Q103 / 160 · TS EAMCET 2008 · Mathematics
If $f(2)=4$ and $f^{\prime}(2)=1$, then
$$
\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}
$$
is equal to
- A$-2$
- B$1$
- Correct answerC$2$
- D$3$
Solution
Given that, $f(2)=4$ and $f^{\prime}(2)=1$
$$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2} \\
= & \lim _{x \rightarrow 2} f(2)-2 \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\
= & f(2)-2 f^{\prime}(2) \\
= & 4-2(1) \\
= & 2
\end{aligned}
$$
Hence, the correct answer is option: 'C'.