Q101 / 160 · WBJEE 2015 · Mathematics
Let $f: R \rightarrow R$ be differentiable at $x=0$. If f(0)=0 and f'(0)=2, then the value of $$
\left.\lim _{x \rightarrow 0} \frac{1}{x} \mid f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)\right] \text{ is}
$$
- A2015
- B0
- Correct answerC$2015 \times 2016$
- D$2015 \times 2014$
Solution
Given, $f(0)=0$ and $f^{\prime}(0)=2$
$\therefore \lim _{x \rightarrow 0} \frac{1}{x}[f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]$
$=\lim _{x \rightarrow 0} \frac{\left[\begin{array}{c}f^{\prime}(x)+2 f^{\prime}(2 x)+3 f^{\prime}(3 x) \\ +\ldots+2015 f^{\prime}(2015 x)\end{array}\right]}{1}$
$=\frac{2+2 \times 2+3 \times 2+\ldots+2015 \times 2}{1}$
$=2[1+2+3+\ldots+2015]$
$=\frac{2(2015)(2015+1)}{2}=\frac{2 \times 2015 \times 2016}{2}$
$=2015 \times 2016$
Hence, the correct answer is option: 'C'.