) = (0, $$\sqrt{85}$$) and (0, –$$\sqrt{85}$$)

Length of the axis:

Major axis = 2m = 22

Minor axis = 2n = 12

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.6^{2}}{11} = \frac{72}{11}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{85}}{11}$$

Q.5: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $$\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1$$ . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{64}$$ > the denominator of $$\frac{y^{2}}{25}$$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{64 – 25} = \sqrt{39}$$

We get:

m = 8, n = 5

The vertices coordinates are (m, 0) and (- m, 0) = (8, 0), (- 8, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = ($$\sqrt{39}$$, 0) and (-$$\sqrt{39}$$, 0)

Length of the axis:

Major axis = 2m = 16

Minor axis = 2n = 10

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.5^{2}}{8} = \frac{25}{4}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{39}}{8}$$

Q.6: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse $$\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1$$ . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{400}$$ > the denominator of $$\frac{y^{2}}{100}$$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{400 – 100} = \sqrt{300} = 10\sqrt{3}$$

We get:

m = 20, n = 10

The vertices coordinates are (m, 0) and (- m, 0) = (20, 0), (- 20, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = ($$10 \sqrt{3}$$, 0) and (-$$10 \sqrt{3}$$, 0)

Length of the axis:

Major axis = 2m = 40

Minor axis = 2n = 20

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.10^{2}}{20} = 10$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{10 \sqrt{3}}{20} = \frac{\sqrt{3}}{2}$$

Q.7: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 49x2 + 9y2 = 441

Given in the question:

49x2 + 9y2 = 441

$$\frac{x^{2}}{9} + \frac{y^{2}}{49} = 1$$ . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{9}$$ < the denominator of $$\frac{y^{2}}{49}$$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

$$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{49 – 9} = \sqrt{40} = 2 \sqrt{10}$$

We get:

m = 7, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 7), (0, – 7)

The foci’s coordinates are (0, c) and (0, – c) = (0, $$2 \sqrt{10}$$) and (0, $$2 \sqrt{10}$$)

Length of the axis:

Major axis = 2m = 14

Minor axis = 2n = 6

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.3^{2}}{7} = \frac{18}{7}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{2 \sqrt{10}}{7}$$

Q.8: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + 16y2 = 64

Answer: Based on formulae given in conic sections

Given in the question:

4x2 + 16y2 = 64

$$\frac{x^{2}}{16} + \frac{y^{2}}{4} = 169$$ . . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{16}$$ > the denominator of $$\frac{y^{2}}{4}$$

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c=\sqrt{m^{2}-n^{2}}=\sqrt{16 – 4}=\sqrt{12}=2\sqrt{3}$$

We get:

m = 4, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (4, 0), (- 4, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = ($$2 \sqrt{3}$$, 0) and (-$$2 \sqrt{3}$$, 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 4

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.4^{2}}{4} = 8$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$$

Q.9: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + y2 = 4

Answer: Based on formulae given in conic sections

Given in the question:

4x2 + y2 = 4

$$\frac{x^{2}}{1} + \frac{y^{2}}{4} = 1$$ . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of $$\frac{x^{2}}{1}$$ < the denominator of $$\frac{y^{2}}{4}$$

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

$$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} – n^{2}} = \sqrt{4 – 1} = \sqrt{3}$$

We get:

m = 2, n = 1

The vertices coordinates are (0, m) and (0, – m) = (0, 2), (0, – 2)

The foci’s coordinates are (0, c) and (0, – c) = (0, $$\sqrt{3}$$) and (0, –$$\sqrt{3}$$)

Length of the axis:

Major axis = 2m = 4

Minor axis = 2n = 2

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.1^{2}}{2} = 1$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{3}}{2}$$

Q.10: For the given condition obtain the equation of the ellipse.

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form $$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . (1)

We get:

m = 6 (semi major axis), c = 3

As we know:

m2 = n2 + c2

62 = n2 + 32

n2 = 62 – 32

n2 = 36 – 9 = 27

n = $$\sqrt{27} = 3\sqrt{3}$$

Hence, $$\frac{x^{2}}{36} + \frac{y^{2}}{27} = 1$$ is the equation of the ellipse.

Q.11: For the given condition obtain the equation of the ellipse.

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

The vertices are represented along y – axis.

The required equation of the ellipse is of the form $$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . . .(1)

We get:

m = 8 (semi major axis), c = 4

As we know:

m2 = n2 + c2

82 = n2 + 42

n2 = 82 – 42

n2 = 64 – 16 = 48

n = $$\sqrt{48} = 4\sqrt{3}$$

Hence, $$\frac{x^{2}}{48} + \frac{y^{2}}{64} = 1$$ is the equation of the ellipse.

Q.12: For the given condition obtain the equation of the ellipse.

(i) Vertices (±5, 0)

(ii) Foci (±3, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Vertices (± 5, 0)

(ii) Foci (± 3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form $$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), c = 3

As we know:

m2 = n2 + c2

52 = n2 + 32

n2 = 52 – 32

n2 = 25 – 9 = 16

n = $$\sqrt{16} = 4$$

Hence, $$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$$ is the equation of the ellipse.

Q.13: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

The major axis is represented along x – axis.

The required equation of the ellipse is of the form

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), n = 3

Hence, $$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$$ is the equation of the ellipse.

Q.14: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±$$\sqrt{3}$$, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±$$\sqrt{3}$$, 0)

The major axis is represented along y – axis.

The required equation of the ellipse is of the form $$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . . . . . (1)

We get:

m = 2 (semi major axis), n = $$\sqrt{3}$$

Hence, $$\frac{x^{2}}{3} + \frac{y^{2}}{4} = 1$$ is the equation of the ellipse.

Q.15: For the given condition obtain the equation of the ellipse.

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form $$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . . . (1)

We get:

2m = 30 (semi major axis),

m = 15

c = 4

As we know:

m2 = n2 + c2

152 = n2 + 42

n2 = 152 – 42

n2 = 225 – 16 = 209

n = $$\sqrt{209}$$

Hence, $$\frac{x^{2}}{225} + \frac{y^{2}}{209} = 1$$ is the equation of the ellipse.

Q.16: For the given condition obtain the equation of the ellipse.

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form

$$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . (1)

We get:

2n = 26 (semi major axis),

n = 13

c = 9

As we know:

m2 = n2 + c2

m2 = 132 + 92

m2 = 169 + 81

m2 = 250

m = $$\sqrt{250}$$ = $$5 \sqrt{10}$$

Hence, $$\frac{x^{2}}{169} + \frac{y^{2}}{250} = 1$$ is the equation of the ellipse.

Q.17: For the given condition obtain the equation of the ellipse.

(i) Coordinates of foci (±4, 0)

(ii) m = 6

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of foci (±4, 0)

(ii) m = 6

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . (1)

We get:

c = 4

m = 6

As we know:

m2 = n2 + c2

62 = n2 + 42

n2 = 36 – 16

n2 = 20

n = $$\sqrt{20}$$ = $$2 \sqrt{5}$$

Hence, $$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$$ is the equation of the ellipse.

Q.18: For the given condition obtain the equation of the ellipse.

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0) i.e., centre is at origin.

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form:

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . (1)

We have:

n = 2, c = 3

As we know:

m2 = n2 + c2

m2 = 22 + 32

m2 = 4 + 9

m2 = 13

m = $$\sqrt{13}$$

Hence, $$\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1$$ is the equation of the ellipse.

Q.19: For the given condition obtain the equation of the ellipse.

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

$$\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1$$ . . . . . . . . . . . . . . . (1)

As the ellipse passes through (2, 1) and (2, 3) points, then

$$\frac{1^{2}}{n^{2}} + \frac{6^{2}}{m^{2}} = 1$$ $$\frac{3^{2}}{n^{2}} + \frac{2^{2}}{m^{2}} = 1$$

$$\frac{1}{n^{2}} + \frac{36}{m^{2}}$$ = 1 . . . . . . . . . . . . . . . (a)

$$\frac{9}{n^{2}} + \frac{4}{m^{2}}$$ = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 40, n2 = 10

Hence, $$\frac{y^{2}}{40} + \frac{x^{2}}{10} = 1$$ is the equation of the ellipse.

Q.20: For the given condition obtain the equation of the ellipse.

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

$$\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . (1)

As the ellipse passes through (6, 2) and (4, 3) points, then

$$\frac{6^{2}}{m^{2}} + \frac{2^{2}}{n^{2}} = 1$$ $$\frac{4^{2}}{m^{2}} + \frac{3^{2}}{n^{2}} = 1$$

$$\frac{36}{m^{2}} + \frac{4}{n^{2}}$$ = 1 . . . . . . . . . . . . . . . . (a)

$$\frac{16}{m^{2}} + \frac{9}{n^{2}}$$ = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 52 and n2 = 13

Hence, $$\frac{y^{2}}{13} + \frac{x^{2}}{52} = 1$$ is the equation of the ellipse.

Exercise 11.4

Q.1: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola $$\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1$$ . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

$$c = \sqrt{m^{2} + n^{2}} = \sqrt{25 + 16} = \sqrt{41}$$

We get:

m = 5 and n = 4

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = ($$\sqrt{41}$$, 0) and (-$$\sqrt{41}$$, 0)

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.4^{2}}{5} = \frac{32}{5}$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{41}}{5}$$

Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola $$\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1$$.

Answer: Based on formulae given in conic sections

Given in the question:

$$\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1$$ . . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

m = 1and n = $$\sqrt{8}$$

As we know:

$$c = \sqrt{m^{2} + n^{2}} = \sqrt{1 + 8} = \sqrt{9}$$ = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 1), (0, – 1)

The foci’s coordinates are (0, c) and (0, – c) = (0, 3), (0, – 3)

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{16}{1} = 16$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{3}}{1}$$ = 3

Q.3: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 25 y2 – 4 x2 = 100

Answer: Based on formulae given in conic sections

Given in the question:

25 y2 – 4 x2 = 100

$$\frac{y^{2}}{4} – \frac{x^{2}}{25} = 1$$ . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

m = 2 and n = 5.

$$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 25} = \sqrt{29}$$

The vertices coordinates are (0, m) and (0, – m) = (0, 2) and (0, – 2)

The foci’s coordinates are (0, c) and (0, – c) = (0, $$\sqrt{29}$$) and (0, –$$\sqrt{29}$$)

Length of the latus rectum = $$\frac{2.n^{2}}{m}$$ = $$\frac{2.5^{2}}{2} = \frac{50}{2}$$ = 25

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{29}}{2}$$

Q.4: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 9 x2 – 4 y2 = 36

Answer: Based on formulae given in conic sections

Given in the question:

9 x2 – 4 y2 = 36

$$\frac{x^{2}}{4} – \frac{y^{2}}{9} = 1$$ . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

m = 2 and n = 3.

$$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 3} = \sqrt{13}$$

The vertices coordinates are (m, 0) and (- m, 0) = (2, 0), (- 2, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = ($$\sqrt{13}$$, 0) and (-$$\sqrt{13}$$, 0)

Length of the latus rectum = $$\frac{2n^{2}}{m}$$ = $$\frac{2.3^{2}}{2} = 9$$

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{13}}{2}$$

Q.5: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 6 y2 – 16 x2 = 96

Answer: Based on formulae given in conic sections

Given in the question:

6 y2 – 16 x2 = 96

$$\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1$$ . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

m = 4 and n = $$\sqrt{6}$$

$$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}$$

The vertices coordinates are (0, m) and (0, – m) = (0, 4) and (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (0, $$\sqrt{10}$$) and (0, –$$\sqrt{10}$$)

Length of the latus rectum = $$\frac{2.n^{2}}{m}$$ = $$\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}$$ = 3

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{10}}{4}$$

Q.6: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 64 y2 – 9 x2 = 96

Answer: Based on formulae given in conic sections

Given in the question:

6 y2 – 16 x2 = 96

$$\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1$$ . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

m = 4 and n = $$\sqrt{6}$$

$$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}$$

The vertices coordinates are (0, m) and (0, – m) = (0, 4) and (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (0, $$\sqrt{10}$$) and (0, –$$\sqrt{10}$$)

Length of the latus rectum = $$\frac{2.n^{2}}{m}$$ = $$\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}$$ = 3

Eccentricity, e = $$\frac{c}{m}$$ = $$\frac{\sqrt{10}}{4}$$

Q.7: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (±3, 0)

(ii) Coordinates of the foci (±4, 0)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the vertices (±3, 0)

(ii) Coordinates of the foci (±4, 0)

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (±3, 0), m = 3

And, coordinates of the foci (±4, 0), c = 4

We know that:

c2 = m2 + n2

42 = 32 + n2

n2 = 16 – 9 = 5

$$\frac{x^{2}}{9} – \frac{y^{2}}{5} = 1$$ is the equation of the hyperbola.

Q.8: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (0, ±7)

(ii) Coordinates of the foci (0, ±9)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the vertices (0, ±7)

(ii) Coordinates of the foci (0, ±9)

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (0, ±7), m = 7

And, coordinates of the foci (0, ±9), c = 9

We know that:

c2 = m2 + n2

92 = 72 + n2

n2 = 81 – 49 = 32

$$\frac{y^{2}}{49} – \frac{x^{2}}{32} = 1$$ is the equation of the hyperbola.

