Question

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

Answer 1

Explanation for the correct option:

Step 1: Finding the value of $a$

Given that

$f:\left[-1,1\right]\to R$

$f\left(x\right)=a{x}^2+bx+c$

$f\left(-1\right)=2$

$f\text{'}\left(-1\right)=1$

$f\text{'}\left(x\right)=\frac{1}{2}$

Consider the given equation,

$\begin{array}{rcl}f\left(x\right)& =& a{x}^2+bx+c\\ f\text{'}\left(x\right)& =& 2ax+b\\ f"\left(x\right)& =& 2a\end{array}$

The maximum value for the constant is $\frac{1}{2}$

$\begin{array}{rcl}2a& =& \frac{1}{2}\\ a& =& \frac{1}{4}\end{array}$

Step 2: Finding the value of $b$and $c$

$\begin{array}{rcl}f\text{'}\left(-1\right)& =& 2a\left(-1\right)+b=1\\ -2a+b& =& 1\\ & \Rightarrow & b=1+2\left(\frac{1}{4}\right)\\ & \Rightarrow & =1+\frac{1}{2}\\ & \Rightarrow & =\frac{3}{2}\end{array}$

Finding the value of $c$

$$\begin{gathered} f\left(-1\right)=a-b+c=2 \\ \Rightarrow \frac{1}{4}-\frac{3}{2}+c=2 \\ \Rightarrow c=2-\frac{1}{4}+\frac{3}{2} \\ \Rightarrow =\frac{8+6-1}{4} \\ \Rightarrow =\frac{13}{4} \end{gathered}$$

Step 3: Finding the least value of $\alpha$

Find the value of $\alpha$ by putting the obtained value.

$\begin{array}{rcl}f\left(x\right)& =& a{x}^2+bx+c\\ & =& \left(\frac{1}{4}\right)x^2+\left(\frac{3}{2}\right)x+\left(\frac{13}{4}\right)\\ & =& \frac{1}{4}\left({\left(x+3\right)}^2+4\right)\end{array}$

For $x\in \left[-1,1\right]$ the minimum value is

$\begin{array}{rcl}f{\left(x\right)}_{min}& =& \frac{1}{4}\left({\left(-1+3\right)}^2+4\right)\\ & =& \frac{1}{4}\left({\left(2\right)}^2+4\right)\\ & =& \frac{1}{4}\left(4+4\right)\\ & =& \frac{1}{4}\left(8\right)\\ & =& 2\end{array}$

For $x\in \left[-1,1\right]$ the maximum value is

$\begin{array}{rcl}f{\left(x\right)}_{max}& =& \frac{1}{4}\left({\left(1+3\right)}^2+4\right)\\ & =& \frac{1}{4}\left({\left(4\right)}^2+4\right)\\ & =& \frac{1}{4}\left(16+4\right)\\ & =& \frac{1}{4}\left(20\right)\\ & =& 5\end{array}$

Since,

$\begin{array}{rcl}2& \le & f\left(x\right)\le 5\\ & \Rightarrow & f\left(x\right)\le \alpha \\ & \Rightarrow & \alpha =5\end{array}$

Hence, then the least value of $\alpha$ is equal to $5$.