Question

If $f(x+y)=f\left(x\right)+f\left(y\right)-xy$ for all $x,y\in R$ and ${lim}_{h\to 0}\frac{f\left(h\right)}{h}=3$, then the area bounded by the curves $y=f\left(x\right)$ and $y=x^2$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Put $x=y=0$
$\Rightarrow f\left(0\right)=0$
Now,

$f{}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)-hx-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}-x=3-x$
$\therefore f\left(x\right)=3x-\frac{x^2}{2}+c$ and $f\left(0\right)=0$
$\Rightarrow c=0$
$\therefore f\left(x\right)=3x-\frac{x^2}{2}$
Required Area$=\left|{\int }_0^2(3x-\frac{x^2}{2}-x^2)dx\right|$
$={(\frac{3x^2}{2}-\frac{3x^3}{6})}_0^2$
$=6-4=2$