Q.9: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (0, ±8)

(ii) Coordinates of the foci (0, ±11)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the vertices (0, ±8)

(ii) Coordinates of the foci (0, ±11)

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (0, ±8), m = 8

And, coordinates of the foci (0, ±11), c = 11

We know that:

c2 = m2 + n2

112 = 82 + n2

n2 = 121 – 64 = 57

$$\frac{y^{2}}{64} – \frac{x^{2}}{57} = 1$$ is the equation of the hyperbola.

Q.10: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (±7, 0),

(ii) The length of the transverse axis is 12

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the foci (±7, 0),

(ii) The length of the transverse axis is 12

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±7, 0), c = 7

As, the transverse axis is of length 12,

2 m = 12

m = 6

We know that:

c2 = m2 + n2

72 = 62 + n2

n2 = 49 – 36 = 13

$$\frac{x^{2}}{36} – \frac{y^{2}}{13} = 1$$ is the equation of the hyperbola.

Q.11: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (0, ±8)

(ii) The length of the conjugate axis is 16

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the foci (0, ±9)

(ii) The length of the conjugate axis is 16

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±9), c = 9

As, the conjugate axis is of length 16,

2 m = 16

m = 8

Therefore, m2 = 64

We know that:

c2 = m2 + n2

92 = 82 + n2

n2 = 81 – 64 = 17

$$\frac{y^{2}}{64} – \frac{x^{2}}{17} = 1$$ is the equation of the hyperbola.

Q.12: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (±$$2\sqrt{3}$$, 0 )

(ii) The length of the latus rectum is 8

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the foci (±$$2\sqrt{3}$$, 0 )

(ii) The length of the latus rectum is 8

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . (1)

And, the coordinates of the foci (±$$2\sqrt{3}$$, 0 ), c = $$2\sqrt{3}$$

As, the latus rectum is of length 8

$$\frac{2n^{2}}{m} = 8 \\ 2n^{2} = 8 m \\ n^{2} = 4 m$$

We know that:

m2 + n2 = c2

m2 + 4m = $$(2\sqrt{3})^{2}$$

m2 + 4m = 12

m2 + 4m – 12 = 0

m2 + 6m – 2m – 12 = 0

m (m + 6) – 2 (m + 6) = 0

m = – 6 and m = 2

m is non – negative so, m = 2 and n2 = 8 [Since, n2 = 12 – 22 = 8]

$$\frac{x^{2}}{4} – \frac{y^{2}}{8} = 1$$ is the equation of the hyperbola.

Q.13: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (±$$2\sqrt{6}$$, 0)

(ii) The length of the latus rectum is 10

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the foci (±$$2\sqrt{3}$$, 0 )

(ii) The length of the latus rectum is 10

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±$$2\sqrt{6}$$, 0 ), c = $$2\sqrt{6}$$

As, the latus rectum is of length 10

Therefore, $$\frac{2n^{2}}{m} = 10 \\ 2n^{2} = 10 m \\ n^{2} = 5m$$

We know that:

m2 + n2 = c2

m2 + 5 m = $$(2\sqrt{6})^{2}$$

m2 + 5 m = 24

m2 + 5 m – 24 = 0

m2 + 8 m – 3 m – 24 = 0

m (m + 8) – 3 (m + 8) = 0

m = – 8 and m = 3

m is non – negative so, m = 3 and n2 = 5m = 15

$$\frac{x^{2}}{9} – \frac{y^{2}}{15} = 1$$ is the equation of the hyperbola.

Q.14: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (±9, 0)

(ii) Eccentricity (e) = $$\frac{9}{4}$$

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the vertices (±9, 0)

(ii) Eccentricity (e) = $$\frac{9}{4}$$

The specific equation for hyperbola is $$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . . . . . . (1)

And, Coordinates of the vertices (±9, 0), m = 9

As, the eccentricity (e) = $$\frac{9}{4}$$,

$$\frac{c}{m} = \frac{9}{4}$$ $$\frac{c}{9} = \frac{9}{4}$$

c = $$\frac{81}{4}$$

We know that:

m2 + n2 = c2

92 + n2 = $$(\frac{81}{4})^{2}$$

n2 = $$(\frac{81}{4})^{2}$$ – 92

n2 = $$\frac{6561}{16} – 81$$

n2 = $$\frac{6561 – 1296}{16} = \frac{5265}{16}$$

$$\frac{x^{2}}{81} – \frac{16 y^{2}}{5265} = 1$$ is the equation of the hyperbola.

Q.15: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (0, ± $$\sqrt{10}$$)

(ii) It passes through (2, 3)

Answer: Based on formulae given in conic sections

Given in the question:

(i) Coordinates of the foci (0, ±$$\sqrt{10}$$)

(ii) It passes through (2, 3)

The specific equation for hyperbola is $$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$$ . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±$$\sqrt{10}$$), c = $$\sqrt{10}$$

We know that:

m2 + n2 = c2

n2 = c2 – m2

n2 = 10 – m2 . . . . . . . . . . . . (2)

As the hyperbola passes through (2, 3)

$$\frac{3^{2}}{m^{2}} – \frac{2^{2}}{n^{2}} = 1$$

$$\frac{9}{m^{2}} – \frac{4}{n^{2}} = 1$$ . . . . . . . . . . . (3)

Putting Equation (2) in equation (3), we get:

$$\frac{9}{m^{2}}-\frac{4}{10 – m^{2}}= 1$$

9 (10 – m2) – 4m2 = m2 (10 – m2)

m4 – 23m2 + 90 = 0

m4 – 18m2 – 5m2 + 90 = 0

(m2 – 5) (m2 – 18) = 0

m2 = 5 or 18

As in hyperbola:

m2 < c2

m2 = 5

n2 = 10 – 5 [From Equation (2)] = 5

$$\frac{y^{2}}{5} – \frac{x^{2}}{5} = 1$$ is the equation of the hyperbola.

DOCTYPE html> Binomial Theorem - Mathematics - UPSCFEVER

#### Chapter-8: Class 11 Binomial Theorem

Formula for Binomial Theorem:

[a + b]n = [ nC0 × an ] + [ nC1 × (an – 1) × b ] + [ nC2 × (a n – 2) × b2 ] + [ nC3 × (an – 3 )× b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × a × (bn – 1) ] + [ nCn × bn ]

$$\Rightarrow$$ Binomial Coefficient:

The coefficients nC0, nC1, nC2 . . . . . . . . . nCn occurring in the Binomial Theorem are known as Binomial coefficients

Some conclusions from Binomial Theorem:

(i) [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

(ii) [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

(iii) [1 – x]n = [ nC0 ] – [ nC1 . x ] + [ nC2 . x2 ] – [ nC3 . x3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn . xn ]

(iv) (a + b)n = $$\sum_{r\;=\;0}^{n}$$ nCr (a)n – r × br

IMPORTANT POINTS:

(1.) nCr = $$\frac{n!}{r!(n-r)!}$$ where, n is a non-negative integer and [0 r n]

(2.) nC0 = nCn = 1

(3.) There are total (n + 1) terms in the expansion of (a + b)n

Exercise 8.1

Q.1: Expand the Expression (1 – 3x)5

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (1 – x)n = [ nC0 ] + [ nC1 × (-x) ] + [ nC2 × (-x)2 ] + [ nC3 × (-x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (-x)n ]

Therefore we get, (1 – 3x)5 = [ 5C0 ] – [ 5C1 × (3x) ] + [ 5C2 × (3x)2 ] – [ 5C3 × (3x)3 ] + [ 5C4 × (3x)4 ] – [ 5C5 × (3x)5 ]

$$\\\Rightarrow$$ 1 – [ 5 × (3x) ] + [ 10 × (9x2) ] – [ 10 × (27x3) ] + [ 5 × (81x4) ] – [ 1 × (243x5) ]

$$\\\Rightarrow$$ 1 – 15x + 90x2 – 270x3 + 405x4 – 243x5

Therefore we get, (1 – 3x)5 = 1 – 15x + 90x2 – 270x3 + 405x4 – 243x5

Q.2: Expand the Expression $$\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}$$

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, $$\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}:$$

[5C0 × $$\left ( \frac{5}{x} \right )^{5}$$ ] – [5C1 × $$\left ( \frac{5}{x} \right )^{4}$$ × $$\left ( \frac{x}{5} \right )^{1}$$ ] + [5C2 × $$\left ( \frac{5}{x} \right )^{3}$$ × $$\left ( \frac{x}{5} \right )^{2}$$ ] – [ 5C3 × $$\left ( \frac{5}{x} \right )^{2}$$ × $$\left ( \frac{x}{5} \right )^{3}$$ ] + [5C4 × $$\left ( \frac{5}{x} \right )^{1}$$ × $$\left ( \frac{x}{5} \right )^{4}$$ ] – [5C5 × $$\left ( \frac{x}{5} \right )^{5}]\\$$

$$\\\Rightarrow \left [ 1\times \frac{3125}{x^{5}}\right ]-\left [ 5\times \frac{625}{x^{4}}\times \frac{x}{5}\right ]+\left [ 10\times \frac{125}{x^{3}}\times \frac{x^{2}}{25}\right ]-\left [ 10\times \frac{25}{x^{2}}\times \frac{x^{3}}{125} \right ]+\left [ 5\times \frac{5}{x}\times \frac{x^{4}}{625} \right ]-\left [ 1\times \frac{x^{5}}{3125} \right ]\\$$ $$\\\Rightarrow \left [\frac{3125}{x^{5}}\right ]-\left [\frac{625}{x^{3}}\right ]+\left [ \frac{50}{x}\right ]-2x+\left [ \frac{3x^{3}}{25} \right ]-\left [\frac{x^{5}}{3125} \right ]\\$$

Therefore we get, $$\\\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}$$:

$$\\\left [\frac{x^{5}}{3125} \right ] +\left [ \frac{3x^{3}}{25} \right ]-2x+\left [ \frac{50}{x}\right ]-\left [\frac{625}{x^{3}}\right ]+ \left [\frac{3125}{x^{5}}\right ]$$

Q.3: Expand the Expression (3x – 2)6

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

[x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, (3x – 2)6 = [ 6C0 × (3x)6 ] – [ 6C1 × (3x)5 × (2) ] + [ 6C2 × (3x)4 × (2)2 ] – [ 6C3 × (3x)3 × (2)3 ] + [ 6C4 × (3x)2 × (2)4 ] – [ 6C5 × (3x)1 × (2)5 ] + [ 6C6 × (2)6 ]

$$\\\Rightarrow$$ [1 × (729x6)] – [6 × (243x5) × 2] + [15 × (81x4) × 4] – [20 × (27x3) × 8 ] + [15 × (9x2) × 16] – [6 × (3x) × 32] + [1 × 64]

$$\\\Rightarrow$$ 729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64

Therefore we get, (3x – 2)6 = 729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64

Q.4: Expand the Expression $$\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}$$

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, $$\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}$$:

[ 5C0 × $$\left ( \frac{4}{x} \right )^{5}$$ ] + [ 5C1 × $$\left ( \frac{4}{x} \right )^{4}$$ × $$\left ( \frac{x}{3} \right )^{1}$$ ] + [ 5C2 × $$\left ( \frac{4}{x} \right )^{3}$$ × $$\left ( \frac{x}{3} \right )^{2}$$ ] + [ 5C3 × $$\left ( \frac{4}{x} \right )^{2}$$ × $$\left ( \frac{x}{3} \right )^{3}$$ ] + [ 5C4 × $$\left ( \frac{4}{x} \right )^{1}$$ × $$\left ( \frac{x}{3} \right )^{4}$$ ] + [ 5C5 × $$\left ( \frac{x}{3} \right )^{5}]\\$$

$$\\\Rightarrow \left [ 1\times \frac{1024}{x^{5}}\;\right ]+\left [ 5\times \frac{256}{x^{4}}\times \frac{x}{3}\;\right ]+\left [ 10\times \frac{64}{x^{3}}\times \frac{x^{2}}{9}\;\right ]+\left [ 10\times \frac{16}{x^{2}}\times \frac{x^{3}}{27} \right ]+\left [ 5\times \frac{4}{x}\times \frac{x^{4}}{81}\; \right ]+\left [ 1\times \frac{x^{5}}{243} \right ]\\\\$$ $$\\\Rightarrow \left [\frac{1024}{x^{5}}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{x^{5}}{243} \right ]\\$$

Therefore we get, $$\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}:$$

$$\\\Rightarrow \left [\frac{x^{5}}{243} \right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [\frac{1024}{x^{5}}\right ]$$

Q.5: Expand the Expression $$\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}$$

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, $$\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}$$:

