Question

Let $f(x)=x^2+ax+b$. If the maximum and the minimum values of $f(x)$ are 3 and 2 respectively for $0\le x\le 2$, then the possible ordered pair(s) of (a, b) is/are

• A. $(-2,3)$
• B. $(-3/2,2)$
• C. $(-5/2,3)$
• D. $(-5/2,2)$

Answer 1

Answer: (A)

$f(x)=x^2+ax+b$
vertex of $f(x)$ is at $-\frac{a}{2}$
f $0\le -\frac{a}{2}\le 2$
i.e. $0\le -a\le 4;-4\le a\le 0$

$f(x){]}_{\min }=f\left(-\frac{a}{2}\right)=\frac{a^2}{4}-\frac{a^2}{2}+b=b-\frac{a^2}{4}=2$ ....(1)
and $f(x){]}_{\max }=f(0)$ or $f(2)$
but $f(0)=b$ and $f(2)=4+2a+b$
let $b=3$;
from(1) $3-\frac{a^2}{4}=2;\frac{a^2}{4}=1\Rightarrow a=2$ or -2
but $a=2$ is rejected hence $a=-2$
$\therefore (-2,3)$ Ans. $\Rightarrow$ (A)
if $f(2)$ is maximum
$4+2x+b=3\Rightarrow 2a+b=-1$
(2) - (1) gives, $2a+\frac{a^2}{4}=-3\Rightarrow a^2+8a+12=0p(a+6)(a+2)=0$ $a=-2$ or $a=-6$ (rejected)
for vertex $<0$ or vertex $>2$ their are no accepted ordered pairs. Hence only $(A)$ is correct