The set of all real values of lambda for whic | vedclass

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(D) Given $f(x) = (1 - \cos^2 x)(\lambda + \sin x) = \sin^2 x(\lambda + \sin x) = \lambda \sin^2 x + \sin^3 x$.Find the derivative: $f'(x) = 2\lambda

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$(-\frac{3}{2}, \frac{3}{2}) - \{0\}$

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(D) Given $f(x) = (1 - \cos^2 x)(\lambda + \sin x) = \sin^2 x(\lambda + \sin x) = \lambda \sin^2 x + \sin^3 x$.Find the derivative: $f'(x) = 2\lambda \sin x \cos x + 3 \sin^2 x \cos x = \sin x \cos x (2\lambda + 3 \sin x)$.For critical points,set $f'(x) = 0$. Since $\cos x 
eq 0$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$,we have $\sin x = 0$ or $\sin x = -\frac{2\lambda}{3}$.$x = 0$ is always a critical point. For the function to have exactly one maxima and one minima,there must be another critical point in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.Thus,$-1 Solving $-1 Excluding $\lambda = 0$ (where the two critical points coincide at $x=0$),the set of values is $(-\frac{3}{2}, \frac{3}{2}) - \{0\}$.

The set of all real values of $\lambda$ for which the function $f(x) = (1 - \cos^2 x)(\lambda + \sin x)$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ has exactly one maxima and exactly one minima is

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