Question

If $\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$, where $\alpha, \beta, \gamma \in \mathbb{R}$ 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 1

Answer: (C)

$\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3 = 0$

Answer 2

$\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$
$⇒ α + β = 0$ (to make indeterminant form) …(i)
Now,
$\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$
(Using L-H Rule)
$⇒ α – β + γ = 0$ (to make indeterminant form) …(ii)
Now,
$\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{6x} = \frac{2}{3}$
(Using L-H Rule)
⇒$$\frac{\alpha - \beta - \gamma}{6} = \frac{2}{3}
⇒$$α – β – γ = 4 …(iii)
$⇒$ $γ = –2$
and eq(i) + eq(ii)
$2α = –γ$
On solving,
$⇒ α = 1\ \text{and}\ β = –1$
and $\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3$
$= 1 – 4 – 2 + 3$
$= –2$
So, the correct option is (C): $\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3 = 0$