Q112 / 200 · WBJEE 2014 · Mathematics

Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous, $f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x) .$ Then,

  • A
    $g^{\prime}(0)$ does not exist
  • B
    $g^{\prime}(0)=0$
  • Correct answer
    C
    $g^{\prime}(0)=1$
  • D
    $g^{\prime}(0)=2$

Solution

Given, $f^{\prime}(0)=1$ and $f^{\prime \prime}$ " (0) does not exist. Also, given $g(x)=x f^{\prime}(x)$ $\therefore \quad g^{\prime}(x)=x f^{\prime \prime}(x)+f^{\prime}(x)$ Put $x=0,$ we get $$ \begin{aligned} g^{\prime}(0) &=0 f^{\prime \prime}(0)+f^{\prime}(0) \\ &=0+1=1 \end{aligned} $$

Hence, the correct answer is option: 'C'.

Relevant Questions