Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f. Then f is continuous at c if
\( \lim_{x \rightarrow c} f(x) = f(c) \)
More elaborately, if the left hand limit, right hand limit and the value of the function at x = c exist and equal to each other, then f is said to be continuous at x = c.
if the right hand and left hand limits at x = c coincide, then we say that the common value is the limit of the function at x = c.
continuity as follows: a function is continuous at x = c if the function is defined at x = c and if the value of the function at x = c equals the limit of the function at x = c. If f is not continuous at c, we say f is discontinuous at c and c is called a point of discontinuity of f.
Q. Check the continuity of the function f given by f (x) = 2x + 3 at x = 1.
First note that the function is defined at the given point x = 1 and its value is 5. Then find the limit of the function at x = 1. Clearly
\( \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} (2x+3) \)
= 2(1) + 3 = 5
Thus, \( \lim_{x \rightarrow 1} f(x) = 5 = f(1) \)
Q. Examine whether the function f given by f (x) = x2 is continuous at x = 0.
First note that the function is defined at the given point x = 0 and its value is 0. Then find the limit of the function at x = 0. Clearly
\( \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} x^2 = 0^2 = 0 \)
\( \lim_{x \rightarrow 0} f(x) = 0 = f(0) \)
Hence, f is continuous at x = 0.
Q. Discuss the continuity of the function f given by f(x) = | x | at x = 0.
f(x) = \( \begin{cases}-x & if x < 0\\ x & if x \ge 0\end{cases} \)
Clearly the function is defined at 0 and f (0) = 0. Left hand limit of f at 0 is
\( \lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (-x) = 0 \)
Similarly, the right hand limit of f at 0 is : \( \lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} (+x) = 0 \)
Thus, the left hand limit, right hand limit and the value of the function coincide at x = 0. Hence, f is continuous at x = 0.
Q. Check the points where the constant function f (x) = k is continuous.
The function is defined at all real numbers and by definition, its value at any real number equals k. Let c be any real number. Then
\( \lim_{x \rightarrow c} f(x) = \lim_{x \rightarrow c} k = k \)
Since f (c) = k = \( \lim_{x \rightarrow c} f(x)\) for any real number c, the function f is continuous at every real number.
Exercise 1
Q1: Prove that the function \(f(x)=5x-3\) is continuous
at \(x=0\), at \(x=-3\) and at \(x=5\).
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=5x-3\)
At \(x=0\), \(f(0)=5\times 0-3=-3\)
\(\lim_{x \to 0 } f(x)=\lim_{x \to 0 }(5x-3)=5\times 0-3=-3\) \(therefore,\; \lim_{x \to 0 } f(x)=f(0)\)So, f is continuous at x=0
At \(x=-3, f(-3)=5\times (-3)-3=-18\)
\(\lim_{x \to -3} f(x)=\lim_{x \to -3 }(5x-3)-3=-18\) \(therefore,\; \lim_{x \to -3} f(x)=f(-3)\)So , f is continuous at x=-3
At x=5,f(x)=f(5)= (5×5) – 3=25-3=22
\(\lim_{x \to 5} f(x)=\lim_{x \to 5}(5x-3)=5\times 5-3=22\) \(therefore,\; \lim_{x \to 0}f(x)=f(5)\)So , f is continuous at x=5
Q2 : Examine the continuity of the function \(f(x)=2x_{2}- 1\) at \(x=3\).
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=2x_{2}-1\)
At \(x=3\), f(x)=f(3) =2x 3^{^{2}}-1=17
\(\lim_{x \to 3}f(x)=\lim_{x \to 3}(2x^{2}-1) =2\times 3^{^{2}}-1=17 \) \(therefore,\; \lim_{x \to 3}f(x)=f(3)\)Thus, f is continuous at \(x=3\) hence solved
Q3: Examine the following function for continuity.
Sol:Based on formulae given in Continuity and Differentiability
It is evident that \(f\) is defined at every real number
\(k\) is k-5.
It is also observed that \(\lim_{x \to k} f(x)=\lim_{x \to k}(x-5)=k-5=f(k)\).
\(therefore,\; \lim_{x \to k} f(x)=f(k)\)Hence, \(f\) is continuous at every real number and
therefore, it is continuous function.
(b) The given function is \(f(x)=\frac{1}{x-5}, x\neq 5\)
For any real number k \(\neq 5\), we obtain
\(\lim_{x \to k} f(x)=\lim_{x \to k}\frac{1}{x-5}=\frac{1}{k-5}\)Also, \(f(k)=\frac{1}{k-5} (as k\neq 5)\)
\(therefore,\;\lim_{x \to k} f(x)=f(k)\)Hence, \(f\) is continuous at every point in the domain of
\(f\) and therefore, it is a continuous function.
(c) The given function is \(f(x)=\frac{x^{2}-25}{x+5}, x\neq -5\)
For any real number c\(\neq -5\), we obtain
\(\lim_{x \to c} f(x)=\lim_{x \to c}\frac{x^{2}-25}{x+5}=\lim_{x \to c}\frac{(x+5)(x-5)}{x+5}=\lim_{x \to c}(x-5)=(c-5)\)Also,\(f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) (as c\neq -5)\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Hence, \(f\) is continuous at every point in the domain of f and therefore, it is a continuous function.
(d) The function is \(f(x)=\left | x-5 \right |\)
\(= \left\{\begin{matrix} 5-x, & if x< 5\\ x-5 ,& if x\geq 5 \end{matrix}\right.\)This function \(f\) is determined at all points of the real line.
Let c be a point on a real line. Then, \(c<5\) or \(c= 5\)
or \(c> 5\)
Case 1: \(c<5\)
Then, \(f(c)=5-c\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(5-x)=5-c\) \(therefore,\; \lim_{x \to c} f(x)=f(c)\)Therefore f is continuous at all real number less than 5.
Case 2: \(c=5\)
Then, \(f(c)=f(5)=(5-5)=0\)
\(\lim_{x \to 5}f(x)=\lim_{x \to 5}(5-x)=(5-5)=0\) \(\lim_{x \to 5}f(x)=\lim_{x \to 5}(x-5)=(5-5)=0\) \(therefore,\; \lim_{x \to c^{-}}f(x)=\lim_{x \to c^{+}}=f(c)\)So \(f\) is continuous at \(x=5\)
Case 3: \(c>5\)
Then, \(f(c)=f(5)=c-5\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(x-5)=c-5\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Based on formulae given in Continuity and Differentiability
Q4: Prove that the function \(f(x)=x^{n}\) is continuous at \(x=n\), where n is a positive integer.
Sol: Based on formulae given in Continuity and Differentiability
Given function \(f(x)=x^{n}\)
f is to be defined at all positive integers, n and its value at n is \(n^{n}\).
Then \(\lim_{x \to n}f(n)=\lim_{x \to n}(x^{n})=n^{n}\)
\(therefore,\; \lim_{x \to n}f(x)=f(n)\)So, f is continuous at n for all positive values of n. Based on formulae given in Continuity and Differentiability
Q5: Is the function \(f\) given by \(f(x)=\begin{cases} x & \text{ if } x\leq 1 \\ 5 & \text{ if } x> 1 \end{cases}\)
continuous at \(x=0,1,2?\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} x & \text{ if } x\leq 1 \\ 5 & \text{ if } x> 1 \end{cases}\)
Case 1: At \(x=0\)
It is evident that f is defined at 0 and its value is 0.
Then,\(\lim_{x \to 0}f(x)=\lim_{x \to 0}x=0\)
\(therefore,\; \lim_{x \to 0}f(x)=f(0)\)Therefore, f is continuous at x=0
Case 2: At \(x=1\),
It is evident that \(f\) is defined at 1 and its value at 1 is 1.
The left hand limit of \(f at x =1 is,\)
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(5)=5\)The right hand limit of f at \(x=1 is,\)
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(5)=5\) \(therefore,\; \lim_{x \to 1^{-}}f(x)\neq \lim_{x \to 1^{+}}f(x)\)So, \(f\) is not continuous at \(x=1\).
Case 3: At \(x=2\)
\(f\) is defined at 2 and its value at 2 is 5.
Then \(\lim_{x \to 2}f(x)=\lim_{x \to 2}(5)=5\)
\(therefore,\; \lim_{x \to 2}f(x)=f(2)\)Therefore, f is continuous at \(x=2\)
Q6: Find all the points of discontinuity of \(f\), where \(f\)
Is defined by
\(f(x)=\begin{cases} 2x+3 & \text{ if } x\leq 2 \\ 2x-3& \text{ if } x> 2 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\begin{cases} 2x+3 & \text{ if } x\leq 2 \\ 2x-3& \text{ if } x> 2 \end{cases}\)
It is necessary that the given function \(f(x)\) is defined at all the points of the real line.
(i) \(c<2\)
(ii) \(c>2\)
(iii) \(c=2\)
Case (i) \(c<2\)
Then,\(f(c)=2c+3\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(2x+3)=2c+3\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points \(x\), such that \(x<2\)
Case (ii) \(c>2\)
Then, \(f(c)=2c+3\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(2c+3)=2c+3\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points \(x\), such that \(x>2\)
Case (iii) \(c=2\)
Then, the left hand limit of \(f\) at \(x=2\) is,
\(\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{-}}(2x+3))=2\times 2+3=7\)The right hand limit of \(f\) at \(x\) =2 is,
\(\lim_{x \to 2^{+}}f(x)=\lim_{x \to 2^{+}}(2x-3))=2\times 2-3=1\)It is observed that the left hand limit of \(f\) at \(x=2\) do not coincide.
Therefore, \(f\) is not continuous at \(x=2\)
Hence, \(x=2\) is the only point of discontinuity of \(f\).
Q7: Find all points of discontinuity of \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} \begin{vmatrix} x \end{vmatrix}+3, & \text{ if } x\leq -3 \\ -2x,& \text{ if } -3< x< 3 \\ 6x+2,& \text{ if } x\geq 3 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} \begin{vmatrix} x \end{vmatrix}+3, & \text{ if } x\leq -3 \\ -2x,& \text{ if } -3< x< 3 \\ 6x+2,& \text{ if } x\geq 3 \end{cases}\)
The given function \(f\) is defined at all the points of the real line.
Let c be a point on the real line.
