Question

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

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

Answer 1

The correct option is B (0,0)

since f is continuous at x = $\frac{\pi }{4}$

$\therefore f\left(\frac{\pi }{4}\right)-f_{h\to 0}(\frac{\pi }{4}+h)-f_{h\to 0}(\frac{\pi }{4}-h)$

$\Rightarrow \frac{\pi }{4}cot\frac{\pi }{4}+b-f_{h\to 0}(\frac{\pi }{4}+h)+a^2\sqrt{2}sin(\frac{\pi }{4}+h)$

$\Rightarrow \frac{\pi }{4}\left(1\right)+b-(\frac{\pi }{4}+0)+a^2\sqrt{2}sin(\frac{\pi }{4}+0)$

$\Rightarrow \frac{\pi }{4}+b-\frac{\pi }{4}+a^2\sqrt{2}sin\frac{\pi }{4}$

$\Rightarrow b-a^2\sqrt{2}\frac{1}{\sqrt{2}}\Rightarrow b-a^2$

Also as f is continuous at x - $\frac{\pi }{4}$

$\therefore f\left(\frac{\pi }{2}\right)=\underset{x\to \frac{\pi }{2}-0}{lim}f\left(x\right)=\underset{x\to -0}{lim}f(\frac{\pi }{2}-h)$

$\Rightarrow bsin2\frac{\pi }{2}-acos2\frac{\pi }{2}=\underset{h\to -0}{lim}\left[\right(\frac{\pi }{2}-h\left)cot\right(\frac{\pi }{2}-h)+b]$

$\Rightarrow b.0-a(-1)=0+b\Rightarrow a=b$

Hence (0,0) satisfy the above relations.