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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
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then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to
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Verified Answer
The correct answer is:
2
Let Let $f(x)=a x+b$
Given $f(2)=4 \& f^{\prime}(2)=1$
$\therefore f(2)=a \cdot 2+b=4 \Rightarrow 2 a+b=4$ ...(1)
$\begin{array}{l}
\& f^{\prime}(x)=a \Rightarrow f^{\prime}(2)=a=1 \Rightarrow a=1 \\
\therefore 2 \times 1+b=4 \Rightarrow b=2 \text { [using equation (1)] } \\
\therefore f(x)=x+2
\end{array}$
Now, $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$
$\begin{array}{l}
=\lim _{x \rightarrow 2} \frac{4 x-2(x+2)}{x-2}=\lim _{x \rightarrow 2} \frac{2 x-4}{x-2} \\
=\lim _{x \rightarrow 2} \frac{2(x-2)}{(x-2)}=2
\end{array}$
Given $f(2)=4 \& f^{\prime}(2)=1$
$\therefore f(2)=a \cdot 2+b=4 \Rightarrow 2 a+b=4$ ...(1)
$\begin{array}{l}
\& f^{\prime}(x)=a \Rightarrow f^{\prime}(2)=a=1 \Rightarrow a=1 \\
\therefore 2 \times 1+b=4 \Rightarrow b=2 \text { [using equation (1)] } \\
\therefore f(x)=x+2
\end{array}$
Now, $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$
$\begin{array}{l}
=\lim _{x \rightarrow 2} \frac{4 x-2(x+2)}{x-2}=\lim _{x \rightarrow 2} \frac{2 x-4}{x-2} \\
=\lim _{x \rightarrow 2} \frac{2(x-2)}{(x-2)}=2
\end{array}$
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