Case 1:
If \(c< -3,\) then \(f(c)=-c+3\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(-x+3)=-c+3\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points \(x\), such that \(x<-3\)
Case 2:
If \(c=-3\), then \(f(-3)=-(-3)=3=6\)
\(\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(-x+3)=-(-3)+3=6\) \(\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(-2x)=-2\times(-3)=6\) \(therefore,\; \lim_{x \to 3^{+}}f(x)=f(-3)\)So, \(f\) is continuous at \(x=-3\).
Case 3:
If \(-3<x<3,\) then \(f(c)=-2c\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(-2x)=-2c\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous in \((-3,3)\).
Case 4:
If \(c=3\), then the left hand limit of \(f\) at \(x=3\) is,
\(\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(-2x)=-2\times 3=-6\)The right hand limit of \(f\) at \(x=3\) is,
\(\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(6x+2)=6\times +2=20\)It is observed that the left hand limit and right hand limit of \(f\) at \(x=3\) do not coincide.
Therefore, \(f\) is not continuous at \(x=3\).
Case 5:
If \(c> 3\), then \(f(c)=5c+2\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(6x+2)=6c+2\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points \(x\), such that \(x>3\)
Hence, \(x=3\) is the only point of discontinuity of \(f\).
Q8: Find all points of discontinuity of \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} \frac{\begin{vmatrix} x \end{vmatrix}}{x} & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} \frac{\begin{vmatrix} x \end{vmatrix}}{x} & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}\)
We know that, \(x<0\Rightarrow \begin{vmatrix} x \end{vmatrix}=-x\) and \(x>0\Rightarrow \begin{vmatrix} x \end{vmatrix}=x\)
Therefore, the given function can be also written as
\(f(x)=\begin{cases} \frac{\begin{vmatrix} x \end{vmatrix}}{x}=\frac{-x}{x}=-1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \frac{\begin{vmatrix} x \end{vmatrix}}{x}=\frac{x}{x}=1, & \text{ if } x>0 \end{cases}\)The function \(f\) is defined at all the points of the real line.
Let \(c\) be a point on a real line.
Case 1:
If \(c<0\), then \(f(c)=-1\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(-1)=-1\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, f is continuous at all points \(x<0\)
Case 2:
If \(c=0\) then the left hand limit of \(f\) at \(x=0\) is,
\(\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}(-1)=-1\)the right hand limit of \(f\) at \(x=0\) is,
\(\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(1)=1\)The left hand limit and right hand limit of \(f\) at \(x=0\) do not coincide.
Therefore, \(f\) is not continuous at \(x=0\).
Case 3:
If \(c>0,\) then \(f(c)=1\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(1)=1\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at \(x>0\)
Hence, \(x=0\) is the only point of discontinuity of \(f\). Based on formulae given in Continuity and Differentiability
Q9: Find all points of discontinuity of \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} \frac{x}{\begin{vmatrix} x \end{vmatrix}}, & \text{ if } x<0 \\ -1,& \text{ if } x\geq 0 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f(x)=\begin{cases} \frac{x}{\begin{vmatrix} x \end{vmatrix}}, & \text{ if } x<0 \\ -1,& \text{ if } x\geq 0 \end{cases}\)
We know that \(x<0\Rightarrow \begin{vmatrix} x \end{vmatrix}=-x\)
Therefore, the given function can be also written as
\(f(x)=\begin{cases} \frac{x}{\begin{vmatrix} x \end{vmatrix}}=\frac{x}{-x}=-1, & \text{ if } x<0 \\ -1,& \text{ if } x\geq 0 \end{cases}\)\(\Rightarrow f(x)=-1\) for all \(x\in \mathbb{R}\)
Let c be any real number. Then, \(\lim_{x \to c}f(x)=\lim_{x \to c}(-1)=-1\)
Also, \(f(c)=-1=\lim_{x \to c}f(x)\)
Therefore, the given function is a continuous function.
Hence the given function has no point of discontinuity .
Q10: Find all the points of discontinuity of f, where \(f\) is defined by
\(f(x)=\begin{cases} x+1 & \text{ if } x\geq 1 \\ x^{2}+1 & \text{ if } x<1 \end{cases}\).
Sol: Based on formulae given in Continuity and DifferentiabilityThe given function \(f\) is \(f(x)=\begin{cases} x+1 & \text{ if } x\geq 1 \\ x^{2}+1 & \text{ if } x<1 \end{cases}\).
The given function \(f\) is defined at all the points of the real line.
Let c be a point on the real line.
Case 1:
If \(c<1,\) then \(f(c)=c^{2}+1\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(x^{2}+1)=(c^{2}+1)\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points \(x\), such that \(x<1\).
Case 2:
If \(c=1,\) then \(f(c)=f(1)=1+1=2\)
The right hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(x^{2}+1)=1^{2}+1=2\)The left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(x^{2}+1)=1^{2}+1=2\) \(therefore,\; \lim_{x \to 1}f(x)=f(c)\)So, \(f\) is continuous at \(x=1\).
Case 3:
If \(c>1\), then \(f(c)=c+1\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(x+1)=c+1\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points of \(x\) such that \(x>1\).
Hence, the given function \(f\) has no point of discontinuity.
Q11: Find all the points of discontinuity of \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} x^{3}-3 & \text{ if } x\leq 2 \\ x^{2}+1 & \text{ if } x>2 \end{cases}\) .
Sol: Based on formulae given in Continuity and DifferentiabilityThe given function \(f\) is \(f(x)=\begin{cases} x^{3}-3 & \text{ if } x\leq 2 \\ x^{2}+1 & \text{ if } x>2 \end{cases}\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case1:
If \(c<2\), then \(f(c)=c^{3}-3\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(x^{3}-3)=c^{3}-3\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points of \(x\), such that \(x<2\).
Case 2:
If \(c=2\), then \(f(c)=f(2)=2^{3}-3=5\)
The right hand limit of \(f\) at \(x=2\) is,
\(\lim_{x \to 2^{+}}f(x)=\lim_{x \to 2^{+}}(x^{2}+1)=2^{2}+1=5\)The left hand limit of \(f\) at \(x=2\) is,
\(\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{-}}(x^{3}-3)=2^{3}-3=5\) \(therefore,\; \lim_{x \to 2}f(x)=f(2)\)So, \(f\) is continuous at \(x=2\).
Case 3:
If \(c>2\), then \(f(c)=c^{2}+1\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(x^{2}+1)=c^{2}+1\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all points of \(x\), such that \(x>2\).
Thus, the given function \(f\) is continuous at every point on the real line.
Hence, the given function \(f\) has no point of discontinuity.
Q12: Find all points of discontinuity of \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} x^{10}-1, & \text{ if } x\leq 1 \\ x^{2}, & \text{ if } x>1 \end{cases}\).
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} x^{10}-1, & \text{ if } x\leq 1 \\ x^{2}, & \text{ if } x>1 \end{cases}\)
The given function \(f\) is defined at all the points of the real line.
Let c be a point on the real line.
Case1:
If \(c<1\), then \(f(c)=c^{10}-1\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(x^{10}-1)=(c^{10}-1)\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at \(x\) such that \(x<1\)
Case 2:
If \(c=1\), then the left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(x^{10}-1)=1^{10}-1=1-1=0\)The right hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(x^{2})=1^{2}=1\)It is observed that the left hand limit and the right hand limit of \(f\) at \(x=1\) do not overlap.
Therefore, \(f\) is not continuous at \(x=1\).
Case 3:
If \(c>1\), then \(f(c)=c^{2}\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(x^{2})=c^{2}\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at \(x\) such that \(x>1\)
Thus, we say that \(x=1\) is the only point of discontinuity of \(f\).
Q13: Is the function defined by
\(f(x)=\begin{cases} x+5, & \text{ if } x\leq 1 \\ x-5, & \text{ if } x> 1 \end{cases}\)
a continuous function?
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\begin{cases} x+5, & \text{ if } x\leq 1 \\ x-5, & \text{ if } x> 1 \end{cases}\)
The function \(f\) is defined at all the points of the real line.
Let \(c\) be a points on the real line.
Case 1:
If \(c<1\), then \(f(c)=c+5\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(x+5)=c+5\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)So, \(f\) is continuous at all the points \(x\), such that \(x<1\)
Case 2:
If \(c=1\), then \(f(1)=1+5=6\)
The left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(x+5)=1+5=6\)The right hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(x-5)=1-5=-4\)It is observed that the left hand limit and the right hand limit of \(f\) at \(x=1\) do not coincide.
Therefore, \(f\) is not continuous at \(x=1\).
Case 3:
If \(c>1\), then \(f(c)=c-5\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(x-5)=c-5\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points \(x\),such that \(x>1\).
Thus, from the above observation, it can be concluded that \(x=1\) is the only point of discontinuity of \(f\).
Q14: Discuss the continuity of the function \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} 3 & \text{ if } 0\leq x\leq 1 \\ 4 & \text{ if } 1< x< 3 \\ 5 & \text{ if } 3\leq x\leq 10 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\begin{cases} 3 & \text{ if } 0\leq x\leq 1 \\ 4 & \text{ if } 1< x< 3 \\ 5 & \text{ if } 3\leq x\leq 10 \end{cases}\)
The given function is defined at all points of the interval \([0,10]\).
Let c be a point in the interval \([0,10]\).
Case 1:
If \(0\leq c\leq 1\), then \(f(c)=3\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(3)=3\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous in the interval \([0,1]\).
Case 2:
If \(c=1\), then \(f(3)=3\)
The left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(3)=3\)The right hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(4)=4\)It is observed that the left hand and right hand limits of \(f\) at \(x=1\) do not coincide.
Therefore, \(f\) is not continuous at \(x=1\).
Case 3:
If \(1<c<3,\), then \(f(c)=4\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(4)=4\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Case 4:
If \(c=3\), then \(f(c)=5\)
The left hand limit of \(f\) at \(x=3\) is,
\(\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(4)=4\)The right hand limit of \(f\) at \(x=3\) is,
\(\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(5)=5\)The left hand and right hand limits of \(f\) at \(x=3\) do not overlap.
Therefore, \(f\) is not continuous at \(x=3\).
Case 5:
If \(3<c\leq10\) then \(f(c)=5\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}f(5)=5\)
therefore, \(f\) is continuous at all points of the interval \((3,10]\).
Hence \(f\) is not continuous at \(x=1\) and \(x=3\).
