The function f(x) = frac{sin(x + a)}{sin(x + | vedclass

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(A) Given the function $f(x) = \frac{\sin(x + a)}{\sin(x + b)}$.Applying the quotient rule for differentiation,$f'(x) = \frac{\cos(x + a)\sin(x + b) -

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

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(A) Given the function $f(x) = \frac{\sin(x + a)}{\sin(x + b)}$.Applying the quotient rule for differentiation,$f'(x) = \frac{\cos(x + a)\sin(x + b) - \sin(x + a)\cos(x + b)}{\sin^2(x + b)}$.Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get $f'(x) = \frac{\sin((x + b) - (x + a))}{\sin^2(x + b)} = \frac{\sin(b - a)}{\sin^2(x + b)}$.For the function to have no maxima or minima,the derivative $f'(x)$ must be zero for all $x$ in its domain,which implies $f(x)$ is a constant function.This occurs when $\sin(b - a) = 0$.The general solution for $\sin 	heta = 0$ is $	heta = n \pi$,where $n \in I$.Thus,$b - a = n \pi$ for $n \in I$.

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

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