If a function f(x) = begin{cases} ax+b, & x l | vedclass

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

Question

(A) Given that $f(x)$ is continuous on $\mathbb{R}$, it must be continuous at $x = -1$ and $x = 1$.At $x = -1$:$\lim_{x 	o -1^-} f(x) = \lim_{x 	o -

Vedclass · Accepted Answer

$(-22, -3)$

Vedclass · Answer

(A) Given that $f(x)$ is continuous on $\mathbb{R}$, it must be continuous at $x = -1$ and $x = 1$.At $x = -1$:$\lim_{x 	o -1^-} f(x) = \lim_{x 	o -1^+} f(x)$$a(-1) + b = 2(-1)^2 + 2b(-1) - \frac{a}{2}$$-a + b = 2 - 2b - \frac{a}{2}$$-\frac{a}{2} + 3b = 2 \quad \dots (i)$At $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x)$$2(1)^2 + 2b(1) - \frac{a}{2} = 7$$2 + 2b - \frac{a}{2} = 7$$-\frac{a}{2} + 2b = 5 \quad \dots (ii)$Subtracting equation $(ii)$ from equation $(i)$:$(-\frac{a}{2} + 3b) - (-\frac{a}{2} + 2b) = 2 - 5$$b = -3$Substituting $b = -3$ into equation $(ii)$:$-\frac{a}{2} + 2(-3) = 5$$-\frac{a}{2} - 6 = 5$$-\frac{a}{2} = 11$$a = -22$Therefore, $(a, b) = (-22, -3)$.

If a function $f(x) = \begin{cases} ax+b, & x \leq -1 \\ 2x^2+2bx-\frac{a}{2}, & -1 < x < 1 \\ 7, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$, then $(a, b) =$

Similar Questions

Examine the following function for continuity: $f(x) = \frac{1}{x-5}, x \neq 5$.

The values of $p$ and $q$ such that the function $f(x) = \begin{cases} (1+|\sin x|)^{\frac{p}{|\sin x|}}, & \frac{-\pi}{6} < x < 0 \\ q, & x = 0 \\ e^{\frac{\sin 2x}{\sin 3x}}, & 0 < x < \frac{\pi}{6} \end{cases}$ is continuous at $x=0$ are:

Let $f : [0,1] \to [0,1]$ be a continuous function,then the equation $f(x) = x$

If $f(x) = x^3 + 7x - 1$,then $f(x)$ has a zero between $x = 0$ and $x = 1$. The theorem which best describes this is:

If $f(x) = \begin{cases} \frac{5}{2} - x, & x < 2 \\ 1, & x = 2 \\ x - \frac{3}{2}, & x > 2 \end{cases}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module