Question

Let D be the domain of a twice differentiable function f. For all $xϵD,f"\left(x\right)+f\left(x\right)=0$ and $f\left(x\right)=\int g\left(x\right)dx+constant$. If $h\left(x\right)=\left(f\right(x){)}^2+(g\left(x\right){)}^2$ and $h\left(0\right)=5$ then $h\left(2015\right)-h\left(2014\right)=$

• A. $5$
• B. $3$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

$f\left(x\right)=\int g\left(x\right)dx+c$

On differentiating

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

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

and $f^{\prime \prime }\left(x\right)+f\left(x\right)=0$ .... (3) (given)

Put value of $f^{\prime \prime }\left(x\right)$ in eqn. (3)

$\Rightarrow f\left(x\right)=-g^{\prime }\left(x\right)$ ....(4)

now,

$h\left(x\right)=f^2\left(x\right)+g^2\left(x\right)$

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

Put value of f'(x) and g'(x) from eqn. (1) and (4)

$h^{\prime }\left(x\right)=2f\left(x\right)g\left(x\right)+2g\left(x\right)[-f(x\left)\right]$

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

This implies $h\left(x\right)$ is constant

$\therefore h\left(x\right)=c$

but $h\left(o\right)=5\Rightarrow c=5$

$\therefore h\left(x\right)=5$

$\therefore h\left(2015\right)-h\left(2014\right)=0$