Question

Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ and $f\left(x\right)$ satisfies the equation $f(x+y)=f\left(x\right)+f\left(y\right)+2xy-1$ for all $x,yϵR$ and $f^{\prime }\left(0\right)=2$ then

• A. $g\left(x\right)$ decreases on $(–\infty ,\infty )$
• B. $g\left(x\right)$ decreases on $(0,\infty )$ and decreases on $(–\infty ,0)$
• C. $g\left(x\right)$ increases on $(0,\infty )$
• D. $g\left(x\right)$ increases on $(0,\infty )$ and decreases on $(–\infty ,–0)$

Answer 1

The correct option is C

$g\left(x\right)$ increases on $(0,\infty )$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+2xh-1-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}+2x$

[putting $x=y=0$ in the given condition, we obtain $f\left(0\right)=1$]

$\therefore f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}+2x$

$\therefore f^{\prime }\left(x\right)=f^{\prime }\left(0\right)+2x=2(x+1)$

$\Rightarrow f\left(x\right)=(x+1{)}^2+C$, Putting $x=0$ we obtain $C=0$
Thus, $f\left(x\right)=(x+1{)}^2$. But $g^{\prime }\left(x\right)=f\left(x\right)=(x+1{)}^2>0$

So, g increase on $(-\infty ,\infty )$ in particular on $(0,\infty )$