Question

If $\mathit{f}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}$ and $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}$, then $\underset{\mathbf{x}\mathbf{\to }\mathbf{5}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{}\mathbf{[}\mathit{x}\mathbf{}\mathit{f}\mathbf{(}\mathbf{5}\mathbf{)}\mathbf{}\mathbf{-}\mathbf{}\mathbf{5}\mathit{f}\mathbf{(}\mathit{x}\mathbf{\left)}\mathbf{\right]}\mathbf{}\mathbf{/}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{5}\mathbf{)}\mathbf{}\mathbf{=}$

• A. $35$
• B. $-35$
• C. $28$
• D. $-28$

Answer 1

The correct option is D

$-28$

Finding the value:

Step1 : Expressing the given data:

$$\begin{gathered} f\left(5\right)=7 \\ f\text{'}\left(5\right)=7 \end{gathered}$$

$\underset{x\to 5}{\lim }\frac{xf\left(5\right)-5f\left(x\right)}{x-5}\mathbf{}\mathbf{}\mathbf{(}\mathbf{}\frac{\mathbf{0}}{\mathbf{0}}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathbf{)}$

Step 2 : Applying L' Hospital's Rule,

$=\underset{x\to 5}{\lim }\frac{f\left(5\right)-5f\text{'}\left(x\right)}{1-0}$

$=f\left(5\right)-5f\text{'}\left(5\right)$

$$\begin{gathered} =7-5\times 7 \\ =7-35 \end{gathered}$$

$=-28$

Hence, option (D) is correct answer.