Q.5. The values of a and b, so that the function
f(x)=\left\{\begin{matrix} x+a\sqrt{2}sin x & 0\leq x\leq \frac{\pi }{4} \\ 2xcot x+b & \frac{\pi }{4}\leq x\leq \frac{\pi }{2} \\ a cos 2x-b sin x& \frac{\pi }{2}< x\leq \pi \ \end{matrix}\right.is continuous for 0 \leq x \leq \pi, are respectively given by
A -\frac{\pi}{12}, \frac{\pi}{6}
B -\frac{\pi}{6}, -\frac{\pi}{12}
C \frac{\pi}{6}, \frac{\pi}{12}
D \frac{\pi}{6}, -\frac{\pi}{12}
Answer:- D \frac{\pi}{6}, -\frac{\pi}{12}
Explanation :- As the given function is continuous at x = \frac{\pi}{4} and x = -\frac{\pi}{2} , we get
\lim\limits_{x \to \frac{\pi}{4}} f(x) = \lim\limits_{x \to \frac{\pi}{4}} g(x) \therefore \lim_{x\to\frac{\pi}{4}}(x+\mathrm{a}\sqrt{2}\sin x)=\lim_{x\to\frac{\pi}{4}}(2x\cot x+\mathrm{b}) \therefore \frac{\pi}{4}+\mathrm{a}=\frac{2\pi}{4}+\mathrm{b} \therefore \mathrm{a-b=\frac{\pi}{4}}\quad...(\mathrm{i}) \text{Also},\lim_{x\to\frac{\pi^{-}}{2}}\mathrm{f}(x)=\lim_{x\to\frac{\pi^{+}}{2}}\mathrm{f}(x) \therefore \lim_{x\to\frac{\pi}{2}}2x\cot x+b=\lim_{x\to\frac{\pi}{2}}a\cos2x-b\sin x \therefore 0+b=-a-b \therefore a+2b=0\quad\mathrm{...(ii)}Solving equations (i) and (ii), we get a = \frac{\pi}{6} and b = -\frac{\pi}{12}