The set of all real values of $\lambda$ for which the
function
$$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right),x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$
has exactly one maxima and exactly one
minima, is :
Solution
$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right)$
<br><br>$\Rightarrow$ f(x) = sin<sup>2</sup> x($\lambda$ + sinx) ....(1)
<br><br>$\therefore$ f'(x) = 2sinx cosx ($\lambda$ +sinx) + sin<sup>2</sup>x (cosx)
<br><br>$\Rightarrow$ f'(x) = sin2x(${{2\lambda + 3\sin x} \over 2}$)
<br><br>For maixma and minima, f'(x) = 0
<br><br>$\therefore$ sin2x = 0 or 2$\lambda$ + 3sinx = 0
<br><br>when sin2x = 0 $\Rightarrow$ x = 0
<br><br>or when 2$\lambda$ + 3sinx = 0
<br><br>$\Rightarrow$ sin x = $- {{2\lambda } \over 3}$
<br><br>As $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$
<br><br>$\therefore$ -1 < sinx < 1
<br><br>$\Rightarrow$ -1 < $- {{2\lambda } \over 3}$ < 1
<br><br>$\Rightarrow$ $- {3 \over 2}$ < $\lambda$ < ${3 \over 2}$
<br><br>$\therefore$ $\lambda$ $\in$ $\left( { - {3 \over 2},{3 \over 2}} \right)$ - {0}
<br><br><b>Note :</b> If $\lambda$ = 0 $\Rightarrow$ f(x) = sin<sup>3</sup>x [from (1)]
<br><br>Which is monotonic. so no maxima/minima.
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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