If $\lim_{x \rightarrow 0} \frac{e^{(a-1)x} + 2\cos bx + (c-2)e^{-x}}{x\cos x - \log_e(1+x)} = 2$, then $a^2 + b^2 + c^2$ is equal to :

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Limits

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

, then

a^{2} + b^{2} + c^{2}

is equal to :

Answer: (C)

Expanding about

x = 0

, the denominator is

x cos x - ln (1 + x) = \frac{x ^{2}}{2} + O (x^{3})

. For a finite limit the constant and linear terms of the numerator must vanish:

1 + 2 + (c - 2) = 0 \Rightarrow c = - 1, (a - 1) - (c - 2) = 0 \Rightarrow a = - 2.

The coefficient of

x^{2}

in the numerator is

\frac{( a - 1 ) ^{2}}{2} - b^{2} + \frac{c - 2}{2} = 3 - b^{2}

, and the limit equals this divided by

\frac{1}{2}

2 (3 - b^{2}) = 2 \Rightarrow b^{2} = 2.

Therefore

a^{2} + b^{2} + c^{2} = 4 + 2 + 1 = 7

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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