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Find the value of \mathrm{ |a / 2 b| }so that the function
\mathrm{f(x)= \begin{cases}x+a \sqrt{2} \sin x, & 0 \leq x<\pi / 4 \\ 2 x \cot x+b, & \pi / 4 \leq x \leq \pi / 2 \\ a \cos 2 x-b \sin x, & \pi / 2<x \leq \pi\end{cases} }
is continuous for \mathrm{0 \leq x \leq \pi. }

Option: 1

0


Option: 2

1


Option: 3

1.3


Option: 4

1.5


Answers (1)

best_answer

 We apply the test of continuity at \mathrm{x=\frac{\pi}{4} and x=\frac{\pi}{2} } to get the values of a and b.
At \mathrm{x=\frac{\pi}{4} }, we have
\mathrm{\begin{aligned} & f\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4} \cot \frac{\pi}{4}+b=\frac{\pi}{2}+b \\ & f\left(\frac{\pi}{4}-0\right)=\lim _{h \rightarrow 0}\left[\frac{\pi}{4}-h+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right)\right] \\ & =\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1}{\sqrt{2}}=a+\frac{\pi}{4} \end{aligned} }

and \mathrm{f\left(\frac{\pi}{4}+0\right)=\lim _{h \rightarrow 0}\left[2\left(\frac{\pi}{4}+h\right) \cot \left(\frac{\pi}{4}+h\right)+b\right] =\frac{\pi}{2} \cdot 1+b=\frac{\pi}{2}+b. }
\mathrm{ \therefore \quad For\, \, continuity\, \, at\, x=\frac{\pi}{4}, }
we have
\mathrm{\frac{\pi}{2}+b=a+\frac{\pi}{4} \Rightarrow a-b=\frac{\pi}{4} . }......................(1)
At  we have
\mathrm{\begin{aligned} & f\left(\frac{\pi}{2}\right)=2 \cdot \frac{\pi}{2} \cot \frac{\pi}{2}+b=b \\ & f\left(\frac{\pi}{2}-0\right)=\lim _{h \rightarrow 0}\left[2 \cdot\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b\right]=b \\ & f\left(\frac{\pi}{2}+0\right)=\lim _{h \rightarrow 0}\left[a \cos 2\left(\frac{\pi}{2}+h\right)-b \sin \left(\frac{\pi}{2}+h\right)\right] \\ & =-a-b . \end{aligned} }
\mathrm{\therefore \quad\, \, For\, \, continuity \, \, at\, \, x=\frac{\pi}{2} }, we have
\mathrm{b=-a-b \Rightarrow a+2 b=0 }.......................(2)

Solving (i) and (ii), we get \mathrm{a=\frac{\pi}{6}, b=-\frac{\pi}{12}. }

Posted by

Deependra Verma

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