Question

Let $f\left(x\right)=a{x}^2+bx+c,a,b,c\in R.$ It is given $|f\left(x\right)|\le 1,|x|\le 1$ then the possible value of $\frac{8}{3}{a}^2+2b^2$ is given by

• A. $32$
• B. $\frac{32}{3}$
• C. $\frac{2}{3}$
• D. $\frac{16}{3}$

Answer 1

Answer: (B)

$\frac{32}{3}$

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

Since,

$\frac{8}{3}{a}^2+2b^2=z\left(say\right)$

$=\frac{8a^2+6b^2}{3}$

$=\frac{2}{3}[4a^2+3b^2]$

$=\frac{2}{3}\left[2\right(a+b{)}^2+2(a-b{)}^2-b^2]-\left(i\right)$

Now,

$f\left(1\right)=a+b+c$

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

$f\left(0\right)=c$

$|f\left(1\right)-f\left(0\right)|\le 2\left(|f\right(x)|⩽1)$

$\Rightarrow |a+b|\le 2\to$ (ii)

Similarly

$|f(-1)-f\left(0\right)|\le 2$

$\Rightarrow |a-b|\le 2-\left(iii\right)$

$z=\frac{2}{3}\left[2\right(a+b{)}^2+2(a-b{)}^2-b^2]$

$z_{max}=\frac{2}{3}[2\times 4+2\times 4-0^2]$ (from ii &iii)

$=\frac{32}{3}(z_{max},$ then $b=0)$