If the function f(x) = begin{cases} x + a^2sq | vedclass

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(D) For $f(x)$ to be continuous at $x = \pi/4$,we must have $\lim_{x 	o \pi/4^-} f(x) = \lim_{x 	o \pi/4^+} f(x) = f(\pi/4)$.$\lim_{x 	o \pi/4^-} (

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$b$ or $c$ both

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(D) For $f(x)$ to be continuous at $x = \pi/4$,we must have $\lim_{x 	o \pi/4^-} f(x) = \lim_{x 	o \pi/4^+} f(x) = f(\pi/4)$.$\lim_{x 	o \pi/4^-} (x + a^2\sqrt{2} \sin x) = \pi/4 + a^2\sqrt{2} \sin(\pi/4) = \pi/4 + a^2\sqrt{2}(1/\sqrt{2}) = \pi/4 + a^2$.$\lim_{x 	o \pi/4^+} (x \cot x + b) = \pi/4 \cot(\pi/4) + b = \pi/4(1) + b = \pi/4 + b$.Equating these,$\pi/4 + a^2 = \pi/4 + b \implies b = a^2$.For $f(x)$ to be continuous at $x = \pi/2$,we must have $\lim_{x 	o \pi/2^-} f(x) = \lim_{x 	o \pi/2^+} f(x) = f(\pi/2)$.$\lim_{x 	o \pi/2^-} (x \cot x + b) = \lim_{x 	o \pi/2^-} (x \frac{\cos x}{\sin x} + b) = 0 + b = b$.$\lim_{x 	o \pi/2^+} (b \sin 2x - a \cos 2x) = b \sin(\pi) - a \cos(\pi) = b(0) - a(-1) = a$.Equating these,$b = a$.Since $b = a^2$ and $b = a$,we have $a^2 = a \implies a^2 - a = 0 \implies a(a - 1) = 0$.Thus,$a = 0$ or $a = 1$.If $a = 0$,then $b = 0$. If $a = 1$,then $b = 1$.Therefore,the possible values for $(a, b)$ are $(0, 0)$ and $(1, 1)$.

If the function $f(x) = \begin{cases} x + a^2\sqrt{2} \sin x, & 0 \le x < \pi/4 \\ x \cot x + b, & \pi/4 \le x < \pi/2 \\ b \sin 2x - a \cos 2x, & \pi/2 \le x \le \pi \end{cases}$ is continuous in the interval $[0, \pi]$,then the values of $(a, b)$ are:

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