JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Application of Derivatives

We have $f^{\prime }(x)=\frac{4}{x-1}-4(x-1)$. For $1<x<2$, $f^{\prime }(x)>0$ (increasing); for $x>2$, $f^{\prime }(x)<0$ (decreasing), so (A) is correct. Since $f$ has a maximum at $x=2$ and $f(2)>-1$ while $f(x)\to -\infty$ as $x\to \infty$ and $x\to 1^{+}$, $f(x)=-1$ has two solutions, so (B) is correct. For (D), $f(e)>0$ and $f(e+1)<0$ by IVT implies a root in $(e,e+1)$, so (D) is correct. For (C), $f^{\prime \prime }(x)=-\frac{4}{(x-1{)}^2}-4$, so $f^{\prime }(e)-f^{\prime \prime }(2)=\frac{4}{e-1}-4(e-1)-(-4-4)=\frac{4}{e-1}-4(e-1)+8>0$, so (C) is NOT correct.