Question

If the function defined by

\(f(x)=\begin{cases} ax^{2}+bx+1, & |2x-3|\geq 2 \\ 3x+2, & \frac{1}{2}\lt x\lt \frac{5}{2} \end{cases}\)

is continuous in its domain, then the values of \(a\) and \(b\) respectively are

• A. \(\frac{-4}{5},\frac{27}{5}\)
• B. \(\frac{5}{4},-\frac{27}{5}\)
• C. \(1,\frac{27}{5}\)
• D. \(3,11\)

Hint:

Solve the modulus inequality to find the domain pieces, then match values at the two boundary points.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(|2x-3|\geq 2\) means \(x\leq \dfrac{1}{2}\) or \(x\geq \dfrac{5}{2}\). So the quadratic applies outside \(\left(\dfrac{1}{2},\dfrac{5}{2}\right)\) and the linear piece inside.

Step 2: Continuity at \(x=\dfrac{1}{2}\)
\(\dfrac{a}{4}+\dfrac{b}{2}+1=3\cdot \dfrac{1}{2}+2=\dfrac{7}{2}\).
This gives \(a+2b=10\).

Step 3: Continuity at \(x=\dfrac{5}{2}\)
\(\dfrac{25a}{4}+\dfrac{5b}{2}+1=\dfrac{15}{2}+2=\dfrac{19}{2}\).
This gives \(25a+10b=34\).

Step 4: Solve
Substitute \(a=10-2b\): \(250-50b+10b=34\), so \(b=\dfrac{27}{5}\) and \(a=-\dfrac{4}{5}\).

Final Answer:
The values are \(a=-\dfrac{4}{5}\) and \(b=\dfrac{27}{5}\). \[ \boxed{\text{(A) }\dfrac{-4}{5},\dfrac{27}{5}} \]