[ 6C0 × $$\left ( \frac{1}{x} \right )^{6}$$ ] – [ 6C1 × $$\left ( \frac{1}{x} \right )^{5}$$ × $$\left ( \frac{x}{2} \right )^{1}$$ ] + [ 6C2 × $$\left ( \frac{1}{x} \right )^{4}$$ × $$\left ( \frac{x}{2} \right )^{2}$$ ] – [ 6C3 × $$\left ( \frac{1}{x} \right )^{3}$$ × $$\left ( \frac{x}{2} \right )^{3}$$ ] + [ 6C4 × $$\left ( \frac{1}{x} \right )^{2}$$ × $$\left ( \frac{x}{2} \right )^{4}$$ ] – [ 6C5 × $$\left ( \frac{1}{x} \right )^{1}$$ × $$\left ( \frac{x}{2} \right )^{5}$$ ] + [ 6C6 ×$$\left ( \frac{x}{2} \right )^{6}]\\$$

$$\\\Rightarrow \left [ 1\times \frac{1}{x^{6}}\;\right ]-\left [ 6\times \frac{1}{x^{5}}\times \frac{x}{2}\;\right ]+\left [ 15\times \frac{1}{x^{4}}\times \frac{x^{2}}{4}\;\right ]-\left [ 20\times \frac{1}{x^{3}}\times \frac{x^{3}}{8} \right ]+\left [ 15\times \frac{1}{x^{2}}\times \frac{x^{4}}{16}\; \right ]-\left [ 6\times \frac{1}{x}\times \frac{x^{5}}{32} \right ]+\left [ 1\times \frac{x^{6}}{64} \right ]\\$$ $$\\\Rightarrow \left [ \frac{1}{x^{6}}\;\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{5}{2} \right ]+\left [\frac{15x^{2}}{16}\; \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{x^{6}}{64} \right ]\\$$

Therefore we get, $$\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}:$$

$$\\\Rightarrow \left [\frac{x^{6}}{64} \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{15x^{2}}{16} \right ]-\left [ \frac{5}{2} \right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{1}{x^{6}}\right ]$$

Q.6: By using Binomial Theorem, Evaluate (98)4

Answer. Based on formula given in Binomial Theorem

(98)4 = (100 – 2)4

Now we get, by using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, (100 – 2)4 = [ 4C0 × (100)4 ] – [ 4C1 × (100)3 × (2) ] + [ 4C2 × (100)2 × (2)2 ] – [ 4C3 × 100 × (2)3 ] + [4C4 × (2)4]

$$\\\Rightarrow$$ (1 × 100000000) – (4 × 1000000 × 2) + (6 × 10000 × 4) – (4 × 100 × 8 ) + (1 × 16)

$$\\\Rightarrow$$ 100000000 – 8000000 + 240000 – 3200 + 16

Therefore we get, (98)4 = (100 – 2)4 = 92236816 = 9.2236816 × 107

Q.7: By using Binomial Theorem, Evaluate (105)5

Answer. Based on formula given in Binomial Theorem

(105)5 = (100 + 5)5

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, (100 + 5)5 = [ 5C0 × (100)5 ] + [ 5C1 × (100)4 × (5) ] + [ 5C2 × (100)3 × (5)2 ] + [ 5C3 × (100)2 × (5)3 ] + [ 5C4 × (100) × (5)4 ] + [ 5C5 × (5)5]

$$\\\Rightarrow$$ (1 × 10000000000) + (5 × 100000000 × 5) + (10 × 1000000 × 52 ) + (10 × 10000 × 53 ) + (5 × 100 × 54 ) + ( 1 × 55 )

$$\\\Rightarrow$$ 10000000000 + 2500000000 + 250000000 + 12500000 + 312500 + 3125

Therefore we get, (105)5 = (100 + 5)5 = 12762815630 = 1.276281563 × 1010

Q.8: By using Binomial Theorem, Evaluate (104)4

Answer. Based on formula given in Binomial Theorem

(104)4 = (100 + 4)4

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, (100 + 4)4 = [4C0 × (100)4 ] + [ 4C1 × (100)3 × (4) ] + [4C2 × (100)2 × (4)2] + [4C3 × 100 × (4)3] + [4C4 × (4)4 ]

$$\\\Rightarrow$$ (1 × 100000000) + (4 × 1000000 × 4) + (6 × 10000 × 16) + (4 × 100 × 64 ) + (1 × 256)

$$\\\Rightarrow$$ 100000000 + 16000000 + 960000 + 25600 + 256

Therefore we get, (104)4 = (100 + 4)4 = 116985856 = 1.16985856 × 108

Q.9: By using Binomial Theorem, Evaluate (95)5

Answer Based on formula given in Binomial Theorem.

(95)5 = (100 – 5)5

Now we get, by using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, (100 – 5)5 = [ 5C0 × (100)5 ] – [ 5C1 × (100)4 × (5) ] + [ 5C2 × (100)3 × (5)2 ] – [ 5C3 × (100)2 × (5)3 ] + [ 5C4 × (100) × (5)4 ] – [ 5C5 × (5)5 ]

$$\\\Rightarrow$$ (1 × 10000000000) – (5 × 100000000 × 5) + (10 × 1000000 × 52 ) – (10 × 10000 × 53 ) + (5 × 100 × 54 ) – ( 1 × 55 )

$$\\\Rightarrow$$ 10000000000 – 2500000000 + 250000000 – 12500000 + 312500 – 3125

Therefore we get, (95)5 = (100 – 5)5 = 7737809375 = 7.737809375 × 109

Q.10: By using Binomial Theorem, determine which number is greater (1.1)10000 or 1000.

Answer. Based on formula given in Binomial Theorem

Now we get, (1.1)10000 = (1 + 0.1)10000

Now we get, on applying Binomial Theorem, the first few terms in the expansion of (1 + 0.1)10000 are:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, (1 + 0.1)10000 = [ 10000C0 × (1)10000 ] + [ 10000C1 × (1)9999 × 0.1 ] + other positive terms

$$\\\Rightarrow$$ [1 × 1] + [10000 × 1 × 0.1] + other positive terms

$$\\\Rightarrow$$ 1 + 1000 + other positive terms > 1000

Therefore we get, (1.1)10000 is greater than 1000.

Q.11: Find (a + b)5 – (a – b)5 . Hence evaluate $$\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}$$

Answer. Based on formula given in Binomial Theorem

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, (a + b)5= [ 5C0 × (a5) ] + [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] + [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] + [ 5C5 × b5 ]

And, [a – b]5 = [ 5C0 × (a5) ] – [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] – [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] – [ 5C5 × b5 ]

Therefore we get, (a + b)5 – (a – b)5 = [ 5C0 a5 + 5C1 a4 b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5 ] – [ 5C0 a55C1 a4b + 5C2 a3b25C3 a2b3 + 5C4 ab4 5C5 b5 ]

$$\\\Rightarrow$$ [ 5C0 a5 + 5C1 a4 b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5 5C0 a5 + 5C1 a4b – 5C2 a3b2 + 5C3 a2b3 5C4 ab4 + 5C5 b5 ]

$$\\\Rightarrow$$ 2[ 5C1 a4b + 5C3 a2b3 + 5C5 b5 ] = 2[ 5a4b + 10a2b3 + b5 ]

Therefore we get, (a + b)5 – (a – b)5 = 2b[ 5a4 + 10a2 b2 + b4 ] . . . . . . (1)

Now we get, on substituting a = $$\sqrt{5}$$ and b = $$\sqrt{3}$$ in equation (1) we will get:

$$\\\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5} = 2\sqrt{3}\times \left [ 5( \sqrt{5})^{4} +10( \sqrt{5})^{2}\times ( \sqrt{3} )^{2}+(\sqrt{3})^{4}\right ]\\$$

$$\\\Rightarrow 2\sqrt{3}\;[125+150+9]=568\sqrt{3}\\$$

Therefore we get, $$\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}$$=$$568\sqrt{3}$$

Q.12: Find (1 + x)5 – (1 – x)5 . Hence evaluate $$\left ( 1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7} \right )^{5}$$

Answer. Based on formula given in Binomial Theorem

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, (1 + x)5= [ 5C0 × (15) ] + [ 5C1 × (14) × (x) ] + [ 5C2 × (13) × (x)2 ] + [ 5C3 × (12) × (x)3 ] + [ 5C4 × (11) × (x)4 ] + [ 5C5 × (x)5 ]

And, [1 – x]5 = [ 5C0 × (15) ] – [ 5C1 × (14) × (x) ] + [ 5C2 × (13) × (x)2 ] – [ 5C3 × (12) × (x)3 ] + [ 5C4 × (11) × (x)4 ] – [ 5C5 × (x)5 ]

Therefore we get, (1 + x)5 – (1 – x)5 = [ 5C0 + 5C1 x + 5C2 x2 + 5C3 x3 + 5C4 x4 + 5C5 x5 ] – [ 5C05C1 x + 5C2 x25C3 x3 + 5C4 x4 5C5 x5 ]

$$\\\Rightarrow$$ [ 5C0 + 5C1 x + 5C2 x2 + 5C3 x3 + 5C4 x4 + 5C5 x5 5C0 + 5C1 x – 5C2 x2 + 5C3 x3 5C4 x4 + 5C5 x5 ]

$$\\\Rightarrow$$ 2[ 5C1x + 5C3 x3 + 5C5 x5 ] = 2[ 5x + 10x3 + x5 ]

Therefore we get, (1 + x)5 – (1 – x)5 = 2x[ 5 + 10x2 + x4 ] . . . . . . (1)

Now we get, on substituting x = $$\sqrt{7}$$ in equation (1) we will get:

$$\\\Rightarrow \left ( 1+\sqrt{7} \right )^{7}-\left ( 1-\sqrt{7} \right )^{5}=2\sqrt{7}\left [ 5+10\left ( \sqrt{7} \right )^{2} + \left ( \sqrt{7} \right )^{4}\right ]\\$$ $$\\\Rightarrow 2\sqrt{7}\;[5+70+49]=248\sqrt{7}\\$$

Therefore we get, $$\left(1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7}\right)^{5}$$=$$\;248\sqrt{7}$$

Q.13 Show that 7n+1 – 6n – 7 is divisible by 36, where n is the positive integer.

Answer. Based on formula given in Binomial Theorem

Equation 7n+1 – 6n – 7 will be divisible by 36 if, 7n+1 – 6n – 7 = 36p [where p is any natural number]

Now we get, by Binomial Theorem:

Since, [1 + x]m = [ mC0 ] + [ mC1 . x ] + [ mC2 . x2 ] + [ mC3 . x3 ] + . . . . . . . . . . . . + [ mCm . xm ]

Therefore we get, for x = 6 and m = n + 1 the equation becomes:

[1 + 6]n+1 = [ n+1C0 ] + [ n+1C1 × 6 ] + [ n+1C2 × (6)2 ] + [ n+1C3 × (6)3 ] + . . . . . . . . . . . . . + [ n+1Cn+1 × (6)n+1 ]

$$\\\Rightarrow$$ [7]n + 1 = 1 + [(n + 1) × 6] + 62 [ n+1C2 + n+1C3 × (6) + . . . . . . . . . + n+1Cn + 1 × (6)n–1 ]

$$\\\Rightarrow$$ [7]n + 1 = 1 + 6n + 6 + 36 [ n+1C2 + n+1C3 × (6) + . . . . . . . . . + n + 1Cn + 1 × (6)n – 1 ]

$$\\\Rightarrow$$ (7)n + 1 – 6 n – 7 = 36p

Where p is any natural number and p = [n + 1C2 + n + 1C3 × (6) + . . . . . . . . . . . . . . . . . . + n + 1Cn + 1 × (6)n – 1]

Therefore we get, 7n+1 – 6n – 7 is divisible by 36 [where n is a positive integer]

Q.14 Prove that $$\sum_{r\;=\;0}^{n}$$ 2r × nCr = 3n

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (a + b)n = $$\sum_{r\;=\;0}^{n}$$ nCr (a)n – r × (b)r

Therefore we get, on substituting the value of a = 1 and b = 2 in the above equation we will get:

(1 + 2)n = $$\sum_{r\;=\;0}^{n}$$ nCr (1)n – r × (2)r

$$\\\Rightarrow$$ 3n = $$\sum_{r\;=\;0}^{n}$$ nCr × (2)r

Therefore we get, $$\\\sum_{r\;=\;0}^{n}$$ 2r × nCr = 3n

Q.15 Show that 5n – 4n always leaves remainder 5 when divided by 16, where n is the positive integer.

Answer. Based on formula given in Binomial Theorem

Equation 5n – 4n will leave remainder 5 when divided by 16 if, 5n – 4n = 16p + 5 [where p is any natural number]

Now we get, by Binomial Theorem:

Since, [1 + x]n = [ nC0 ] + [ nC1 . x ] + [ nC2 . x2 ] + [ nC3 . x3 ] + . . . . . . . . . . . . . . . . . . + [ nCn . xn ]

Therefore we get, for x = 4 the equation becomes:

[1 + 4]n = [ nC0 ] + [ nC1 × 4 ] + [ nC2 × (4)2 ] + [ nC3 × (4)3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × (4)n ]

$$\\\Rightarrow$$ [5]n = 1 + [(n + 1) × 4] + 42 [ nC2 + nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n ]

$$\\\Rightarrow$$ [5]n = 1 + 4n + 4 + 16 [ nC2 + nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n–1 ]

$$\\\Rightarrow$$ 5n – 4n – 5 = 16p

Where p is any natural number and p = [ nC2 + nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n–1]

i.e. 5n – 4n = 16 + 5

Therefore we get, 5n – 4n will leave remainder 5 when divided by 16, where n is any natural number.

