f(x)={(36)x−9x−4x+12sin2(x4) if x≠0k, if x=0is continuous at x=0 then 'k'=

f(x)={(36)x−9x−4x+12sin2(x4) if x≠0k, if x=0 is continuous at x=0⇒ltx→0f(x)=f(0)⇒k=ltx→09x(4x−1)−1⋅(4x−1)2sin2(x4)×x216×x216⇒k=ltx→0(9x−1)(4x−1)2×(x4)2(sinx4)2×1x216⇒k=ltx→0(9x−1x)(4x−1x)×(ltx→0x4sin(x4))2×162⇒k=8⋅loge9×loge4⇒k=32⋅loge3⋅loge2

f(x)={(36)x−9x−4x+12sin2(x4) if x≠0k, if x=0is continuous at x=0 then 'k'=

f(x)={(36)x−9x−4x+12sin2(x4) if x≠0k, if x=0 is continuous at x=0⇒ltx→0f(x)=f(0)⇒k=ltx→09x(4x−1)−1⋅(4x−1)2sin2(x4)×x216×x216⇒k=ltx→0(9x−1)(4x−1)2×(x4)2(sinx4)2×1x216⇒k=ltx→0(9x−1x)(4x−1x)×(ltx→0x4sin(x4))2×162⇒k=8⋅loge9×loge4⇒k=32⋅loge3⋅loge2

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$f (x) = {\begin{cases} \frac{{(36)}^{x} - 9^{x} - 4^{x} + 1}{2 \sin^{2} (\frac{x}{4})} i f x \neq 0 \\ k, i f x = 0 \end{cases}$ is continuous at $x = 0$ then $' k' =$

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$8 \log_{e} 5$

$32 \log_{e} 2. \log_{e} 3$

$\log_{e} 2. \log_{e} 3$

$8 \log_{e} (6)$

answer is C.

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Detailed Solution

$f (x) = {\begin{cases} \frac{{(36)}^{x} - 9^{x} - 4^{x} + 1}{2 \sin^{2} (\frac{x}{4})} i f x \neq 0 \\ k, i f x = 0 \end{cases}$ is continuous at $x = 0$

$\begin{array}{l} \Rightarrow \underset{x \to 0}{l t} f (x) = f (0) \\ \Rightarrow k = \underset{x \to 0}{l t} \frac{9^{x} (4^{x} - 1) - 1 \cdot (4^{x} - 1)}{2 \sin^{2} (\frac{x}{4}) \times \frac{x^{2}}{16}} \times \frac{x^{2}}{16} \\ \Rightarrow k = \underset{x \to 0}{l t} \frac{(9^{x} - 1) (4^{x} - 1)}{2} \times \frac{{(\frac{x}{4})}^{2}}{{(\sin \frac{x}{4})}^{2}} \times \frac{1}{\frac{x^{2}}{16}} \\ \Rightarrow k = \underset{x \to 0}{l t} (\frac{9^{x} - 1}{x}) (\frac{4^{x} - 1}{x}) \times {(\underset{x \to 0}{l t} \frac{\frac{x}{4}}{\sin (\frac{x}{4})})}^{2} \times \frac{16}{2} \\ \Rightarrow k = 8 \cdot \log_{e} 9 \times \log_{e} 4 \\ \Rightarrow k = 32 \cdot \log_{e} 3 \cdot \log_{e} 2 \end{array}$