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Continuity

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Question

$If the function g (x) = \{\begin{matrix} \frac{e^{p x} + \log_{e} (1 + 4 x) + q}{x^{3}}, x \neq 0 is continuous at x = 0, \\ r, x = 0 \end{matrix}$ $then the value of 6 p q r is$

Moderate

Solution

$Since g (x) is continuous at x = 0 then$
$\begin{array}{l} lim_{x \to 0} \frac{e^{p x} + \log_{e} (1 + 4 x) + q}{x^{3}} = r \\ ∴ lim_{x \to 0} \{\frac{(1 + (p x) + \frac{p^{2} x^{2}}{2!} + \frac{p^{3} x^{3}}{3!} + \dots \dots) + (4 x - \frac{16 x^{2}}{2} + \frac{64}{3} x^{3} + \dots \dots) + q}{x^{3}}\} = r \end{array}$
$\begin{array}{l} \Rightarrow lim_{x \to 0} \{\frac{(1 + q) + (p + 4) x + (\frac{p^{2}}{2} - 8) x^{2} + (\frac{p^{3}}{6} + \frac{64}{3}) x^{3} + \dots . .}{x^{3}}\} = r \\ ∴ 1 + q = 0, p + 4 = 0 and \frac{p^{2}}{2} - 8 = 0 and \frac{p^{3}}{6} + \frac{64}{3} = r \\ ∴ q = - 1, p = - 4, r = \frac{32}{3} \\ ∴ 6 p q r = 6 (- 4) (- 1) (\frac{32}{3}) = 256 \end{array}$

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