Question

If $f\left(x\right)=\{\begin{array}{c}\frac{{20}^x+3^x-6^x-{10}^x}{1-\cos 8x}\text{; for}x\ne 0\\ \\ \left(\frac{k}{16}\right)\log \left(\frac{10}{3}\right).\log 2\text{; for}x=0\end{array}$ is continous at $x=0$, then the value of $k$ is

• A. ${\sin }^2{30}^{\circ }$
• B. $3^{{\log }_3\left(\frac{1}{2}\right)}$
• C. $\sqrt[3]{3\frac{1}{4}}$
• D. $\frac{\log 2^2}{3}$

Answer 1

The correct option is B $3^{{\log }_3\left(\frac{1}{2}\right)}$

$f\left(x\right)=\frac{{20}^x+3^x-6^x-{10}^x}{1-\cos 8x}$ when $x\ne 0$

$=\left(\frac{k}{16}\right)\log \left(\frac{10}{3}\right)\log 2$ for $x=0$

Since the value at $x=0$ is of the form $\frac{0}{0}$, we apply L'Hospital rule.

Differentiating,

$\underset{x\to 0}{lim}f\left(x\right)=\frac{{20}^x\ln 20+3^x\ln 3-6^x\ln 6-{10}^x\ln 10}{8\sin 8x}$

Differentiating again,

$\underset{x\to 0}{lim}f\left(x\right)=\frac{{20}^x{\ln }^{2}20+3^x{\ln }^{2}3-6^x{\ln }^{2}6-{10}^x{\ln }^{2}10}{64\cos 8x}$

Substituting $x=0$, we get

$\underset{x\to 0}{lim}f\left(x\right)=\frac{{\ln }^{2}20+{\ln }^{2}3-{\ln }^{2}6-{\ln }^{2}10}{64}$

$\underset{x\to 0}{lim}f\left(x\right)=\frac{{\ln }^{2}10+{\ln }^{2}2+2\left(\ln 2\right)\left(\ln 10\right)+{\ln }^{2}3-{\ln }^{2}2-{\ln }^{2}3-2\left(\ln 2\right)\left(\ln 3\right)-{\ln }^{2}10}{64}$

$=\frac{\left(\ln 2\right)\left(\ln 10\right)-\left(\ln 2\right)\left(\ln 3\right)}{32}$

$=\frac{\ln \frac{10}{3}\times \ln 2}{32}$

$\therefore k=\frac{1}{2}$, which can also be written as $3^{{\log }_3\frac{1}{2}}$