Q.16: Find (a + b)6 – (a – b)6 . Hence evaluate: $$\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}$$

Answer. Based on formula given in Binomial Theorem

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, [a + b]6= [ 6C0 × (a6) ] + [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] + [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] + [ 6C5 × a × b5 ] + [6C6 × b6]

And, [a – b]6 = [ 6C0 × (a6) ] – [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] – [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] – [ 6C5 × a × b5 ] + [6C6 × b6]

Therefore we get, (a + b)6 – (a – b)6 = [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b6] – [ 6C0 a66C1 a5b + 6C2 a4b2 6C3 a3b3 + 6C4 a2b4 6C5 ab5 + 6C6 b6 ]

$$\\\Rightarrow$$ [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b66C0 a6 + 6C1 a5b – 6C2 a4b2 + 6C3 a3b3 6C4 a2b4 + 6C5 ab56C6 b6 ]

$$\\\Rightarrow$$ 2[ 6C1 a5b + 6C3 a3b3 + 6C5 ab5 ] = 2[ 6a5b + 20a3b3 + 6ab5 ]

Therefore we get, (a + b)6 – (a – b)6 = 4ab[ 3a4 + 10a2 b2 + 3b4 ] . . . (1)

Now we get, on substituting a = $$\sqrt{2}$$ and b = $$\sqrt{3}$$ in equation (1) we will get:

$$\\(\sqrt{2}+\sqrt{3} )^{6}-(\sqrt{2}-\sqrt{3})^{6}=4 (\sqrt{2}. \sqrt{3})\left [ 3( \sqrt{2})^{4} +10 ( \sqrt{2})^{2}\times( \sqrt{3} )^{2}+3(\sqrt{3})^{4} \right ]\\$$ $$\\\Rightarrow 4\sqrt{6}\;[12+60+27]=396\sqrt{6}\\$$

Therefore we get, $$\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}$$=$$\;396\sqrt{6}$$

Q.17: By using Binomial Theorem, Evaluate (91)4

Answer. Based on formula given in Binomial Theorem

(91)4 = (100 – 9)4

Now we get, by using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, (100 – 9)4 = [ 4C0 × (100)4 ] – [ 4C1 × (100)3 × (9) ] + [ 4C2 × (100)2 × (9)2 ] – [ 4C3 × 100 × (9)3 ] + [4C4 × (9)4]

$$\\\Rightarrow$$ (1 × 100000000) – (4 × 1000000 × 9) + (6 × 10000 × 81) – (4 × 100 × 729) + (1 × 6561)

$$\\\Rightarrow$$ 100000000 – 36000000 + 4860000 – 291600 + 6561

Therefore we get, (91)4 = (100 – 9)4 = 68574961 = 6.8574961 × 107

Q.18: By using Binomial Theorem, Evaluate (107)5

Answer. Based on formula given in Binomial Theorem

(107)5 = (100 + 7)5

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, (100 + 7)5 = [ 5C0 × (100)5 ] + [ 5C1 × (100)4 × (7) ] + [ 5C2 × (100)3 × (7)2 ] + [ 5C3 × (100)2 × (7)3 ] + [ 5C4 × (100) × (7)4 ] + [ 5C5 × (7)5]

$$\\\Rightarrow$$ (1 × 10000000000) + (5 × 100000000 × 7) + (10 × 1000000 × 72 ) + (10 × 10000 × 73 ) + (5 × 100 × 74 ) + ( 1 × 75 )

$$\\\Rightarrow$$ 10000000000 + 3500000000 + 490000000 + 34300000 + 1200500 + 16807

Therefore we get, (107)5 = (100 + 7)5 = 1402551731 = 1.402551731 × 1010

Q.19: Expand the Expression $$\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}$$

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, $$\\\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\$$

[6C0 × $$\left ( \frac{3}{2x} \right )^{6}$$ ] – [6C1 × $$\left ( \frac{3}{2x} \right )^{5}$$ × $$\left ( \frac{x}{5} \right )^{1}$$ ] + [6C2 × $$\left ( \frac{3}{2x} \right )^{4}$$ × $$\left ( \frac{x}{5} \right )^{2}$$ ] – [6C3 × $$\left ( \frac{3}{2x} \right )^{3}$$ × $$\left ( \frac{x}{5} \right )^{3}$$ ] + [6C4 × $$\left ( \frac{3}{2x} \right )^{2}$$ × $$\left ( \frac{x}{5} \right )^{4}$$ ] – [6C5 × $$\left ( \frac{3}{2x} \right )^{1}$$ × $$\left ( \frac{x}{5} \right )^{5}$$ ] + [6C6 ×$$\left ( \frac{x}{5} \right )^{6}]\\$$

$$\\\Rightarrow \left [ 1\times \frac{729}{64\times x^{6}}\;\right ]-\left [ 6\times \frac{243}{32\times x^{5}}\times \frac{x}{5}\;\right ]+\left [ 15\times \frac{81}{16\times x^{4}}\times \frac{x^{2}}{25}\;\right ]-\left [ 20\times \frac{27}{8x^{3}}\times \frac{x^{3}}{125} \right ]+\left [ 15\times \frac{9}{4x^{2}}\times \frac{x^{4}}{625}\; \right ]-\left [ 6\times \frac{3}{2x}\times \frac{x^{5}}{3125} \right ]+\left [ 1\times \frac{x^{6}}{15625} \right ]\\$$ $$\\\Rightarrow \left [\frac{729}{64\; x^{6}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{x^{6}}{15625} \right ]\\$$

Therefore we get, $$\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\$$

$$\Rightarrow \left [\frac{x^{6}}{15625} \right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{729}{64\; x^{6}}\;\right ]$$

Q.20: Expand the Expression (5x – 3)6

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore we get, (5x – 3)6 = [ 6C0 × (5x)6 ] – [ 6C1 × (5x)5 × (3) ] + [ 6C2 × (5x)4 × (3)2 ] – [ 6C3 × (5x)3 × (3)3 ] + [ 6C4 × (5x)2 × (3)4 ] – [ 6C5 × (5x)1 × (3)5 ] + [ 6C6 × (3)6 ]

$$\\\Rightarrow$$ [1 × (15625 x6)] – [6 × (3125 x5) × 3] + [15 × (625 x4) × 9] – [20 × (125 x3) × 27 ] + [15 × (25 x2) × 81] – [6 × (5x) × 243] + [1 × 729]

$$\\\Rightarrow$$ 15625 x6 – 56250 x5 + 84375 x4 – 67500 x3 + 30375 x2 – 7290 x + 729

Therefore we get, (5x – 3)6 = 15625 x6 – 56250 x5 + 84375 x4 – 67500 x3 + 30375 x2 – 7290 x + 729

Exercise 8.2

Formulas:

1. The general term in the expansion of (a + b)n :

Tr + 1 = nCr × (a)n – r × br

2. The middle term in the expansion of (a + b)n :

(a). If n is even:

The middle term = $$\left ( \frac{n}{2}+1 \right )^{th}term$$

(b). If n is odd:

The middle term = $$\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term$$

Q.1: Find the Coefficient of x6 in the expansion of (x + 2)9

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore we get, on substituting n = 9, a = x and b = 2 in the above expression we will get:

Tr + 1 = 9Cr × (x)9 – r × 2r . . . . . . . (1)

Now we get, on comparing the coefficient of x in equation (1) with x6, we will get:

i.e. (x)9 – r = x6

(or) 9 – r = 6

Therefore we get, r = 3

Now we get, on substituting the value of r in equation (1) we will get:

T4 = 9C3 × (x)9 – 3 × 23

$$\\\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times x^{6}\times 8\\\\$$ $$\\\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 8\;x^{6}= 672\;x^{6}\\\\$$

Therefore we get, the Coefficient of x6 in the expansion of (x + 2)9 = 672

Q.2: Find the Coefficient of a6 b8 in the expansion of (2a – 3b)14

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × yr

Therefore we get, on substituting n = 14, x = 2a and y = (-3b) in the above expression we will get:

Tr + 1 = 14Cr × (2a)14 – r × (-3b)r

i.e. Tr + 1 = [14Cr × (2)14 – r (-3)r ] (a)14 – r × (b)r . . . . . . . . . . (1)

Now we get, on comparing the coefficients of a and b in equation (1) with a6 b8, we will get:

i.e. a6 b8 = a14 – r br

Therefore we get, r = 8

Now we get, on substituting the value of ‘r’ in equation (1) we will get:

T9 = [ 14C8 × (2)14 – 8 (-3)8 ] (a)14 – 8 × (b)8

i.e. T9 = [ 14C8 × (2)6 (-3)8 ] a6 b8

$$\\\\\Rightarrow T_{9}=\frac{14!}{8!\;6!}\times 64\times 729\times a^{6}\;b^{8}\\\\$$ $$\\\\\Rightarrow T_{9}=\frac{14\times 13\times 12\times 11\times 10\times 9\times 8! }{6\times 5\times 4\times 3\times 2\times 1\times 8!}\times 64\times 729\times a^{6}\;b^{8}\\\\$$ $$\\\\\Rightarrow T_{9} = 3003\times 64\times 729\times a^{6}\;b^{8}=140107968\;\;a^{6}b^{8}\\\\$$

Therefore we get, the Coefficient of a6 b8 in the expansion of (2a – 3b)14 = 140107968

Q.3: Write the general term in the expansion of (x3 – y2)5

Answer. Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now we get, on substituting n = 5, a = x3 and b = y2 we will get:

Tr+1 = 5Cr × (x3)5 – r × (y2)r

Therefore we get, the general term in the expansion of (x3 – y2)5:

Tr+1 = 5Cr × (x)15 – 3 r × (y)2 r

Q.4: Write the general term in the expansion of (x4 – xy2)9

Answer.Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now we get, on substituting n = 9, a = x4 and b = xy2 we will get:

Tr+1 = 9Cr × (x4)9 – r × (xy2)r

$$\Rightarrow$$ Tr+1 = 9Cr × (x)36 – 4 r × (x)2 r × (y)2 r

$$\Rightarrow$$ Tr+1 = 9Cr × (x)36 – 4 r + 2 r × (y)2 r

$$\Rightarrow$$ Tr+1 = 9Cr × (x)36 – 2 r × (y)2 r

Therefore we get, the general term in the expansion of (x4 – xy2)9:

Tr+1 = 9Cr × (x)36 – 2 r × (y)2r

Q.5: Find the 4th term in the expansion of (2x + 3y)10

Answer.Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore we get, from the above equation the value of r should be 3; for finding out the values of 4th Term (T4)

Therefore we get, T3 + 1 = nC3 × (a)n – 3 × b3

Now we get, on substituting n =10, a = 2x and b = 3y we will get:

$$\\\Rightarrow$$ T3 + 1 = 10C3 × (2x)10 – 3 × (3y)3

$$\\\Rightarrow$$ T4 = 10C3 × (2x)7 × (3y)3

$$\\\\\Rightarrow T_{4}=\frac{10!}{3!\; 7!}\times 128\;x^{7}\times 27\;y^{3}\\\\$$ $$\\\\\Rightarrow T_{4}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\times 3456\;x^{7}y^{3}\\\\$$ $$\\\\\Rightarrow T_{4} = 120\times 3456\;x^{7}y^{3}=414720\;x^{7}y^{3}\\\\$$

Therefore we get, 4th term in the expansion of (2x + 3y)10 = 414720 x7y3

Q.6 Find the 11th term in the expansion of $$\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}$$

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore we get, from the above equation the value of r should be 10 for finding out the values of 11th Term (T11)

Therefore we get, T10 + 1 = nC10 × (a)n – 10 × b10

Now we get, on substituting n =15, a = 8x and b = $$\left (-\frac{1}{2\sqrt{x}} \right )$$ we will get:

$$\\\Rightarrow$$ T10 + 1 = 15C10 × (8x)15 – 10 × $$\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\$$

$$\\\Rightarrow$$ T11 = 15C10 × (8x)5 × $$\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\$$

$$\\\\\Rightarrow T_{11}=\frac{15!}{10!\; 5!}\times 8\times 8\times 8\times 8\times 8\times x^{5}\times \frac{1}{2^{10}}\times \left ( \frac{1}{\sqrt{x}} \right )^{10}\\\\$$ $$\\\\\\\Rightarrow T_{11}=\frac{15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times 1\times 10!}\times 32\times x^{5}\times \frac{1}{x^{5}}\\\\$$ $$\\\\\Rightarrow T_{11} = 3003\times 32\boldsymbol{=96096}\\\\$$

Therefore we get, 11th term in the expansion of $$\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}$$= 96096

Q.7: In the expansion of $$\left ( 5-\frac{x^{4}}{10} \right )^{9}$$, Find the middle terms.

