Question

If $f^2\left(x\right)+2g^2\left(x\right)+3h^2\left(x\right)\le 1$ and $u\left(x\right)=2f\left(x\right)-2g\left(x\right)+3h\left(x\right),$ then the maximum value of $u^2\left(x\right)$ is equal to

Answer 1

Let two vectors $\overrightarrow{P}=f\left(x\right)\hat{i}+\sqrt{2}g\left(x\right)\hat{j}+\sqrt{3}h\left(x\right)\hat{k}$ and $\overrightarrow{Q}=2\hat{i}-\sqrt{2}\hat{j}+\sqrt{3}\hat{k}$
if $\theta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q},$
then
$\cos \theta =\frac{\overrightarrow{P}\cdot \overrightarrow{Q}}{|\overrightarrow{P}||\overrightarrow{Q}|}=\frac{2f\left(x\right)-2g\left(x\right)+3h\left(x\right)}{\sqrt{f^2\left(x\right)+2g^2\left(x\right)+3h^2\left(x\right)}\times \sqrt{4+2+3}}\Rightarrow \frac{\left[2f\right(x)-2g(x)+3h(x){]}^2}{\left(f^2\left(x\right)+2g^2\left(x\right)+3h^2\left(x\right)\right)\times 9}\le 1[\because {\cos }^2\theta \le 1]\Rightarrow u^2\left(x\right)\le 9\left(f^2\right(x)+2g^2(x)+3h^2(x\left)\right)\Rightarrow u^2\left(x\right)\le 9\times 1\Rightarrow u^2\left(x\right)\le 9$