Q15: Discuss the continuity of the function \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} 2x, & \text{ if } x<0 \\ 0,& \text{ if } 0\leq x\leq 1\\ 4, & \text{ if } x>1 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\begin{cases} 2x, & \text{ if } x<0 \\ 0,& \text{ if } 0\leq x\leq 1\\ 4, & \text{ if } x>1 \end{cases}\)
The given function is defined at all the points in the real line.
Let \(c\) be a point on the real line.
Case 1:
If c<0, then \(f(c)=2c\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(2x)=2c\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points \(x\) such that \(x<0\).
Case 2:
If \(c=0\), then \(f(c)f(0)=0\)
The left hand limit of \(f\) at \(x=0\) is,
\(\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}(2x)=2\times0=0\)The right hand limit of \(f\) at \(x=0\) is,
\(\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(0)=0\) \(therefore,\; \lim_{x \to 0}f(x)=f(0)\)Therefore, \(f\) is continuous at \(x=0\)
Case 3:
If \(0<c<1\), then \(f(x)=0\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(0)=0\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points of the interval \((0,1)\).
Case 4:
If \(c=1\), then \(f(c)=f(1)=0\)
The left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(0)=0\)The left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(4x)=4\times 1=4\)The left hand and right hand limits of \(f\) at \(x=1\) do not overlap.
Therefore, \(f\) is not continuous at \(x=1\).
Case 5:
If c<1, then \(f(c)=4c\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(4x)=4c\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points \(x\) such that \(x>1\).
Hence, \(f\) is discontinuous only at \(x=1\).
Q16: Discuss the continuity of the function \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} -2, & \text{ if } x\leq -1\\ 2x, & \text{ if } -1<x\leq 1 \\ 2, & \text{ if } x>1 \end{cases}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} -2, & \text{ if } x\leq -1\\ 2x, & \text{ if } -1<x\leq 1 \\ 2, & \text{ if } x>1 \end{cases}\)
The given function is defined at all points of the real line.
Let \(c\) be a point on the real line.
Case 1:
If \(c<-1\), then \(f(c)=-2\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(-2)=-2\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points \(x\) such that \(x<-1\).
Case 2:
If \(c=-1\), then \(f(c)=f(-1)=-2\)
The left hand limit of \(f\) at \(x=-1\) is,
\(\lim_{x \to -1^{-}}f(x)=\lim_{x \to -1^{-}}(-2)=-2\)The right hand limit of \(f\) at \(x=-1\) is,
\(\lim_{x \to -1^{+}}f(x)=\lim_{x \to -1^{+}}(2x)=2\times (-1)=-2\) \(therefore,\;\lim_{x \to -1}f(x)=f(-1)\)Therefore, \(f\) is continuous at \(x=-1\).
Case 3:
If \(-1<c<1\), then \(f(c)=2c\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(2x)=2c\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points of the interval \((-1,1)\).
Case 4:
If \(c=1\), then \(f(c)=f(1)=2\times 1=2\)
The left hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(2x)=2\times 1=2\)The right hand limit of \(f\) at \(x=1\) is,
\(\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(2)=2\) \(therefore,\; \lim_{x \to 1}f(x=f(c))\)Therefore, f is continuous at \(x=2\)
Case 5:
If \(c>1\), then \(f(c)=2\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(2)=2\) \(\lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all points \(x\) such that \(x>1\).
Thus, from the above observation it can be concluded that \(f\) is continuous at all points of the real line.
Q17: Find the relationship between a and b so that the function \(f\) is defined by
\(f(x)=\begin{cases} ax-1, & \text{ if } x\leq 3 \\ bx+3, & \text{ if } x>3 \end{cases}\)
is continuous at \(x=3\).
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} ax-1, & \text{ if } x\leq 3 \\ bx+3, & \text{ if } x>3 \end{cases}\)
If \(f\) is continuous at \(x-3\), then
\(\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}f(x)=f(3)\; \; \; \; \; \; …..(1)\)Also,
The left hand limit of \(f\) at \(x=3\) is,
\(\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(ax+1)=3a+1\)The left hand limit of \(f\) at \(x=3\) is,
\(\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(bx+3)=3b+3\) \(f(3)=3a+1\)Therefore, from (1) we obtain
3a+1-3b+3-3a+1
\(\Rightarrow 3a+1-3b-3\) \(\Rightarrow 3a=3b+2\) \(\Rightarrow a-b=\frac{2}{3}\)Therefore, the required relationship is given by, \(\Rightarrow a-b=\frac{2}{3}\)
Q18: For what value of \(\lambda\), is the function defined by
\(f(x)=\begin{cases} \lambda (x^{2}-2x), & \text{ if } x\leq 0 \\ 4x+1, & \text{ if } x> 0 \end{cases}\)
continuous at x=0? What about continuity at \(x=1\)?
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} \lambda (x^{2}-2x), & \text{ if } x\leq 0 \\ 4x+1, & \text{ if } x> 0 \end{cases}\)
If \(f\) is continuous at \(x=0\), then
\(\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{+}}f(x)=f(0)\) \(\Rightarrow \lim_{x \to 0^{-}}\lambda(x^{2}-2x)=\lim_{x \to 0^{+}}(4x+1)\) \(\Rightarrow \lambda (0^{2}-2\times 0)=(4\times 0-1)\)\(\Rightarrow 0=1\) which is not possible.
Therefore, there is no value of \(\lambda\) for which the function \(f\) is continuous at \(x=0\).
At \(x=1\),
\(f(1)=4x+1=4\times 1+1=5\) \(\lim_{x \to 1}f(1)=4x+1=4\times 1+1=5\) \(therefore,\; \lim_{x \to 1}f(x)=f(1)\)Therefore, for any values of \(\lambda\), \(f\) is continuous at \(x=1\).
Q19: Show that the function defined by \(g(x)=x-[x]\) is discontinuous at all integral points. Hence \([x]\) denotes the greatest integer less than or equal to \(x\).
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(g(x)=x-[x]\)
It is evident that \(g\) is defined at all integral points.
Let \(n\) be an integer.
Then,
\(g(n)=n-[n]=n-n=0\)Then left hand limit of \(f\) at \(x=n\) is,
\(\lim_{x \to n^{-}}g(x)=\lim_{x \to n^{-}}(x-[x])=\lim_{x \to n^{-}}(x)-\lim_{x \to n^{-}}[x]=n-(n-1)=1\)Then right hand limit of \(f\) at \(x=n\) is,
\(\lim_{x \to n^{+}}g(x)=\lim_{x \to n^{+}}(x-[x])=\lim_{x \to n^{+}}(x)-\lim_{x \to n^{+}}[x]=n-n=0\)It is observed that the left and right hand limits of \(f\) at \(x=n\) do not coincide.
Therefore, \(f\) is not continuous at \(x=n\)
Hence, \(g\) is discontinuous at all integral points.
Q20: Is the function defined by \(f(x)=x^{2}-\sin x+5\) is continuous at x?
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=x^{2}-\sin x+5\)
It is evident that \(f\) is defined at x
At \(x=\pi\), \(f(x)=f(\pi )=\pi ^{2}-\sin \pi +5=\pi ^{2}-0+5=\pi ^{2}+5\)
Consider \(\lim_{x \to \pi }f(x)=\lim_{x \to \pi }(x^{2}-sin x +5)\)
Put \(x=\pi +h\)
If \(x\rightarrow \pi\), then it is evident that \(h\rightarrow 0\)
\(therefore,\; \lim_{x \to \pi }f(x)=\lim_{x \to \pi }(x^{2}-\sin x+5)\) \(=\lim_{x \to 0 }[(\pi +h)^{2}-\sin (\pi +h)+5]\) \(=\lim_{x \to 0 }(\pi +h)^{2}-\lim_{x \to 0 }\sin (\pi +h)+\lim_{x \to 0 }5\) \(=(\pi +0)^{2}-\lim_{h \to 0}[\sin\pi\cosh+\cos\pi\sinh]+5\) \(=\pi ^{2}-\sin\pi\cos0-\cos\pi\sin0+5\) \(=\pi ^{2}-0\times 1-(-1)\times 0+5\) \(=\pi ^{2}+5\) \(=therefore,\; \lim_{x \to \pi }f(x)=f(\pi )\)Therefore, the given function is continuous at \(x=\pi\)
Q21: Discuss the continuity of the following functions.
Sol: Based on formulae given in Continuity and Differentiability
It is known that if \(g\) and \(g\) are two continuous functions, then
\(g+h, g-h, g.h\) are also continuous functions.
It has to be proved first that \(g(x)=\sin x\) and \(h(x)=\cos x\) are continuous functions.
Let \(g(x)=\sin x\)
It is evident that \(g(x)=\sin x\) is defined for every real number.
Let \(c\) be a real number.
Put \(x=c+h\)
If \(X\;\tilde{A}\;C\;\hat{a}\;\epsilon \;c,\;then\;h\;\tilde{A}\;C\;\hat{a}\;\epsilon \;0\)
\(h(c)-\cos c\) \(\lim_{x \to c}h(x)=\lim_{x \to c}\cos x\) \(=\lim_{h \to 0}\cos (c+h)\) \(=\lim_{h \to 0}\cos [\cos c\cos h-\sin c\sin h]\) \(=\lim_{h \to 0}\cos \cos c\cos h-\lim_{h \to 0}\sin c\sin h\) \(=\cos \cos c\cos 0-\sin c\sin 0\) \(=\cos c\times 1-\sin c\times 0\) \(=\cos c\) \(therefore,\; \lim_{x \to c}h(x)=h(c)\)Therefore, \(h\) is continuous function.
Therefore, it can be said that
Q22: Discuss the continuity of the cosine, cosecant, secant and cotangent function.
Sol: Based on formulae given in Continuity and Differentiability
It is known that if \(g\) and \(h\) are two continuous functions, then
(i) \(\frac{h(x)}{g(x)},g(x)\neq 0\) is continuous
(ii) \(\frac{1}{g(x)},g(x)\neq 0\) is continuous
(iii) \(\frac{1}{h(x)},h(x)\neq 0\) is continuous
It has to be proved first that \(g(x)=\sin x\) and \(h(x)=\cos x\) are continuous functions
Let \(g(x)=\sin x\)
Clearly \(g(x)=\sin x\) is defined for every real number.
