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Question: \(f ( x ) = \sin ^ { p } x \cdot \cos ^ { q } x\) \(\left( p,q > 0,0 < x < \frac{\pi}{2} \right)\) h...

f(x)=sin⁡px⋅cos⁡qxf ( x ) = \sin ^ { p } x \cdot \cos ^ { q } x (p,q>0,0<x<π2)\left( p,q > 0,0 < x < \frac{\pi}{2} \right) has point

of maximum at

A

x=tan⁡−1(pq)x = \tan^{- 1}\left( \frac{\sqrt{p}}{q} \right)

B

x=tan⁡−1(qp)x = \tan^{- 1}\left( \frac{\sqrt{q}}{p} \right)

C

No such point exist

D

) None of these

Answer

x=tan⁡−1(qp)x = \tan^{- 1}\left( \frac{\sqrt{q}}{p} \right)

Explanation

Solution

f(x)=sin⁡px.cos⁡qxf(x) = \sin^{p}x.\cos^{q}x

f′(x)=sin⁡px.q.cos⁡q−1x.(−sin⁡x)+cos⁡qx.psin⁡p−1x.cos⁡xf'(x) = \sin^{p}x.q.\cos^{q - 1}x.\left( - \sin x \right) + \cos^{q}x.p\sin^{p - 1}x.\cos x= sin⁡p−1x.cos⁡q−1x(pcos⁡2x−qsin⁡2x)\sin^{p - 1}x.\cos^{q - 1}x\left( p\cos^{2}x - q\sin^{2}x \right).

For maxima & minima f′(x)=0f'(x) = 0

⇒ sin⁡p−1x.cos⁡q−1x(pcos⁡2x−qsin⁡2x)=0\sin^{p - 1}x.\cos^{q - 1}x\left( p\cos^{2}x - q\sin^{2}x \right) = 0

⇒ pcos⁡2x−qsin⁡2x=0p\cos^{2}x - q\sin^{2}x = 0

tanx = pq\sqrt{\frac{p}{q}}

f′(x)=q⋅sin⁡p−1x⋅cos⁡q+1x⋅(pq−tan⁡2x)f ^ { \prime } ( x ) = q \cdot \sin ^ { p - 1 } x \cdot \cos ^ { q + 1 } x \cdot \left( \frac { p } { q } - \tan ^ { 2 } x \right)

f′(x−h)=q+ve[pq−(tan⁡(x−h)2)]=+vef ^ { \prime } ( x - h ) = q + v e \left[ \frac { p } { q } - \left( \tan ( x - h ) ^ { 2 } \right) \right] = + v e f′(x+h)=+ve[pq−(tan⁡(x+h))2]=−vef ^ { \prime } ( x + h ) = + v e \left[ \frac { p } { q } - ( \tan ( x + h ) ) ^ { 2 } \right] = - v e

sign changes from +ve to -ve;

so x = tan-1 (pq)\left( \sqrt { \frac { p } { q } } \right) is point of local maxima.

So, 'a' is correct.