JEE Advanced (One or More Correct Option): Functions Free MCQ Practice

Let us check each option one by one.

(b) f (1) ≠ 3 as function is not defined for x = 1
∴ (b) is not correct.

As (0, 0) and (x, g (x)) are two vertices of an equilateral triangle; therefore, length of the side of D is

∴ (b), (c) are the correct answers as (a) is not a function
(∴ image of x is not unique)

f(x) = cos [π²] x + cos [– π²] x
We know 9 < π² < 10 and – 10 < – π² < – 9
⇒ [π²] = 9 and [–π²] = – 10
⇒ ∴ f (x) = cos 9x + cos (–10x)
f (x) = cos 9x  + cos 10 x
Let us check each option one by one.

Thus (a) and (c) are the correct options.

f (x) = 3x – 5 (given), which is strictly increasing on R,
so f –1 (x) exists.
Let y = f (x) = 3x – 5

    ...(1)
and y = f (x)  ⇒ x = f ^–1(y)   ...(2)

Let us check each option one by one.

∴ f is neither onto nor invertible

Hence a and b are the correct options.

the critical points can be obtained by solving |x| = 0

The graph of y = f(x) is as follows

From graph f(x) has local minimum at –2 and 0 and 

∴ f  is an odd function.
(a) is correct and (d) is not correct.
Also
We know that strictly increasing function is one one.
∴ f  is one one
∴ (b) is correct

∴ Range of f = (–∞, ∞) = R
∴ f  is an onto function.
∴ (c) is correct.

f (x) = x⁵ - 5 x+a
f (x) = 0  ⇒ x⁵ - 5 x +a = 0
⇒ a = 5x – x⁵ = g(x)
⇒ g(x) = 0 when x = 0, 5^1/4 - 5^1/4 and g' (x ) = 0 ⇒ x = 1, – 1
Also g (– 1) = – 4 and g(1) = 4
∴ graph of g(x) will be as shown below.
From graph

if a ∈ ( -4,-4)
then g(x) = a or f (x) = 0 has 3 real roots If a > 4 or a < – 4
then f(x) = 0 has only one real root.
∴ (b) and (d) are the correct options.