If lim x ? 0 a x 2 e x - b log e ( 1 + x ) + c x e - x x 2 sin x = 1 , then 16 ( a 2 + b 2 + c 2 ) is equal to _ _ _ _ _ . [2024]

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Solution

Q.
$If \lim_{x \to 0} \frac{a x^{2} e^{x} - b \log_{e} (1 + x) + c x e^{- x}}{x^{2} \sin x} = 1, then 16 (a^{2} + b^{2} + c^{2}) is equal to_____.$ [2024]

Ans.
(81)

Given, $\lim_{x \to 0} \frac{a x^{2} e^{x} - b \log_{e} (1 + x) + c x e^{- x}}{x^{2} \sin x} = 1$

$\Rightarrow \lim_{x \to 0} \frac{a x^{2} (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots) - b (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} \dots) + c x (1 - \frac{x}{1!} + \frac{x^{2}}{2!} \dots)}{x^{2} (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} \dots)} = 1$

$\Rightarrow \lim_{x \to 0} (a x^{2} + a x^{3} + \frac{a x^{4}}{2!} + \dots) - b x + \frac{b x^{2}}{2} - \frac{b x^{3}}{3} \dots + c x - c x^{2} + \frac{c x^{3}}{2!} \dots = \lim_{x \to 0} x^{3} - \frac{x^{5}}{3!} + \frac{x^{7}}{5!} \dots$

Comparing the coefficient of $x^{3},$ we get

$a - \frac{b}{3} + \frac{c}{2} = 1$ ...(i)

Comparing the coefficient of $x^{2},$ we get

$a + \frac{b}{2} - c = 0$ ...(ii)

Comparing the coefficient of $x,$ we get

$- b + c = 0$ ...(iii)

On solving (i), (ii) and (iii), we get

$a = \frac{3}{4}, b = c = \frac{3}{2}$

$∴ 16 (a^{2} + b^{2} + c^{2}) = 16 [\frac{9}{16} + \frac{9}{4} + \frac{9}{4}] = 16 [\frac{9 + 36 + 36}{16}] = 81$

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