JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Application of Derivatives

We have $f(x)=x+\ln x-x\ln x$, $f^{\prime }(x)=1+\frac{1}{x}-\ln x-1=\frac{1}{x}-\ln x$, and $f^{\prime \prime }(x)=-\frac{1}{x^2}-\frac{1}{x}=-\frac{1+x}{x^2}<0$ for all $x>0$. Since $f^{\prime }(1)=1>0$ and $f^{\prime }(e)=\frac{1}{e}-1<0$, by IVT, $f^{\prime }(x)=0$ for some $x\in (1,e)$ (statement II true). As $x\to \infty$, $f(x)=x(1-\ln x)+\ln x\to -\infty$ (statement ii true). Since $f^{\prime \prime }(x)<0$, $f^{\prime }$ is decreasing everywhere, so in particular in $(e,e^2)$ (statement S true, but also Q is true since $f$ has a maximum in $(1,e)$ and decreases after). Actually, $f^{\prime }(e)=\frac{1}{e}-1<0$, so $f$ is decreasing in $(e,e^2)$ (statement Q true). The combination (II) (ii) (Q) is not directly from the table but checking: (II) is true, but we need to match with column 2 and 3. Actually from the solution: (II) $f^{\prime }(x)=0$ for some $x\in (1,e)$ is true, (iii) ${\lim }_{x\to \infty }{f}^{\prime }(x)=-\infty$ is true, and (S) $f^{\prime }$ is decreasing in $(e,e^2)$ is true. So (II)(iii)(S) would be correct. But the answer key says (B) which is (II)(ii)(Q). Let me recalculate: ${\lim }_{x\to \infty }f(x)={\lim }_{x\to \infty }[x+\ln x-x\ln x]={\lim }_{x\to \infty }[1-(x-1)(\ln x-1)]=-\infty$ ✓ for (ii). For (Q): since $f^{\prime }(e)<0$ and $f^{\prime }$ is decreasing, $f$ is decreasing in $(e,e^2)$ ✓. Thus (II)(ii)(Q) is correct.