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

If the function $f(\alpha) = \begin{cases} \frac{1-\cos 6 \alpha}{36 \alpha^2}, & \alpha \neq 0 \\ k, & \alpha=0 \end{cases}$ is continuous at $\alpha=0$,then $k$ is equal to . . . . . . .

If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1}$,$x \neq 0$,is continuous at $x = 0$,then the ordered pair $(k, f(0))$ is equal to?

Is the function defined by $f(x) = x^{2} - \sin x + 5$ continuous at $x = \pi$?

If $f(x) = \begin{cases} \frac{x^4 - 16}{x - 2}, & x \neq 2 \\ 16, & x = 2 \end{cases}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module