A function is a rule that attributes to every number x from D one definite number y where y belongs to the set of real numbers. Here, x is called the independent variable and y is called the dependent variable
The set D is referred to as the domain of definition of the function and the set of all values attained by the variable y is called the range of the function
In other words, a variable y is said to be the value of function of a variable x in the domain of definition D if to each value of x belonging to this domain there corresponds a definite value of the variable y. This is symbolised as y = f(x) where f denotes the rule by which y varies with x.
For representing functions through a table, we simply write down a sequence of values of the independent variable x and then write down the corresponding values of the dependent variable y.
The process is: on the coordinate xy plane for every value of x from the domain D of the function, a point P(x, y) is constructed whose abscissa is x and whose ordinate y is got by putting the particular value of x in the formula representing the function.
For example, for plotting the function y = x^{2} , we first decide on the values of x for which we need to plot the graph. Thus we can take x = 0 and get y = 0 (means the point (0, 0) is on the graph). Then for x = 1, y = 1 ; for x = 2, y = 4; for x = 3, y = 9 and for x = –1 y = 1; for x = –2 , y = 4, and so on
Let a function y = f(x) be given in a certain interval. The function is said to be even if for any value of x , f(x) = f(–x)
Properties of even functions: (a) The sum, difference, product and quotient of an even function is also an even function. (b) The graph of an even function is symmetrical about the y-axis.
However, when y is the independent variable, it is symmetrical about the x-axis. In other words, if x = f(y) is an even function, then the graph of this function will be symmetrical about the x-axis. Example: x = y^{2}
Examples of even functions: y = x^{2}, y = x^{4}, y = –3x^{8}, y = x^{2} + 3 , y = x^{4}/5, y = | x | are all even functions.
Let a function y = f(x) be given in a certain interval. The function is said to be odd if for any value of x; f(x) = – f(–x)
Properties of odd functions. (a) The sum and difference of an odd function is an odd function. (b) The product and quotient of an odd function is an even function. (c) The graph of an odd function is symmetrical about the origin.
Examples of odd functions y = x^{3}, y = x^{5}, y = x^{3} + x, y = x/(x^{2} + 1).
Not all functions need be even or odd. However, every function can be represented as the sum of an even function and an odd function.
Let there be a function y = f(x), which is defined for the domain D and has a range R.
Then, by definition, for every value of the independent variable x in the domain D, there exists a certain value of the dependent variable y. In certain cases the same value of the dependent variable y can be got for different values of x. For example, if y = x^{2} , then for x = 2 and x = –2 give the value of y as 4.
In such a case, the inverse function of the function y = f(x) does not exist
However, if a function y = f(x) is such that for every value of y (from the range of the function R) there corresponds one and only one value of x from the domain D, then the inverse function of y = f(x) exists and is given by x = g(y). Here it can be noticed that x becomes the dependent variable and y becomes the independent variable. Hence, this function has a domain R and a range D
Under the above situation, the graph of y = f(x) and x = g(y) are one and the same
However, when denoting the inverse of the function, we normally denote the independent variable by y and, hence, the inverse function of y = f(x) is denoted by y = g(x) and not by x = g(y).
The graphs of two inverse functions when this change is used are symmetrical about the line y = x (which is the bisector of the first and the third quadrants)
The relationship between the graph of y = f(x) and y = f(x) + c (where c represents a positive constant.): The shape of the graph of y = f(x) + c will be the same as that of the y = f(x) graph. The only difference would be in terms of the fact that f(x) + c is shifted c units up on the x – y plot.
Relationship between y = x^{2} and y = x^{2} + 2.
The relationship between y = f(x) and y = f(x) – c: In this case while the shape remains the same, the position of the graph gets shifted c units down.
The relationship between y = f(x) and y = f(x + c): In this case the graph will get shifted c units to the left. (Remember, c was a positive constant)
The relationship between y = f(x) and y = f(x – c): In this case the graph will get shifted c units to the right on the x – y plane.
