For x in mathbb{R},f(x) = |log 2 - sin x| and | vedclass

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(D) Given $f(x) = |\log 2 - \sin x|$. Since $\log 2 \approx 0.693  0$. Thus,$f(x) = \log

Vedclass · Accepted Answer

$g'(0) = \cos(\log 2)$

Vedclass · Answer

(D) Given $f(x) = |\log 2 - \sin x|$. Since $\log 2 \approx 0.693  0$. Thus,$f(x) = \log 2 - \sin x$ in the neighborhood of $x=0$.Then $g(x) = f(f(x)) = \log 2 - \sin(f(x)) = \log 2 - \sin(\log 2 - \sin x)$.Since $f(x)$ is differentiable at $x=0$ and the composition of differentiable functions is differentiable,$g(x)$ is differentiable at $x=0$.Now,$g'(x) = -\cos(\log 2 - \sin x) \cdot (-\cos x) = \cos(\log 2 - \sin x) \cdot \cos x$.Evaluating at $x=0$,$g'(0) = \cos(\log 2 - \sin 0) \cdot \cos 0 = \cos(\log 2) \cdot 1 = \cos(\log 2)$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

For $x \in \mathbb{R}$,$f(x) = |\log 2 - \sin x|$ and $g(x) = f(f(x))$,then which of the following is true?

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