Answer. Based on formula given in Binomial Theorem

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)n are given by:

middle terms = $$\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term$$

Therefore we get, the middle terms in the expansion of $$\left ( 5-\frac{x^{4}}{10} \right )^{9}$$are:

$$\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term$$

$$\Rightarrow$$ 5th term and 6th term

Now we get, 5th and 6th terms in the expansion of $$\left ( 5-\frac{x^{4}}{10} \right )^{9}$$ are:

T5 = T4 + 1 = 9C4 × (5)9 – 4 × $$\left ( -\frac{x^{4}}{10} \right )^{4}$$

$$\\\\\Rightarrow$$ T5 = 9C4 × (5)5 × $$\left ( -\frac{x^{4}}{10} \right )^{4}\\\\$$

$$\\\\\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (5)^{5}\times \left ( -\frac{x^{4}}{10} \right )^{4}\\\\$$ $$\\\\\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 5\times 5\times 5\times 5\times 5\times \frac{1}{10^{4}}\times x^{16}\\\\$$ $$\\\\\Rightarrow T_{5} = 126\times \frac{5}{16}\;x^{16} \boldsymbol{=\frac{315}{8}\;x^{16}}\\\\$$

Therefore we get, 5th term $$=\frac{315}{8}\;x^{16}$$

Now we get, T6 = T5 + 1 = 9C5 × (5)9 – 5 × $$\left ( -\frac{x^{4}}{10} \right )^{5}$$

$$\\\\\Rightarrow$$ T6 = 9C5 × (5)4 × $$\left ( -\frac{x^{4}}{10} \right )^{4}\\\\$$

$$\\\\\Rightarrow T_{6}=\frac{9!}{5!\; 4!}\times (5)^{4}\times \left ( -\frac{x^{4}}{10} \right )^{5}\\\\$$ $$\\\\\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 5\times 5\times 5\times 5\times \frac{-1}{10^{5}}\times x^{20}\\\\$$ $$\\\\\Rightarrow T_{6} = -126\times \frac{1}{320}\;x^{20}\boldsymbol{=\frac{-63}{160}\;x^{20}}\\\\$$

Therefore we get, 6th term $$=\frac{63}{160}\;x^{20}$$

Therefore we get, 5th and 6th terms are the middle terms in the expansion of $$\left ( 5-\frac{x^{4}}{10} \right )^{9}$$

And also, 5th term $$=\frac{315}{8}\;x^{16}$$ and 6th term $$=\frac{-63}{160}\;x^{20}$$

Q.8: In the expansion of $$\left ( \frac{x}{2}+8y \right )^{10}$$, Find the middle term.

Answer. Based on formula given in Binomial Theorem

Here, n = 10

When n is even, the middle term in the expansion of (a + b)n is given by:

$$\left ( \frac{n}{2}+1 \right )^{th}term$$

Therefore we get, the middle term in the expansion of $$\left ( \frac{x}{2}+8y \right )^{10}$$ is:

$$\Rightarrow \left ( \frac{10}{2}+1 \right )^{th}term$$ = 6th term

Now we get, 6th term in the expansion of $$\left ( \frac{x}{2}+8y \right )^{10}$$ is:

T6 = T5 + 1 = 10C5 × $$\left ( \frac{x}{2} \right )^{10-5}$$ × (8y)5

$$\\\Rightarrow$$ T6 = 10C5 × $$\left ( \frac{x}{2} \right )^{5}$$ × (8y)5

$$\\\Rightarrow T_{6}=\frac{10!}{5!\; 5!}\times \left ( \frac{x}{2} \right )^{5}\times (8y)^{5}\\$$ $$\\\Rightarrow T_{6}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{x^{5}}{32}\times 8\times 8\times 8\times 8\times 8\times y^{5}\\$$ $$\\\Rightarrow T_{6} = 252\times 1024\;x^{5}y^{5}\boldsymbol{=258048\;x^{5}y^{5}}\\$$

Therefore we get, 6th term = 258048 x5 y5

Hence, the middle term in the expansion of $$\left ( \frac{x}{2}+8y \right )^{10}$$ = 258048 x5 y5

Q.9: Prove that the coefficients of pa and pb are equal in the expansion of (1 + p)a + b

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

Now we get, on substituting n = (a + b), x =1 and y = p, we will get:

Tr + 1 = (a + b)Cr × (1)(a + b) – r × (p)r

$$\Rightarrow$$ Tr + 1 = (a + b)Cr × (p)r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of p in equation (1) with pa, we will get r = a

Therefore we get, from equation (1):

T a + 1 = (a + b)Ca × (p)a

$$\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;(a+b-a)!}\times p^{a}\\$$ $$\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;b!}\times p^{a}\\$$

Therefore we get, the coefficient of (p)a = $$\frac{(a+b)!}{a!\;b!}$$ . . . . . . . . . . (2)

Now we get, on comparing the coefficient of p in equation (1) with pb, we will get r = b

Therefore we get, from equation (1):

T b + 1 = (a + b)Cb × (p)b

$$\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;(a+b-b)!}\times p^{b}\\\\$$ $$\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;a!}\times p^{b}\\\\$$

Therefore we get, the coefficient of (p)b = $$\frac{(a+b)!}{b!\;a!}$$ . . . . . . . . . . (3)

Now we get, on comparing equation (2) and equation (3) we conclude that the coefficients of pa and pb are equal in the expansion of (1 + p)a + b.

Hence, Proved

Q.10: The coefficients of the (k + 1)th, kth and (k – 1)th terms in the expansion of (x + 1)n are in the ratio 5 : 3 : 1. Find the values of n and k.

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now we get on substituting a = x and b = 1 in the above equation:

Tr + 1 = nCr × (x)n – r × (1)r

Now we get, (k + 1)th term in the expansion of (x + 1)n :

T(k) + 1 = nCk × (x)n – k × 1(k )

i.e. T(k + 1) = nCk × (x)n – k

$$\\\Rightarrow T_{k+1} = \frac{n!}{k!\;(n-k)!}\times (x)^{n-k}\\$$

Therefore we get the coefficient of T(k + 1)th term = $$\frac{n!}{k!\;(n-k)!}\\\\$$

Now we get, (k)th term in the expansion of (x + 1)n :

T(k – 1) + 1 = nCk – 1 × (x)n – (k – 1) × 1(k – 1)

i.e. Tk = nCk – 1 × (x)n – k + 1

$$\\\\\Rightarrow T_{k} = \frac{n!}{(k-1)!\;(n-k+1)!}\times (x)^{n-k+1}\\\\$$

Therefore we get the coefficient of T(k)th term = $$\frac{n!}{(k-1)!\;(n-k+1)!}$$

And, (k – 1)th term in the expansion of (x + 1)n :

T(k – 2 )+ 1 = nCk – 2 × (x)n – (k – 2) × 1(k – 2)

i.e. Tk – 1 = nCk – 2 × (x)n – k + 2

$$\\\\\Rightarrow T_{k-1} = \frac{n!}{(k-2)!\;(n-k+2)!}\times (x)^{n-k+2}\\\\$$

Therefore we get the coefficient of T(k – 1)th term = $$\frac{n!}{(k-2)!\;(n-k+2)!}$$

Now we get, according to the given conditions: Coefficients of T(k + 1), Tk and Tk – 1 are in the ratio of 5 : 3 : 1

Therefore we get, $$\frac{T_{k-1}}{T_{k}}=\frac{1}{3}\\$$

$$\\\\\Rightarrow \left [ \frac{n!}{(k-2)!\;(n-k+2)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{1}{3}\\\\$$ $$\\\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{(k-2)!\;(n-k+2)!}=\frac{1}{3}\\\\$$ $$\\\\\Rightarrow \frac{(k-1)(k-2)!\;(n-k+1)!}{(k-2)!\;(n-k+2)(n-k+1)!}=\frac{1}{3}\\\\$$ $$\\\Rightarrow \frac{k-1}{n-k+2}=\frac{1}{3}\\$$ $$\\\Rightarrow 3k-3=n-k+2\\$$

(or) n – 4k + 5 = 0 . . . . . . . . . . . . . . . . (1)

And, $$\frac{T_{k+1}}{T_{k}}=\frac{5}{3}\\$$

$$\\\Rightarrow \left [ \frac{n!}{k!\;(n-k)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{5}{3}\\$$ $$\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{k!\;(n-k)!}=\frac{5}{3}\\$$ $$\\\Rightarrow \frac{(k-1)!\;[(n-k+1)\;(n-k)!]}{[k\;(k-1)!]\;(n-k)!}=\frac{5}{3}\\$$ $$\\\Rightarrow \frac{n-k+1}{k}=\frac{5}{3}\\$$ $$\\\Rightarrow 3n-3k+3=5k\\$$

(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)

On multiplying equation (1) by 3 and subtracting it to equation (2) we will get:

(3n – 8k + 3) – 3(n – 4k + 5) = 0

$$\\\Rightarrow$$ 3n – 8k + 3 – 3n + 12k – 15 = 0

$$\\\Rightarrow$$ 4k = 12

Therefore we get, k = 3

Now we get on substituting the value of k in equation (2) we will get:

3n – 8(3) + 3 = 0

Therefore we get, n = 7

Q.11: Prove that the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

For (1 + 3a)2p:

On putting n = 2p, x =1 and y = 3a, we will get:

Tr + 1 = (2p)Cr × (1)(3p) – r × (3a)r

$$\\\\\Rightarrow$$ Tr + 1 = (2p)Cr × (3a)r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of a in equation (1) with ap, we will get r = p

Therefore we get, from equation (1):

T p + 1 = (2p)Cp × (3a)p

$$\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;(2p-p)!}\times (3a)^{p}\\\\$$

$$\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;p!}\times 3^{p}\times (a)^{p}\\\\$$

Therefore we get, the coefficient of ap $$=\frac{(2p)!}{p!\;p!}\times 3^{p}$$ . . . . . . . . . . . . . . . . (2)

Now we get, for (1 + 3a)2p – 1 :

On substituting n = (2p – 1), x =1 and y = 3a, we will get:

Tr + 1 = (2p – 1)Cr × (1)(2p – 1) – r × (3a)r

$$\Rightarrow$$ Tr + 1 = (2p – 1)Cr × (3a)r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of ‘a’ in equation (1) with ‘ap ’, we will get r = p

Therefore we get, from equation (1):

T p + 1 = (2p – 1)Cp × (3a)p

$$\\\\\Rightarrow T_{p+1}=\frac{(2p-1)!}{p!\;(2p-1-p)!}\times (3a)^{p}\\\\$$ $$\\\\\Rightarrow T_{p+1}= \frac{(2p-1)!}{p!\;(p-1)!}\times (3)^{p}\times (a)^{p}\\\\$$ $$\\\\\Rightarrow T_{p+1}=\frac{\frac{(2p)!}{2p}}{p!\times \frac{p!}{p}}\times (3)^{p}.(a)^{p}=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}.a^{p}\\$$

Therefore we get, the coefficient of ap = $$=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}$$ . . . . . . . . . . (3)

Now we get, on comparing equation (2) and equation (3):

$$\\\\\Rightarrow \frac{(2p)!}{p!\;p!}\times 3^{p}=\frac{(2p)!}{p!\;p!}\times 3^{p}\\\\$$

$$\\\Rightarrow$$ (2p)Cp × 3p = $$\frac{1}{2}$$ (2p – 1)Cp × 3p

Therefore we get, the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

Q.12: For what values of ‘m’ the coefficient of x2 in the expansion of (1+ 2x)k is 140.

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × (b)r

Now we get, on substituting n = m, a =1 and b = 2x we will get:

Tr + 1 = mCr × (1)m – r × (2x)r

$$\Rightarrow$$ Tr + 1 = mCr × (2)r × (x)r . . . . . . . . . . . . (1)

Now we get, on comparing the coefficient of x in equation (1) with x2, we will get r = 2

Therefore we get, from equation (1):

T 2 + 1 = mC2 × (2)2 × (x)2

Since, the coefficient of x2 = 140 [GIVEN]

Therefore we get, 140 = mC2 × (2)2

$$\Rightarrow 84=\left [ \frac{m!}{2!\;\;(m-2)!} \right ]\times 4\\\\$$ $$\Rightarrow 21=\frac{m\times (m-1)\;\;(m-2)!}{[2\times 1]\times (m-2)!}\\\\$$ $$\Rightarrow 42 = m^{2}-m$$

Therefore we get, m2 – m – 42 = 0

Now we get, by splitting of middle term method, the roots of this quadratic equation are:

$$\Rightarrow$$m2 – (7 – 6)m – 42 = 0

$$\Rightarrow$$m2 – 7m + 6m – 42 = 0

$$\Rightarrow$$ m(m – 7) +6(m – 7) = 0

i.e. (m + 6) (m – 7) = 0

Therefore we get, m = 7 or m = -6

Hence, for the coefficient of x2 in the expansion of (1+ 2x)k to be 140, the value of ‘m’ should be 7.

Q.13: In the expansion of (2x+3y)9, Find the middle terms.

