If f(x) = begin{cases} x^2-1, & 0

Question

(B) Given $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \ 2x+3, & 2 \leq x < 3 \end{cases}$.First,we find the left-hand limit at $x=2$:$\lim_{x ightarro

Vedclass · Accepted Answer

$x^2-10x+21=0$

Vedclass · Answer

(B) Given $f(x) = \begin{cases} x^2-1, & 0 First,we find the left-hand limit at $x=2$:$\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{-}} (x^2-1) = 2^2 - 1 = 3$.Next,we find the right-hand limit at $x=2$:$\lim_{x ightarrow 2^{+}} f(x) = \lim_{x ightarrow 2^{+}} (2x+3) = 2(2) + 3 = 7$.The roots of the required quadratic equation are $\alpha = 3$ and $\beta = 7$.The quadratic equation is given by $x^2 - (\alpha + \beta)x + (\alpha 	imes \beta) = 0$.Substituting the values: $x^2 - (3 + 7)x + (3 	imes 7) = 0$.Thus,the equation is $x^2 - 10x + 21 = 0$.

If $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \\ 2x+3, & 2 \leq x < 3 \end{cases}$,the quadratic equation whose roots are $\lim_{x \rightarrow 2^{-}} f(x)$ and $\lim_{x \rightarrow 2^{+}} f(x)$ is

Similar Questions

$\lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}} = $

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - {b^x}}}{x}} \right) = $

$\mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sum_{k=1}^{n} {k^2}}}{{{n^3}}}} \right] = $

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module