The function $f(x) = a\cos x + b an x + x$ has extreme values at $x = 0$ and $x = \frac{\pi}{6}$, then

Question

The function $f(x) = a\cos x + b	an x + x$ has extreme values at $x = 0$ and $x = \frac{\pi}{6}$, then

Dr. Abubacker · Accepted Answer

(a)
$f^{'}(x) = - {a\ sin}x + b\sec^{2}x + 1$
Now, $f^{'}(0) = 0$ and $f^{'}\left( \frac{\pi}{6} \right) = 0$
$\Rightarrow \ \ b + 1 = 0$ and $–\frac{a}{2} + \frac{4b}{3} + 1 = 0$
$\Rightarrow \ \ b = - 1,\ a = - \frac{2}{3}$

Q. The function $f(x) = a\cos x + b\tan x + x$ has extreme values at $x = 0$ and $x = \frac{\pi}{6}$, then

Explanation

Reviewed by Dr. Abubacker

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