Let \(c\) be a real number. Put \(x\rightarrow c+h\)
If \(x\rightarrow c\), then \(h\rightarrow 0\)
\(g(x)=\sin x\) \(\lim_{x \to c}g(x) = \lim_{x \to c}\sin x\\ =\lim_{h \to 0}\sin (c+h)\\ =\lim_{h \to 0}[\sin c\cos h+\cos c\sin h]\\ = \sin c\cos 0+\cos c\sin 0\\ =\sin c+0 \\ =\sin c\) \(therefore,\; \lim_{x \to c}g(x)=g(c)\) Based on formulae given in Continuity and DifferentiabilityTherefore, \(g\) is a continuous function.
Let \(h(x)=\cos x\)
It is evident that \(h(x)=\cos x\) is defined for every real number.
Let \(c\) be a real number. Put \(x\rightarrow c+h\)
\(x\;\hat{A}\;c,\;then\;h\;\hat{A}\) \(h(c)=\cos x\) \(\lim_{x \to c}h(x) = \lim_{x \to c}\cos x\\ =\lim_{h \to 0}\cos (c+h)\\ =\lim_{h \to 0}[\cos c\cos h-\sin c\sin h]\\ = \cos c\cos 0+\sin c\sin 0\\ =\cos c+0 \\ =\cos c\) \(therefore,\; \lim_{x \to c}h(x)=h(c)\)Therefore, \(h\) is a continuous function.
It can be concluded that,
\(\csc x-\frac{1}{sinx}, \sin x\neq 0 is continuous\) \(\Rightarrow \csc x, x\neq n\pi (n\in \mathbb{Z})\)Based on formulae given in Continuity and Differentiability
Q23: Find the points of discontinuity of \(f\), where
\(f(x)=\begin{cases} \frac{\sin x}{x}, & \text{ if } x<0 \\ x-1, & \text{ if } x\geq 0 \end{cases}\)
Sol:
The given function \(f\) is \(f(x)=\begin{cases} \frac{\sin x}{x}, & \text{ if } x<0 \\ x-1, & \text{ if } x\geq 0 \end{cases}\)
It is evident that \(f\) is defined at all the points of the real line.
Let \(c\) be a real number.
Case 1:
If \(c<0, then f(c)=\frac{\sin c }{c}\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(\frac{\sin x}{x})=(\frac{\sin c }{c})\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Based on formulae given in Continuity and DifferentiabilityTherefore, \(f\) is continuous at all the points \(x\), such that \(x<0\).
Case 2:
If \(c>0, then f(c)=c+1\) and \(\lim_{x \to c}f(x)=\lim_{x \to c}(x+1)=(c+1)\)
\(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all the points \(x\), such that \(x>0\).
Case 3:
If \(c=0, then f(c)=f(0)=0+1=1\)
The left hand limit of \(f\) at \(x\) is,
\(\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}\frac{\sin x}{x}=1\)The right hand limit of \(f\) at \(x\) is,
\(\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(x+1)=1\) \(therefore,\; \lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{+}}f(x)=f(0)\)Therefore, \(f\) is continuous at \(x=0\)
Based on formulae given in Continuity and DifferentiabilityIt can be concluded that \(f\) is continuous at all points of the real line.
Thus, \(f\) has no point of discontinuity.
Q24: Determine if \(f\) defined by
\(f(x)=\begin{cases} x^{2}\sin \frac{1}{x}, & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}\)
Sol:
The given function \(f\) is \(f(x)=\begin{cases} x^{2}\sin \frac{1}{x}, & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}\)
It is evident that \(f\) is defined at all points of the real line.
Let \(c\) be a real number.
Case 1:
If \(c\neq 0, then f(c)=c^{2\sin\ frac{1}{c}}\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}\left ( x^{2}\sin \frac{1}{x}\right )=\left ( \lim_{x \to c}x^{2} \right )\left ( \lim_{x \to c}\sin \frac{1}{x} \right )=c^{2}\sin \frac{1}{c}\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all the points \(x\), such that \(x\neq 0\).
Case 2:
If \(c= 0, then f(0)=0\)
\(\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}\left ( x^{2}\sin \frac{1}{x} \right )=\lim_{x \to 0^{-}}\left ( x^{2} \sin \frac{1}{x}\right )\)It is known that, \(1\leq \sin \frac{1}{x}\leq 1, x\neq 0\)
\(\Rightarrow -x^{2}\leq \sin \frac{1}{x}\leq x^{2}\) \(\Rightarrow \lim_{x \to 0}(-x^{2})\lim_{x \to 0}\leq \lim_{x \to 0}(\sin \frac{1}{x})\leq \lim_{x \to 0}(x^{2})\) \(\Rightarrow 0\lim_{x \to 0}\leq \lim_{x \to 0}(\sin \frac{1}{x})\leq 0\) \(\Rightarrow \lim_{x \to 0}(\sin \frac{1}{x})= 0\) \(therefore,\; \lim_{x \to 0^{-}}f(x)= 0\)Similarly, \(\lim_{x \to 0^{+}}f(x)= \lim_{x \to 0^{+}}x^{2}\sin \frac{1}{x} =0 \)
Based on formulae given in Continuity and Differentiability \(therefore,\; \lim_{x \to 0^{-}}f(x)=f(0)=\lim_{x \to 0^{+}}f(0)\)Therefore, \(f\) is continuous at \(x=0\)
It can be concluded that \(f\) is continuous at all points of the real line.
Thus, \(f\) has no point of discontinuity.
Q25: Determine the continuity of \(f\), where \(f\) is defined by
\(f(x)=\begin{cases} \sin x- \cos x,& \text{ if } x\neq 0 \\ -1,& \text{ if } x= 0 \end{cases}\)
Sol:
The given function \(f\) is \(f(x)=\begin{cases} \sin x- \cos x,& \text{ if } x\neq 0 \\ -1,& \text{ if } x= 0 \end{cases}\)
It is evident that \(f\) is defined at all points of the real line.
Let \(c\) be a real number.
Case 1:
If \(c\neq 0, then f(c)= \sin c- \cos c\)
\(\lim_{x \to c}f(x)=\lim_{x \to c}(\sin x- \cos x)=\sin c- \cos c\) \(therefore,\; \lim_{x \to c}f(x)=f(c)\)Therefore, \(f\) is continuous at all the points \(x\), such that \(x\neq 0\).
Case 2:
If \(c= 0, then f(0)=-1\)
\(\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}(\sin x- \cos x)=\sin 0- \cos 0=0-1=-1\) \(\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(\sin x- \cos x)=\sin 0- \cos 0=0-1=-1\) \(therefore,\; \lim_{x \to 0^{-}}f(x)=f(0)=\lim_{x \to 0^{+}}f(0)\)Therefore, \(f\) is continuous at \(x=0\)
It can be concluded that \(f\) is continuous at all points of the real line.
Thus, \(f\) is a continuous function.
Q26: Find the values of \(k\) so that the function \(f\) is continuous at the indicated points
\(f(x)=\begin{cases} \frac{k\cos x}{\pi -2x} ,& \text{ if } x\neq \frac{\pi }{2} \\ 3,& \text{ if} x= \frac{\pi }{2} \end{cases}\)
At \(x=\frac{\pi }{2}\).
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f\) is \(f(x)=\begin{cases} \frac{k\cos x}{\pi -2x} ,& \text{ if } x\neq \frac{\pi }{2} \\ 3,& \text{ if} x= \frac{\pi }{2} \end{cases}\)
The given function \(f\) is continuous at \(x=\frac{\pi }{2}\), if \(f\) is defined at \(x=\frac{\pi }{2}\) and the value of \(f\) at \(x=\frac{\pi }{2}\) equals the limit of \(f\) at \(x=\frac{\pi }{2}\).
It is evident that the limit of \(f\) is defined at \(x=\frac{\pi }{2}\) and \(f(\frac{\pi }{2})=3\).
\(\lim_{x \to \frac{\pi }{2}}f(x)=\lim_{x \to \frac{\pi }{2}}\frac{k\cos x}{\pi -2x}\) \(\lim_{x \to \frac{\pi }{2}}f(x)=\lim_{x \to \frac{\pi }{2}}\frac{k\cos x}{\pi -2x}\)Put \(x=\frac{\pi }{2}+h\)
Then,\(x\rightarrow \frac{\pi }{2}\Rightarrow h\rightarrow 0\)
\(therefore,\; \lim_{x \to \frac{\pi }{2}}f(x)=\lim_{x \to \frac{\pi }{2}}\frac{k \cos x}{\pi -2x}=\lim_{h \to 0}\frac{k \cos (\frac{\pi }{2}+h)}{\pi -2(\frac{\pi }{2}+h)}\) \(=k\lim_{h \to 0}\frac{-\sin h}{-2h}=\frac{k}{2}\lim_{h \to 0}\frac{\sin h}{h}=\frac{k}{2}\) \(therefore,\; \lim_{x \to \frac{\pi }{2}}f(x)=f(\frac{\pi }{2})\) \(\Rightarrow \frac{k}{2}=3\) \(\Rightarrow k=6\)therefore , the required value of \( k\) is 6 Based on formulae given in Continuity and Differentiability.
Q27: Find the values of \(k\) so that the function \(f\) is continuous at the indicated points
\(f(x)=\begin{cases} kx^{2,} & \text{ if } x=2 \\ 3, & \text{ if } x>2 \end{cases}\)
At \(x=2\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is\(f(x)=\begin{cases} kx^{2,} & \text{ if } x=2 \\ 3, & \text{ if } x>2 \end{cases}\)
The given function \(f\) is continuous at \(x=2\). If \(f\) is defined at \(x=2\) and if the value of \(f\) at \(x=2\) equals the limit of \(f at x=2\)
It is evident that \(f\) is defined at \(x=2\) and \(f(2)=k\left ( 2 \right )^{2}=4k\)
\(\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{+}}f(x)=f(2)\) \(\Rightarrow \lim_{x \to 2^{-}}(kx^{2})=\lim_{x \to 2^{+}}3=4k\) \(\Rightarrow (k\times 2^{2})=3=4k\) \(\Rightarrow 4k=3\) \(\Rightarrow k=\frac{3}{4}\)Therefore, the required value of \(k=\frac{3}{4}\).
