A function \( f(x) \) is defined by \( f(x)=\begin{cases}x^2+\alpha x+\beta & :\ 0\le x\lt 2\\3x+2 & :\ 2\le x\le 4\\2\alpha x+5\beta & :\ 4\lt x\le 8\end{cases} \). If function \( f(x) \) is continuous on [0, 8] , then value of \( 2\alpha+3\beta \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Each piece is a polynomial, so f can fail to be continuous only at the joining points x = 2 and x = 4.
At x = 2: the left-hand limit must equal f(2) = 3(2) + 2 = 8:
\( \displaystyle 4+2\alpha+\beta=8 \)
\( \displaystyle 2\alpha+\beta=4 \)
At x = 4: f(4) = 3(4) + 2 = 14 must equal the right-hand limit:
\( \displaystyle 8\alpha+5\beta=14 \)
Putting \( \beta=4-2\alpha \) in the second equation:
\( \displaystyle 8\alpha+20-10\alpha=14 \)
\( \displaystyle \alpha=3,\ \beta=-2 \)
\( \displaystyle 2\alpha+3\beta=6-6=0 \)