Given, f:[−1,1]→R and f(x)=ax2+bx+c f(−1)=a−b+c=2 (given)...(i) f′(−1)=−2a+b=1 (given)...(ii) f′′(x)=2a ∴‌‌fmax′′(x)=2a Also, given maximum value of f′′(x)=‌
1
2
i.e. 2a=‌
1
2
⇒a=‌
1
4
From Eq. (ii), b=‌
3
2
From Eq. (i), c=‌
13
4
∴‌‌f(x)=‌
x2
4
+‌
3
2
x+‌
13
4
Here, f(−1)=‌
1
4
−‌
3
2
+‌
13
4
=2 and f(1)=‌
1
4
+‌
3
2
+‌
13
4
=5
For x∈[−1,1] f(x)∈[2,5] ∴ Least value of α is 5 .