Answer. Based on formula given in Binomial Theorem

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)n are given by:

middle terms = $$\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term$$

Therefore we get, the middle terms in the expansion of (2x+3y)9 are:

$$\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term$$

$$\Rightarrow$$ 5th term and 6th term

Now we get, 5th and 6th terms in the expansion of (2x+3y)9 are:

T5 = T4 + 1 = 9C4 × (2x)9 – 4 × (3y)4

$$\Rightarrow$$ T5 = 9C4 × (2x)5 × (3y)4

$$\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (2x)^{5}\times (3y)^{4}$$ $$\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 32x^{5}\times 81y^{4}\\\\$$

$$\Rightarrow T_{5} = 126\times 2592\;x^{5}y^{4}\boldsymbol{=414720\;x^{5}y^{4}}\\$$

Therefore we get, 5th term = 414720 x5 y4

Now we get, T6 = T5 + 1 = 9C5 × (2x)9 – 5 × (3y)5

$$\Rightarrow$$ T6 = 9C5 × (2x)4 × (3y)5

$$\Rightarrow T_{6}=\frac{9!}{5!\times 4!}\times (2x)^{4}\times (3y)^{5}$$ $$\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 16x^{4}\times 243y^{5}\\\\$$ $$\Rightarrow T_{6} = 126\times 3888\;x^{4}y^{5}\boldsymbol{=489888\;x^{4}y^{5}}\\$$

Therefore we get, 6th term = 489888 x4 y5

Hence, 5th and 6th terms are the middle terms in the expansion of (2x+3y)9

And also, 5th term = 414720 x5 y4 and 6th term = 489888 x4 y5

Q.14: Find the Coefficient of x6 in expansion of (x + 3)11

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore we get, on substituting n = 11, a = x and b = 3 in the above expression we will get:

Tr + 1 = 11Cr × (x)11 – r × 3r . . . . . . . (1)

Now we get, on comparing the coefficient of x in equation (1) with x7, we will get:

i.e. (x)11 – r = x6

(or) 11 – r = 6

Therefore we get, r = 5

Now we get, on substituting r in equation (1) we will get:

T5 + 1 = 11C5 × (x)11 – 5 × 35

$$\Rightarrow T_{6}=\frac{11!}{5!\times 6!}\times x^{6}\times 3^{5}$$ $$\Rightarrow T_{6}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{5\times 4\times 3\times 2\times 1\times 6!}\times 243\;x^{6}\\\\$$

$$\Rightarrow T_{6} = 462\times 243\;x^{6}\boldsymbol{=112266\;x^{6}}$$

Therefore we get, the Coefficient of x6 in the expansion of (x + 3)11 = 112266

Q.15: Write the general term in the expansion of (x2y3 – x3 y2)7

Answer. Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now we get, on substituting n = 7, a = x2y3 and b = x3 y2 we will get:

Tr+1 = 7Cr × (x2y3)7 – r × (x3 y2)r

$$\Rightarrow$$ Tr + 1 = 7Cr × (x2)7 – r × (y3)7 – r × (x)3 r × (y)2 r

$$\Rightarrow$$ Tr + 1 = 7Cr × (x)14 – 2 r × (x)3 r× (y)2 r× (y)21 – 3r

$$\Rightarrow$$ Tr + 1 = 7Cr × (x)14 – 2 r + 3r × (y)21 – 3r + 2 r

i.e. Tr + 1 = 7Cr × (x)14 + r × (y)21 – r

Therefore we get, the general term in the expansion of (x2y3 – x3 y2)7:

Tr + 1 = 7Cr × (x)14 + r × (y)21 – r

Miscellaneous Exercise:

Q.1: If 1st term, 2nd term and 3rd term in the expansion of (a + b)n are 729, 7290 and 30375 respectively then find the value of ‘a’, ‘b’ and ‘c’.

Answer. Based on formula given in Binomial Theorem

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now we get, according to the given conditions:

T(0 + 1) = nC0 × (a)n – 0 × b0

T1 = an

i.e. an = 729 . . . . . . . . . . . . . . . . . . (1)

Also, T(1 + 1) = nC1 × (a)n – 1 × b1

T2 = nC1 × (a)n – 1 × b

i.e. 7290 = nC1 × (a)n – 1 × b

$$\\\Rightarrow 7290=\frac{n!}{1!\;\;(n-1)!}\times a^{n}\times a^{-1}\times b\\\\$$ $$\\\\\Rightarrow 7290=\frac{n(n-1)!}{(n-1)!}\times a^{n}\times \frac{b}{a}\;\;\;$$

Therefore we get, 7290 = n × an × $$\frac{b}{a}$$ . . . . . . . . . . . . . . . . (2)

And, T(2 + 1) = nC2 × (a)n – 2 × b2

T3 = nC2 × (a)n – 2 × b2

i.e. 30375 = nC2 × (a)n – 2 × b2

$$\\\Rightarrow 30375=\frac{n!}{2!\;\;(n-2)!}\times a^{n}\times a^{-2}\times b^{2}\\$$ $$\\\Rightarrow 30375=\frac{n\;(n-1)\;(n-2)!}{(2\times 1)\;\;(n-2)!}\times a^{n}\times \left ( \frac{b}{a} \right )^{2}\\$$

Therefore we get, 60750 = n (n – 1) × an × $$\left ( \frac{b}{a} \right )^{2}$$ . . . . . . . . . . . . . . . . (3)

Now we get, on substituting equation (1) in equation (2) we will get:

$$\\\Rightarrow 7290 = n\times (729)\times \frac{b}{a}\\$$ $$\\\Rightarrow \frac{10a}{b}=n\\$$

Therefore we get, n = $$\frac{10a}{b}$$ . . . . . . . . . . . . . . . . . (4)

Now we get, on dividing equation (2) and equation (3) we will get:

$$\\\Rightarrow \frac{n \times a^{n} \times \frac{b}{a}}{n\times (n-1)\times a^{n}\times \left ( \frac{b}{a} \right )^{2} }=\frac{7290}{60750}\\$$ $$\\\Rightarrow \frac{3}{25}=\frac{1}{(n-1)\times \frac{b}{a}}\\$$ $$\\\Rightarrow 3n-3=\frac{25a}{b}\\$$

Now we get, from equation (4):

$$\frac{a}{b}$$ = $$\frac{n}{10}$$

$$\Rightarrow 3n-3=25\times \left ( \frac{n}{10} \right )$$ $$\Rightarrow 30n-30=25n$$

i.e. 30n – 30 = 25n

Therefore we get, n = 6

Now we get, on substituting the value of ‘n’ in equation (1) we will get:

$$\Rightarrow$$ a6 = 729

i.e. a6 = 36

Therefore we get, a = 3

Now we get, from equation (4):

$$\\\Rightarrow$$ 6 = $$\frac{30}{b}\\$$

Therefore we get, b = 5

Q.2: If the coefficients of z2 and z3 are equal in the expansion of (3 + bz)9. Find the value of ‘b’

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × yr

Therefore we get, on substituting n = 9, x = 3 and y = bz in the above expression we will get:

Tr + 1 = 9Cr × (3)9 – r × (bz)r

Tr + 1 = 9Cr × (3)9 – r × br × zr . . . . . (1)

Now we get, on comparing the coefficient of z in equation (1) with z2, we will get: r = 2

On substituting the value of r in equation (1) we will get:

T(2 + 1) = 9C2 × (3)9 – 2 × b2 × z2

$$\\\Rightarrow T_{3}=\frac{9!}{2!\;7!}\times 3^{7}\times b^{2}\times z^{2}\\$$ $$\\\Rightarrow T_{3}=\frac{9\times 8\times 7!}{2\times 1\times 7!}\times 3^{7}\times b^{2}\times z^{2}\\$$

$$\\ \Rightarrow T_{3} = 36\times 3^{7}\times b^{2}\times z^{2}$$ . . . . . . . . . . . . (2)

Now we get, on comparing the coefficient of z in equation (1) with z3, we will get: r = 3

On substituting the value of ‘r’ in equation (1) we will get:

T(3 + 1) = 9C3 × (3)9 – 3 × b3 × z3

$$\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times 3^{6}\times b^{3}\times z^{3}\\$$ $$\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 3^{6}\times b^{3}\times z^{3}\\$$

$$\\\Rightarrow T_{4} = 84\times 3^{6}\times b^{3}\times z^{3} . . . . . . . . . (3)$$

Now we get, according to the given conditions:

Coefficient of z2 = Coefficient of z3

Therefore we get from equation (2) and equation (3):

$$\\\Rightarrow 36\times 3^{7}\times b^{2}=84\times 3^{6}\times b^{3}\\$$ $$\\\Rightarrow b = \frac{36\times 3^{7}}{84\times 3^{6}}\\$$

Therefore we get, the value of b = $$\frac{9}{7}$$

Q.3: Find the coefficient of x6 in the product of (1 + 3x)7 (1 – 2x)6 by using binomial theorem.

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (1 + x)n = [ nC0 ] + [ nC1 × (x) ] + [ nC2 × (x)2 ] + [ nC3 × (x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (x)n ]

Therefore we get, (1 + 3x)7 = [ 7C0 ] + [ 7C1 × (3x) ] + [ 7C2 × (3x)2 ] + [ 7C3 × (3x)3 ] + [ 7C4 × (3x)4 ] + [ 7C5 × (3x5) ] + [ 7C6 × (3x)6 ] + [ 7C7 × (3x)7 ]

$$\\\Rightarrow$$ 1 + [ 7 × (3x) ] + [ 21 × (9x2) ] + [ 35 × (27x3) ] + [ 35 × (81x4) ] + [ 21 × (243x5) ] + [ 7 × 729x6 ] + [ 1 × 2187x7 ]

$$\\\Rightarrow$$ 1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7

Therefore we get, (1 + 3x)7 = 1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7

Now we get, (1 – 2x)6:

By using Binomial Theorem:

(1 – 2x)6 = [ 6C0 ] – [ 6C1 × (2x) ] + [ 6C2 × (2x)2 ] – [ 6C3 × (2x)3 ] + [ 6C4 × (2x)4 ] – [ 6C5 × (2x5) ] + [ 6C6 × (2x)6 ]

$$\\\Rightarrow$$ 1 – [ 6 × (2x) ] + [ 15 × (4x2) ] – [ 20 × (8x3) ] + [ 15 × (16x4) ] – [ 6 × (32x5) ] + [ 1 × 64x6 ]

$$\\\Rightarrow$$ 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6

Therefore we get, (1 – 2x)6 = 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6

Now we get, (1 + 3x)7 (1 – 2x)6:

=(1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7) × (1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6)

Now we get, it not required to multiply the above equation, only terms with x5 are required and that can be identified by analyzing the above equation.

$$\\\Rightarrow$$ [{1 × (- 192x5)} + {(21x) × (240x4)} + {(189x2) × (- 160x3)} + {(945x3) × (60x2)} + {(2835x4) × (- 12x)} + {(5103x5) × (1)}]

$$\\\Rightarrow$$ [ -192x5 + 5040x5 – 30240x5+ 56700x5 – 34020x5 + 5103x5 ]

$$\\\Rightarrow$$ 2391 x5

Therefore we get, the coefficient of x5 = 2391

Q.4: Show that (a – b) is a factor of (an – bn), where ‘n’ is a positive integer and ‘a’ and ‘b’ are distinct integers.

Answer. Based on formula given in Binomial Theorem

Now we get, for (a – b) to be factor of (an – bn), (an – bn) = p(a – b) [where p is any natural number]

We can write an = (a – b + b)n = [(a – b) + b]n

Now we get, by Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, for x = (a – b) and y = b the equation becomes:

[(a – b) + b]n = [ nC0 × (a – b)n ] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 × (a – b)n – 2 × b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × (a – b) × bn – 1 ] + [ nCn × bn ]

$$\\\Rightarrow$$ [a]n = [(a – b)n] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 × (a – b)n – 2 × b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × (a – b) × bn – 1 ] + [ bn ]

Now we get, an – bn = [(a – b)n] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 × (a – b)n – 2 × b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × (a – b) × bn – 1 ] + [ bn ] – [ bn ]

$$\\\Rightarrow$$ an – bn = (a – b) × [ (a – b)n – 1 + nC1 (a – b)n – 2 b + nC2 (a – b)n – 3 b2 + nC3 (a – b)n – 4 b3 + . . . . . . . . . . . . . + nCn – 1 bn – 1 ]

$$\\\Rightarrow$$ (an – bn) = (a – b) × p

Where, p = [ (a – b)n – 1 + nC1 (a – b)n – 2 b + nC2 (a – b)n – 3 b2 + nC3 (a – b)n – 4 b3 + . . . . . . . . . . . . . + nCn – 1 bn – 1 ] is any natural number.

Therefore we get, (a – b) is a factor of (an – bn), where ‘n’ is a positive integer.