Q28: Find the values of \(k\) so that the function \(f\) is continuous at the indicated points
\(f(x)=\begin{cases} kx+1, & \text{ if } x\leq \pi \\ \cos x,& \text{ if } x>\pi \end{cases}\)
At \(x=5\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\begin{cases} kx+1, & \text{ if } x\leq \pi \\ \cos x,& \text{ if } x>\pi \end{cases}\)
The given function \(f\) is continuous at \(x=c\), If \(f\) is defined at \(x=c\) and if the value of \(f\) at \(x=c\) equals the limit of \(f at x=c\)
It is evident that \(f\) is defined at \(x=c\) and \(f(\pi)=k\pi +1\)
\(\lim_{x \to \pi^{-}}f(x)=\lim_{x \to \pi^{+}}f(x)=f(\pi )\) \(\Rightarrow \lim_{x \to \pi^{-}}(kx+1)=\lim_{x \to \pi^{+}}\cos x=k\pi +1\) \(\Rightarrow (k\pi +1)=\cos \pi =k\pi +1\) \(\Rightarrow (k\pi +1)=-1 =k\pi +1\) \(\Rightarrow k=-\frac{2}{\pi }\)Therefore, the required value of \( k is -\frac{2}{\pi }\)
Q29: Find the values of \(k\) so that the function \(f\) is continuous at the indicated points
\(f(x)=\begin{cases} kx+1, & \text{ if } x\leq 5\\ 3x-5, & \text{ if } x>5 \end{cases}\)
At \(x=5\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(k\) is \(f(x)=\begin{cases} kx+1, & \text{ if } x\leq 5\\ 3x-5, & \text{ if } x>5 \end{cases}\)
The given function \(f\) is continuous at \(x=5\), If \(f\) is defined at \(x=5\) and if the value of \(f\) at \(x=5\) equals the limit of \(f at x=5\)
It is evident that \(f\) is defined at \(x=5\) and \(f(5)=kx +1=5k+1\)
\(\lim_{x \to 5^{-}}f(x)=\lim_{x \to 5^{+}}f(x)=f(5 )\) \(\Rightarrow \lim_{x \to 5^{-}}(kx+1)=\lim_{x \to 5^{+}}(3x-5)=5k+1\) \(\Rightarrow (5k +1)=15-5 =5k+1\) \(\Rightarrow 5k+1=10\) \(\Rightarrow 5k=9\) \(\Rightarrow k=\frac{9}{5 }\)Therefore, the required value of \( k is \frac{9}{5 }\)
Q30: Find the values a and b such that the function defined by
\(f(x)=\begin{cases} 5, & \text{ if } x\leq 2 \\ ax+b, & \text{ if } 2<x<10 \\ 21, & \text{ if } x\geq 10 \end{cases}\)
Is a continuous function.
Sol: Based on formulae given in Continuity and Differentiability
The given function \(k\) is \(f(x)=\begin{cases} 5, & \text{ if } x\leq 2 \\ ax+b, & \text{ if } 2<x<10 \\ 21, & \text{ if } x\geq 10 \end{cases}\)
It is evident that the given function \(f\) is defined at all points of the real line.
If \(f\) is continuous function, then \(f\) is continuous at all real numbers.
In particular, \(f\) is continuous at \(x=2 and x=10\)
Since \(f\) is continuous at \(x=2\), we obtain
\(\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{+}}f(x)=f(2)\) \(\Rightarrow \lim_{x \to 2^{-}}(5)=\lim_{x \to 2^{+}}(ax+b)=5\) \(\Rightarrow 5=2a+b=5\) \(\Rightarrow 2a+b=5 \; \; \; \; \; \; \; \; \; \; ……(1))\)Since \(f\) is continuous at \(x=10\), we obtain
\(\lim_{x \to 10^{-}}f(x)=\lim_{x \to 10^{+}}f(x)=f(10)\) \(\Rightarrow \lim_{x \to 10^{-}}(ax+b)=\lim_{x \to 10^{+}}(21)=21\) \(\Rightarrow 10a+b=21=21\) \(\Rightarrow 10a+b=21 \; \; \; \; \; \; \; \; \; \; ……(2))\)On subtracting equation (1) from equation (2), we obtain
\(8a=16\) \(a=2\)By putting \(a=2\) in equation (1), we obtain
\(2\times 2+b=5\) \(\Rightarrow 4+b=5\) \(\Rightarrow b=1\)Therefore the values of a and b for which \(f\) is continuous function are 2 and 1 respectively.
Q31: Show that the function defined by \(f(x)=\cos (x^{2})\) is a continuous function.
Sol: Based on formulae given in Continuity and Differentiability
The given function is\(f(x)=\cos (x^{2})\).
This function \(f\) is defined for every real number and \(f\) be written as the composition of two function as,
\(f-g\, o\, h \; where \; g(x)=\cos x \; and\; h(x)=x^{2}\) \([because \; (goh)(x)= g(h(x))=g(x^{2})=\cos (x^{2})=f(x)]\)It has to be first proved that \(g(x)=\cos x\; \; and \; \; h(x)=x^{2}\) are continuous functions.
Let \(c\) be a real number.
Then, \(g(x)=\cos c\)
Put \(x=c+h\)
If \(x\rightarrow c, then h\rightarrow 0\)
\(\lim_{x \to c}g(x)=\lim_{x \to c}\cos x\) \(\lim_{x \to c}g(x)=\lim_{x \to c}\cos x\\ =\lim_{h \to 0}\cos(c+h)\\ =\lim_{h \to 0}[\cos c\cos h- \sin c\sin h]\\ =\lim_{h \to 0}\cos c\cos h-\lim_{h \to 0} \sin c\sin h\\ =\cos c\cos 0-\sin c\sin 0\\ =\cos c\times 1-\sin c\times 0\\ =\cos c\) \(therefore,\; \lim_{x \to c}=g(c)\)Therefore, \(g(x)=\cos x\) is continuous function.
Clearly, \(h\) is defined for every real number
Let \(k\) be a real number, then \(h(k)=k^{2}[\)
\(\lim_{x \to k}h(x)=x^{2}=k^{2}\) \(therefore,\; \lim_{x \to k}h(x)=h(k)\)Therefore, \(h\) is continuous function.
It is known that for real valued function \(g\) and \(h\), such that \((g\; o\; h)\) is defined at \(c\), if \(g\) is continuous at \(c\) and \(f\) is continuous at \(g(c)\), then \((f\; o\; g)\) is continuous at \(c\).
Therefore, \(f(x)=(g\; o\; h)(x)=cos(x^{2})\) is a continuous function.
Q32: Show that the function defined by \(f(x)=\left | \cos x \right |\) is a continuous function.
Sol: Based on formulae given in Continuity and Differentiability
The given function\(f(x)=\left | \cos x \right |\)
This function \(f\) is defined for every real number and \(f\) can be written as the composition of two function as,
\(f=g\, o\, h \; where \; g(x)=\left | x \right |; and\; h(x)=\cos x\) \([because \; (goh)(x)= g(h(x))=g(\cos x)= \left |\cos x \right |=f(x)]\)It has to be first proved that \(g(x)=\left |x \right |\; \; and \; \; h(x)=\left |\cos x \right |\) are continuous functions.
\(g(x)=\left |x \right |\) can be written as
\(g(x)=\begin{cases} -x, & \text{ if } x<0 \\ x, & \text{ if } x\geq 0 \end{cases}\)Clearly \(g\) is for all real numbers.
Let \(c\) be a real number.
Case 1:
If \(c< 0, then g(c)=-c\) and \(\lim_{x \to c}g(x)=\lim_{x \to c}(-x)=-c\)
\(therefore,\; \lim_{x \to c}g(x)=g(c)\)Therefore, \(g\) is continuous at all the points \(x\), such that \(x< 0\).
Case 2:
If \(c> 0, then g(c)=c\) and \(\lim_{x \to c}g(x)=\lim_{x \to c}(x)=c\)
\(therefore,\; \lim_{x \to c}g(x)=g(c)\)Therefore, \(g\) is continuous at all the points \(x\), such that \(x> 0\).
Case 3:
If \(c= 0, then g(c)=g(0)=0\)
\(\lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{-}}(-x)=0\) \(\lim_{x \to 0^{+}}g(x)=\lim_{x \to 0^{+}}(x)=0\) \(therefore,\; \lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{+}}g(x)=g(0)\)Therefore, \(g\) is continuous at \(x=0\)
It can be concluded that \(g\) is continuous at all points.
\(h(x)=\cos x\)Clearly \(h(x)=\cos x\) is defined for every real number.
Let \(c\) be a real number. Put \(x=c+h\)
If \(x\: \tilde{A}\: c\: \hat{a}\: \in \; ‘c\), then \(h\: \tilde{A}\: c\: \hat{a}\: \in \; ‘0\)
\(h(c)=\cos c\) \(\lim_{x \to c}h(x)=\lim_{x \to c}\cos x\) \(\lim_{x \to c}h(x)=\lim_{x \to c}\cos x \\ =\lim_{h \to 0} \cos (c+h)\\ =\lim_{h \to 0}[ \cos c \cos h- \sin c \sin h]\\ =\lim_{h \to 0} \cos c \cos h- \lim_{h \to 0} \sin c \sin h\\ =\cos c \cos 0-\sin c \sin 0\\ =\cos c \times 1-\sin c\times 0\\ =\cos c\) \(therefore,\; \lim_{x \to c}h(x)=h(c)\)Therefore, \(h(x)=\cos x\) is a continuous function.
We know that for every real valued function \(g\) and \(h\), such that \((g\: o\: h)\) is defined at \(c\), if \(g\) is continuous at \(c\) and if \(f\) is continuous at \(g(c)\), then \((f\: o\: g)\) is continuous at \(c\).
Therefore,
\(f(x)= (goh)(x)= g(h(x))=g(\cos x)= \left |\cos x \right |\) is a continuous function.
Q33: Examine that \(\ sin\left | x \right |\) is a continuous function.
Sol: Based on formulae given in Continuity and Differentiability
Let \(f(x)=\ sin\left | x \right |\)
The function \(f\) is defined for every real number and \(f\) can be written as the composition of two function as,
\(f=g\, o\, h \; where \; g(x)=\left | x \right |; and\; h(x)=\sin \left | x \right |\) \([because \; (goh)(x)= g(h(x))=g(\sin x)= \left |\sin x \right |=f(x)]\)It has to be first proved that \(g(x)=\left |x \right |\; \; and \; \; h(x)=\left |\sin x \right |\) are continuous functions.
