Q89 / 155 · WBJEE 2019 · Mathematics

Let $f(x)$ be a derivable function, $f^{\prime}(x)>f(x)$ and $f(0)=0 .$ Then

  • Correct answer
    A
    $f(x)>0$ for all $x>0$
  • B
    $f(x) < 0$ for all $x>0$
  • C
    no sign of $f(x)$ can be ascertained
  • D
    $f(x)$ is a constant function

Solution

Let $g(x)=e^{-x} f(x)$ $\therefore$ $g^{\prime}(x)=e^{-x} f^{\prime}(x)-e^{-x} f(x)$ $\quad=e^{-x}\left(f^{\prime}(x)-f(x)\right)$ Since, $f^{\prime}(x)>f(x),$ so $f^{\prime}(x)-f(x)>0$ $\therefore e^{-x}\left(f^{\prime}(x)-f(x)\right)>0$ $\Rightarrow \quad g^{\prime}(x)>0$ $\therefore g(x)$ is an increasing function. Now, $g(x)>g(0) \forall x>0$ $\begin{array}{lr}\Rightarrow e^{-x} f(x)>e^{0} f(0) & \\ \Rightarrow e^{-x} f(x)>0 & {[\because f(0)=0]}\end{array}$ $\Rightarrow \quad f(x)>0, \forall x>0$

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

Relevant Questions