If the function $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,x + {a^2}\sqrt 2 \sin x,\,0 \le x < \pi /4\\\,\,\,\,\,\,\,\,\,\,\,\,\,x\cot x + b,\,\pi /4 \le x < \pi /2\\b\sin 2x - a\cos 2x,\,\pi /2 \le x \le \pi \end{array} \right.$ is continuous in the interval $[0,\,\pi ]$, then the values of $(a,\,b)$ are
- A
$(-1, -1)$
- B
$(0,0)$
- C
$(1,1)$
- ✓
$b$ or $c$ both
✓
Answer
Correct option: D.$b$ or $c$ both
d
(d) Since $f$ is continuous at $x = \frac{\pi }{4}$
$\therefore \,\,\,f\left( {\frac{\pi }{4}} \right) = \mathop f\limits_{h \to 0} \,\left( {\frac{\pi }{4} + h} \right) = \mathop f\limits_{h \to 0} \,\left( {\frac{\pi }{4} - h} \right)$
$ \Rightarrow \,\,\,\frac{\pi }{4}\cot \frac{\pi }{4} + b = \mathop f\limits_{h \to 0} \,\left( {\frac{\pi }{4} + h} \right) + {a^2}\sqrt 2 \sin \,\left( {\frac{\pi }{4} + h} \right)$
$ \Rightarrow \,\,\frac{\pi }{4}(1) + b = \left( {\frac{\pi }{4} + 0} \right) + {a^2}\sqrt 2 \sin \,\left( {\frac{\pi }{4} + 0} \right)$
$ \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 $x = \frac{\pi }{2}$
$\therefore$ $f\left( {\frac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to \frac{\pi }{2} - 0} f(x) = \mathop {\lim }\limits_{h \to 0} f\left( {\frac{\pi }{2} - h} \right)$
$ \Rightarrow \,\,b\sin 2\frac{\pi }{2} - a\cos 2\frac{\pi }{2} = \mathop {\lim }\limits_{h \to 0} \,\left[ {\left( {\frac{\pi }{2} - h} \right)\,\cot \,\left( {\frac{\pi }{2} - h} \right) + b} \right]$
$ \Rightarrow \,\,b.0 - a\,( - 1) = 0 + b\,\, \Rightarrow \,\,a = b$
Hence $(0, 0), (1, 1)$ satisfy the above relations.
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