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

Question

Dr. Abubacker · Accepted Answer

(b)
Since, $h^{'}(x) = 2f(x)f^{'}(x) + 2g(x)g^{'}(x)$
Now, $f^{'}(x) = g(x)\$and$\ f^{''}(x) = - f(x)$
$\Rightarrow \ \ \ f^{''}(x) = g^{'}(x)$and$\ f^{''}(x) = - f(x)$
$\Rightarrow \  - f(x) = g^{'}(x)$
Thus, ${\ f}^{'}(x) = g(x)\$and$\ g^{'}(x) = - f(x)\$
$	herefore\ \ \ \ \ h^{'}(x) = \  - 2g(x){\ g}^{'}(x) + 2g(x)g^{'}(x)$
$= 0,\forall x$
$\Rightarrow \ \ h(x) =$constant for all$\ x\ \ \ \$
But$\ \ h(5) = 11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
Hence, $h(x) = \ 11\ for\ all\ x\ \ \ \ \ \ \ \ \ \ \ \ \$

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

Explanation

Reviewed by Dr. Abubacker

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