Let f(x)=lim _{theta rightarrow 0}left(frac{c | vedclass

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

Question

(A) Given $f(x) = \lim_{	heta ightarrow 0} \frac{\cos \pi x - x^{(2/	heta)} \sin(x-1)}{1 + x^{(2/	heta)}(x-1)}$.Case $1$: $|x| < 1$. As $	heta \

Vedclass · Accepted Answer

Neither $(I)$ nor $(II)$ is True

Vedclass · Answer

(A) Given $f(x) = \lim_{	heta ightarrow 0} \frac{\cos \pi x - x^{(2/	heta)} \sin(x-1)}{1 + x^{(2/	heta)}(x-1)}$.Case $1$: $|x| Case $2$: $|x| > 1$. As $	heta ightarrow 0$, $x^{(2/	heta)} ightarrow \infty$. Dividing numerator and denominator by $x^{(2/	heta)}$, we get $f(x) = \frac{-\sin(x-1)}{x-1}$.At $x=1$: $LHL = \lim_{x ightarrow 1^-} \cos \pi x = -1$. $RHL = \lim_{x ightarrow 1^+} \frac{-\sin(x-1)}{x-1} = -1$. Since $LHL = RHL = f(1) = -1$, $f(x)$ is continuous at $x=1$. Statement $(I)$ is False.At $x=-1$: $LHL = \lim_{x ightarrow -1^-} \frac{-\sin(x-1)}{x-1} = \frac{-\sin(-2)}{-2} = \frac{-\sin 2}{2}$. $RHL = \lim_{x ightarrow -1^+} \cos \pi x = \cos(-\pi) = -1$. Since $LHL 
eq RHL$, $f(x)$ is discontinuous at $x=-1$. Statement $(II)$ is False.Therefore, neither $(I)$ nor $(II)$ is true.

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in R$. Consider the following two statements: $(I)$ $f(x)$ is discontinuous at $x=1$. $(II)$ $f(x)$ is continuous at $x=-1$. Then,

Similar Questions

If $f(x) = \frac{4}{x^4} \left[ 1 - \cos \frac{x}{2} - \cos \frac{x}{4} + \cos \frac{x}{2} \cdot \cos \frac{x}{4} \right]$ is continuous at $x = 0$,then $f(0)$ is

The number of points of discontinuity of $f(x)$ where $f(x) = | | |x + [x]| - 3[x] | - 5[x] |$ on $[-2, 2]$ is (where $[ \cdot ]$ denotes the greatest integer function).

If the function $f(x) = \begin{cases} \frac{\sqrt{1+px}-\sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $p = $

If the function $f(x) = \begin{cases} 1+\cos x, & x \leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{cases}$ is continuous everywhere,then $a^2+b^2=$

If a function $f(x) = \begin{cases} \frac{\sqrt[3]{1+ax^2+bx^3}-\sqrt[3]{1-ax^2-bx^3}}{x^2}, & x < 0 \\ 5, & x=0 \\ \frac{\tan 3x - \sin 3x}{bx^3}, & x > 0 \end{cases}$ is continuous at $x=0$,then the geometric mean of $a$ and $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module