JEE Main 2021 — Mathematics Calculus

Given $f:[-1,1]\rightarrow R,$f:[−1,1]→R, $f(x)=a{x}^{2}+bx+c,{f}^{'}(x)=2ax+b$f(x)=ax2+bx+c,f′(x)=2ax+b and ${f}^{''}(x)=2a$f′′(x)=2a

$$\Rightarrow f(-1)=a-b+c=2...(1),$$⇒f(−1)=a−b+c=2...(1),

$\Rightarrow {f}^{'}(-1)=-2a+b=1...(2)$⇒f′(−1)=−2a+b=1...(2) and

$${f}^{''}(x)=2a$$f′′(x)=2a

Given the maximum value of ${f}^{''}(x)=2a=\frac{1}{2}$f′′(x)=2a=21​ $\Rightarrow a=\frac{1}{4},$⇒a=41​,

From the equations $(1)$(1) and $(2)$(2) we get $b=\frac{3}{2}$b=23​ and $c=\frac{13}{4}.$c=413​.

$$\therefore f(x)=\frac{{x}^{2}}{4}+\frac{3x}{2}+\frac{13}{4}$$∴f(x)=4x2​+23x​+413​

We know that, the vertex of the quadratic $A{x}^{2}+Bx+C$Ax2+Bx+C is at $(-\frac{B}{2A},-\frac{{B}^{2}-4AC}{4A})$(−2AB​,−4AB2−4AC​)

Thus, the vertex of $\frac{{x}^{2}}{4}+\frac{3x}{2}+\frac{13}{4}$4x2​+23x​+413​ is at $(-3,1),$(−3,1), hence both the numbers $\pm 1$±1 are on the same side of the vertex and the graph of $f(x)$f(x) is given below.

For, $x\in [-1,1],$x∈[−1,1], we have f(-1)=2&f(1)=5

$$\Rightarrow 2\leq f(x)\leq 5$$⇒2≤f(x)≤5

$\therefore$∴ Least value of $\alpha$α is$5.$5.