If f(x)= 36^x-9^x-4x+1 2- 1+ ,&x 0 ,&x=0 is continuous at x = 0, these k equals.

If f(x)= 36^x-9^x-4x+1 2- 1+ ,&x 0 ,&x=0 is continuous at x = 0, …

MCQ

If $\text{f(x)}=\begin{cases}\frac{36^\text{x}-9^\text{x}-4\text{x}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at $x = 0$, these $k$ equals.

A
$16\sqrt{2}\log2\log3$
B
$16\sqrt{2}\text{ in }6$
✓
$16\sqrt{2}\text{ in }6\text{ in }3$
D
None of these

✓

Answer

Correct option: C.

$16\sqrt{2}\text{ in }6\text{ in }3$

$\text{k}=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{ \text{x}}+1}{\sqrt2-\sqrt{1+\cos\text{x}}}$
consider,
$=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(4\times9)^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(4^{\text{x}}\times9^{\text{x}})-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{1-\cos^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\\cos\text{x})}{\sin^2\text{x}}$
dividing by $x^2$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{9^{\text{x}}-1}{\text{x}}{\times\frac{4^{\text{x}}-1}{\text{x}}\times\big(\sqrt{2}+\sqrt{1+\cos\text{x}}\big)(1+\cos\text{x})}}{\frac{\sin^2\text{x}}{\text{x}^2}}$
$=(\log9)(\log4)\big(\sqrt{2}+\sqrt{1+1}\big)(1+1)$
$=4\sqrt{2}(\log9)(\log4)$
$=4\sqrt{2}(2\log3)(2\log2)$
$=16\sqrt{2}(\log3)(\log2)$