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Local Maxima

The function ...

Question

The function $y = a (1 - cos x)$ is maximum when $x$ is equal to

π

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\frac{π}{2}

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- \frac{π}{2}

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- \frac{π}{6}

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Solution

The correct option is A $π$
$y = a (1 - cos x)$ .
On differentiating, we get $y^{'} = a sin x$
Put $y^{'} = 0$ for maxima or minima
$sin x = 0 \Rightarrow x = 0, π$
Again differentiating, we get
$y^{''} = a cos x \Rightarrow y^{''} (0) = a$ and $y^{''} (π) = - a$
Hence, $y$ is maximum when $x = π$ .

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} 0, & - \frac{π}{2} < x < - \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = - \frac{π}{6} \end{matrix} \begin{matrix} x, & - \frac{π}{6} < x < \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = \frac{π}{6} \end{matrix} \begin{matrix} 0, & \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{matrix}

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