\(g(x)=\left |x \right |\) can be written as
\(g(x)=\begin{cases} -x, & \text{ if } x<0 \\ x, & \text{ if } x\geq 0 \end{cases}\)Clearly \(g\) is defined for all real numbers.
Let \(c\) be a real number.
Case 1:
If \(c< 0, then g(c)=-c\) and \(\lim_{x \to c}g(x)=\lim_{x \to c}(-x)=-c\)
\(therefore,\; \lim_{x \to c}g(x)=g(c)\)Therefore, \(g\) is continuous at all the points \(x\), such that \(x< 0\).
Case 2:
If \(c> 0, then g(c)=c\) and \(\lim_{x \to c}g(x)=\lim_{x \to c}(x)=c\)
\(therefore,\; \lim_{x \to c}g(x)=g(c)\)Therefore, \(g\) is continuous at all the points \(x\), such that \(x> 0\).
Case 3:
If \(c= 0, then g(c)=g(0)=0\)
\(\lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{-}}(-x)=0\) \(\lim_{x \to 0^{+}}g(x)=\lim_{x \to 0^{+}}(x)=0\) \(therefore,\; \lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{+}}g(x)=g(0)\)Therefore, \(g\) is continuous at \(x=0\)
From the above three observation, it can be concluded that \(g\) is continuous at all points.
\(h(x)=\sin x\)It is evident that \(h(x)=\sin x\) is defined for every real number.
Let \(c\) be a real number. Put \(x=c+h\)
If \(x\: \tilde{A}\: c\: \hat{a}\: \in \; ‘c\), then \(h\: \tilde{A}\: c\: \hat{a}\: \in \; ‘0\)
\(h(c)=\sin c\) \(\lim_{x \to c}h(x)=\lim_{x \to c}\sin x\) \(\lim_{x \to c}h(x)=\lim_{x \to c}\sin x \\ =\lim_{h \to 0} \sin (c+h)\\ =\lim_{h \to 0}[ \sin c \cos h- \cos c \sin h]\\ =\lim_{h \to 0} \sin c \cos h- \lim_{h \to 0} \cos c \sin h\\ =\sin c \cos 0-\cos c \sin 0\\ =\sin c \times 1-\cos c\times 0\\ =\sin c\) \(therefore,\; \lim_{x \to c}h(x)=h(c)\)Therefore, \(h(x)=\sin x\) is a continuous function.
We know that for every real valued function \(g\) and \(h\), such that \((g\: o\: h)\) is defined at \(c\), if \(g\) is continuous at \(c\) and if \(f\) is continuous at \(g(c)\), then \((f\: o\: g)\) is continuous at \(c\).
Therefore,\(f(x)= (goh)(x)= g(h(x))=g(\sin x)= \left |\sin x \right |\) is a continuous function.
Q34: Find all the points of discontinuity of \(f\) defined by \(f(x)=\left | x \right |-\left | x+1 \right |\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\left | x \right |-\left | x+1 \right |\)
The two functions, \(g and h\), are defined as
\(g(x)=\left | x \right | and h(x)=\left | x+1 \right |\)Then, \(f=g\, o\, h \;\)
The continuity of \(g \;and\; h\) is examined first.
\(g(x)=\left |x \right |\) can be written as
\(g(x)=\begin{cases} -x, & \text{ if } x<0 \\ x, & \text{ if } x\geq 0 \end{cases}\)Clearly \(g\) is defined for all real numbers.
Let \(c\) be a real number.
Case 1:
If \(c< 0, then g(c)=-c\) and \(\lim_{x \to c}g(x)=\lim_{x \to c}(-x)=-c\)
\(therefore,\; \lim_{x \to c}g(x)=g(c)\)Therefore, \(g\) is continuous at all the points \(x\), such that \(x< 0\).
Case 2:
If \(c> 0, then g(c)=c\) and \(\lim_{x \to c}g(x)=\lim_{x \to c}(x)=c\)
\(therefore,\; \lim_{x \to c}g(x)=g(c)\)Therefore, \(g\) is continuous at all the points \(x\), such that \(x> 0\).
Case 3:
If \(c= 0, then g(c)=g(0)=0\)
\(\lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{-}}(-x)=0\) \(\lim_{x \to 0^{+}}g(x)=\lim_{x \to 0^{+}}(x)=0\) \(therefore,\; \lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{+}}g(x)=g(0)\)Therefore, \(g\) is continuous at \(x=0\)
From the above three observation, it can be concluded that \(g\) is continuous at all points.
\(h(x)=\left | x+1 \right |\)Which can be written as also second form:
\(h(x)=\begin{cases} -(x+1), & \text{ if } x<-1 \\ x+1, & \text{ if } x\geq -1 \end{cases}\)It is evident that \(h(x)=\sin x\) is defined for every real number.
Let \(c\) be a real number.
Case 1:
If \(c< -1, then h(c)=-(c+1)\) and \(\lim_{x \to c}h(x)=\lim_{x \to c}[-(x+1)]=-(c+1)\)
\(therefore,\; \lim_{x \to c}h(x)=h(c)\)Therefore, \(h\) is continuous at all the points \(x\), such that \(x< -1\).
Case 2:
If \(c> -1, then h(c)=-c+1\) and \(\lim_{x \to c}h(x)=\lim_{x \to c}(x+1)=c+1\)
\(therefore,\; \lim_{x \to c}h(x)=h(c)\)Therefore, \(h\) is continuous at all the points \(x\), such that \(x> -1\).
Case 3:
If \(c= -1, then h(c)=h(-1)=-1+1=0\)
\(\lim_{x \to -1^{-}}h(x)=\lim_{x \to -1^{-}}[-(x+1)]=(-1+1)=0\) \(\lim_{x \to -1^{+}}h(x)=\lim_{x \to -1^{+}}(x+1)=(-1+1)=0\) \(therefore,\; \lim_{x \to -1^{-}}h(x)=\lim_{x \to -1^{+}}h(x)=h(-1)\)Therefore, \(h\) is continuous at \(x=-1\)
It can be concluded that \(h\) is continuous at all the real points.
Exercise 5.2 :
Q1: Differentiate the function with respect to \(x\)
\(f(x)=\ sin (x^{2}+5)\)
Sol:The given function is \(f(x)=\ sin (x^{2}+5)=y\)
Let \(t=(x^{2}+5)\)
\(So f(t)=\sin t\)\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\) ……..(i)
\(\frac{\mathrm{d} y}{\mathrm{d} t}=\cos t\) ……..(ii)
And \(\frac{\mathrm{d} t}{\mathrm{d} x}=2x\) …….(iii)
Substituting equation (ii) and (iii) in (i) we have,
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\cos t \times (2x)\)We know \(t=x^{2}+5\)
Thus \(\frac{\mathrm{d} y}{\mathrm{d} x}=(2x)\times \cos x\)
Q2: Differentiate the function with respect to \(x\)
\(f(x)=\cos (\sin x)\)
Sol: Based on formulae given in Continuity and Differentiability
Let \(f(x)=\cos (\sin x)\)
Here \(f\) is a composite function which can be written in the form of two composite function \(u and v\).
\(u(x)=\sin x\; \; and\; \; v(t)=\cos t\) \((vou)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)\)Put \(t=u(x)=\sin x\)
\(therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\cos t)}{\mathrm{d} x}=-\sin t = -\sin(\sin x)\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (\sin x)}{\mathrm{d} x}=\cos x\)By chain rule ,
\(\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\) \(\frac{\mathrm{d} f}{\mathrm{d} x}=-\sin (\sin x)\times \ cos x=-\cos x\sin (\sin x)\)
Q3: Differentiate the function with respect to \(x\)
\(f(x)=\sin (ax+b)\)
Sol: Based on formulae given in Continuity and Differentiability
Let \(f(x)=\sin (ax+b)\)
Here \(f\) is a composite function which can be written in the form of two composite function \(u and v\).
\(u(x)=ax+b\; \; and\; \; v(t)=\sin t\) \((vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)=f(x)\)Put \(t=u(x)=ax+b\)
\(therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} x}=\cos t = \cos(ax+b)\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a\)By chain rule ,
\(\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\) \(\frac{\mathrm{d} f}{\mathrm{d} x}=a\times \ cos (ax+b)\)
Q4: Differentiate the function with respect to \(x\)
\(\sec (\tan (\sqrt{x}))\).
Sol: Based on formulae given in Continuity and Differentiability
Let \(\sec (\tan (\sqrt{x}))\)
Here \(f\) is a composite function which can be written in the form of three composite function \(u,v and w\).
\(u(x)=\sqrt{x}\; \; v(t)=\tan t\; \; and w(s)=\sec s\) \((wovou)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w[\tan( \sqrt{x})]=\sec (\tan (\sqrt{x}))=f(x)\)Put \(s=v(t)=\tan t and t=u(x)=\sqrt{x}\)
then ,\(\frac{\mathrm{d} w}{\mathrm{d} s}=\frac{\mathrm{d} (\sec s)}{\mathrm{d} s}=\sec s\tan s =\sec(\tan t). \tan (\tan t)\) ( as \(s=\tan t\)
\(=\sec(\tan \sqrt{x}). \tan (\tan \sqrt{x})\) \(\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{\mathrm{d} (\tan t)}{\mathrm{d} t}=\sec ^{2}t=\sec ^{2}\sqrt{t}\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} \sqrt{x}}{\mathrm{d} x}=\frac{\mathrm{d} (x^{\frac{1}{2}})}{\mathrm{d} x}=\frac{1}{2}.x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}\)By chain rule ,
\(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} w}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\)=\(\sec (\tan\sqrt{x}).\tan (\tan\sqrt{x})\times \sec ^{2} \sqrt{x}\times \frac{1}{2\sqrt{x}}\)
=\(\frac{\sec ^{2}\sqrt{x}.\sec (\tan \sqrt{x}.\tan (\tan \sqrt{x}))}{2\sqrt{x}}\)
Q5: Differentiate the function with respect to \(x\)
\(f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}= \frac{g(x)}{h(x)}\),
where \(g(x)= \sin (ax+b) \) and \(h(x)=\cos (cx+d)\)
Consider \(g(x)=\sin (ax+b)\)
Here \(g\) is a composite function which can be written in the form of two composite function \(u and v\).
