Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{r}$ is a differentiable function such that $\mathrm{f}\left(3\right)=3,\mathrm{f}\text{'}\left(3\right)=\frac{1}{2}$. Then the value of $\underset{\mathrm{x}\to 3}{\lim }\frac{\left[{\int }_3^{\mathrm{f}\left(\mathrm{x}\right)}2{\mathrm{t}}^3\mathrm{dt}\right]}{\mathrm{x}-3}$ is

• A. $25$
• B. $26$
• C. $27$
• D. None of these

Answer 1

The correct option is C

$27$

Explanation for the correct answer:

Step 1: Given functions $\mathrm{f}\left(3\right)=3,\mathrm{f}\text{'}\left(3\right)=\frac{1}{2}$

To find, $\underset{\mathrm{x}\to 3}{\lim }\frac{\left[{\int }_3^{\mathrm{f}\left(\mathrm{x}\right)}2{\mathrm{t}}^3\mathrm{dt}\right]}{\mathrm{x}-3}$

At, $x=3$ the value of limit $\frac{0}{0}$

Clearly, the limit is of the form $\frac{0}{0}$

Step 2: Applying the L-hospitals rule,

Differentiating the numerator and denominator with respect to $x$

$$\begin{gathered} \therefore \underset{\mathrm{x}\to 3}{\lim }\frac{2\mathrm{f}{\left(\mathrm{x}\right)}^3\mathrm{f}\text{'}\left(\mathrm{x}\right)-2\left(3\right){\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(3\right)}{1} \\ =\frac{2\mathrm{f}{\left(3\right)}^3\mathrm{f}\text{'}\left(3\right)-0}{1} \\ =\frac{2{\left(3\right)}^3\left({\displaystyle \frac{1}{2}}\right)-0}{1} \\ =\frac{2\times 27}{2} \\ =27 \end{gathered}$$

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