SATHEE: Application of Derivatives 2 Question 32

Application of Derivatives 2 Question 32

####32. Given $A = x : \frac{π}{6} \leq x \leq \frac{π}{3}$ and $f (x) = \cos x - x (1 + x)$ . Find $f (A)$ .

$(1980, 2 M)$

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Answer:

Solution:

Given, $A = x : \frac{π}{6} \leq x \leq \frac{π}{3}$

and $f (x) = \cos x - x - x^{2}$

$\Rightarrow f^{'} (x) = - \sin x - 1 - 2 x = - (\sin x + 1 + 2 x)$

which is negative for $x \in \frac{π}{6}, \frac{π}{3}$

$∴ f^{'} (x) < 0$

or $f (x)$ is decreasing.

Hence, $f (A) = f \frac{π}{3}, f \frac{π}{6}$

$= \frac{1}{2} - \frac{π}{3} 1 + \frac{π}{3}, \frac{\sqrt{3}}{2} - \frac{π}{6} 1 + \frac{π}{6}$

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