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← JEE Main 2022 · 29 Jul · Shift 1
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JEE Main 2022 · 29 Jul · Shift 1, Mathematics question 5

If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\displaystyle\lim_{x\to0}\frac{\alpha e^x+\beta e^{-x}+\gamma\sin x}{x\sin^2x}=\frac23, where α, β, γ ∈ R, then which of the following is NOT correct ?

Choose one option
Answer and worked solution

Official answer: C

Expand. Constant and xx terms vanish: α+β=0\alpha+\beta=0, α−β+γ=0\alpha-\beta+\gamma=0. The x3x^3 term gives α−β−γ6=23\frac{\alpha-\beta-\gamma}6=\frac23.

Values γ=−2\gamma=-2, α=1\alpha=1, β=−1\beta=-1. Then αβ2+βγ2+γα2+3=1−4−2+3=−2≠0\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2+3=1-4-2+3=-2\ne0; the other three hold.

(C)

Topics:Limits
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