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Question:

Let f1: R → R, f2: (-π/2, π/2) → R, f3: ( -1, eπ/2 - 1) → R and f4: R → R be functions defined by
(i) f1(x) = sin(√(1 - e-x²))
(ii) f2(x) = {|sin x| tan-1(x) if x ≠ 0, 1 if x = 0}, where the inverse trigonometric function tan-1(x) 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, 0 if x = 0}
List - I
List - II
P. The function f1 is

  1. NOT continuous at x = 0
    Q. The function f2 is
  2. continuous at x = 0 and NOT differentiable at x = 0
    R. The function f3 is
  3. differentiable at x = 0 and its derivative is NOT continuous at x = 0
    S. The function f4 is
  4. differentiable at x = 0 and its derivative is continuous at x = 0
    The correct option is
    P → 2; Q → 3; R → 1; S → 4
    P → 4; Q → 1; R → 2; S → 3
    P → 4; Q → 2; R → 1; S → 3
    P → 2; Q → 1; R → 4; S → 3

P → 4; Q → 2; R → 1; S → 3

P → 2; Q → 3; R → 1; S → 4

P → 4; Q → 1; R → 2; S → 3

P → 2; Q → 1; R → 4; S → 3

Solution:

The correct option is D
P → 2; Q → 1; R → 4; S → 3
(i) f1(x) = sin√(1 - e-x²)
f1'(x) = cos√(1 - e-x²) * 1/(2√(1 - e-x²)) * (0 - e-x² * (-2x))
at x = 0 f1'(x) does not exist
So. P → 2
(ii) f2(x) = {|sin x| tan-1(x) if x ≠ 0, 1 if x = 0}
limx→0+ (sin x / x) tan-1 x = 1
⇒ f2(x) is not continuous at x = 0
So Q → 1
(iii) f3(x) = [sin(ln(x + 2))] = 0
1 < x + 2 < eπ/2 ⇒ 0 < ln(x + 2) < π/2 ⇒ 0 < sin(ln(x + 2)) < 1 ⇒ f3(x) = 0
So R → 4
(iv) f4(x) = {x²sin(1/x) if x ≠ 0, 0 if x = 0}
So S → 3