Question

Select the correct answer from the given alternatives:

f(x) ${\displaystyle \begin{array}{ll}=\frac{{32}^x-8^x-4^x+1}{4^x-2^{x+1}+1}\text{,}& \text{for} x\ne 0\\ =\text{k}\text{,}& \text{for} x=0\end{array}}$ is continuous at x = 0, then value of ‘k’ is

Options

• 6

• 4

• (log 2)(log 4)

• 3 log 4

Answer 1

6

Explanation;

f(x) is continuous at x = 0

∴ f(0) = ${\displaystyle \underset{x\to 0}{lim}\text{f}\left(x\right)}$

∴ k = ${\displaystyle \underset{x\to 0}{lim}\frac{{32}^x-8^x-4^x+1}{4^x-2^{x+1}+1}}$

= ${\displaystyle \underset{x\to 0}{lim}\frac{(4^x-1)(8^x-1)}{{(2^x-1)}^2}}$

= ${\displaystyle \frac{\underset{x\to 0}{lim}\left(\frac{4^x-1}{x}\right)\left(\frac{8^x-1}{x}\right)}{\underset{x\to 0}{lim}{\left(\frac{2^x-1}{x}\right)}^2}}$

= ${\displaystyle \frac{\underset{x\to 0}{lim}\left(\frac{4^x-1}{x}\right)\cdot \underset{x\to 0}{lim}\left(\frac{8^x-1}{x}\right)}{{\left(\underset{x\to 0}{lim}\frac{2^x-1}{x}\right)}^2}}$

= ${\displaystyle \frac{\log 4\times \log 8}{{\left(\log 2\right)}^2}}$

= ${\displaystyle \frac{2\log 2\times 3\log 2}{{\left(\log 2\right)}^2}}$

= 6