Question

The Function $f\left(x\right)=\frac{1-\cos x\left(\cos 2x{)}^{\frac{1}{2}}\right(\cos 3x{)}^{\frac{1}{3}}}{x^2}$ is not defined at $x=0$. If $f\left(x\right)$ is continuous at $x=0$ then $f\left(0\right)$ equals

• A. $1$
• B. $3$
• C. $6$
• D. $-6$

Answer 1

Answer: (B)

$3$

Applying L'Hospital's Rule,
$=\underset{x\to 0}{lim}\left(\frac{3sin3x.cosx(cos2x{)}^{\frac{1}{2}}.\frac{1}{3}(cos3x{)}^{\frac{-2}{3}}+2sin2xcosx\left(cos3x{)}^{\frac{1}{3}}\frac{1}{2}\right(cos2x{)}^{\frac{-1}{2}}+\left(cos2x{)}^{\frac{1}{2}}\right(cos3x{)}^{\frac{1}{3}}sinx}{2x}\right)$
Applying L'Hospitals Rule again, here we have ignored the terms containing $sinx$ as $x\to 0$ when $sinx\to 0$
$=\underset{x\to 0}{lim}\left(\frac{9cos3x.cosx(cos2x{)}^{\frac{1}{2}}.\frac{1}{3}(cos3x{)}^{\frac{-2}{3}}+4cos2xcosx\left(cos3x{)}^{\frac{1}{3}}\frac{1}{2}\right(cos2x{)}^{\frac{-1}{2}}+\left(cos2x{)}^{\frac{1}{2}}\right(cos3x{)}^{\frac{1}{3}}cosx}{2}\right)$
$=\frac{9.\frac{1}{3}+4.\frac{1}{2}+1}{2}$$=\frac{3+2+1}{2}$
$=3$