Visualizing a graph for a function like x^{2} – 4x + 7. First convert x^{2} – 4x + 7 into (x – 2)^{2} + 3
[Note: In order to do this conversion, the key point of your thinking should be on the –4x. Your first focus has to be to put down a bracket (x – a)^{2} which on expansion gives – 4x as the middle term.
When you think this way, you will get (x – 2)^{2} . On expansion (x – 2)^{2} = x^{2} – 4x + 4. But you wanted x^{2} – 4x + 7. Hence add +3 to (x – 2)^{2} . Hence the expression x^{2} – 4x + 7 is equivalent to (x – 2)^{2} + 3.]
To visualize (x – 2)^{2} + 3 shift the x^{2} graph two units right [to account for (x – 2)^{2}] and 3 units up [to account for the +3] on the x – y plot. This will give you the required plot.
Use options for solving. If a function is even it should satisfy the equation f(x) = f(–x) f (x) = e^{2x} + e^{–2x} Putting (–x) at the place of x. f (–x) = e^{–2x} + e^{–(-2x)} = e^{–2x} + e^{2x} = f(x)
The graph is symmetrical about the y-axis. This is the definition of an even function.
A graph is symmetrical about origin. This is the definition of an odd function. B graph is neither symmetrical about the y-axis nor about origin.
For two functions to be identical, their domains should be equal. Checking the domains of f(x), g(x) and h(x), f(x) = x^{2}/x, x should not be equal to zero. So, domain will be all real numbers except at x = 0. g(x) = (√2)^{2}, x should be non-negative. So, domain will be all positive real numbers. h(x) = x, x is defined every where. So, we can see that none of them have the same domain. Hence, (d) is the correct option.
fogoh (2) is the same as f(g(h(2))) To solve this, open the innermost bracket first. This means that we first resolve the function h(2). Since h(2) = 4 we will get f(g(h(2))) = f(g(4)) = f(–1/3) = –3. Hence, the option (d) is the correct answer.
For solving questions containing a function in the question as well as a function in the options (where values are absent), the safest process for students weak at math is to assume certain convenient values of the variables in the expression and checking for the correct option that gives us equality with the expression in the question. The advantages of this process of solution is that there is very little scope for making mistakes. Besides, if the expression is not simple and directly visible, this process takes far less time as compared to simplifying the expression from one form to another. This process will be clear after perusing the following solution to the above problem. Solution The problem statement above defines the expression: (1/x) + (1/y) + 1 and asks us to find out which of the four options is equal to this expression. If we try to simplify, we can start from the problem expression and rewrite it to get the correct option. However, in the above case this will become extremely complicated since the symbols are indirect. Hence, if we have to solve through simplifying, we should start from the options one by one and try to get the problem expression. However, this is easier said than done and for this particular problem, going through this approach will take you at least two minutes plus. Hence, consider the following approach: Take the values of x and y as 1 each. Then, (1/x) + (1/y) + 1 = 3 Put the value of x and y as 1 each in each of the four options that we have to consider. Option (a) will give a value of –1 π 3. Hence, option (a) is incorrect. Opton (b) will give a value of 0. 0 π 3. Hence, option (b) is incorrect. Option (c) gives an answer of 3. Hence, option (c) could be the answer. Option (d) gives an answer of 0. 0 π 3. Hence, option (d) is incorrect. Now since options (a), (b) and (d) are incorrect and option (c) is the only possibility left, it has to be the answer. [Note that in case there is a fourth option of ‘none of these’ then we have to be careful before marking the answer.]
y = |x| will be defined for all values of x. From = – ∞ to + ∞
For y = √x; to be defined, x should be non-negative. i.e. x ≥ 0.
Since the function contains a √x in it, x ≥ 0 would be the domain.
For (x – 2)^{1/2} to be defined x ≥ 2. For (8 – x)^{1/2} to be defined x <= 8. Thus, 2 <= x <= 8 would be the required domain
(9 – x^{2}) ≥ 0 then –3 <= x <= 3
The function would be defined for all values of x except where the denominator viz: x^{2} – 4x + 3 becomes equal to zero. The roots of x^{2} – 4x + 3x = 0 being 1, 3, it follows that the domain of definition of the function would be all values of x except x = 1 and x = 3.
f(x) = x and g(x) = (√x)^{2} would be identical if is defined. Hence, x ≥ 0 would be the answer
f(x) = x is defined for all values of x. g(x) = x^{2}/x also returns the same values as f(x) except at x = 0 where it is not defined. Hence. option (a).
