If f(x) = begin{cases} frac{a sin x - b x + c | vedclass

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

Question

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$. The Taylor expansion of the numerator $N(x)$ is: $N(x) = a(x -

Vedclass · Accepted Answer

$a=b$

Vedclass · Answer

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$. The Taylor expansion of the numerator $N(x)$ is: $N(x) = a(x - \frac{x^3}{6} + O(x^5)) - bx + cx^2 + x^3 = (a-b)x + cx^2 + (1 - \frac{a}{6})x^3 + O(x^4)$. The Taylor expansion of the denominator $D(x)$ is: $D(x) = 2(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + O(x^5)) - 2x + x^2 - \frac{2}{3}x^3 = -\frac{1}{2}x^4 + O(x^5)$. For the limit to exist and equal $0$,the coefficients of $x, x^2,$ and $x^3$ in the numerator must be zero. Thus,$a-b = 0 \implies a=b$,$c=0$,and $1 - \frac{a}{6} = 0 \implies a=6$. Since the question asks for the relation between $a, b, c$ based on the provided options,$a=b$ is the correct relation.

If $f(x) = \begin{cases} \frac{a \sin x - b x + c x^2 + x^3}{2 \log(1+x) - 2x + x^2 - \frac{2}{3}x^3} &, x \neq 0 \\ 0 &, x=0 \end{cases}$ is continuous at $x=0$,then find the relation between $a, b, c$.

Similar Questions

If the function $f(x) = \frac{4\sqrt{2}(\sin 3x + \sin x)}{2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2} - \cos \frac{3x}{2}}$ for $x \neq \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$, then the value of $f(\frac{\pi}{2})$ is equal to

Is the function defined by $f(x) = |x|$ a continuous function?

If $f(x) = \begin{cases} \frac{x^2}{a} - a, & x < a \\ 0, & x = a \\ a - \frac{x^2}{a}, & x > a \end{cases}$ then:

Number of points of discontinuity of the function $f(x) = \sin(\{2^x + [2^x] + [3^{-x}]\})$ for $x \in [0, 4]$ is (where $[.]$ and $\{.\}$ denote the greatest integer and fractional part functions,respectively).

Let $f(x) = \begin{cases} \frac{(1 + \tan x)^{\frac{1}{x}} - e}{x} & x \neq 0 \\ k & x = 0 \end{cases}$ be continuous at $x = 0$,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module