Let $f:
Question:

Let $f:[-1,1] \rightarrow R$ be defined as $f(x)=a x^{2}+b x+c$

for all $\mathrm{x} \in[-1,1]$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$ such that $\mathrm{f}(-1)=2, \mathrm{f}^{\prime}(-1)=1$ and

for $x \in(-1,1)$ the maximum value of $f^{\prime \prime}(x)$ is $\frac{1}{2}$. If $f(x) \leq \alpha$

$x \in[-1,1]$, then the least value of $\alpha$ is equal to

 

Solution:

$\mathrm{f}:[-1,1] \rightarrow \mathrm{R}$

$f(x)=a x^{2}+b x+c$

$f(-1)=a-b+c=2$

$f^{\prime}(-1)=-2 a+b=1$

$\mathrm{f}^{\prime \prime}(\mathrm{x})=2 \mathrm{a}$

$\Rightarrow$ Max. value of $\mathrm{f}^{\prime \prime}(\mathrm{x})=2 \mathrm{a}=\frac{1}{2}$

$\Rightarrow a=\frac{1}{4} ; \quad b=\frac{3}{2} ; \quad c=\frac{13}{4}$

$\therefore \quad f(x)=\frac{x^{2}}{4}+\frac{3}{2} x+\frac{13}{4}$

For, $x \in[-1,1] \Rightarrow 2 \leq f(x) \leq 5$

$\therefore \quad$ Least value of $\alpha$ is 5

Administrator

Leave a comment

Please enter comment.
Please enter your name.