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The set of all real values of \lambda for which the function f(x)=(1-\cos^{2}x).\left ( \lambda +\sin\; x \right ), x\epsilon \left ( -\frac{\pi }{2},\frac{\pi }{2} \right ), has exactly one maxima and exactly one minima, is :
Option: 1 \left ( -\frac{1}{2},\frac{1}{2} \right )-\left \{ 0 \right \}
 
Option: 2 \left ( -\frac{3}{2},\frac{3}{2} \right )
Option: 3 \left ( -\frac{1}{2},\frac{1}{2} \right )  
Option: 4 \left ( -\frac{3}{2},\frac{3}{2} \right )-\left \{ 0 \right \}

Answers (1)

best_answer

\\f(x)=\left(1-\cos ^{2} x\right)(\lambda+\sin x) \\\\ x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

\\ f(x)=\lambda \sin ^{2} x+\sin ^{3} x \\ \\f^{\prime}(x)=2 \lambda \sin x \cos x+3 \sin ^{2} x \cos x \\ \\f^{\prime}(x)=\sin x \cos x(2 \lambda+3 \sin x)

\sin x=0, \frac{-2 \lambda}{3},(\lambda \neq 0)

for exactly one maxima & minima

\\\frac{-2 \lambda}{3} \in(-1,1) \Rightarrow \lambda \in\left(\frac{-3}{2}, \frac{3}{2}\right) \\ \lambda\in \left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}

Posted by

himanshu.meshram

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