MHT CET · Maths · Continuity and Differentiability
If
\(f(x)= \begin{cases}a x^2+b x+1 & \text { if }|2 x-3| \geq 2 \ 3 x+2 & \text {; if } \end{cases}\) \(\frac{1}{2} < x < \frac{5}{2}\) is continuous on its domain, then \(a+b\) has the value
- A \(\frac{23}{5}\)
- B \(\frac{1}{5}\)
- C \(\frac{13}{5}\)
- D \(\frac{31}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{23}{5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & f(x)=\left\{\begin{array}{ccc}a x^2+b x+1 & ; & |2 x-3| \geq 2 \\ 3 x+2 & ; & \frac{1}{2} < x < \frac{5}{2}\end{array}\right. \\ & =\left\{\begin{array}{ccc}a x^2+b x+1 & ; & x £ \frac{1}{2} \\ 3 x+2 & ; & \frac{1}{2} < x < \frac{5}{2} \\ a x^2+b x+1 & ; & x^3 \frac{5}{2}\end{array}\right.\end{aligned}\)
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 \(x=\frac{1}{2}\)
for continuity at \(x=\frac{5}{2}\)
from (1) and (2)
\(a=-\frac{4}{5}\) and \(b=\frac{27}{5}\)
\(a+b=\frac{-4+27}{5}=\frac{23}{5}\)
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