**CBSE Class 11 Maths – MCQ and Online Tests – Unit 2 – Relations and Functions**

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**CBSE Class 11 Maths – MCQ and Online Tests – Unit 2 – Relations and Functions**

Question 1.

The number of binary operations on the set {a, b} are

(a) 2

(b) 4

(c) 8

(d) 16

## Answer

Answer: (d) 16

Hint:

Let S is a finite set containing n elements.

Since binary operation on S is a function from SŨS to S, therefore total number of

binary operations on S is the

total number of functions from SŨS to S = (n^{n})ē

Given Set = {a, b}

Total number of elements = 2

Total number of binary operations = (2ē)ē = 2^{4} = 16

Question 2.

If f is an even function and g is an odd function the fog is a/an

(a) Even function

(b) Odd function

(c) Either even or odd function

(d) Neither even nor odd function

## Answer

Answer: (a) Even function

Hint:

Given, f is an even function and g is an odd function.

Now, fog(-x) = f{g(-x)}

= f{-g(x)} {since g is an odd function}

= f{g(x)} for all x {since f is an even function}

So, fog is an even function.

Question 3.

If f(x) = e^{x} and g(x) = log_{e} x then the value of fog(1) is

(a) 0

(b) 1

(c) -1

(d) None of these

## Answer

Answer: (b) 1

Hint:

Given, f(x) = e^{x}

and g(x) = log x

fog(x) = f(g(x))

= f (log x)

= e^{log x}

= x

So, fog(1) = 1

Question 4.

Two functions f and g are said to be equal if f

(a) the domain of f = the domain of g

(b) the co-domain of f = the co-domain of g

(c) f(x) = g(x) for all x

(d) all of above

## Answer

Answer: (d) all of above

Hint:

Two functions f and g are said to be equal if f

1. the domain of f = the domain of g

2. the co-domain of f = the co-domain of g

3. f(x) = g(x) for all x

Question 5.

A function f(x) is said to be an odd function if

(a) f(-x) = f(x)

(b) f(-x) = -f(x)

(c) f(-x) = k * f(x) where k is a constant

(d) None of these

## Answer

Answer: (b) f(-x) = -f(x)

Hint:

A function f(x) is said to be an odd function if

f(-x) = -f(x) for all x

Question 6.

If f(x) is an odd differentiable function on R, then df(x)/dx is a/an

(a) Even function

(b) Odd function

(c) Either even or odd function

(d) Neither even nor odd function

## Answer

Answer: (a) Even function

Hint:

Given, f(x) is an odd differentiable function on R

? f(-x) = -f(x) for all x ? R

differentiate on both side, we get

? -df(-x)/dx = -df(x)/dx for all x ? R

? df(-x)/dx = df(x)/dx for all x ? R

? df(x)/dx is an even function on R.

Question 7.

The function f(x) = sin (?px/2) + cos (px/2) is periodic with period

(a) 4

(b) 6

(c) 12

(d) 24

## Answer

Answer: (a) 4

Hint:

Period of sin (?px/2) = 2p/(p/2) = 4

Period of cos (px/2) = 2p/(p/2) = 4

So, period of f(x) = LCM (4, 4) = 4

Question 8.

If f(x) = log_{3} x and A = (3, 27) then f(A) =

(a) (1, 1)

(b) (3, 3)

(c) (1, 3)

(d) (2, 3)

## Answer

Answer: (c) (1, 3)

Hint:

Since f(x) = log_{3} x is an increasing function

So, f(A) = (log_{3} 3, log_{3} 27) = (1, 3)

Question 9.

If f(x) = (a – x)^{1/n}, a > 0 and n ? N, then the value of f(f(x)) is

(a) 1/x

(b) x

(c) xē

(d) x^{1/2}

## Answer

Answer: (b) x

Hint:

Given, f(x) = (a – x)^{1/n}

Now, f(f(x)) = [(a – f(x))^{n}]^{1/n}

? f(f(x)) = [(a – {(a – x^{n})^{1/n} }^{n} ]^{1/n}

? f(f(x)) = [a – (a – x^{n})]^{1/n}

? f(f(x)) = [a – a + x^{n})]^{1/n}

? f(f(x)) = (x^{n})^{1/n}

? f(f(x)) = x

Question 10.

The domain of the definition of the real function f(x) = v(log_{12} xē ) of the real variable x is

(a) x > 0

(b) |x| = 1

(c) |x| > 4

(d) x = 4

## Answer

Answer: (b) |x| = 1

Hint:

We have f(x) = v(log_{12} xē)

Since, log_{a} k = 0 if a > 1, k = 1

or 0 < a < 1 and 0 < k = 1

So, the function f(x) exists if

log_{12} xē = 0

? xē = 1

? |x| = 1

Question 11.

Let A = {-2, -1, 0} and f(x) = 2x – 3 then the range of f is

(a) {7, -5, -3}

(b) {-7, 5, -3}

(c) {-7, -5, 3}

(d) {-7, -5, -3}

## Answer

Answer: (d) {-7, -5, -3}

Hint:

Given, A = {-2, -1, 0}

and f(x) = 2x – 3

Now, f(-2) = 2 Ũ (-2) – 3 = -4 – 3 = -7

f(-1) = 2 Ũ (-1) – 3 = -2 – 3 = -5

f(0) = 2 Ũ 0 – 3 = -3

So, range of f = {-7, -5, -3}

Question 12.

