Let gx - fx sin x where fx is a twice differentiable function on - - - such that f- - 1- The value of g- equals

The correct option is C 2 g(x) = f(x) sin x ⇒ g'(x) = f(x) cos x+f'(x)sin x ⇒ g #34;(x)= f(x) sin x + f'(x) cos x + sin x f #34;(x) + ...

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Byju's Answer

Standard XII

Mathematics

Product Rule of Differentiation

Let gx = fx...

Question

Let $g (x) = f (x) sin x$ where f(x) is a twice differentiable function on $(- \infty, \infty)$ such that $f^{'} (- π) = 1$ . The value of $g " (- π)$ equals

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Solution

The correct option is C $- 2$
$g (x) = f (x) sin x$
$\Rightarrow g^{'} (x) = f (x) cos x + f^{'} (x) sin x$
$\Rightarrow g " (x) = - f (x) sin x + f^{'} (x) cos x + sin x f " (x) + f^{'} (x) cos x$
$∴ g " (- π) = - 2 f^{'} (- π) = - 2 (1) = - 2$

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Let g(x)=f(x)sinx where f(x) is a twice differentiable function on (−∞,∞) such that f′(−π)=1. The value of g"(−π) equals

The correct option is C −2g(x)=f(x)sinx⇒g′(x)=f(x)cosx+f′(x)sinx⇒g"(x)=−f(x)sinx+f′(x)cosx+sinxf"(x)+f′(x)cosx∴g"(−π)=−2f′(−π)=−2(1)=−2

Let $g (x) = f (x) sin x$ where f(x) is a twice differentiable function on $(- \infty, \infty)$ such that $f^{'} (- π) = 1$ . The value of $g " (- π)$ equals

The correct option is C $- 2$
$g (x) = f (x) sin x$
$\Rightarrow g^{'} (x) = f (x) cos x + f^{'} (x) sin x$
$\Rightarrow g " (x) = - f (x) sin x + f^{'} (x) cos x + sin x f " (x) + f^{'} (x) cos x$
$∴ g " (- π) = - 2 f^{'} (- π) = - 2 (1) = - 2$