If $f(x)= \begin{cases}a x^2+b x+1 & \text { if }|2 x-3| \geq 2 \\ 3 x+2 & \text {; if } \frac{1}{2} < x <…
If $f(x)= \begin{cases}a x^2+b x+1 & \text { if }|2 x-3| \geq 2 \\ 3 x+2 & \text {; if } \frac{1}{2} < x < \frac{5}{2}\end{cases}$ is continuous on its domain, then $a+b$ has the value
$\frac{23}{5}$
$\frac{1}{5}$
$\frac{13}{5}$
$\frac{31}{5}$
Solution
\(f(x)=\left\{\begin{array}{ccc}a x^2+b x+1 & ; & |2 x-3| \geq 2 \\ 3 x+2 & ; & \frac{1}{2} < x < \frac{5}{2}\end{array}\right.\)
$=\left\{\begin{array}{ccc}a x^2+b x+1 & ; & x £ \frac{1}{2} \\ 3 x+2 & ; & \frac{1}{2} < x < \frac{5}{2} \\ a x^2+b x+1 & ; & x^3 \frac{5}{2}\end{array}\right.$
for continuity at $x=\frac{1}{2}$
for continuity at $x=\frac{5}{2}$
from (1) and (2)
$a=-\frac{4}{5}$ and $b=\frac{27}{5}$
$a+b=\frac{-4+27}{5}=\frac{23}{5}$
for continuity at $x=\frac{5}{2}$
from (1) and (2)
$a=-\frac{4}{5}$ and $b=\frac{27}{5}$
$a+b=\frac{-4+27}{5}=\frac{23}{5}$