Let f(x) = begin{cases} x^2 left| cos frac{pi | vedclass

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

Question

(B) $(i)$ Check for differentiability at $x=0$: $LHD = f'(0^-) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{(-h)^2 |\cos(-\pi/

Vedclass · Accepted Answer

differentiable at $x=0$ but not differentiable at $x=2$

Vedclass · Answer

(B) $(i)$ Check for differentiability at $x=0$: $LHD = f'(0^-) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{(-h)^2 |\cos(-\pi/h)| - 0}{-h} = \lim_{h 	o 0^+} -h |\cos(\pi/h)| = 0$. $RHD = f'(0^+) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h^2 |\cos(\pi/h)| - 0}{h} = \lim_{h 	o 0^+} h |\cos(\pi/h)| = 0$. Since $LHD = RHD = 0$,$f(x)$ is differentiable at $x=0$. $(ii)$ Check for differentiability at $x=2$: $f(2) = 2^2 |\cos(\pi/2)| = 0$. $RHD = f'(2^+) = \lim_{h 	o 0^+} \frac{f(2+h) - f(2)}{h} = \lim_{h 	o 0^+} \frac{(2+h)^2 |\cos(\pi/(2+h))|}{h}$. Since $\cos(\pi/(2+h)) > 0$ for small $h > 0$,$|\cos(\pi/(2+h))| = \cos(\pi/(2+h)) = \sin(\pi/2 - \pi/(2+h)) = \sin(\pi h / (2(2+h)))$. $RHD = \lim_{h 	o 0^+} \frac{(2+h)^2 \sin(\pi h / (2(2+h)))}{h} = 4 \cdot \frac{\pi}{4} = \pi$. $LHD = f'(2^-) = \lim_{h 	o 0^+} \frac{f(2-h) - f(2)}{-h} = \lim_{h 	o 0^+} \frac{(2-h)^2 |\cos(\pi/(2-h))|}{-h}$. Since $\cos(\pi/(2-h))  0$,$|\cos(\pi/(2-h))| = -\cos(\pi/(2-h)) = -\sin(\pi/2 - \pi/(2-h)) = -\sin(-\pi h / (2(2-h))) = \sin(\pi h / (2(2-h)))$. $LHD = \lim_{h 	o 0^+} \frac{(2-h)^2 \sin(\pi h / (2(2-h)))}{-h} = 4 \cdot (-\pi/4) = -\pi$. Since $LHD 
eq RHD$,$f(x)$ is not differentiable at $x=2$.

Let $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x=0 \end{cases}$,$x \in \mathbb{R}$,then $f$ is

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} x^{5} \sin \left(\frac{1}{x}\right) + 5x^{2} & , x < 0 \\ 0 & , x = 0 \\ x^{5} \cos \left(\frac{1}{x}\right) + \lambda x^{2} & , x > 0 \end{cases}$. The value of $\lambda$ for which $f''(0)$ exists is:

Identify the correct statement about the function $f(x) = \max(x^2 - 1, 7 - x^2, 5)$.

If $[t]$ denotes the greatest integer $\leq t$,then the number of points at which the function $f(x) = 4|2x + 3| + 9[x + \frac{1}{2}] - 12[x + 20]$ is not differentiable in the open interval $(-20, 20)$ is:

If the function $g(x)=\begin{cases} K \sqrt{x+1} &, 0 \leq x \leq 3 \\ mx+2 &, 3 < x \leq 5 \end{cases}$ is differentiable, then $K+m=$

Let $f: R \rightarrow R$ be a continuous function such that for any two real numbers $x$ and $y$,$|f(x)-f(y)| \leq 10|x-y|^{201}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module