If lim _{x rightarrow 0} frac{alpha e^{x}+bet | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $\lim _{x \rightarrow 0} \frac{\alpha e^{x}+\beta e^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$.Since $\sin x \approx x$ as $x \rightarr

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $\lim _{x ightarrow 0} \frac{\alpha e^{x}+\beta e^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$.Since $\sin x \approx x$ as $x ightarrow 0$,the limit becomes $\lim _{x ightarrow 0} \frac{\alpha e^{x}+\beta e^{-x}+\gamma \sin x}{x^{3}}=\frac{2}{3}$.Expanding the series: $\alpha(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots) + \beta(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\dots) + \gamma(x-\frac{x^3}{6}+\dots) = \frac{2}{3}x^3$.For the limit to exist and be finite,the coefficients of $x^0, x^1, x^2$ must be zero.$x^0: \alpha + \beta = 0 \implies \beta = -\alpha$.$x^1: \alpha - \beta + \gamma = 0 \implies \alpha - (-\alpha) + \gamma = 0 \implies 2\alpha + \gamma = 0 \implies \gamma = -2\alpha$.$x^2: \frac{\alpha}{2} + \frac{\beta}{2} = 0$,which is satisfied since $\beta = -\alpha$.Now,the coefficient of $x^3$ is $\frac{\alpha}{6} - \frac{\beta}{6} - \frac{\gamma}{6} = \frac{2}{3}$.$\frac{\alpha}{6} - \frac{-\alpha}{6} - \frac{-2\alpha}{6} = \frac{2}{3} \implies \frac{4\alpha}{6} = \frac{2}{3} \implies \alpha = 1$.Thus,$\alpha = 1, \beta = -1, \gamma = -2$.Checking options:$A: 1^2 + (-1)^2 + (-2)^2 = 1 + 1 + 4 = 6$ (Correct).$B: (1)(-1) + (-1)(-2) + (-2)(1) + 1 = -1 + 2 - 2 + 1 = 0$ (Correct).$C: (1)(-1)^2 + (-1)(-2)^2 + (-2)(1)^2 + 3 = 1 - 4 - 2 + 3 = -2 
eq 0$ (Incorrect).$D: 1^2 - (-1)^2 + (-2)^2 = 1 - 1 + 4 = 4$ (Correct).Therefore,option $C$ is not correct.

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?

Similar Questions

The integral value of $n$ for which $\lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is a finite non-zero real number is

If $\mathop {\lim }\limits_{x \to 1} {\sin ^{ - 1}}\left( {\frac{k}{{\ln x}} - \frac{k}{{x - 1}}} \right)$ exists, then the number of integers in the range of $k$ is:

If $\alpha, \beta$ are the roots of the equation $ax^2+bx+c=0$,then $\lim_{x \rightarrow \alpha} \frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2} =$

Define $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$. If $\lim_{x \rightarrow 0} f(x)$ exists,then $p =$

If $\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=500$,then the value of $k$,where $k \in N$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module