Question

Let f₁ : R → R, f₂ : (-π/2, π/2) → R, f₃:(-1, e^π/2 - 2) → R and f₄: R → R be functions defined by

(i) f₁(x) = sin(√(1 - e^{-x^2}))

(ii) f₂(x) = {(|sinx|/tan^-1x if x ≠ 0), (1, if x = 0), where the inverse trigonometric function tan^–1x assumes values in (-π/2, π/2),

(iii) f₃(x) = [sin(log_e(x + 2))], where, for t ∈ R, [t]  denotes the greatest integer less than or equal to t,

(iv) f₄(x) = {(x²sin(1/x) if x ≠ 0), (1, if x = 0) 

LIST–I	LIST–II
P. The function f 4 is	1. NOT continuous at x = 0
Q. The function f 2 is	2. continuous at x = 0 and NOT differentiable at x = 0
R. The function f 3 is	3. differentiable at x = 0 and its derivative is NOT continuous at x = 0
S. The function f 4 is	4. differentiable at x = 0 and its derivative is continuous at x = 0

The correct option is: 

(A) P→2; Q→3; R→1; S→4 

(B) P→4; Q→1; R→2; S→3 

(C) P→4; Q→2; R→1; S→3 

(D) P→2; Q→1; R→4; S→3

Answer 1

Answer is (D) P→2; Q→1; R→4; S→3

Let us check for each given item:

(i) It is given that

At x = 0: f₁(x) does not exist. Therefore, f₁(x)  is continuous at x = 0 and it is not differentiable at x = 0.

(ii) It is given that

Therefore, f₃(x)  is differentiable at x = 0 and its derivative is continuous at x = 0.

This is oscillating. Thus, f₄(x)  is differentiable at x = 0 and its derivative is not continuous at x = 0.

Therefore, the correct mapping is P→2; Q→1; R→4; S→3.