JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Continuity and Differentiability

For option (A): Let $h(x)=x^9-f(x)$. Since $f:(0,1)\to (0,1)$, we have $h(0)=-f(0)<0$ and $h(1)=1-f(1)>0$. By the Intermediate Value Theorem, $h(x)=0$ for some $x\in (0,1)$. For option (B): Let $h(x)=x-{\int }_0^{\frac{\pi }{2}-x}f(t)\cos tdt$. Then $h(0)=-{\int }_0^{\pi /2}f(t)\cos tdt<0$ and $h(1)=1-{\int }_0^{\pi /2-1}f(t)\cos tdt>0$. By IVT, $h(x)=0$ for some $x\in (0,1)$. Option (C): $e^x>1$ for $x>0$ while the integral is less than 1, so no solution. Option (D): both terms are positive, so no solution.