Question

If the range of the function $f\left(x\right)=8({\sin }^{4}x+{\cos }^{4}x-\sin x\cos x)\forall x\in R$ is $[a,b],$ then the value of ${\left(\underset{x\to a}{lim}\frac{{(3x+b-1)}^{1/3}-{(b-1)}^{1/3}}{x-a}\right)}^{-1}$ is

Answer 1

$f\left(x\right)=8({\sin }^{4}x+{\cos }^{4}x-\sin x\cos x)=8\left(\right({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x-\sin x\cos x)=8(1-\frac{1}{2}{\sin }^{2}2x-\frac{1}{2}\sin 2x)f\left(x\right)=9-{(2\text{sin}2x+1)}^2$
Range of $f\left(x\right)$ is $[0,9]$
$\Rightarrow a=0,b=9$

$\therefore L=\underset{x\to 0}{lim}\frac{{(3x+8)}^{1/3}-2}{x}\frac{0}{0}$ form
Applying L'Hospital rule
$L=\underset{x\to 0}{lim}\left(\frac{\frac{1}{3}{(3x+8)}^{-2/3}\cdot 3}{1}\right)=\frac{1}{4}$
$\therefore L^{-1}=4$