Question

Let $f:[-1,1]\to R$ be defined as $f\left(x\right)=a{x}^2+bx+c$ for all $x\in [-1,1],$ where $a,b,c\in R$ such that $f(-1)=2,f^{\prime }(-1)=1$ and for $x\in (-1,1)$ the maximum value of $f^{\prime \prime }\left(x\right)$ is $\frac{1}{2}.$ If $f\left(x\right)\le \alpha ,x\in [-1,1],$ then the least value of $\alpha$ is equal to

Answer 1

$f\left(x\right)=a{x}^2+bx+c{f}^{\prime }\left(x\right)=2ax+b{f}^{\prime \prime }\left(x\right)=2a$
We know
$f(-1)=2\Rightarrow a-b+c=2\cdots \left(1\right)f^{\prime }(-1)=1\Rightarrow b-2a=1\cdots \left(2\right)f^{\prime \prime }\left(x\right)\le \frac{1}{2}\Rightarrow a\le \frac{1}{4}\cdots \left(3\right)$
From equations $\left(1\right)$ and $\left(2\right)$, we get
$b=1+2a,c=3+a\Rightarrow f\left(x\right)=a{x}^2+(1+2a)x+(3+a)\Rightarrow f\left(x\right)=a(x+1{)}^2+(x+3)$
To get the maximum value of $f\left(x\right)$, $a$ should be maximum, so
$a=\frac{1}{4}f\left(x\right)=\frac{(x+1{)}^2}{4}+(x+3)\Rightarrow f\left(x\right)\in [2,5],x\in [-1,1]$
As $f\left(x\right)\le \alpha ,x\in [-1,1]$
$\therefore \alpha =5$