If the function f(x) = frac{sin 3x + alpha si | vedclass

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

Question

(D) For the function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x 	o 0} f(x)$ must exist and be equal to $f(0)$.Using the Taylor series expa

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(D) For the function $f(x)$ to be continuous at $x = 0$,the limit $\lim_{x 	o 0} f(x)$ must exist and be equal to $f(0)$.Using the Taylor series expansions:$\sin 3x = 3x - \frac{(3x)^3}{3!} + \dots = 3x - \frac{27x^3}{6} + \dots$$\sin x = x - \frac{x^3}{6} + \dots$$\cos 3x = 1 - \frac{(3x)^2}{2!} + \dots = 1 - \frac{9x^2}{2} + \dots$Substituting these into the expression for $f(x)$:$f(x) = \frac{(3x - \frac{27x^3}{6} + \dots) + \alpha(x - \frac{x^3}{6} + \dots) - \beta(1 - \frac{9x^2}{2} + \dots)}{x^3}$$f(x) = \frac{-\beta + x(3 + \alpha) + x^2(\frac{9\beta}{2}) + x^3(-\frac{27}{6} - \frac{\alpha}{6}) + \dots}{x^3}$For the limit to exist as $x 	o 0$,the coefficients of $x^0$,$x^1$,and $x^2$ must be zero:$1$) $-\beta = 0 \implies \beta = 0$$2$) $3 + \alpha = 0 \implies \alpha = -3$$3$) $\frac{9\beta}{2} = 0$ (which is satisfied since $\beta = 0$)Now,the limit is the coefficient of $x^3$:$f(0) = -\frac{27}{6} - \frac{\alpha}{6} = \frac{-27 - (-3)}{6} = \frac{-24}{6} = -4$.

If the function $f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$,$x \in R$,is continuous at $x = 0$,then $f(0)$ is equal to:

Similar Questions

In order that the function $f(x) = (x + 1)^{\cot x}$ is continuous at $x = 0$,$f(0)$ must be defined as

Discuss the continuity of the function $f$ given by $f(x) = x^{3} + x^{2} - 1$.

If $f(x) = \begin{cases} \frac{x}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then

The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

Let $[t]$ represent the greatest integer not more than $t$. Then the number of discontinuous points of $f(x) = [x^{1/x}]$ in $(0, \infty)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module