JEE Mains · Maths · STD 11 - 12. limits
If \(\lim _{x \rightarrow 0} \frac{\alpha e^{x}+\beta e^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}\), where \(\alpha, \beta, \gamma \in R\), then which of the following is \(NOT\) correct ?
- A \(\alpha^{2}+\beta^{2}+\gamma^{2}=6\)
- B \(\alpha \beta+\beta \gamma+\gamma \alpha+1=0\)
- C \(\alpha \beta^{2}+\beta \gamma^{2}+\gamma \alpha^{2}+3=0\)
- D \(\alpha^{2}-\beta^{2}+\gamma^{2}=4\)
Answer & Solution
Correct Answer
(C) \(\alpha \beta^{2}+\beta \gamma^{2}+\gamma \alpha^{2}+3=0\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{\alpha\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots\right)+\beta\left(1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots\right)+\gamma\left(x-\frac{x^{3}}{3 !}+\ldots\right)}{x^{3}}\) constant terms should be zero \(a+\beta=0\) coeff of \(x\)…
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