If maximum value of $f(x)=\frac{3\sqrt{2x^2+x}}{2x^2+x+2\sin^2\theta}$ (where $x\gt0$) is $\frac3{√2}$ then $\theta$ may lie in

Question

Question:

If maximum value of $f(x)=\frac{3\sqrt{2x^2+x}}{2x^2+x+2\sin^2\theta}$ (where $x\gt0$) is $\frac3{√2}$ then $\theta$ may lie in

A) $(-\fracπ6,\fracπ6)$

B) $(-\fracπ4,\fracπ4)$

C) $(-\fracπ3,\fracπ3)$

D) $(-π,π)$

My Attempt:

$\frac{3\sqrt{2x^2+x}}{2x^2+x+2(1)} \le f(x)\le\frac{3\sqrt{2x^2+x}}{2x^2+x+2(0)}$

$\frac{3}{\sqrt{2x^2+x}+\frac2{\sqrt{2x^2+x}}} \le f(x)\le\frac{3}{\sqrt{2x^2+x}}$

Not sure what to do next.

Edit:

$f'(x)=\frac{(2x^2+x+2\sin^2\theta)\frac{3(4x+1)}{2\sqrt{2x^2+x}}-3\sqrt{2x^2+x}(4x+1)}{(2x^2+x+2\sin^2\theta)^2}$

Numerator is $3(4x+1)(2x^2+x+2\sin^2\theta-2(2x^2+x))$

i.e. $(2\sin^2\theta-2x^2-x))$

Answer 1

Let $t=\sqrt{2x^2+x}.$ Then $t>0$ and $t$ admits all the values from $(0,\infty).$ The function takes the form $${3t\over t^2+2\sin^2\theta}\qquad (*)$$ The inequality $t^2+a^2\ge 2|a|\,t$ implies ($a=\sqrt{2}\sin\theta$) $$t^2+2\sin^2\theta\ge 2\sqrt{2}|\sin \theta|\, t$$ and the equality occurs when $t=|a|=\sqrt{2}|\sin\theta|.$ Hence the minimal value of $(*)$ is equal ${3\over 2\sqrt{2}|\sin\theta|}.$ Hence $|\sin \theta|={1\over 2}.$ In the interval $(-\pi,\pi)$ the only solutions are $\theta=\pm {\pi\over 6}$ and $\theta=\pm {5\pi\over 6}.$ Thus B, C and D may occur unlike A.