Question

If the function $f\left(x\right)=\frac{1-\cos x\cos 2x\cos 3x}{{\sin }^{2}2x}$ is continuous at $x=0$, then $f\left(0\right)$must be equal to

• A. $\frac{7}{4}$
• B. $\frac{3}{4}$
• C. $\frac{7}{8}$
• D. $-\frac{7}{4}$

Answer 1

Answer: (A)

$\frac{7}{4}$

For the function to be continuous at $x=0$, $f\left(0\right)={lim}_{x\to 0}f\left(x\right)$
$f\left(x\right)=\frac{1-\cos x\cos 2x\cos 3x}{{\sin }^{2}2x}=\frac{1-\left(\cos x\cos 3x\right)\cos 2x}{{\sin }^{2}2x}=\frac{1-\frac{1}{2}(\cos 4x+\cos 2x)\cos 2x}{{\sin }^{2}2x}=\frac{1-\frac{1}{2}(2{\cos }^{2}2x-1+\cos 2x)\cos 2x}{1-{\cos }^{2}2x}=\frac{1-\frac{1}{2}(2y^2-1+y)y}{1-y^2}$
Where $y=\cos 2x$.
As $x\to 0,y\to 1$
This expression can be rewritten as:
$\frac{2-(2y^2-1+y)y}{2(1-y^2)}=\frac{(2y^3-y+y^2-2)}{2(y^2-1)}=\frac{(y-1)(2y^2+3y+2)}{2(y-1)(y+1)}$
$f\left(0\right)={lim}_{x\to 0}f\left(x\right)={lim}_{y\to 1}\frac{(y-1)(2y^2+3y+2)}{2(y-1)(y+1)}=\frac{(2\cdot 1^2+3\cdot 1+2)}{2(1+1)}=\frac{7}{4}$