If f(x) =(3x – 2)/(2x – 3) then the value of f(f(x)) is

(a) x

(b) xē

(c) xģ

(d) None of these

## Answer

Answer: (a) x

Hint:

Given, f(x) = (3x – 2)/(2x – 3)

Now, f(f(x)) = f{(3x – 2)/(2x – 3)}

= {(3Ũ(3x – 2)/(2x – 3) – 2)}/{(2(3x – 2)/(2x – 3) – 3)}

= {(9x – 6)/(2x – 3) – 2)}/{((6x – 4)/(2x – 3) – 3)}

= [{(9x – 6) – 2(2x – 3)}/(2x – 3)]/[{(6x – 4) – 3(2x – 3)}/(2x – 3)]

= {(9x – 6) – 2(2x – 3)}/{(6x – 4) – 3(2x – 3)}

= (9x – 6 – 4x + 6)/(6x – 4 – 6x + 9)

= 5x/5

= x

So, f(f(x)) = x

Question 13.

Let R be the set of real numbers. If f(x) = xē and g(x) = 2x + 1, then fog(x) is equal to

(a) 2x + 1

(b) 2xē + 1

(c) (2x + 1)ē

(d) None of these

## Answer

Answer: (b) 2xē + 1

Hint:

Given, f(x) = xē and g(x) = 2x + 1

Now gof(x) = g(f(x)) = f(xē) = 2xē + 1

Question 14.

A relation R is defined from the set of integers to the set of real numbers as (x, y) = R if xē + yē = 16 then the domain of R is

(a) (0, 4, 4)

(b) (0, -4, 4)

(c) (0, -4, -4)

(d) None of these

## Answer

Answer: (b) (0, -4, 4)

Hint:

Given that:

(x, y) ? R ? xē + yē = 16

? y = ąv(16 – xē )

when x = 0 ? y = ą4

(0, 4) ? R and (0, -4) ? R

when x = ą4 ? y = 0

(4, 0) ? R and (-4, 0) ? R

Now for other integral values of x, y is not an integer.

Hence R = {(0, 4), (0, -4), (4, 0), (-4, 0)}

So, Domain(R) = {0, -4, 4}

Question 15.

The range of the function ^{7-x}P_{x-3} is

(a) {1, 2, 3, 4, 5}

(b) {3, 4, 5}

(c) None of these

(d) {1, 2, 3}

## Answer

Answer: (d) {1, 2, 3}

Hint:

The function f(x) = ^{7-x}P_{x-3} is defined only if x is an integer satisfying the following inequalities:

1. 7 – x = 0

2. x – 3 = 0

3. 7 – x = x – 3

Now, from 1, we get x = 7 ……… 4

from 2, we get x = 3 ……………. 5

and from 2, we get x = 5 ………. 6

From 4, 5 and 6, we get

3 = x = 5

So, the domain is {3, 4, 5}

Now, f(3) = ^{7-3}P_{3-3} = ^{4}P_{0} = 1

? f(4) = ^{7-4}P_{4-3} = ^{3}P_{1} = 3

? f(5) = ^{7-5}P_{5-3} = ^{2}P_{2} = 2

So, the range of the function is {1, 2, 3}

Question 16.

The period of the function f(x) = sin^{4} 3x + cos^{4} 3x is

(a) p/2

(b) p/3

(c) p/4

(d) p/6

## Answer

Answer: (d) p/6

Hint:

Since g(x) = sin^{4} x + cos^{4} x is periodic with period p/2

So, f(x) = sin^{4} 3x + cos^{4} 3x is periodic with period (p/2)/3 = p/6

Question 17.

The domain of the function f(x) = 1/(xē – 3x + 2) is

(a) {1, 2}

(b) R

(c) R – {1, 2}

(d) R – {1, -2}

## Answer

Answer: (c) R – {1, 2}

Hint:

Given, function is f(x) = 1/(xē – 3x + 2)

Clearly, f(x) is not defined when xē – 3x + 2 = 0

? (x – 1)Ũ(x – 1) = 0

? x = 1, 2

So, f(x) is not defined when x = 1, 2

So, domain of function is R – {1, 2}

Question 18.

The domain of the function f(x) = sin^{-1} (tan x) is

(a) -p/4 = x = p/4

(b) np – p/4 = x = np + p/4

(c) np – p/3 = x = np + p/3

(d) -p/3 = x = p/3

## Answer

Answer: (b) np – p/4 = x = np + p/4

Hint:

sin^{-1} (tan x) is defined for -1 = tan x = 1

= -p/4 = x = p/4

The general solution of the above inequality is

np -p/4 = x = np + p/4

Question 19.

The domain of tan^{-1} (2x + 1) is

(a) R

(b) R -{1/2}

(c) R -{-1/2}

(d) None of these

## Answer

Answer: (a) R

Hint:

Since tan^{-1} x exists if x ? (-8, 8)

So, tan^{-1} (2x + 1) is defined if

-8 < 2x + 1 < 8

? -8 < x < 8

? x ? (-8, 8)

? x ? R

So, domain of tan^{-1} (2x + 1) is R.

Question 20.

the function f(x) = x – [x] has period of

(a) 0

(b) 1

(c) 2

(d) 3

## Answer

Answer: (b) 1

Hint:

Let T is a positive real number.

Let f(x) is periodic with period T.

Now, f(x + T) = f(x), for all x ? R

? x + T – [x + T] = x – [x], for all x ? R

? [x + T] – [x] = T, for all x ? R

Thus, there exist T > 0 such that f(x + T) = f(x) for all x ? R

Now, the smallest value of T satisfying f(x + T) = f(x) for all x ? R is 1

So, f(x) = x – [x] has period 1