If the function f(x) = begin{cases} frac{cos | vedclass

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

Question

(B) For the function to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = -1$. Evaluating the limit: $\lim_{x 	o 0} \frac{\cos ax -

Vedclass · Accepted Answer

Arithmetic progression

Vedclass · Answer

(B) For the function to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = -1$. Evaluating the limit: $\lim_{x 	o 0} \frac{\cos ax - \cos bx}{\cos cx - \cos bx}$. Using the expansion $\cos 	heta \approx 1 - \frac{	heta^2}{2}$ for small $	heta$: $\lim_{x 	o 0} \frac{(1 - \frac{a^2x^2}{2}) - (1 - \frac{b^2x^2}{2})}{(1 - \frac{c^2x^2}{2}) - (1 - \frac{b^2x^2}{2})} = \lim_{x 	o 0} \frac{\frac{b^2x^2}{2} - \frac{a^2x^2}{2}}{\frac{b^2x^2}{2} - \frac{c^2x^2}{2}} = \frac{b^2 - a^2}{b^2 - c^2}$. Setting this equal to $-1$: $\frac{b^2 - a^2}{b^2 - c^2} = -1$. $b^2 - a^2 = -(b^2 - c^2) \implies b^2 - a^2 = c^2 - b^2$. $2b^2 = a^2 + c^2$. This condition implies that $a^2, b^2, c^2$ are in Arithmetic progression.

If the function $f(x) = \begin{cases} \frac{\cos ax - \cos bx}{\cos cx - \cos bx} & , x \neq 0 \\ -1 & , x = 0 \end{cases}$ is continuous at $x = 0$,then $a^2, b^2, c^2$ are in

Similar Questions

If $f(x) = \begin{cases} |x - 3|, & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & x < 1 \end{cases}$,then $f(x)$ is:

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.

If $f(x) = \begin{cases} x+a, & x \leq 0 \\ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$,then $(g \circ f)(2) + (f \circ g)(-2)$ is equal to.

The function $f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$, where $[x]$ denotes the greatest integer $\leq x$, is

If the function $f(x)$ defined by $f(x) = \begin{cases} ax + 1 & \text{if } x \leq 3 \\ bx + 3 & \text{if } x > 3 \end{cases}$ is continuous at $x = 3$, then $(a - b) = ..........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module