If the function f(x) = begin{cases} k_{1}(x-p | vedclass

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

Question

(A) For $f(x)$ to be twice differentiable,it must be continuous and differentiable at $x = \pi$.$1$. Continuity at $x = \pi$:$f(\pi^{-}) = f(\pi) = f(

Vedclass · Accepted Answer

$(\frac{1}{2}, 1)$

Vedclass · Answer

(A) For $f(x)$ to be twice differentiable,it must be continuous and differentiable at $x = \pi$.$1$. Continuity at $x = \pi$:$f(\pi^{-}) = f(\pi) = f(\pi^{+})$$k_{1}(\pi-\pi)^{2} - 1 = k_{2} \cos(\pi)$$-1 = -k_{2} \implies k_{2} = 1$.$2$. First derivative $f'(x)$:$f'(x) = \begin{cases} 2k_{1}(x-\pi), & x \leq \pi \ -k_{2} \sin x, & x > \pi \end{cases}$At $x = \pi$,$f'(\pi^{-}) = 2k_{1}(\pi-\pi) = 0$ and $f'(\pi^{+}) = -k_{2} \sin(\pi) = 0$.Since $0 = 0$,the function is differentiable at $x = \pi$ for any $k_{1}, k_{2}$.$3$. Second derivative $f''(x)$:$f''(x) = \begin{cases} 2k_{1}, & x \leq \pi \ -k_{2} \cos x, & x > \pi \end{cases}$For $f''(x)$ to be continuous at $x = \pi$:$f''(\pi^{-}) = f''(\pi^{+})$$2k_{1} = -k_{2} \cos(\pi)$$2k_{1} = -k_{2}(-1) = k_{2}$Since $k_{2} = 1$,we have $2k_{1} = 1 \implies k_{1} = \frac{1}{2}$.Thus,$(k_{1}, k_{2}) = (\frac{1}{2}, 1)$.

If the function $f(x) = \begin{cases} k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi \end{cases}$ is twice differentiable,then the ordered pair $(k_{1}, k_{2})$ is equal to

Similar Questions

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}$. Is $f$ a continuous function?

Prove that the function defined by $f(x) = \tan x$ is a continuous function.

If $a$ and $b$ $(a > b)$ are points of discontinuity of the function $f(x) = \begin{cases} 3-2x^2, & \text{for } x \leq 0 \\ 2x+3, & \text{for } 0 < x \leq 1 \\ 2x^2-3x, & \text{for } 1 < x < 2 \\ 2x-3, & \text{for } 2 \leq x < 3 \\ |x|, & \text{for } x \geq 3 \end{cases}$,then $3a-b = $

If $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $a$ will be

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by $f(x)= \begin{cases} \frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0 \end{cases}$ is continuous at $x=0$, then the ordered pair $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module