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The function \frac{\sin (x+a)}{\sin (x+b)} has no maxima or minima if

 

Option: 1

b-a=n \pi, n \in I


Option: 2

b-a=(2 n+1) \pi, n \in I


Option: 3

b-a=2 n \pi, n \in I


Option: 4

None of these


Answers (1)

best_answer

\begin{aligned} f(x) & =\frac{\sin (x+a)}{\sin (x+b)} \\ f^{\prime}(x) & =\frac{\sin (x+b) \times \cos (x+a)-\sin (x+a) \cos (x+b)}{\sin ^2(x+b)} \\ & =\frac{\sin (b-a)}{\sin ^2(x+b)} \end{aligned}
If \sin (b-a)=0, then f^{\prime}(x)=0
\Rightarrow f(x) will be constant.
i.e., b-a=n \pi \ \text{or} \ b-a=(2 n+1) \pi \text{ or } b-a=2 n \pi, then f(x) has no minima.

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seema garhwal

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