Question

If the function $f\left(x\right)=\{\begin{array}{c}x+a^2\sqrt{2}sinx,0\le x<\frac{\pi }{4}\\ xcotx+b,\frac{\pi }{4}\le x<\frac{\pi }{2}\\ bsin2x-acos2x,\frac{\pi }{2}\le x\le \pi \end{array}$ (a, b) are is continuos in the interval $[0,\pi ]$, then the values of

• A. (-1, -1)
• B. (1, 0)
• C. (-1, 1)
• D. (1,1)

Answer 1

The correct option is D (1,1)

Since f is continuous at $x=\frac{\pi }{4}$
$\therefore f\left(\frac{\pi }{4}\right)=\underset{h\to 0}{f}(\frac{\pi }{4}+h)=\underset{h\to 0}{f}(\frac{\pi }{4}-h)\Rightarrow \frac{\pi }{4}cot\frac{\pi }{4}+b=\underset{h\to 0}{f}(\frac{\pi }{4}+h)+a^2\sqrt{2}sin(\frac{\pi }{4}+h)\Rightarrow \frac{\pi }{4}\left(1\right)+b=(\frac{\pi }{4}+0)+a^2\sqrt{2}sin(\frac{\pi }{4}+0)\Rightarrow \frac{\pi }{4}+b=\frac{\pi }{4}+a^2\sqrt{2}sin\frac{\pi }{4}\Rightarrow b=a^2\sqrt{2}\frac{1}{\sqrt{2}}\Rightarrow b=a^2$
Also as f is continuous at $x=\frac{\pi }{2}$
$\therefore f\left(\frac{\pi }{2}\right)=\underset{x\to \frac{\pi }{2}-0}{lim}f\left(x\right)=\underset{lim}{h\to 0}f(\frac{\pi }{2}-h)\Rightarrow bsin2\frac{\pi }{2}-acos2\frac{\pi }{2}=\underset{h\to 0}{lim}\left[\right(\frac{\pi }{2}-h\left)cot\right(\frac{\pi }{2}-h)+b]$
$\Rightarrow b.0-a(-1)=0+b\Rightarrow a=b$
Hence (0, 0), (1, 1) satisfy the above relations.