1: g f is differentiable at x = 0 and its derivative is continuous at that point, where f(x) = x|x| and g(x) = x . 2: g f is twice differentiable at x - Mathematics MCQ AIEEE 2009 2026 | ExamTest.live

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1: g \circ f is differentiable at x = 0 and its derivative is continuous at that point, where f (x) = x ∣ x ∣ and g (x) = sin x . 2: g \circ f is twice differentiable at x = 0 .

A $Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1$
B $Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1$
C $Statement-1 is true, Statement-2 is false$
D $Statement-1 is false, Statement-2 is true$

Solution & Step-by-step Explanation

h (x) = g (f (x)) = sin (x ∣ x ∣) .For x \geq 0, h (x) = sin (x^{2}); for x < 0, h (x) = sin (- x^{2}) = - sin (x^{2}) . h^{'} (x) = 2 x cos (x^{2}) for x > 0 and h^{'} (x) = - 2 x cos (x^{2}) for x < 0 .At x = 0, h^{'} (0) = 0 from both sides. Continuous. h^{''} (x) = 2 cos (x^{2}) - 4 x^{2} sin (x^{2}) for x > 0 . h^{''} (0^{+}) = 2 . h^{''} (x) = - 2 cos (x^{2}) + 4 x^{2} sin (x^{2}) for x < 0 . h^{''} (0^{-}) = - 2 .Since h^{''} (0^{+}) \neq = h^{''} (0^{-}), it is not twice differentiable. Statement-2 is false.

Try it yourself before checking the explanation above.

1: g \circ f is differentiable at x = 0 and its derivative is continuous at that point, where f (x) = x ∣ x ∣ and g (x) = sin x . 2: g \circ f is twice differentiable at x = 0 .

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

Statement-1 is true, Statement-2 is false

Statement-1 is false, Statement-2 is true

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