If f(x)=72x−9x−8x+12−1+cosxx≠0=k log2.log3x=0 is a continuous function then k =

f(x)=72x−9x−8x+12−1+cosxf(0)=Ltx→0f(x)f(0)=Ltx→0 (9x−1)(8x−1)2−1+cosx×(2+1+cosx)(2+1+cosx)f(0)=Ltx→0 (9x−1)(8x−1)(2+1+cosx)1−cosxf(0)=Ltx→0 9x−1x8x−1x×x22sin2x2x24×x24 Ltx→0(2+1+cosx)f(0)=klog2.log3=2log9.log8(2+2)k.log2.log3=2.22 .3(log2).2(log3)k=242

If f(x)=72x−9x−8x+12−1+cosxx≠0=k log2.log3x=0 is a continuous function then k =

f(x)=72x−9x−8x+12−1+cosxf(0)=Ltx→0f(x)f(0)=Ltx→0 (9x−1)(8x−1)2−1+cosx×(2+1+cosx)(2+1+cosx)f(0)=Ltx→0 (9x−1)(8x−1)(2+1+cosx)1−cosxf(0)=Ltx→0 9x−1x8x−1x×x22sin2x2x24×x24 Ltx→0(2+1+cosx)f(0)=klog2.log3=2log9.log8(2+2)k.log2.log3=2.22 .3(log2).2(log3)k=242

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If $f (x) = \begin{matrix} \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} & x \neq 0 \end{matrix}$
$= \begin{matrix} k \log 2. \log 3 & x = 0 \end{matrix}$ is a continuous function then k =

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$18 \sqrt{3}$

$24 \sqrt{2}$

$\sqrt{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{72^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} \\ f (0) = \underset{x \to 0}{L t} f (x) \\ f (0) = \underset{x \to 0}{L t} \frac{(9^{x} - 1) (8^{x} - 1)}{\sqrt{2} - \sqrt{1 + \cos x}} \times \frac{(\sqrt{2} + \sqrt{1 + \cos x})}{(\sqrt{2} + \sqrt{1 + \cos x})} \end{array}$

$f (0) = \underset{x \to 0}{L t} \frac{(9^{x} - 1) (8^{x} - 1) (\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$

$f (0) = \underset{x \to 0}{L t} \frac{(\frac{9^{x} - 1}{x}) (\frac{8^{x} - 1}{x}) × x^{2}}{\frac{2 \sin^{2} \frac{x}{2}}{\frac{x^{2}}{4}} \times \frac{x^{2}}{4}}$ $\underset{x \to 0}{L t} (\sqrt{2} + \sqrt{1 + \cos x})$

$\begin{array}{l} f (0) = k \log 2. \log 3 = 2 \log 9. \log 8 (\sqrt{2} + \sqrt{2}) \\ k . \log 2. \log 3 = 2.2 \sqrt{2} .3 (\log 2) .2 (\log 3) \\ k = 24 \sqrt{2} \end{array}$