The values of a and b,so that the function f( | vedclass

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(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit at these point

Vedclass · Accepted Answer

$\frac{\pi}{6}, -\frac{\pi}{12}$

Vedclass · Answer

(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit at these points.At $x = \frac{\pi}{4}$:$\lim_{x 	o \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x 	o \frac{\pi}{4}^+} (2x \cot x + b)$$\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2(\frac{\pi}{4}) \cdot 1 + b$$\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4} \quad (i)$At $x = \frac{\pi}{2}$:$\lim_{x 	o \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x 	o \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$Using $\lim_{x 	o \frac{\pi}{2}} x \cot x = \lim_{x 	o \frac{\pi}{2}} x \frac{\cos x}{\sin x} = 0$:$0 + b = a \cos(\pi) - b \sin(\frac{\pi}{2})$$b = -a - b \implies a + 2b = 0 \quad (ii)$From $(ii)$,$a = -2b$. Substituting into $(i)$:$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.Thus,$a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$.

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x, & 0 \le x \le \frac{\pi}{4} \\ 2x \cot x + b, & \frac{\pi}{4} < x \le \frac{\pi}{2} \\ a \cos 2x - b \sin x, & \frac{\pi}{2} < x \le \pi \end{cases}$ is continuous for $0 \le x \le \pi$,are respectively:

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