JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Application of Derivatives

For $x<0$: $f(x)=(x+1{)}^5-2x$, so $f^{\prime }(x)=5(x+1{)}^4-2$ which changes sign, making $f$ non-monotonic on $(-\infty ,0)$. ✗ for option (C)

As $x\to -\infty$, $f(x)\to -\infty$ and ${\lim }_{x\to 0^{-}}f(x)=1$, so $(-\infty ,1)\subset$ Range.

For $x\ge 3$: ${\lim }_{x\to \infty }f(x)=\infty$ and $f(3)=\frac{1}{3}$, so $[\frac{1}{3},\infty )\subset$ Range.

Combining all intervals, Range$(f)=\mathbb{R}$, so $f$ is onto. ✓

For $0\le x<3$:


At $x=1$: $f^{\prime }(1^{-})=1$ and $f^{\prime }(1^{+})=1$, but the left derivative of $f^{\prime }$ at $x=1$ is 2 while the right derivative is $-4$, so $f^{\prime }$ is not differentiable at $x=1$. ✓

From the graph of $f^{\prime }$, it has a local maximum at $x=1$. ✓