7 f(x) = 7 e^{x}
While log x^{2} is defined for – ∞ < x < ∞ , 2 log x is only defined for 0 < x < ∞. Thus, the two functions are identical for 0 < x < ∞.
y – axis by definition.
Origin by definition.
x^{–8} is even since f(x) = f(–x) in this case
(x + 1)^{3} is not odd as f(x) is -f(–x).
dy/dx = 2x + 10 = 0 when x = –5.
Required value = (–5)^{2} + 10(–5) + 11 = 25 – 50 + 11 = –14
Since the denominator x^{2} – 3x + 2 has real roots, the maximum value would be infinity
The minimum value of the function would occur at the minimum value of (x^{2} – 2x + 5) as this quadratic function has imaginary roots. For y = x^{2} – 2x + 5 dy/dx = 2x – 2 = 0 fi x = 1 fi x^{2} – 2x + 5 = 4. Thus, minimum value of the argument of the log is 4. So minimum value of the function is log_{2} 4 = 2
y = 1/x + 1 Hence, y – 1 = 1/x fi x = 1/(y – 1) Thus f^{–1}(x) = 1/(x – 1).
f(1) = 0, f(2) = 1, f(3) = f(1) – f(2) = –1 f(4) = f(2) – f(3) = 2 f(5) = f(3) – f(4) = –3 f(6) = f(4) – f(5) = 5 f(7) = f(5) – f(6) = –8 f(8) = f(6) – f(7) = 13 f(9) = f(7) – f(8) = –21
–8 + 2 = –6
0 + 1 – 1 + 2 – 3 + 5 – 8 + 13 – 21 = –12.
For any ^{n}C_{r} , n should be positive and r ≥ 0. Thus, for positive x, 5 – x ≥ 0 fi x = 1, 2, 3, 4, 5.
Mark a if f(–x) = f(x);
Mark b if f(–x) = – f(x)
Mark c if neither a nor b is true
Mark d if f(x) does not exist at at least one point of the domain
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd. Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things. Either the function is returning two values for one value of x. or the function has a break in between (not seen in any of these questions). We see even functions in [Symmetry about the y axis]. We see odd functions in Symmetry about the Origin by definition.
a @ b = \( \frac{a+b}{2} \)
a # b = a^{2} - b^{2}
a ! b = \( \frac{a-b}{2} \)
{[(3@4)! (3 #2)] @ [(4!3) @ (2 # 3)]} {[(3.5) ! (5)] @ [ (0.5) @ (–5)]} {[–0.75] @ [–2.25]} = –1.5.
(7) @ (–0.5) = 3.25.
0 @ 0.5 = 0.25. Thus, a
b = (1) (4) = 4. C = 16/1 * 4 = 4 Hence, both (b) and (c).
(a) will always be true because (a + b)/2 would always be greater then (a – b)/2 for the given value range. Further, a^{2} – b^{2} would always be less than a^{3} – b^{3} . Thus, option (d) is correct.
(a) (a M b) = a – b (b) (a D b) = a + b (c) (a H b) = (ab) (d) (a P b) = a/b
Option a = (a – b) (a + b) = a^{2} – b^{2}
Option a = (a^{2} – b^{2}) + b^{2} = a^{2}
3 – 4 × 2 + 4/8 – 2 = 3 – 8 + 0.5 – 2 = – 6.5 (using BODMAS rule)
The maximum would depend on the values of a and b. Thus, cannot be determined
The minimum would depend on the values of a and b. Thus, cannot be determined
Any of (a + b) or a/b could be greater and thus we cannot determine this.
Again (a + b) or a/b can both be greater than each other depending on the values we take for a and b. E.g. for a = 0.9 and b = 0.91, a + b > a/b. For a = 0.1 and b = 0.11, a + b < a/b