Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ be a differentiable function having $\mathrm{f}\left(2\right)=6,\mathrm{f}\text{'}\left(2\right)=\left(\frac{1}{48}\right)$. Then $\underset{\mathrm{x}\to 2}{\lim }{\int }_6^{\mathrm{f}\left(\mathrm{x}\right)}\frac{4{\mathrm{t}}^3\mathrm{dt}}{\left[\mathrm{x}-2\right]}=$

• A. $18$
• B. $12$
• C. $36$
• D. $24$

Answer 1

The correct option is A

$18$

Explanation for the correct answer:

Step 1: Using Leibnitz's Theorem,

According to the theorem: $\frac{\mathrm{d}}{\mathrm{dx}}\left[{\int }_{\mathrm{v}\left(\mathrm{x}\right)}^{\mathrm{u}\left(\mathrm{x}\right)}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\right]=\mathrm{f}\mathrm{u}\left(\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{u}\left(\mathrm{x}\right)-\mathrm{f}\mathrm{v}\left(\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{v}\left(\mathrm{x}\right)$

Given, $\mathrm{f}\left(2\right)=6,\mathrm{f}\text{'}\left(2\right)=\frac{1}{48}$

To find, $\underset{\mathrm{x}\to 2}{\lim }{\int }_6^{\mathrm{f}\left(\mathrm{x}\right)}\frac{4{\mathrm{t}}^3\mathrm{dt}}{\mathrm{x}-2}$

Now it is clear from Leibnitz's Theorem,

The form of the limit is $\frac{0}{0}$

Step 2: Using L-hospital's Rule,

L-hospital's Rule: $\underset{\mathrm{x}\to \mathrm{c}}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{c}}{\lim }\frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{g}\text{'}\left(\mathrm{x}\right)}$

$\underset{\mathrm{x}\to 2}{\lim }\frac{{\int }_6^{\mathrm{f}\left(\mathrm{x}\right)}4{\mathrm{t}}^3\mathrm{dt}}{\mathrm{x}-2}$

Applying, L-hospital's rule,

$=\underset{\mathrm{x}\to 2}{\lim }\frac{4\mathrm{f}{\left(\mathrm{x}\right)}^3{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\mathrm{f}\left(\mathrm{x}\right)-4\times 6^3{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(6\right)}{1}$

$$\begin{gathered} =\underset{\mathrm{x}\to 2}{\lim }\frac{4(\mathrm{f}{\left(\mathrm{x}\right))}^3\mathrm{f}\text{'}(\mathrm{x})-0}{1} \\ =\frac{4\times 6^3\times {\displaystyle \frac{1}{48}}}{1} \\ =\frac{4\times 216}{48} \\ =18 \end{gathered}$$

Hence, the correct answer is an option (A).