Question

If $f\left(x\right)$ & $g\left(x\right)$ be two functions such that $f\left(a\right)=g\left(a\right)$ is zero also $f$ and $g$ both are differentiable every where in the immediate neighborhood of point $c$ except possibly $c$.

Then

$\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\left(\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}\right)$ provided $f^{\prime }\left(c\right)$ and $g^{\prime }\left(c\right)$ are not both zero.

If $\underset{x\to 1}{lim}\frac{{\int }_2^{x^2(1+x)}f\left(t\right)dt}{x-1}=1$ then $f\left(2\right)$ is equal to?

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (A)

$\frac{1}{5}$

Let $h\left(x\right)={\int }_2^{x^2(1+x)}f\left(t\right)dt$

$g\left(x\right)=n-1$

Using Leibuitz Rule,

$h^{\prime }\left(x\right)=f\left(x^2\right(1+x\left)\right).(2x+3x^2)$

$\stackrel{lim}{x\to 1}\frac{h\left(x\right)}{g\left(x\right)}=\stackrel{lim}{x\to 1}\frac{h^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}{g}^{\prime }\left(x\right)=1\stackrel{lim}{x\to 1}\frac{h\left(x\right)}{g\left(x\right)}=1\stackrel{lim}{x\to 1}\frac{h^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=f\left(2\right).5\Rightarrow 1=f\left(2\right).5\Rightarrow f\left(2\right)=\frac{1}{5}$

Hence correct answer is $f\left(2\right)=\frac{1}{5}$