If $\lim_{x \rightarrow 0} \frac{e^{ax}-\cos(bx)-\frac{cxe^{-cx}}{2}}{1-\cos(2x)}=17$, then $5a^{2}+b^{2}$ is equal to

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Limits

lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17

, then

5 a^{2} + b^{2}

is equal to

Answer: (C)

Using Taylor expansions around

x = 0

e^{a x} = 1 + a x + \frac{a ^{2} x ^{2}}{2} + \dots

cos (b x) = 1 - \frac{b ^{2} x ^{2}}{2} + \dots

e^{- c x} = 1 - c x + \frac{c ^{2} x ^{2}}{2} + \dots

1 - cos (2 x) = 2 x^{2} + \dots

Substituting:

x \to 0 lim \frac{( 1 + a x + \frac{a ^{2} x ^{2}}{2} ) - ( 1 - \frac{b ^{2} x ^{2}}{2} ) - \frac{c x}{2} ( 1 - c x )}{2 x ^{2}}

= x \to 0 lim \frac{( a - \frac{c}{2} ) x + ( \frac{a ^{2} + b ^{2} + c ^{2}}{2} ) x ^{2}}{2 x ^{2}}

For the limit to exist, the coefficient of

x

must be zero:

a = \frac{c}{2}

Then:

\frac{a ^{2} + b ^{2} + c ^{2}}{4} = 17 \Rightarrow a^{2} + b^{2} + 4 a^{2} = 68 \Rightarrow 5 a^{2} + b^{2} = 68

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option

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