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A function f(x) is defined by f(x)={x2+αx+β:0≤x<23x+2:2≤x≤42αx+5β:4<x≤8f(x)=\begin{cases}x^2+\alpha x+\beta: & 0\leq x<2\\3x+2 : & 2\leq x\leq4\\2\alpha x+5\beta :& 4<x\leq8\end{cases}. If function f(x) is continuous on [0, 8], then value of 2α+3β2\alpha+3\beta is

Solution

✅ Correct Option: 1

Continuity at x=2x=2: 4+2α+β=84+2\alpha+\beta=8, so 2α+β=42\alpha+\beta=4. Continuity at x=4x=4: 14=8α+5β14=8\alpha+5\beta. Solving, α=3\alpha=3 and β=−2\beta=-2, so 2α+3β=6−6=02\alpha+3\beta=6-6=0.

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