\(u(x)=ax+b\; \; v(t)=\sin t\) \((vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)]=g(x)\)Put \(t=u(x)=ax+b\)
\(\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} t}=\cos t=\cos (ax+b)\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a\)By chain rule ,
\(g^{‘}=\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}=\cos (ax+b).a=a \cos (ax+b)\)Consider \(h(x)=\cos (cx+d)\)
Here \(h\) is a composite function which can be also written in the form of two composite function \(p and q\) of type.
\(p(x)=cx+d\; \; q(y)=\cos y\) \((qop)(x)=q(p(x))=q(cx+d)=\cos (cx+d)]=h(x)\)Put \(y=p(x)=cx+d\)
\(\frac{\mathrm{d} q}{\mathrm{d} y}=\frac{\mathrm{d} (\cos y)}{\mathrm{d} y}=-\sin y=-\sin (cx+d)\) \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} (cx+d)}{\mathrm{d} x}=c\)By chain rule ,
\(h^{‘}=\frac{\mathrm{d} h}{\mathrm{d} x}=\frac{\mathrm{d} q}{\mathrm{d} y}\times \frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (cx+d).c=-c \sin (cx+d)\)Therefore by chain rule , we obtain
\(f^{‘}=\frac{a\cos (ax+b).\ cos (cx+d)-\sin (ax+b)(-\sin(cx+d))}{[\cos(cx+d)]^{2}}\)=\(=\frac{a\cos (ax+b)}{\ cos (cx+d)}+c\sin (ax+b).\frac{\sin (cx+d)}{\cos (cx+d)}\times \frac{1}{\cos (cx+d)}\)
=\(=a\cos (ax+b)\ sec (cx+d)+c\sin (ax+b).\tan (cx+d).\sec (cx+d)\)
Q6: Differentiate the function with respect to \(x\)
\(f(x)=\cos x^{3}.\sin ^{2}(x^{5})\)
Sol:
The given function is \(f(x)=\cos x^{3}.\sin ^{2}(x^{5})\)
\(\frac{\mathrm{d} }{\mathrm{d} x}\left [ \cos x^{3}.\sin ^{2}(x^{5}) \right ]=\sin ^{2}(x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\cos x^{3}+\cos x^{3}\times \frac{\mathrm{d} }{\mathrm{d} x}\sin ^{2}(x^{5})\)=\(\sin ^{2}(x^{5})\times (-\sin x^{3})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{3})+\cos x^{3}\times 2\sin (x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \sin (x^{5}) \right ]\)
\(=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\frac{\mathrm{d} }{\mathrm{d} x} (x^{5})\) \(=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\times 5x^{4}\) \(=10x^{4}\sin x^{5}.\cos x^{3}.\cos x^{5}\times -3x^{2}\sin x^{3}.\sin ^{2}(x^{5} )\)
Q7: Differentiate the functions with respect to x.
\(2\sqrt{\cot (x^{2})}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(2\sqrt{\cot (x^{2})}\)
\(\frac{\mathrm{d} }{\mathrm{d} x}2\sqrt{\cot (x^{2})}\) \(=2.\frac{1}{2\sqrt{\cot (x^{2})}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \cot (x^{2}) \right ]\) \(=\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times -cosec^{2} (x^{2})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})\) \(=-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times \frac{1}{\sin ^{2(x^{2})}}\times(2x)\) \(=\frac{-2x}{\sqrt{\cos (x^{2})}\sqrt{\sin (x^{2})}\sin (x^{2})}\) \(=\frac{-2\sqrt{2}x}{\sqrt{2\cos (x^{2})\sin (x^{2})}\sin (x^{2})}\) \(=\frac{-2\sqrt{2}x}{\sin (x^{2})\sqrt{\sin 2(x^{2})}}\)
Q8: Differentiate the functions with respect to x.
\(\cos\sqrt{x}\)
Sol: Based on formulae given in Continuity and Differentiability
The given function \(f(x)\) is \(\cos\sqrt{x}\).
Let \(u(x)=\sqrt{x}\)
And \(v(t)=\cos t\)
\((vou)(x)=v(u(x))\) \(=v(\sqrt{x})\) \(=\cos (\sqrt{x})\)=\(f(x)\)
Clearly, \(f\) is a composite function of two functions, \(u\; and \; v\), such that
\(t=u(x)=\sqrt{x}\)Then, \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}\)
\(=\frac{1}{2\sqrt{x}}\)And, \(\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}(\cos t)=-\sin t\)
\(=-\sin (\sqrt{x})\)By chain rule we have,
\(\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}.\frac{\mathrm{d} t}{\mathrm{d} x}\).
\(=-\sin (\sqrt{x}).\frac{1}{2\sqrt{x}}\) \(=-\frac{1}{2\sqrt{x}}\sin (\sqrt{x})\) \(=-\frac{\sin (\sqrt{x})}{2\sqrt{x}}\)
Q9: Prove that the function f given by \(f(x)=\left | x-1 \right |,x\in \mathbb{R}\), is not differentiable at \(x=1\).
Sol: Based on formulae given in Continuity and Differentiability
The given function is \(f(x)=\left | x-1 \right |,x\in \mathbb{R}\).
We have known that a function \(f\) is differentiable at a point \(x=c\) in its domain if the right hand limit and the left hand limit are finite and equal.
To check the differentiability of the given function at x=1,
The right hand and the left hand limits where x=c are
\(\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}\) and \(\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}\)
Considering the right hand limit of the given function at x=1
\(\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{+}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}\) \(=\lim_{h \to 0^{+}}\frac{\left | h \right |-\left | 0 \right |}{h}\) \(=\lim_{h \to 0^{+}}\frac{h}{h}\) \(=1\)Considering the left hand limit of the given function at x=1
\(\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{-}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}\) \(=\lim_{h \to 0^{-}}\frac{\left | h \right |-\left | 0 \right |}{h}\) \(=\lim_{h \to 0^{-}}\frac{-h}{h}\) \(=-1\)Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1
Q10: Prove that the greatest integer function defined by \(f(x)=\left [ x \right ], 0<x<3\) is not differentiable at x = 1 and x = 2.
Sol:
The function f is \(f(x)=\left [ x \right ], 0<x<3\)
It is known that a function f is differentiable at a point \(x=c\) in its domain if both the left hand and the left hand limit are equal
\(\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}\) and \(\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}\) are finite and equal.
To check the differentiability of the given function at \(x=1\), consider the right hand limit of f at \(x=1\)
\(\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{+}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}\) \(=\lim_{h \to 0^{+}}\frac{1-1}{h}\) \(=\lim_{h \to 0^{+}}\frac{1-1}{h}=\lim_{h \to 0^{+}}(0)=0\)Now consider the left hand limit of f at \(x=1\)
\(\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{-}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}\) \(=\lim_{h \to 0^{-}}\frac{0-1}{h}\) \(=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty \)Since left hand and the right hand limit of f at x=1 are not equal, f is not differentiable at x=1.
Now to check the differentiability of the given function at x=2,
consider the left hand limit at x=2.
\(\lim_{h \to 0^{-}}\frac{f(2+h)-f(2)}{h}\) \(\lim_{h \to 0^{-}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}\) \(=\lim_{h \to 0^{-}}\frac{1-2}{h}\) \(=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty \)Now consider the right hand limit of f at \(x=2\)
\(\lim_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}\) \(\lim_{h \to 0^{+}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}\) \(=\lim_{h \to 0^{+}}\frac{2-2}{h}\) \(=\lim_{h \to 0^{+}}\frac{0}{h}=\lim_{h \to 0^{+}}(0)=0\)Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2
Exercise 5.3
Q1: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(2x+3y=\sin x\)
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(2x+3y=\sin x\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)\) \(\Rightarrow 2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x\) \(\Rightarrow 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x-2\)\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\cos x-2}{3}\).
Q2: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(2x+3y=\sin y\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(2x+3y=\sin y\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)\) \(\Rightarrow 2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos y \frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow 2=(\cos y-3) \frac{\mathrm{d} y}{\mathrm{d} x}\)\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\cos y -3}\).
Q3: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(ax+by^{2}=\cos y\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(ax+by^{2}=\cos y\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(ax+by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (ax) + \frac{\mathrm{d} }{\mathrm{d} x}(by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)\) \(\Rightarrow a + 2by\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin y \frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow a=-(\sin y+2by) \frac{\mathrm{d} y}{\mathrm{d} x}\)\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-a}{(\sin y+2by )}\).
Q4: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(xy+y^{2}=\tan x+ y\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(xy+y^{2}=\tan x+ y\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x + y)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x)+ \frac{\mathrm{d} y}{\mathrm{d} x} \) \(\Rightarrow y.\frac{\mathrm{d} }{\mathrm{d} x} (x)+x .\frac{\mathrm{d} y}{\mathrm{d} x} +2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x} \) \(\Rightarrow y.1 + x.\frac{\mathrm{d} y}{\mathrm{d} x}+2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow (x+2y-1)\frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x-y\)\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sec ^{2}x-y}{(x+2y-1 )}\).
