Let $f$ be twice differentiable function such that $f''\left( x \right) = - f\left( x \right),$ and $f'\left( x \right) = {\rm{g}}\left( x \right),$ $h\left( x \right) = {\left\{ {f\left( x \right)} \right\}^2} + {\left\{ {{\rm{g}}\left( x \right)} \right\}^2}.$ If $h\left( 5 \right) = 11,$ then $h\left( {10} \right)$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Let

f

be twice differentiable function such that

f''\left( x \right) = - f\left( x \right),

and

f'\left( x \right) = \mathrm{g}\left( x \right),

h\left( x \right) = {\left\{ {f\left( x \right)} \right\}^2} + {\left\{ {\mathrm{g}\left( x \right)} \right\}^2}.

h\left( 5 \right) = 11,

then

h\left( {10} \right)

is equal to

View answer and explanation

Correct option: B

Option B (11) is correct because h(x) is a constant function, so h(5) = h(10) = 11.