Question

If $y=\frac{\sin (x+a)}{\sin (x+b)}$, $a\ne b$, then $y$ is

• A. minimum at $x=0$
• B. maxima at $x=0$
• C. neither minima nor maxima at $x=0$
• D. None of the above

Answer 1

The correct option is C

neither minima nor maxima at $x=0$

Explanation for the correct option:

$y=\frac{\sin (x+a)}{\sin (x+b)}$, $a\ne b$

Using Quotient Rule, differentiating with respect to $x$ both sides:

$\frac{dy}{dx}=\frac{\sin (x+b)\cos (x+a)-\sin (x+a)\cos (x+b)}{{\sin }^2(x+b)}$

Using $\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{B}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}$.We get,

$$\begin{gathered} \frac{dy}{dx}=\frac{\sin \left[(x+b)-(x+a)\right]}{{\sin }^2(x+b)} \\ \Rightarrow \frac{dy}{dx}=\frac{\sin (b-a)}{{\sin }^2(x+b)} \end{gathered}$$

For maxima or minima, Put,

$$\begin{gathered} \frac{dy}{dx}=0 \\ \Rightarrow \frac{\sin (b-a)}{{\sin }^2(x+b)}=0 \\ \Rightarrow \sin (b-a)=0 \end{gathered}$$

This is independent of $x$.

Therefore, $y$ is neither minima nor maxima at $x=0$.

Hence, Option (C) is the correct answer.