Q5: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(x^{2}+xy+y^{2}=100\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(x^{2}+xy+y^{2}=100\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}+xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(100)\)\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (x^{2})+\frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=0 \) (derivatives of constant function is 0)
\(\Rightarrow 2x+ y.1+x.\frac{\mathrm{d} y}{\mathrm{d} x} +2y.\frac{\mathrm{d} y}{\mathrm{d} x}=0 \) \(\Rightarrow 2x+ y+(x+2y)\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2x+y}{x+2y}\)
Q6: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(x^{3}+x^{2}y+xy^{2}+y^{3}\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(x^{3}+x^{2}y+xy^{2}+y^{3}=81\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(x^{3}+x^{2}y+xy^{2}+y^{3})=\frac{\mathrm{d} }{\mathrm{d} x}(81)\)\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (x^{3})+\frac{\mathrm{d} }{\mathrm{d} x} (x^{2}y) + \frac{\mathrm{d} }{\mathrm{d} x}(xy^{2})+\frac{\mathrm{d} }{\mathrm{d} x}(y^{3})=0 \) (derivatives of constant function is 0)
\(\Rightarrow 3x^{2}+ y.\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}\frac{\mathrm{d} }{\mathrm{d} x}(x)+x\frac{\mathrm{d} }{\mathrm{d} x}(y^{2}+3y^{2}.\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow 3x^{2}+ y.2x+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}.1+x.2y.\frac{\mathrm{d} y}{\mathrm{d} x}+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow (x^{2}+2xy+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+(3x^{2}+2xy+y^{2}=0\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-(3x^{2}+2xy+y^{2})}{x^{2}+2xy+3y^{2}}\)
Q7: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(\sin ^{2}y +\cos xy=\Pi\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(\sin ^{2}y+\cos xy=\Pi\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y+\cos xy)=\frac{\mathrm{d} }{\mathrm{d} x}(\Pi)\)\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos xy) =0 \) ……(i) (derivatives of constant function is 0)
Using chain rule,we get
\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y=2 \sin y \frac{\mathrm{d} }{\mathrm{d} x}(\sin y) =2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}\) …….(ii)
\(\frac{\mathrm{d} }{\mathrm{d} x}(\cos xy)=-\sin xy \frac{\mathrm{d} }{\mathrm{d} x}(xy)= -\sin xy (y. \frac{\mathrm{d} }{\mathrm{d} x}(x)+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -\sin xy (y. 1+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -y \sin xy -x\sin xy \frac{\mathrm{d} y}{\mathrm{d} x}\) …….(iii)
From (i), (ii) and (iii) we have
\( 2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}- y\sin xy- x \sin xy \frac{\mathrm{d} y}{\mathrm{d} x}=0\) \(\Rightarrow (2\sin y \cos y- x \sin xy)\frac{\mathrm{d} y}{\mathrm{d} x}= y \sin xy\) \(\Rightarrow (\sin 2y-x\sin xy )\frac{\mathrm{d} y}{\mathrm{d} x}=y \sin xy\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y \sin xy}{\sin 2y -x \sin xy}\)
Q8: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(\sin ^{2}x +\cos ^{y}=1\).
Sol:
The given relationship is \(\sin ^{2}x +\cos ^{y}=1\)
Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}x+\cos ^{y})=\frac{\mathrm{d} }{\mathrm{d} x}(1)\)\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos ^{y}) =0 \) (derivatives of constant function is 0)
\(\Rightarrow 2\sin x \frac{\mathrm{d} }{\mathrm{d} x} (\sin x)+2\cos y.\frac{\mathrm{d} }{\mathrm{d} x} (\cos y) =0 \) \(\Rightarrow 2\sin x \cos x+2\cos y(-\sin y).\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow \sin 2x-\sin 2y.\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin 2x}{\sin 2y}\)
Q9: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )\).
Sol:
The given relationship is \(y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )\)
\(\sin y=\left ( \frac{2x}{1+x^{2}} \right )\)Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )\)\(\Rightarrow \cos y \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{2x}{1+x^{2}} \right ) \) ………..(i)
The right side function \( \left ( \frac{2x}{1+x^{2}} \right ) \) is of the form \(\frac{u}{v}\)
So, by quotient rule, Based on formulae given in Continuity and Differentiability
\(\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )=\frac{(1+x^{2}).\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}\)\(=\frac{(1+x^{2}).2-2x.(0+2x)}{(1+x^{2})^{2}}=\frac{2+2x^{2}-4x^{2}}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}\) ………(ii)
Also \(\sin y=\frac{2x}{1+x^{2}}\)
\(\Rightarrow \cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}=\sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}\)\(=\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\frac{1-x^{2}}{1+x^{2}}\) ……….(iii)
From (i) , (ii) and (iii) we obtain
\(=\frac{1-x^{2}}{1+x^{2}}\times \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x^{2}}\)
Q10: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right ),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )\)
\(\tan y=\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )\) ………(i)
Differentiating the equation with respect to x, we have
We know that, \(\tan y=\frac{3\tan \frac{y}{3}-\tan ^{3}\frac{y}{3}}{1-3\tan ^{2}\frac{y}{3}}\) …………(ii)
Comparing equation (i) and (ii), we have
\(x=\tan \frac{y}{3}\)Differentiating this relationship w.r.t. x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \tan \frac{y}{3} \right )\) \(\Rightarrow 1=\sec ^{2}\frac{y}{3}.\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{y}{3} \right )\) \(\Rightarrow 1=\sec ^{2}\frac{y}{3}.\frac{1}{3}\frac{\mathrm{d} y}{\mathrm{d} x} \) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{\sec ^{2}\frac{y}{3}}=\frac{3}{1+\tan ^{2}\frac{y}{3}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{1+x^{2}}\)
Q11: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\)
\(\cos y= \left ( \frac{1-x^{2}}{1+x^{2}} \right )\) \(\frac{1-\tan ^{2}\frac{y}{2}}{1+\tan ^{2}\frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}}\)Comparing both sides equation
\(\tan \frac{y}{2}=x\)Differentiating the equation with respect to x, we have
\(\sec ^{2} \frac{y}{2}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )=\frac{\mathrm{d} }{\mathrm{d} x}(x)\) \(\sec ^{2} \frac{y}{2}\times \frac{1}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}=1\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sec ^{2}\frac{y}{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+\tan ^{2}\frac{y}{2}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x ^{2}}\)
Q12: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\)
\(\sin y= \left ( \frac{1-x^{2}}{1+x^{2}} \right )\)Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )\) ……..(i)
Using chain rule
\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\cos y \frac{\mathrm{d}y }{\mathrm{d} x}\) \(\cos y=\sqrt{1-\sin ^{2}y}=\sqrt{1-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}\) \(=\sqrt{({1+x^{2}})^{2}-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}\) \(=\sqrt{ \frac{({1+x^{2}})^{2}-(1-x^{2})}{(1+x^{2})^{2}} ^{2}}=\sqrt{\frac{4x^{2}}{(1+x^{2})^{2}}}=\frac{2x}{1+x^{2}}\)\(therefore,\; \frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{2x}{1+x^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}\) …..(ii)
\(\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )=\frac{(1+x^{2}).(1-x^{2})-(1-x^{2}).(1+x^{2})}{(1+x^{2})^{2}}\) [using quotient rule]
\(=\frac{(1+x^{2}).(-2x)-(1-x^{2}).(2x)}{(1+x^{2})^{2}}\) \(=\frac{-2x-2x^{3}-2x+2x^{3}}{(1+x^{2})^{2}}\)\(=\frac{-4x}{(1+x^{2})^{2}}\) ……..(iii)
From (i),(ii) and (iii) we have
\(\frac{2x}{(1+x^{2})}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-4x}{(1+x^{2})^{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{(1+x^{2})}\)
Q13: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right ), -1<x<1\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right )\)
\(\cos y= \left ( \frac{2x}{1+x^{2}} \right )\)Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )\) \(-\sin y \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{(1+x^{2}.\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}\) \(\Rightarrow -\sqrt{1-\cos^{2}y}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{(1+x^{2}.2-2x.2x)}{(1+x^{2})^{2}}\) \(\Rightarrow \sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]\) \(\Rightarrow \sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]\) \(\Rightarrow \sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]\) \(\Rightarrow \frac{(1-x^{2})}{(1+x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{1+x^{2}}\)
Q14: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
\(\sin y= (2x\sqrt{1-x^{2}})\)Differentiating the equation with respect to x, we have
\(\cos y{\mathrm{d} y}{\mathrm{d} x}=2\left [ x.\frac{\mathrm{d} }{\mathrm{d} x}\sqrt{1-x^{2}} +\sqrt{1-x^{2}}\frac{\mathrm{d} x}{\mathrm{d} x}\right ]\) \(\sqrt{1-\sin^{2}y} \frac{\mathrm{d}y }{\mathrm{d} x}=2\left [ \frac{x}{2}.\frac{-2x}{\sqrt{1-x^{2}}} +\sqrt{1-x^{2}}\right ]\) \(\Rightarrow \sqrt{1-(2x\sqrt{1-x^{2}})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}} \right ]\) \(\Rightarrow \sqrt{1-4x^{2}(1-x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]\) \(\Rightarrow \sqrt{(1-2x^{2})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sqrt{1-x^{2}}}\)
Q15: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)
\(y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}\).
Sol: Based on formulae given in Continuity and Differentiability
The given relationship is \(y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}\)
\(\sec y= \left ( \frac{1}{2x^{2}-1} \right )\) \(\Rightarrow \cos y=2x^{2}-1\) \(\Rightarrow 2x^{2}=1+\cos y\) \(\Rightarrow 2x^{2}=2\cos {2}\frac{y}{2}\) \(\Rightarrow x=cos \frac{y}{2}\)Differentiating the equation with respect to x, we have
\(\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \frac{y}{2} \right )\) \(\Rightarrow 1=-\sin \frac{y}{2}.\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )\) \(\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2}\frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos^{2}\frac{y}{2}}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sqrt{1-x^{2}}}\)
Exercise 5.4
Q1: Differentiate the following w.r.t. x
\(\frac{e^{x}}{\sin x}\)
Sol: Based on formulae given in Continuity and Differentiability
Let y=\(\frac{e^{x}}{\sin x}\)
Using quotient rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin x.\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})-e^{x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)}{\sin^{2} x}\) \(=\frac{\sin x.(e^{x})-e^{x}.(\cos x)}{\sin^{2} x}\) \(=\frac{e^{x}(\sin x-\cos x)}{\sin^{2} x},x\neq n\pi ,n\in \mathbb{Z}\)
Q2: Differentiate the following w.r.t. x
\(e^{\sin ^{-1}x}\)
Sol: Based on formulae given in Continuity and Differentiability
Let y=\(y=e^{\sin ^{-1}x}\)
Using chain rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{\sin ^{-1}x})\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)\) \(=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}\) \(=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)\)
Q3: Differentiate the following w.r.t. x
\(e^{x^{3}}\)
Sol: Based on formulae given in Continuity and Differentiability
Let y=\(e^{x^{3}}\)
Using chain rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{x^{3}})=e^{x^{3}}.3x^{2}=3x^{2}.e^{x^{3}}\)
Q4: Differentiate the following w.r.t. x
\(\sin (\tan^{-1} e^{-x})\)
Sol: Based on formulae given in Continuity and Differentiability
Let y=\(\sin (\tan^{-1} e^{-x})\)
Using chain rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[\sin (\tan^{-1} e^{-x})]\) \(=\cos (\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1} e^{-x})\) \(=\cos (\tan^{-1} e^{-x}).\frac{1}{1+(e^{-x})^{2}}(\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(e^-{x})\) \(=\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{\mathrm{d} }{\mathrm{d} x}(-x)\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)\) \(=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}\) \(=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)\)