MCQs Of Functions, Past Year Questions JEE Mains, Class 12, Maths

I accept the Terms & Conditions.

Description

This mock test of MCQs Of Functions, Past Year Questions JEE Mains, Class 12, Maths for Class 12 helps you for every Class 12 entrance exam. This contains 17 Multiple Choice Questions for Class 12 MCQs Of Functions, Past Year Questions JEE Mains, Class 12, Maths (mcq) to study with solutions a complete question bank. The solved questions answers in this MCQs Of Functions, Past Year Questions JEE Mains, Class 12, Maths quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this MCQs Of Functions, Past Year Questions JEE Mains, Class 12, Maths exercise for a better result in the exam. You can find other MCQs Of Functions, Past Year Questions JEE Mains, Class 12, Maths extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.

QUESTION: 1

The domain of sin^-1 [log₃ (x/3)] is

A.
[1, 9]
B.
[–1, 9]
C.
[–9, 1]
D.
[–9, –1]

Solution:

Domain of sin^-1x is [-1,1]

Therefore, -1 ≤ log₃(x/3) ≤ 1

3^-1 ≤ x/3 ≤ 3

1 ≤ x ≤ 9

Therefore, the domain of sin^-1[log₃(x/3)] is [1,9]

QUESTION: 2

The function

A.
neither an even nor an odd function
B.
an even function
C.
an odd function
D.
a periodic function.

Solution:

QUESTION: 3

Domain of definiti on of the function

A.
(-1,0) ∪ (1,2) ∪ (2, ∞)
B.
(a, 2)
C.
(-1,0) ∪ ( a,2)
D.
(1,2) ∪ (2, ∞) .

Solution:

QUESTION: 4

If f : R → R satisfies f (x + y) = f ( x) + f (y) , for all x,

A.
B.
C.
D.
7n + (n + 1) .

Solution:

QUESTION: 5

A function f from the set of natural numbers to integers defined by

A.
neither one -one nor onto
B.
one-one but not onto
C.
onto but not one-one
D.
one-one and onto both.

Solution:

We have f: N →I
If x and y are two even natural numbers,

Again if x and y are two odd natural numbers then

Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.

∴ f is onto.
Hence f is one one and onto both.

QUESTION: 6

The range of the function f (x) =^{7- x} P_{x -3} is

A.
{1, 2, 3, 4, 5}
B.
{1, 2, 3, 4, 5, 6}
C.
{1, 2, 3, 4,}
D.
{1, 2, 3,}

Solution:

QUESTION: 7

If f : R → S, defined by

f (x) = sin x - √3 cosx+ 1, is onto, then the interval of S is

A.
[ –1, 3]
B.
[–1, 1]
C.
[ 0, 1]
D.
[0, 3]

Solution:

f (x) is onto ∴ S = range of f (x)

QUESTION: 8

The graph of the function y = f(x) is symmetrical about the line x = 2, then

A.
f (x) = - f (-x)
B.
f (2 +x) = f (2-x)
C.
f (x) = f (-x)
D.
f (x +2) = f (x-2)

Solution:

Let us consider a graph symm. with respect to line x = 2 as shown in the figure.

From the figure

f (x₁) = f (x₂), where x₁ = 2-x and x₂ = 2+x
∴ f (2 - x) = f (2+x)

QUESTION: 9

The domain of the function

A.
[1, 2]
B.
[2, 3)
C.
[1, 2 ]
D.
[2, 3]

Solution:

Taking common solution of (i) and (ii), we get 2 < x < 3 ∴ Domain = [2, 3)

QUESTION: 10

L et f : (– 1, 1) → B , be a function defined by then f is both one - one and onto when B is the interval

Solution:

Clearly, range of f (x)

For f to be onto, codomain = range
∴ Co-domain of function

QUESTION: 11

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

A.
a
B.
b
C.
c
D.
d

Solution:

Clearly function f (x) = 3x² - 2x+1 is increasing when
f ' (x) = 6x – 2 > 0 ⇒ x ∈[1 / 3,∞)

∴ f (x) is incorrectly matched with

QUESTION: 12

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

A.
– f(x)
B.
f(x)
C.
f(a) + f (a – x)
D.
f(– x)

Solution:

f(2a – x) = f (a – (x – a))
= f(a) f(x – a) – f(0) f(x) = f(a) f(x –a) – f(x)
= – f(x)
[∴ x = 0, y = 0, f(0) = f²(0)- f²(a)
⇒ f²(a) = 0
⇒ f(a) = 0]
⇒ f(2a - x) = - f(x)

QUESTION: 13

The largest interval lying in for which the function, , is defined, is

Solution:

QUESTION: 14

Let f: N→Y be a function defined as f(x) = 4x + 3 where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}.

Show that f is invertible and its inverse is

Solution:

Clearly f is one one and onto, so invertible

QUESTION: 15

Let f(x) = ( x + 1)² – 1,x > –1

Statement -1 : The set {x : f(x) = f ^–1(x) = {0, –1}

Statement-2 : f is a bijection.

A.
Statement-1 is true, Statement-2 is true.Statement-2 is not a correct explanation for Statement-1.
B.
Statement-1 is true, Statement-2 is false.
C.
Statement-1 is false, Statement-2 is False.
D.
Statement-1 is False, Statement-2 is true.Statement-2 is not a correct explanation for Statement-1.

Solution:

Given that f (x) = (x + 1)² –1, x > –1 Clearly D_f = [–, ∞) but co-demain is not given.
Therefore f (x) need not be necessarily onto.
But if f (x) is onto then as f (x) is one one also, (x + 1) being something +ve, f^–1(x) will exist where (x + 1)² –1 = y