Test: JEE Main 35 Year PYQs: Functions - Airforce X Y / Indian Navy SSR MCQ

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]

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.

f (x) is onto ∴ S = range of f (x)

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)

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

Clearly, range of f (x) 

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

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

∴ f (x) is incorrectly matched with 

Clearly f is one one and onto, so invertible

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

∴ The statement-1 is correct but statement-2 is false.

Given that f (x) = x³ + 5x + 1

⇒ f (x) is strictly increasing on R
⇒ f (x) is one one
∴ Being a polynomial f (x) is cont. and inc.

Hence f is onto also. So, f is one one and onto R.