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

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

$\therefore \underset{x\to a}{lim}\frac{(3x+b-1{)}^{1/3}-(b-1{)}^{1/3}}{x-a}$
$=\underset{x\to 0}{lim}\frac{(3x+8{)}^{1/3}-2}{x}=2\underset{x\to 0}{lim}\frac{{(1+\frac{3x}{8})}^{1/3}-1}{x}=\frac{1}{4}$