JEE Main 2016 · 3 Apr · Shift 1 · Mathematics Q71

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MathematicsSingle correct+4 / -1Question 71 of 90

For $x\in\mathbb R$ , $f(x)=|\log2-\sin x|$ and $g(x)=f(f(x))$ , then :

g is not differentiable at x = 0

g′(0) = cos(log2)

g′(0) = −cos(log2)

g is differentiable at x = 0 and g′(0) = −sin(log2)

Answer and worked solution

Official answer: B

Near 0, $\log2-\sin x>0$ , so $f(x)=\log2-\sin x$ , and $\log2-\sin(f(x))>0$ too. So $g(x)=\log2-\sin(\log2-\sin x)$ and
$g'(0)=\cos(\log2-0)\cdot\cos0=\cos(\log2).$

Topics:Differentiability Chain rule and implicit differentiation

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JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 71