F x -27 x -9 x -3 x -1-2-1-cos x is continuous at x -0 and f 0- k -2log 32 then k -A- 4B- 8C- 1D- 2

The correct option is B 8 lim_x→10(9^x 1)(3^x 1)/(1 cosx)(√(2)+√(1+cosx)) = 4(log3)(log3).2√(2) = 8√(()log3)^2 ...

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Square Root of a Complex Number

f x =27 x -9 ...

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$f (x) = \frac{27^{x} - 9^{x} - 3^{x} + 1}{\sqrt{2} - \sqrt{1 + c o s x}}$ is continuous at x = 0 and f(0)= $k \sqrt{2} (l o g 3)^{2}$ then k =

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I f (x) = {\begin{matrix} \frac{36^{x} - 9^{x} - 4^{x} + 1}{\sqrt{2} - \sqrt{1 + c o s x}}, x \neq 0 k, x = 0 \end{matrix}

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16 \sqrt{2}

16 \sqrt{2}

16 \sqrt{2}

f

f (x) = \frac{1}{x} - \frac{k - 1}{e^{2 x} - 1}

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(k, f (0))

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