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## Sunday 21 February 2021

### CBSE Class 11 Maths - MCQ and Online Tests - Unit 2 - Relations and Functions

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

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.

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

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 = (nn
Given Set = {a, b}
Total number of elements = 2
Total number of binary operations = (2ē)ē = 24 = 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

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) = ex and g(x) = loge x then the value of fog(1) is
(a) 0
(b) 1
(c) -1
(d) None of these

Hint:
Given, f(x) = ex
and g(x) = log x
fog(x) = f(g(x))
= f (log x)
= elog 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

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

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

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

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) = log3 x and A = (3, 27) then f(A) =
(a) (1, 1)
(b) (3, 3)
(c) (1, 3)
(d) (2, 3)

Hint:
Since f(x) = log3 x is an increasing function
So, f(A) = (log3 3, log3 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) x1/2

Hint:
Given, f(x) = (a – x)1/n
Now, f(f(x)) = [(a – f(x))n]1/n
? f(f(x)) = [(a – {(a – xn)1/n }n ]1/n
? f(f(x)) = [a – (a – xn)]1/n
? f(f(x)) = [a – a + xn)]1/n
? f(f(x)) = (xn)1/n
? f(f(x)) = x

Question 10.
The domain of the definition of the real function f(x) = v(log12 xē ) of the real variable x is
(a) x > 0
(b) |x| = 1
(c) |x| > 4
(d) x = 4

Hint:
We have f(x) = v(log12 xē)
Since, loga k = 0 if a > 1, k = 1
or 0 < a < 1 and 0 < k = 1
So, the function f(x) exists if
log12 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}

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

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

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

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-xPx-3 is
(a) {1, 2, 3, 4, 5}
(b) {3, 4, 5}
(c) None of these
(d) {1, 2, 3}

Hint:
The function f(x) = 7-xPx-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) = 7-3P3-3 = 4P0 = 1
? f(4) = 7-4P4-3 = 3P1 = 3
? f(5) = 7-5P5-3 = 2P2 = 2
So, the range of the function is {1, 2, 3}

Question 16.
The period of the function f(x) = sin4 3x + cos4 3x is
(a) p/2
(b) p/3
(c) p/4
(d) p/6

Hint:
Since g(x) = sin4 x + cos4 x is periodic with period p/2
So, f(x) = sin4 3x + cos4 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: (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: (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

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

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

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