all values of $ \tan^2 3x \cos^2 x - 4\tan 3x \sin 2x + 16\sin^2 x$ lie in interval

Question

For $0 < x < \frac{\pi}{6}$ , all values of

$ \tan^2 3x \cos^2 x - 4\tan 3x \sin 2x + 16\sin^2 x$ lie in interval

A. $(0 , \frac{121}{36})$

B.$(1,\frac{121}{9})$

C.$(-1,0)$

D.None of above

Now i completed square and write as $(\tan 3x \cos x - 4\sin x)^2$. How do i proceed? Thanks

Answer 1

The answer is $D$ since you have a perfect square and hence the value has to be positive. Indeed, let $f(x)=(\tan 3x \cos x - 4\sin x)^2$.

For $x\to 0$ the value of $f(x)$ tends to $0$. This leaves you with possibilities: A) or D).

Now remark that $f(x)$ is continuous in the interval you are considering and that for $x\to\frac{\pi}{6}$ the function $f(x)\to\infty$

Hence the only option is D).