JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Inverse Trigonometric Functions

Differentiating, $f^{\prime }(x)=-\frac{2}{\sqrt{1-x^2}}-\frac{4}{1+x^2}-6x-2$. On $[-1,1]$ the first two terms are always negative, and even near $x=-1$, where $-6x-2$ is at its largest value of about $4$, the term $-\frac{2}{\sqrt{1-x^2}}$ grows without bound in the negative direction, so $f^{\prime }(x)<0$ throughout and $f$ is strictly decreasing. The maximum therefore occurs at $x=-1$ and the minimum at $x=1$. At $x=1$, with ${\cos }^{-1}1=0$ and ${\cot }^{-1}1=\frac{\pi }{4}$, $f(1)=0+4\cdot \frac{\pi }{4}-3-2+10=\pi +5$. At $x=-1$, with ${\cos }^{-1}(-1)=\pi$ and ${\cot }^{-1}(-1)=\frac{3\pi }{4}$ since the inverse cotangent takes values in $(0,\pi )$, $f(-1)=2\pi +4\cdot \frac{3\pi }{4}-3+2+10=2\pi +3\pi +9=5\pi +9$. Hence the range is $[a,b]=[\pi +5,5\pi +9]$ and $4a-b=4(\pi +5)-(5\pi +9)=4\pi +20-5\pi -9=11-\pi$.