F x - 4 x -1 3 - sin x - p log 1- x 2 - 3 is continous at x-0 and f0-12 log4 3 then p- -

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Trigonometric Ratios

f x = 4 x -1 ...

Question

$f (x) = \frac{{(4^{x} - 1)}^{3}}{s i n (\frac{x}{p}) l o g (1 + \frac{x^{2}}{3})}$ is continous at x=0 and f(0)=12 ${(l o g 4)}^{3}$ then p= __________.

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p

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(4^{x} - 1)^{3}}{sin \frac{x}{p} log (1 + \frac{x^{2}}{3})} & ; x \neq 0 12 (log 4)^{3} & ; x = 0 \end{matrix}

x = 0

Determine f(0) so that the function f(x) defined by f(x) = (4x-1)³ / [sinx/4 log(1+x²/3) becomes continuous at x=0.

f (0)

f (x) = \frac{{(4^{x} - 1)}^{3}}{sin \frac{x}{4} log (1 + x^{2} / 3) {(log 2)}^{3}}, x \neq 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(4^{x} - 1)^{3}}{sin (\frac{x}{a}) ln (1 + \frac{x^{2}}{3})}, & x \neq 0 9 (ln 4)^{3}, & x = 0 \end{matrix}

x = 0

a

f (x) = \frac{(4^{x} - 1)^{3}}{s i n (\frac{x}{p}) l n (1 + \frac{x^{2}}{3})}, x \neq 0 = 12 (l n 4)^{3}, x = 0

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