If f(x) = begin{cases} x^2 & text{if } x leqs | vedclass

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

Question

(A) For $f(x)$ to be derivable at $x = x_0$,it must be continuous at $x = x_0$. Continuity at $x = x_0$ implies $\lim_{x 	o x_0^-} f(x) = \lim_{x 	o

Vedclass · Accepted Answer

$2x_0, -x_0^2$

Vedclass · Answer

(A) For $f(x)$ to be derivable at $x = x_0$,it must be continuous at $x = x_0$. Continuity at $x = x_0$ implies $\lim_{x 	o x_0^-} f(x) = \lim_{x 	o x_0^+} f(x) = f(x_0)$. $\lim_{x 	o x_0^-} x^2 = x_0^2$ and $\lim_{x 	o x_0^+} (ax + b) = ax_0 + b$. So,$x_0^2 = ax_0 + b$  (Equation $1$). For differentiability at $x = x_0$,the left-hand derivative must equal the right-hand derivative. $f'(x) = \begin{cases} 2x & 	ext{if } x  x_0 \end{cases}$. Equating derivatives at $x = x_0$: $2x_0 = a$. Substitute $a = 2x_0$ into Equation $1$: $x_0^2 = (2x_0)x_0 + b$. $x_0^2 = 2x_0^2 + b \implies b = -x_0^2$. Thus,$a = 2x_0$ and $b = -x_0^2$.

If $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively:

Similar Questions

In the interval $[0, 3]$,the function $f(x) = |x - 1| + |x - 2|$ is

$\text{The domain of the derivative of the function } f(x) = \begin{cases} \tan^{-1} x, & \text{if } |x| \le 1 \\ \frac{1}{2}(|x|-1), & \text{if } |x| > 1 \end{cases} \text{ is given by:}$

Let $f(x) = x^{2} - x + k - 2$,where $k \in R$. Find the complete set of values of $k$ for which $y = |f(|x|)|$ is non-derivable at $5$ distinct points.

If $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \\ ax^2 + b, & -1 < x < 1 \end{cases}$ is differentiable $\forall x \in \mathbb{R}$,then one of the values of $a$ and $b$ is-

Which of the following statements is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module