Question

If $\underset{x\to \frac{1}{2}}{\lim }$$\frac{32x^5-1}{2x-1}$ $=\underset{\theta \to 0}{\lim }$$\frac{tan\left(\theta /m\right)}{\theta }$, then $m$ is equal to

• A. $2$
• B. $5$
• C. $\frac{1}{5}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\underset{x\to \frac{1}{2}}{\lim }$$\frac{32x^5-1}{2x-1}$ $=\underset{\theta \to 0}{\lim }$$\frac{tan\left(\theta /m\right)}{\theta }$
$\Rightarrow \left(\frac{32}{2}\right)$ $\underset{x\to \frac{1}{2}}{\lim }$$\frac{x^5-\frac{1}{32}}{x-\frac{1}{2}}=\frac{1}{m}$ $\underset{\theta \to 0}{\lim }$$\left(\frac{tan\frac{\theta }{m}}{\frac{\theta }{m}}\right)$
$\Rightarrow 16$ $\underset{x\to \frac{1}{2}}{\lim }$$\frac{x^5-{\left(\frac{1}{2}\right)}^5}{x-\frac{1}{2}}=\frac{1}{m}\times 1$
$\Rightarrow 16\times 5\times {\left(\frac{1}{2}\right)}^{5-1}=\frac{1}{m}$
$\Rightarrow 16\times 5\times \frac{1}{16}=\frac{1}{m}$ or $m=\frac{1}{5}$