Question

Let $f$ be a continuous functions satisfying $f^{\prime}(\ln x )=\left[\begin{array}{ll}1 & \text { for } 0< x \leq 1 \\ x & \text { for } x >1\end{array}\right.$ and $f (0)=0$ then $f( x )$ can be defined as

• A. $f(x)=\left[\begin{array}{ll}1 & \text { if } x \leq 0 \\ 1-e^x & \text { if } x>0\end{array}\right.$
• B. $f(x)=\left[\begin{array}{ll}1 & \text { if } x \leq 0 \\ e^x-1 & \text { if } x>0\end{array}\right.$
• C. $f(x)=\left[\begin{array}{ll}x & \text { if } x<0 \\ e^x & \text { if } x>0\end{array}\right.$
• D. $f(x)=\left[\begin{array}{ll}x & \text { if } x \leq 0 \\ e^x-1 & \text { if } x>0\end{array}\right.$

Answer 1

Answer: (D)

put $ \ln x=t \Rightarrow x=e^t$
for $ x >1 ; f^{\prime}(t)=e^t $ for $t >0$
integrating $f(t)=e^t+C ; f(0)=e^0+C \Rightarrow C=-1 $ (given $f(0)=0$ )
$\therefore f(t)=e^t-1$ for $t >0$ (corresponding to $x>1$ )
hence $f(x)=e^x-1$ for $x>0$....(1)
again for $0< x \leq 1$
$f ^{\prime}(\ln x )=1 \left( x = e ^{ t }\right)$
$f^{\prime}(t)=1$ for $t \leq 0$
$f ( t )= t + C$
$f(0)=0+C \Rightarrow C=0 \Rightarrow f(t)=t$ for $t \leq 0 \Rightarrow f(x)=x$ for $x \leq 0$