Question

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

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

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 $0<x\le 1$
$f^{\prime }(\ln x)=1\left(x=e^t\right)$
$f^{\prime }(t)=1$ for $t\le 0$
$f(t)=t+C$
$f(0)=0+C\Rightarrow C=0\Rightarrow f(t)=t$ for $t\le 0\Rightarrow f(x)=x$ for $x\le 0$