Question

Assume $f$ is differentiable function satisfying $f(x+y)=f\left(x\right)+f\left(y\right)+xy$ and $\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}=3$ then $f\left(1\right)$ equals

• A. $\frac{5}{2}$
• B. $\frac{7}{2}$
• C. $-\frac{5}{2}$
• D. $-\frac{7}{2}$

Answer 1

Answer: (B)

$\frac{7}{2}$

Given $f$ is differentiable, so

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

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

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

$\therefore f^{\prime }\left(x\right)=3+x$

$\therefore f\left(x\right)=3x+\frac{x^2}{2}+c$

$[\text{Using}x=0,y=0\therefore f\left(0\right)=0]$

$\therefore c=0$

$\therefore f\left(x\right)=3x+\frac{x^2}{2}$

$\therefore f\left(1\right)=3+\frac{1}{2}$

$\Rightarrow f\left(1\right)=\frac{7}{2}$