Let f(x) = begin{cases} x^2 - a & x

Question

(B) For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 3^-} (x^2 - a) = \lim_{x 	o 3^+} (b\sqrt

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 3^-} (x^2 - a) = \lim_{x 	o 3^+} (b\sqrt{x - 2} + a) \implies 9 - a = b(1) + a \implies 2a + b = 9$. For $f(x)$ to be continuous at $x = 6$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 6^-} (b\sqrt{x - 2} + a) = \lim_{x 	o 6^+} (2x + b) \implies b\sqrt{4} + a = 12 + b \implies 2b + a = 12 + b \implies a + b = 12$. Solving the system of equations: $2a + b = 9$ and $a + b = 12$. Subtracting the second from the first gives $a = -3$. Substituting $a = -3$ into $a + b = 12$ gives $b = 15$. Now,$f(1) = 1^2 - (-3) = 4$ and $f(3) = b\sqrt{3 - 2} + a = 15(1) - 3 = 12$. Finally,$\frac{f(1) - f(3)}{4} = \frac{4 - 12}{4} = \frac{-8}{4} = -2$.

Let $f(x) = \begin{cases} x^2 - a & x < 3 \\ b\sqrt{x - 2} + a & 3 \leqslant x < 6 \\ 2x + b & x \geqslant 6 \end{cases}$. If $f(x)$ is continuous $\forall x \in R$,then find the value of $\frac{f(1) - f(3)}{4}$.

Similar Questions

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

Show that the function $f$ defined by $f(x) = |1 - x + |x||$,where $x$ is any real number,is a continuous function.

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

If $f(x) = \begin{cases} [x] + [-x], & x \neq 2 \\ \lambda, & x = 2 \end{cases}$ is continuous at $x = 2$,then $\lambda = $ (where $[.]$ denotes the greatest integer function).

The function $f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$, where $[x]$ denotes the greatest integer $\leq x$, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module