Q.5: Evaluate $$\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}$$

Answer. Based on formula given in Binomial Theorem

Find (a + b)6 – (a – b)6

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, [a + b]6= [ 6C0 × (a6) ] + [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] + [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] + [ 6C5 × a × b5 ] + [6C6 × b6]

And, [a – b]6 = [ 6C0 × (a6) ] – [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] – [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] – [ 6C5 × a × b5 ] + [6C6 × b6]

Therefore we get, (a + b)6 – (a – b)6 = [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b6] – [ 6C0 a66C1 a5b + 6C2 a4b2 6C3 a3b3 + 6C4 a2b4 6C5 ab5 + 6C6 b6 ]

$$\\\Rightarrow$$ [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b66C0 a6 + 6C1 a5b – 6C2 a4b2 + 6C3 a3b3 6C4 a2b4 + 6C5 ab56C6 b6 ]

$$\\\Rightarrow$$ 2[ 6C1 a5b + 6C3 a3b3 + 6C5 ab5 ] = 2[ 6a5b + 20a3b3 + 6ab5 ]

Therefore we get, (a + b)6 – (a – b)6 = 4ab[ 3a4 + 10a2 b2 + 3b4 ] . . . . . . (1)

Now we get, on substituting a = $$\sqrt{7}$$ and b = $$\sqrt{5}$$ in equation (1) we will get:

$$\\\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}:\\$$

$$\\=4( \sqrt{7}\times \sqrt{5} )\left [ (3 \sqrt{7} )^{4} +10\times (\sqrt{7} )^{2}\times ( \sqrt{5})^{2}+3(\sqrt{5})^{4} \right ]\\$$

$$\\\Rightarrow 4\sqrt{35}\;[\;147+350+75\;]\\$$

Therefore we get, $$\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}$$= $$2288\sqrt{35}$$

Q.6: Find the expansion of $$\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}$$

Answer. Based on formula given in Binomial Theorem

The above given expression can be assumed as: (x + y)4 + (x – y)4

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, [x + y]4 = [4C0 × (x4)] + [4C1 × (x3) × (y)] + [4C2 × (x2) × (y)2] + [4C3 × (x1) × (y)3] + [4C4 × (x0) × (y)4]

And, [x – y]4 = [4C0 × (x4) ] – [4C1 × (x3) × (y) ] + [4C2 × (x2) × (y)2 ] – [4C3 × (x1) × (y)3 ] + [4C4 × (x0) × (y)4 ]

Therefore we get, (x + y)4 + (x – y)4 = [ 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 ] + [ 4C0 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4 y4 ]

$$\\\Rightarrow$$ [ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x44C1 x3y + 4C2 x2y24C3 xy3 + 4C4 y4 ]

$$\\\Rightarrow$$ 2[ 4C0 x4 + 4C2 x2 y2 + 4C4 y4] = 2[ x4 + 6x2 y2 + y4 ]

Therefore we get, (x + y)4 + (x – y)4 = 2[ x4 + y4 + 6 x2 y2 ] . . . . . . . . . . . . . . . . . . . (1)

Now we get, on substituting x = a3 and y = $$\sqrt{a^{3}-2}$$ in equation (1) we will get:

$$\\\Rightarrow \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}:\\$$

$$\\=2\times \left [ (a^{3})^{4}+ (\sqrt{a^{3}-2})^{4}+[6\times (a^{3})^{2}\times (\sqrt{a^{3}-2})^{2}] \right ]\\$$

$$\\=2\left [ a^{12}+ (a^{3}-2)^{2}+[6\times a^{6}\times (a^{3}-2)] \right ]\\$$

$$\\=2\left [\; a^{12}+ a^{6}+4-4a^{3}+6a^{9}-12a^{6} \;\right ]\\$$

$$\\\Rightarrow 2a^{12}+12a^{9}-22a^{6}-8a^{3}+8$$

Therefore we get, the expansion of $$\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}$$:

= $$2a^{12}+12a^{9}-22a^{6}-8a^{3}+8$$

Q.7: Find the approximate value of (0.98)5 using the first four terms of its expansion.

Answer. Based on formula given in Binomial Theorem

(0.98)5 = (1 – 0.02)5

Now we get, by using Binomial Theorem:

Since, [ x – y ]n = [ nC0 × (x)n ] – [ nC1 × (x)n – 1 × y ] + [ nC2 × (x)n – 2 × y2 ] – [ nC3 × (x)n – 3 × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore we get, (1 – 0.02)5 = [ 5C0 × (1)5 ] – [ 5C1 × (1)4 × (0.02) ] + [ 5C2 × (1)3 × (0.02)2 ] – [ 5C3 × (1)2 × (0.02)3 ]

= 1 – [5 × 1 × 0.02] + [5 × 1 × 0.0004] – [5 × 1 × 0.000008]

= 1 – 0.10 + 0.0020 + 0.000040

= 0.90204

Therefore we get, (0.98)5 = 0.90204

Q.8: The coefficient of the 5th term from end and the 5th term from the beginning in the expansion of $$\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}$$ are in the ratio $$1:\sqrt{6}$$.Find the value of ‘n’.

Answer. Based on formula given in Binomial Theorem

Since, The general term in the expansion of (a + b)n is given by: Tk + 1 = nCk × (a)n – k × bk

Now we get on substituting a = $$\left ( 2\right )^{{\frac{1}{4}}}$$ and b = $$\frac{1}{(3)^{\frac{1}{4}}}$$ in the above equation:

Tk + 1 = nCk ×$$\left ( \sqrt[4]{2} \right )^{n-k} \times \left ( \frac{1}{\sqrt[4]{3}}\right )^{k}\\$$

Now we get, 5th term from beginning in the expansion of $$\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}$$:

T(4 + 1) = nC4 × $$\left ( \sqrt[4]{2} \right )^{n-4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}\\$$

i.e. T5 = nC4 × $$\left ( 2 \right )^{\frac{n}{4}-1} \times \left ( \frac{1}{3} \right )\\$$

$$\\\Rightarrow T_{5} = \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times (2)^{-1}\times \frac{1}{3}\\$$

Therefore we get the coefficient of 5th term from beginning $$\boldsymbol{= \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6}}\\$$

Now we get, 5th term from end in the expansion of $$\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}$$:

1st term, 2nd term, 3rd term . . . . . . . . . . . (n – 5)th term, (n – 4)th term, (n – 3)th term, (n – 2)th term, (n – 1)th term, nth term.

Therefore we get, 5th term from end will be (n – 4)th term from beginning.

i.e. T(n – 4) + 1 = nCn – 4 × $$\left ( \sqrt[4]{2} \right )^{n-(n-4)} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}\\$$

i.e. T(n – 3) = nCn – 4 × $$\left ( \sqrt[4]{2} \right )^{4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n}\times \left ( \frac{1}{\sqrt[4]{3}} \right )^{-4}\\$$

$$\\\Rightarrow T_{(n-3)} = \frac{n!}{(n-4)!\;[(n-(n-4)]!}\times 2 \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 3\\$$

Therefore we get the coefficient of 5th term from end [ T(n – 4)th term ] $$= \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6\\$$

Now we get, according to the given conditions:

The ratio of coefficient of 5th term from end and the coefficient of 5th term from beginning is $$1:\sqrt{6}\\$$

Therefore we get, $$\\ \frac{T_{n-3}}{T_{5}}=\frac{1}{\sqrt{6}}\\$$

$$\\\Rightarrow \left ( \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6 \right )\div\left ( \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6} \right )=\frac{1}{\sqrt{6}}\\$$ $$\\\Rightarrow \frac{4!\;(n-4)!}{(n-4)!\;4!}\times \frac{1}{(2)^{\frac{n}{4}}}\times \frac{1}{(3)^{\frac{n}{4}}}\times 6\times 6=\frac{1}{\sqrt{6}}\\$$ $$\\\Rightarrow \frac{1}{(6)^{\frac{n}{4}}}\times 36=\frac{1}{\sqrt{6}}\\$$ $$\\\Rightarrow 36\sqrt{6}=(6)^{\frac{n}{4}}\\$$ $$\\\Rightarrow (6)^{2}\times (6)^{\frac{1}{2}}=6^{\frac{n}{4}}\\$$ $$\\\Rightarrow (6)^{\frac{5}{2}}=(6)^{\frac{n}{4}\\}$$

On comparing RHS and LHS:

$$\\\Rightarrow \frac{5}{2}=\frac{n}{4}\\$$

$$\\\Rightarrow n = 10$$

Therefore we get, the value of n = 10.

Q.9: By using the Binomial Theorem, find the expansion of $$\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}$$; where 0

Answer. Based on formula given in Binomial Theorem

The given expression $$\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}$$ can be expanded as $$\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}$$

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, $$\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}$$ = [4C0 × $$\left (1-\frac{3}{x} \right )^{4}$$] + [4C1 × $$\left (1-\frac{3}{x} \right )^{3}$$ × $$\left ( \frac{x}{3} \right )$$] + [4C2 × $$\left (1-\frac{3}{x} \right )^{2}$$ × $$\left ( \frac{x}{3} \right )^{2}$$] + [4C3 × $$\left (1-\frac{3}{x} \right )^{1}$$ × $$\left ( \frac{x}{3} \right )^{3}$$ + [4C4 × $$\left ( \frac{x}{3} \right )^{4}$$]

$$\\=\left [ 1\times \left (1-\frac{3}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{3}\times \left ( \frac{x}{3} \right ) \right ]+\left [ 6\times \left (1-\frac{3}{x} \right )^{2} \times \left ( \frac{x}{3} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{1} \times \left ( \frac{x}{3} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{3} \right )^{4} \right ]$$ . . . . . . . . . . . . . . . . . . . (1)

Now we get, $$\left (1-\frac{3}{x} \right )^{4}$$:

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore we get, $$\left (1-\frac{3}{x} \right )^{4}$$ = [4C0] – [4C1 × $$\left ( \frac{3}{x} \right )^{1}$$] + [4C2 × $$\left ( \frac{3}{x} \right )^{2}$$] – [4C3 × $$\left ( \frac{3}{x} \right )^{3}$$] + [4C4 × $$\left ( \frac{3}{x} \right )^{4}$$]

$$\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{3}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{27}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{81}{x^{4}} \right ) \right ]\\$$

$$\\=\left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}} \right ]$$ . . . . . . . . (2)

And, $$\left (1-\frac{3}{x} \right )^{3}$$ = [ 3C0 ] – [ 3C1 × $$\left ( \frac{3}{x} \right )^{1}$$] + [3C2 × $$\left ( \frac{3}{x} \right )^{2}$$] – [3C3 × $$\left ( \frac{3}{x} \right )^{3}$$]

$$\\= [1]-\left [ 3\times \left ( \frac{3}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{27}{x^{3}} \right ) \right ]\\$$

$$\\=\left [1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right ]$$ . . . . . . . . . . . . (3)

Now we get, from equation (1), equation (2) and equation (3):

$$\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ 4 \times \left ( 1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right )\times \frac{x}{3} \right ]+\left [ 6\times \left (1+\frac{9}{x^{2}}-\frac{6}{x} \right )\times \frac{x^{2}}{9}\right ] +\left [ \left ( 4-\frac{12}{x} \right )\times \frac{x^{3}}{27} \right ]+\left [\frac{x^{4}}{81}\right ]\\$$ $$\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ \frac{4x}{3}-12+\frac{36}{x}-\frac{36}{x^{2}}\right ]+\\\\\\\\+\left [ \frac{2x^{2}}{3}+6-4x\right ] +\left [ \frac{4x^{3}}{27}-\frac{4x^{2}}{9} \right ]+\left [\frac{x^{4}}{81}\right ]\\$$ $$\\\boldsymbol{\Rightarrow \left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\\$$

Therefore we get, the expansion of $$\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}$$:

= $$\boldsymbol{\left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}$$

Q.10: By using the Binomial Theorem, find the expansion of (6x2 – 4ax + 6a2)3.

