Question

Let $f\left(x\right)$ be a function defined on $R$ such that $f\left(1\right)=2,f\left(2\right)=8$ and $f(u+v)=f\left(u\right)+kuv-2v^2$ for $u,v\in$ $R$ $(k$ is a fixed constant$)$, then?

• A. $f\left(x\right)=g\left(x\right)$
• B. $f^{\prime }\left(x\right)=4x$
• C. $f\left(x\right)=x$
• D. $f\left(x\right)=0$

Answer 1

The correct option is B $f^{\prime }\left(x\right)=4x$

Given, $f(x+h)=f\left(x\right)+kxh-2h^2$
$\Rightarrow \underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}(kx-2h)$
$\Rightarrow f^{\prime }\left(x\right)=kx$
$\Rightarrow f\left(x\right)=\frac{k{x}^2}{2}+c$ ..... [By integrating]
$\Rightarrow f\left(1\right)=\frac{k}{2}+c$
$\Rightarrow 2=\frac{k}{2}+c...\left(i\right)$
Also, $f\left(2\right)=2k+c$
$\Rightarrow 8=2k+c...\left(ii\right)$
From (i) and (ii), we get
$k=4$ & $c=0$
Then, $f\left(x\right)=2x^2$
$\Rightarrow f^{\prime }\left(x\right)=4x$
Hence, option 'B' is correct.