If f(x) = int_{-1}^x |t| , dt,x ge -1,then | vedclass

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

Question

(A) Let us divide the interval into two sub-intervals: $I_1: -1 \le x < 0$ and $I_2: x \ge 0$.For $I_1$,$f(x) = \int_{-1}^x (-t) \, dt = -\frac{1}{2}[

Vedclass · Accepted Answer

$f$ and $f'$ are continuous for $x + 1 > 0$

Vedclass · Answer

(A) Let us divide the interval into two sub-intervals: $I_1: -1 \le x For $I_1$,$f(x) = \int_{-1}^x (-t) \, dt = -\frac{1}{2}[t^2]_{-1}^x = -\frac{1}{2}(x^2 - 1) = \frac{1}{2}(1 - x^2)$.For $I_2$,$f(x) = \int_{-1}^0 (-t) \, dt + \int_0^x (t) \, dt = -\frac{1}{2}[t^2]_{-1}^0 + \frac{1}{2}[t^2]_0^x = -\frac{1}{2}(0 - 1) + \frac{1}{2}(x^2 - 0) = \frac{1}{2}(1 + x^2)$.Thus,$f(x) = \begin{cases} \frac{1}{2}(1 - x^2), & -1 \le x Taking the derivative,$f'(x) = \begin{cases} -x, & -1  0 \end{cases}$.At $x = 0$,$\lim_{x 	o 0^-} f(x) = \frac{1}{2}(1 - 0) = \frac{1}{2}$ and $\lim_{x 	o 0^+} f(x) = \frac{1}{2}(1 + 0) = \frac{1}{2}$. Since $f(0) = \frac{1}{2}$,$f$ is continuous at $x = 0$.For $f'(x)$,$\lim_{x 	o 0^-} f'(x) = 0$ and $\lim_{x 	o 0^+} f'(x) = 0$. Since $f'(0) = 0$,$f'$ is continuous at $x = 0$.Therefore,both $f$ and $f'$ are continuous for all $x > -1$,i.e.,$x + 1 > 0$.

If $f(x) = \int_{-1}^x |t| \, dt$,$x \ge -1$,then

Similar Questions

If $f(x) = \frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}, x \neq 0$ is continuous at $x=0$, then the value of $f(0)$ is

Let $f(x) = \begin{cases} 0, & x=0 \\ 2-x, & 0 < x < 1 \\ 2, & x=1 \\ \frac{1}{2}-x, & 1 < x < 2 \\ \frac{-3}{2}, & x \geq 2 \end{cases}$ then which of the following is true?

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

Which of the following function$(s)$ not defined at $x = 0$ has/have a removable discontinuity at $x = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module