Answer. Based on formula given in Binomial Theorem

The given expression (6x2 – 4ax + 6a2)3 can be expanded as [(6x2 – 4ax) + 6a2)3]

By using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, [(6x2 – 4ax) + 6a2)3] = [3C0 × (6x2 – 4ax)3] + [3C1 × (6x2 – 4ax)2 × 6a2] + [3C2 × (6x2 – 4ax)1 × (6a2)2] + [3C3 × (6x2 – 4ax)0 × (6a2)3]

= [1× (6x2 – 4ax)3] + [3 × (6x2 – 4ax)2 × 6a2] + [3 × (6x2 – 4ax) × (6a2)2] + [1 × (6a2)3]

= [(6x2 – 4ax)3] + [(6x2 – 4ax)2 × 18a2] + [(6x2 – 4ax) × 108a4 ] + [216a6] . . . . . . . . . . . . (1)

Now we get, [(6x2 – 4ax)3]:

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore we get, [(6x2 – 4ax)3] = [3C0 × (6x2)3] – [3C1 × (6x2)2) × 4ax] + [3C2 × 6x2 × (4ax)2] – [3C3 × (4ax)3]

= [1 × (216x6)] – [3 × (36x4) × 4ax] + [3 × (6x2 16a2x2] – [1 × (64a3x3)]

$$\Rightarrow$$ [(6x2 – 4ax)3] = [216x6 – 432 ax5 + 288 a2x4 – 64a3x3] . . . . . . . . . . (2)

From equation (1) and equation (2)

=[216x6 – 432 ax5 + 288 a2x4 – 64a3x3] + [(36x4 + 16a2x2 – 48ax3) × 18a2] + [(6x2 – 4ax) × 108a4 ] + [ 216a6 ]

=[216x6 – 432 ax5 + 288 a2x4 – 64a3x3 + 648a2x4 + 288a4x2 – 864a3x3 + 648x2a4 – 432xa5 + 216a6]

=8[27x6 – 54 ax5 + 36 a2x4 – 8a3x3 + 81a2x4 + 36a4x2 – 108a3x3 + 81x2a4 – 54xa5 + 27a6]

=8[27x6 – 54 ax5 + 117a2 x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6]

Therefore we get, the expansion of (6x2 – 4ax + 6a2)3 :

= 8[27x6 – 54 ax5 + 117a2 x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6]

Q.11: By using the Binomial Theorem, find the expansion of $$\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}$$; where 0

Answer. Based on formula given in Binomial Theorem

The given expression $$\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}$$ can be expanded as $$\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}$$

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, $$\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}$$

= [4C0 × $$\left (1-\frac{5}{x} \right )^{4}$$] + [4C1 × $$\left (1-\frac{5}{x} \right )^{3}$$ × $$\left ( \frac{x}{5} \right )$$] + [4C2 × $$\left (1-\frac{5}{x} \right )^{2}$$ × $$\left ( \frac{x}{5} \right )^{2}$$] + [4C3 × $$\left (1-\frac{5}{x} \right )^{1}$$ × $$\left ( \frac{x}{5} \right )^{3}$$ + [4C4 × $$\left ( \frac{x}{5} \right )^{4}$$]

$$\\=\left [ 1\times \left (1-\frac{5}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{3}\times \left ( \frac{x}{5} \right ) \right ]+\left [ 6\times \left (1-\frac{5}{x} \right )^{2} \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{1} \times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{5} \right )^{4} \right ]$$ . . . . . . . . . . . . . . . . . . . (1)

Now we get, $$\left (1-\frac{5}{x} \right )^{4}:\\$$

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore we get, $$\left (1-\frac{5}{x} \right )^{4}$$ = [4C0] – [4C1 × $$\left ( \frac{5}{x} \right )^{1}$$] + [4C2 × $$\left ( \frac{5}{x} \right )^{2}$$] – [4C3 × $$\left ( \frac{5}{x} \right )^{3}$$] + [4C4 × $$\left ( \frac{5}{x} \right )^{4}$$]

$$\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{5}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{125}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{625}{x^{4}} \right ) \right ]\\$$

$$\\=\left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]$$ . . . . . . . . (2)

And, $$\left (1-\frac{5}{x} \right )^{3}:\\$$

[ 3C0 ] – [ 3C1 × $$\left ( \frac{5}{x} \right )^{1}$$] + [3C2 × $$\left ( \frac{5}{x} \right )^{2}$$] – [3C3 × $$\left ( \frac{5}{x} \right )^{3}$$]

$$\\= [1]-\left [ 3\times \left ( \frac{5}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{125}{x^{3}} \right ) \right ]\\$$

$$\\=\left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ]$$ . . . . . . . . . . . . . . . . . . . (3)

Now we get, from equation (1), equation (2) and equation (3):

$$\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ 4 \times \left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ] \times \left ( \frac{x}{5} \right ) \right ]+\\\\\\+\left [ 6\times \left (1+\frac{25}{x^{2}} -\frac{10}{x}\right ) \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )\times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ \left ( \frac{x}{5} \right )^{4} \right ]\\$$ $$\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ \frac{4x}{5}-12+\frac{60}{x}-\frac{100}{x^{2}} \right ]+\left [ \frac{6x^{2}}{25}+6-\frac{12x}{5}\right ]+\left [ \frac{4x^{3}}{125}-\frac{x^{2}}{25} \right ]+\left [ \frac{x^{4}}{625} \right ]\\$$ $$\\\\\Rightarrow \left [\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}-\frac{8x}{5}-5+\frac{40}{x}+ \frac{x^{2}}{5}+\frac{4x^{3}}{125}+\frac{x^{4}}{625} \right ]\\\\$$

Therefore we get, the expansion of $$\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}$$:

= $$\left [\frac{x^{4}}{625}+\frac{4x^{3}}{125}+ \frac{x^{2}}{5}-\frac{8x}{5}-5+\frac{40}{x}+\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}\right ]$$

Q.12: By using the Binomial Theorem, find the expansion of (5x3 – 2ax + 3a3)3.

Answer. Based on formula given in Binomial Theorem

The given expression (5x3 – 2ax + 3a3)3 can be expanded as [(5x3 – 2ax) + 3a3)3]

By using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore we get, [ (5x3 – 2ax) + 3a3)3 ] = [ 3C0 × (5x3 – 2ax)3 ] + [ 3C1 × (5x3 – 2ax)2 × 3a3 ] + [ 3C2 × (5x3 – 2ax)1 × (3a3)2 ] + [ 3C3 × (5x3 – 2ax)0 × (3a3)3 ]

= [1× (5x3 – 2ax)3] + [3 × (5x3 – 2ax)2 × 3a3] + [3 × (5x3 – 2ax) × (3a3)2] + [1 × (3a3)3]

= [(5x3 – 2ax)3] + [(5x3 – 2ax)2 × 9a3] + [(5x3 – 2ax) × 27a6] + [27a9] . . . . . . . . . . . . (1)

Now we get, [(5x3 – 2ax)3]:

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore we get, [(5x2 – 2ax)3] = [3C0 × (5x3)3] – [3C1 × (5x3)2) × 2ax] + [3C2 × 5x3 × (2ax)2] – [3C3 × (2ax)3]

= [1 × 125x9] – [3 × (25x6)× 2ax] + [3 × (5x3) × 4a2x2] – [1 × 8a3x3]

[(5x3 – 2ax)3] = [125x6 – 150ax5 + 60a2x4 – 8a3x3] . . . . . . . . . . (2)

From equation (1) and equation (2):

=[125x6 – 150ax5 + 60a2x4 – 8a3x3] + [(25x6 + 4a2x2 – 20ax4) × 9a3] + [(5x3 – 2ax) × 27a6 ] + [ 27a9 ]

=[125x6 – 150ax5 + 60a2x4 – 8a3x3] + [225a3x6 + 36a5x2 – 180a4x4] + [135a6x3 – 54a7x ] + [ 27a9 ]

=[125x6 – 150ax5 + 60a2x4 – 8a3x3 + 225a3x6 + 36a5x2 – 180a4x4 + 135a6x3 – 54a7x + 27a9 ]

=[125x6 + 225a3x6 – 150ax5 + 60a2x4 – 180a4x4 – 8a3x3 + 135a6x3 + 36a5x2 – 54a7x + 27a9 ]

Therefore we get, the expansion of (5x3 – 2ax + 3a3)3 :

=[125x6 + 225a3x6 – 150ax5 + 60a2x4 – 180a4x4 – 8a3x3 + 135a6x3 + 36a5x2 – 54a7x + 27a9]

Q.13: Find the coefficient of x7 in the product of (1 + x)7 (1 + 3x)6 by using binomial theorem.

Answer. Based on formula given in Binomial Theorem

By using Binomial Theorem:

Since, (1 + x)n = [ nC0 ] + [ nC1 × (x) ] + [ nC2 × (x)2 ] + [ nC3 × (x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (x)n ]

Therefore we get, (1 + x)7 = [ 7C0 ] + [ 7C1 × (x) ] + [ 7C2 × (x)2 ] + [ 7C3 × (x)3 ] + [ 7C4 × (x)4 ] + [ 7C5 × (x5) ] + [ 7C6 × (x)6 ] + [ 7C7 × (x)7

$$\\\Rightarrow$$ 1 + [ 7x ] + [ 21x2 ] + [ 35x3 ] + [ 35x4 ] + [ 21x5 ] + [ 7x6 ] + [ x7 ]

Therefore we get, (1 + x)7 = 1+7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7

Now we get, (1 + 3x)6:

By using Binomial Theorem:

(1 + 3x)6 = [ 6C0 ] + [ 6C1 × (3x) ] + [ 6C2 × (3x)2 ] + [ 6C3 × (3x)3 ] + [ 6C4 × (3x)4 ] + [ 6C5 × (3x5) ] + [ 6C6 × (3x)6 ]

$$\\\Rightarrow$$ 1 + [ 6 × (3x) ] + [ 15 × (9x2) ] + [ 20 × (27x3) ] + [ 15 × (81x4) ] + [ 6 × (243x5) ] + [ 1 × 729x6 ]

$$\\\Rightarrow$$ 1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6

Therefore we get, (1 + 3x)6 = 1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6

Now we get, (1 + x)7 (1 + 3x)6:

(1+7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7) × (1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6)

Now we get, it not required to multiply the above equation, only terms with x5 are required and that can be identified by analysing the above equation.

$$\\\Rightarrow$$ [{(7x) × (729x6)} + {(21x2) × (1458x5)} + {(35x3) × (1215x4)} + {(35x4) × (540x3)} + {(21x5) × (135x2)} + {(7x6) × (18x)} + {(x7) × (1)}]

$$\\\Rightarrow$$ [5103x5 + 30618x5 + 42525x5+ 18900x5 + 2835x7 + 126x7 + x7]

$$\\\Rightarrow\\$$ 72551 x7

Therefore we get, the coefficient of x7 = 72551

Q.14: Evaluate $$\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}$$

Answer. Based on formula given in Binomial Theorem

Find (a + b)5 – (a – b)5

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, [a + b]5 = [ 5C0 × (a5) ] + [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] + [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] + [ 5C5 × b5 ]

And, [a – b]5 = [ 5C0 × (a5) ] – [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] – [ 5C3 × (a2) × b3 ] + [ 5C4 × (a) × b4 ] – [ 5C5 × b5 ]

Therefore we get, (a + b)5 – (a – b)5 = [5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5] – [5C0 a55C1 a4 b + 5C2 a3b2 5C3 a2b3 + 5C4 ab4 5C5 ab5]

$$\\\Rightarrow$$ [5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b55C0 a5 + 5C1 a4 b – 5C2 a3b2 + 5C3 a2b3 5C4 ab4 + 5C5 ab5]

$$\\\Rightarrow$$ 2[ 5C1 a4b + 5C3 a2b3 + 5C5 b5 ] = 2[ 5a4b + 10a2b3 + b5 ]

Therefore we get, (a + b)5 – (a – b)5 = 2b[ 5a4 + 10a2 b2 + b4 ] . . . . . . (1)

Now we get, on substituting a = $$\sqrt{3}$$ and b = $$\sqrt{7}$$ in equation (1) we will get:

$$\\\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}:\\$$

$$\\=2(\sqrt{7})\times \left [ 5( \sqrt{3})^{4} +10\times ( \sqrt{3})^{2}\times (\sqrt{7})^{2}+(\sqrt{7})^{4} \right ]\\$$

$$\\\Rightarrow 2\sqrt{7}\left [\;45+210+49\; \right ]\\$$

Therefore we get, $$\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}$$= $$608\sqrt{7}$$

Q.15: Find the expansion of $$\left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}$$

Answer. Based on formula given in Binomial Theorem

The above given expression can be assumed as: (x + y)4 + (x – y)4

Now we get, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]

Therefore we get, [x + y]4 = [4C0 × (x4)] + [4C1 × (x3) × (y)] + [4C2 × (x2) × (y)2] + [4C3 × (x1) × (y)3] + [4C4 × (x0) × (y)4]

And, [x – y]4 = [4C0 × (x4) ] – [4C1 × (x3) × (y) ] + [4C2 × (x2) × (y)2 ] – [4C3 × (x1) × (y)3 ] + [4C4 × (x0) × (y)4 ]

Therefore we get, (x + y)4 + (x – y)4 = [ 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 ] + [ 4C0 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4 y4 ]

$$\\\Rightarrow$$ [ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x44C1 x3y + 4C2 x2y24C3 xy3 + 4C4 y4 ]

$$\\\Rightarrow$$ 2[ 4C0 x4 + 4C2 x2 y2 + 4C4 y4] = 2[ x4 + 6x2 y2 + y4 ]

Therefore we get, (x + y)4 + (x – y)4 = 2[ x4 + y4 + 6 x2 y2 ] . . . . . . . . . . . . . . . . . . . (1)

Now we get, on substituting x = a2 and y = $$\sqrt{a^{2}-5}$$ in equation (1) we will get:

$$\\\Rightarrow \left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}:\\$$

$$\\=2\times \left [ (a^{2})^{4}+ \left ( \sqrt{a^{2}-5} \right )^{4}+6\times (a^{2})^{2}\times \left ( \sqrt{a^{2}-5} \right )^{2} \right ]\\$$ $$\\\Rightarrow 2\left [ a^{8}+ (a^{2}-5)^{2}+[6a^{4}\times (a^{2}-5)] \right ]\\$$ $$\\\Rightarrow 2 \left [ a^{8}+a^{4}+ 25 + 10a^{2}+6a^{6}-30a^{4} \right ]\\$$

Therefore we get, the expansion of $$\left [ a^{2} +\sqrt{a^{2}-5}\right ]^{4}+ \left [ a^{2} -\sqrt{a^{2}-5}\right ]^{4}:$$

=$$2 \left [ a^{8}+6a^{6}-29a^{4}+10a^{2}+25 \right ]$$