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 $|a+b|$, if $\frac{8}{3}{a}^2+2b^2$ is maximum, is given by

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

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

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

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

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

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

Now,

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

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

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

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

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

Similarly

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

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

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

$z_{max}=\frac{2}{3}[2\times 4+2\times 4-0^2]\left(\begin{array}{c}\text{from}\left(ii\right)\\ \text{and}\left(11\right)\end{array}\right)$

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

$\text{Now,}b=0$

$|a|\le 2$

a can be-2 or+2

$\text{Now,}a=2$

$f\left(1\right)=12+c1\le 1$

C=-1

and $a=-2$

$f\left(1\right)=1-2+c1\le 1$

$c=1$

Hence, Two possibility:

a=2 or -2

b= 0 or 0

c= -1 or 1

Now,

$|a+b|=|2+0|=|-2+0|=2$