Question

A function $y$ $=$ $f\left(x\right)$ satisfying. $f^{\prime \prime }\left(x\right)$$=x^{-3/2},f^{\prime }\left(4\right)=2$ and $f\left(0\right)=0$ is :

• A. $-2\sqrt{x}+3x$
• B. $2\sqrt{x}+3x$
• C. $4\sqrt{x}+3x$
• D. $-4\sqrt{x}+3x$

Answer 1

The correct option is D $-4\sqrt{x}+3x$

Given, $y=f\left(x\right),f"\left(x\right)=x^{\frac{-3}{2}},f^{\prime }\left(4\right)=2,f\left(0\right)=0$

Consider,$f"\left(x\right)=x^{\frac{-3}{2}}$

On integrating both sides,

$\int f"\left(x\right)=\int x^{\frac{-3}{2}}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+k$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x^{\frac{-1}{2}}}{\frac{-1}{2}}+k$

$\Rightarrow f^{\prime }\left(x\right)=-2x^{\frac{-1}{2}}+k---->A$

But $f^{\prime }\left(4\right)=2$,

on substituting$\Rightarrow f^{\prime }\left(4\right)=-2\times 4^{\frac{-1}{2}}+k$

$\Rightarrow 2=-2\times \frac{1}{2}+k$

$\Rightarrow k=3$

Equation A becomes,

$\Rightarrow f^{\prime }\left(x\right)=-2x^{\frac{-1}{2}}+3$

On integrating once again,

$\Rightarrow \int f^{\prime }\left(x\right)=\int (-2x^{\frac{-1}{2}}+3)$

$\Rightarrow f\left(x\right)=(-4x^{\frac{1}{2}}+3x+k_1)$

But f(0)=0

$\Rightarrow f\left(0\right)=(-4\times 0^{\frac{1}{2}}+3\times 0+k_1)$

$\Rightarrow 0=(0+0+k_1)$

$\Rightarrow k_1=0$

$f\left(x\right)=-4\sqrt{x}+3x$

Option D is correct