JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1
If for non-zero distinct real numbers $a,b$ and $c$, $f(x)=\begin{cases}x\alpha+\sqrt{2}a\sin x, & 0\le x<\frac{\pi}{4}\\ 2x\alpha\cot x-c\sin2x, & \frac{\pi}{4}\le x<\frac{\pi}{2}\\ -8ac\cos2x-\pi\alpha b, & \frac{\pi}{2}\le x\le\pi\end{cases}$ is continuous on $[0,\pi]$, then
Limits & Continuity
Solution
At $x=\pi/4$: $\frac{\alpha\pi}{4}+a=\frac{\alpha\pi}{2}-c\Rightarrow a+c=\frac{\alpha\pi}{4}$. At $x=\pi/2$: left limit $=0$ (as $x\cot x\to0$, $\sin\pi=0$); right $=8ac-\pi\alpha b$. So $8ac=\pi\alpha b$. Substituting $\pi\alpha=4(a+c)$: $8ac=4(a+c)b\Rightarrow b=\frac{2ac}{a+c}$, i.e. $\frac2b=\frac1a+\frac1c$. So $\frac1a,\frac1b,\frac1c$ are in A.P. Option (B).