f(x)={2αx2+2βx−4α(α+3)x+α−β;x<1;x≥1
f(1−)=2α−β+3, f(1)=−2α+2β
2α−β+3=2β−2α⇒4α−3β+3=0 ...(1)
f′(1+)=4α+2β, f′(1−)=α+3
4α+2β=α+3⇒3α+2β−3=0 ...(2)
Solving (1) & (2)
We get α=173, β=1721
⇒34(α+β)=34×1727=48
JEE Main 2026 — Mathematics Calculus
Let α,β∈R be such that the function f(x)={2α(x2−2)+2βx(α+3)x+(α−β),x<1,x≥1
be differentiable at all x∈R. Then 34(α+β) is equal to
Held on 24 Jan 2026 · Verified 6 Jul 2026.
36
24
84
48
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