If lim x 0 e a x - cos ( b x ) - c x e - c x 2 1 - cos ( 2 x ) = 17 , then 5 a 2 + b 2 is equal to [2023]

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Solution

Q.
$If \lim_{x \to 0} \frac{e^{a x} - \cos (b x) - \frac{c x e^{- c x}}{2}}{1 - \cos (2 x)} = 17, then 5 a^{2} + b^{2} is equal to$ [2023]

1 72

2 64

3 68

4 76

Ans.
(3)

$\lim_{x \to 0} \frac{a e^{a x} + b \sin b x - \frac{c}{2} e^{- c x} + \frac{c^{2} x}{2} e^{- c x}}{2 \sin 2 x} = 17$

$\Rightarrow a - \frac{c}{2} = 0$ $∴$ $a = \frac{c}{2}$

Again, applying (L'Hospital's rule)

$\lim_{x \to 0} \frac{a^{2} e^{a x} + b^{2} \cos b x + \frac{c^{2}}{2} e^{- c x} - \frac{c^{3} x}{2} e^{- c x} + \frac{c^{2}}{2} e^{- c x}}{4 \cos 2 x} = 17$

$\Rightarrow \frac{a^{2} + b^{2} + \frac{c^{2}}{2} + \frac{c^{2}}{2}}{4} = 17$

$\Rightarrow a^{2} + b^{2} + c^{2} = 68 \Rightarrow a^{2} + b^{2} + 4 a^{2} = 68$

$\Rightarrow 5 a^{2} + b^{2} = 68$

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