Question

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

• A. $22$
• B. $11$
• C. $0$
• D. $1$

Answer 1

The correct option is B $11$

Given,$f^{\prime }\left(x\right)=g^{\prime }\left(x\right)$ ......$\left(1\right)$

And $f^{\prime }\left(x\right)=g\left(x\right)$

Differentiating w.r.t $x$, we get

$f^{\prime \prime }\left(x\right)=g^{\prime }\left(x\right)$

$g^{\prime }\left(x\right)=-f\left(x\right)$ ......$\left(2\right)$(Given)

Differentiating w.r.t $x$ we get

$\Rightarrow h^{\prime }\left(x\right)=2f\left(x\right)\times f^{\prime }\left(x\right)+2g\left(x\right)\times g^{\prime }\left(x\right)$

$\Rightarrow h^{\prime }\left(x\right)=2f\left(x\right)\times g\left(x\right)+2g\left(x\right)\times f\left(x\right)$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow h^{\prime }\left(x\right)=0$

So, $h\left(x\right)$ is a constant function.

i.e., $h\left(x\right)=c$(a constant) for all real $x$.

It is given that $h\left(5\right)=11$

$\therefore h\left(10\right)=11$

$\therefore c=11$

Hence $h\left(10\right)=11$