Question

The maximum value of the expression $\frac{3}{9co{s}^2\Theta +8sin\Theta cos\Theta +3si{n}^2\Theta }$ is

Answer 1

Consider the given function

$f\left(x\right)=\frac{3}{9{\cos }^2\theta +8\sin \theta \cos \theta +3{\sin }^2\theta }$

$f\left(x\right)=\frac{3}{8{\cos }^2\theta +8\sin \theta \cos \theta +2{\sin }^2\theta +{\sin }^2\theta +{\cos }^2\theta }$

$f\left(x\right)=\frac{3}{8{\cos }^2\theta +8\sin \theta \cos \theta +2{\sin }^2\theta +1}$

$f\left(x\right)=\frac{3}{8{\cos }^2\theta +4\sin \theta \cos \theta +4\sin \theta \cos \theta +2{\sin }^2\theta +1}$

$f\left(x\right)=\frac{3}{4\cos \theta (2\cos \theta +\sin \theta )+2\sin \theta (2\cos \theta +\sin \theta )+1}$

$f\left(x\right)=\frac{3}{(2\cos \theta +\sin \theta )(4\cos \theta +2\sin \theta )+1}$

For the maximum value of expression the denominator must be minimum and for denominator to be minimum,

$(2\cos \theta +\sin \theta )(4\cos \theta +2\sin \theta )$ can be zero.

When

$(2\cos \theta +\sin \theta )(4\cos \theta +2\sin \theta )=0$

$\Rightarrow 2\cos \theta +\sin \theta =0,4\cos \theta +2\sin \theta =0$

$\Rightarrow 2\cos \theta =-\sin \theta ,4cos\theta =-2sin\theta$

$\Rightarrow tan\theta =-2,tan\theta =-2$

Therefore, the maximum value of expression

$=\frac{3}{(2\cos \theta +\sin \theta )(4\cos \theta +2\sin \theta )+1}$

$=\frac{3}{0\times 0+1}$

$=3$

Hence, this is the answer.