Test Index

UPSC NDA Math Model Paper 3

© examsnet.com

Question : 10 of 120

Marks: +1, -0

find the value of

k

f(x)=k \sin 2x + \frac{2}{3} \cos 3x

x=\frac{\pi}{6}?

-3
$\frac{1}{3}$
2
$\frac{2}{3}$

Solution:

Given that,

\Rightarrow f(x)=k \sin 2x + \frac{2}{3} \cos 3x

has maxima at

x=\frac{\pi}{6}

To find the stationary point

\Rightarrow f'(x)=k \times 2 \times \cos 2x - \left(\frac{2}{3}\right) \times 3

\times \sin 3x

\Rightarrow f'(x)=2k \cos 2x - 2 \sin 3x

\Rightarrow f'(x)=0

Given that stationary point

x=\frac{\pi}{6}

so, it will satisfy the

f'(x)=0

\Rightarrow 2k \cos\left(2 \times \frac{\pi}{6}\right) - 2 \sin\left(3 \times \frac{\pi}{6}\right)=0

\Rightarrow 2k \left(\frac{1}{2}\right) - 2(1)=0

\left[\because \cos \left(\frac{\pi}{3}\right)=\frac{1}{2}, \sin \left(\frac{\pi}{2}\right)=1\right]

\Rightarrow K=2

© examsnet.com

Go to Question:

Prev Question

Next Question