Question

Let f1 : R → R, f2 : (-π/2, π/2) → R, f3:(-1, eπ/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(loge(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 |
| *. 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.