Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{{36}^x-9^x-4^x+1}{\sqrt{2}-\sqrt{1+cosx}}& ,x\ne 0\\ k& ,x=0\end{array}$ Is continuous at x =0, then k equals:

• A. $\sqrt{2}ln2ln3$
  
• B. $16\sqrt{2}ln6$
  
• C. $16\sqrt{2}ln2ln3$
• D. None of these

Answer 1

The correct option is C $16\sqrt{2}ln2ln3$

$f\left(x\right)=\{\begin{array}{c}\frac{{36}^x-9^x-4^x+1}{\sqrt{2}-\sqrt{1+cosx}}:x\ne 0\end{array}(cos\frac{x}{2}>0about0)$
$=\underset{x\to 0}{lim}\left(fx\right)=\underset{x\to 0}{lim}\frac{(9^x-1)(4^x-1)}{\sqrt{2}(1-cos\frac{x}{2})}(cos\frac{x}{2}>0about0)$
$=\underset{x\to 0}{lim}\frac{(9^x-1)(4^x-1)}{2\sqrt{2}si{n}^2\left(\frac{x}{4}\right)}$
$=\underset{x\to 0}{lim}\frac{1}{2\sqrt{2}}\times \frac{9^x-1}{x}\times \frac{4^x-1}{x}\times {\left(\frac{\frac{x}{4}}{sin\frac{x}{4}}\right)}^2\times 16$
$=4\sqrt{2}ln9\times ln4\times 1=16\sqrt{2}ln3ln2.$
Hence, for f(x) to be continuous at x =0,
$\underset{x\to 0}{lim}f\left(x\right)$=k i.e., $k=16\sqrt{2}ln3ln2$