Question

${\cos }^{3}x-{\cos }^3({60}^0-x)-{\cos }^3({60}^0+x)\in$ $[-\frac{k}{4}\frac{k}{4}]$, $k=$

• A. $-3$
• B. $-5$
• C. $3$
• D. $5$

Answer 1

The correct option is C $3$

Algebric identity,
$a^3-b^3=(a-b{)}^3+3a^{2}b-3a{b}^2$
$a^3+b^3=(a+b{)}^3-3ab(a+b)$
$={\cos }^{3}x-\left[\right(\cos (60+x)+\cos (60-x){)}^3-3\cos (60+x)\cos 60-x\left(\cos \right(60+x)\cos 60-x)]$
$={\cos }^{3}x-\left[\right(2\times \frac{1}{2}\times \cos x{)}^3-\frac{3}{2}[\cos 2x+\cos 120]\left(\cos x\right)]$
$={\cos }^{3}x-\left[\right(\cos x{)}^3-\frac{3}{2}[2{\cos }^{2}x-\frac{3}{2}]\cos x]$
$=(3{\cos }^{2}x-\frac{9}{4})\cos x[\cos A\cos B=\frac{1}{2}(\cos (A-B)+\cos (A+B)\left)\right]$
$=3\frac{(4{\cos }^{3}x-3\cos x)}{4}$
$f\left(x\right)=\frac{3\cos 3x}{4}$
$\therefore$ range of $f\left(x\right)ϵ[-\frac{3}{4},\frac{3}{4}]=[-\frac{k}{4},\frac{k}{4}]$
$k=3$