Question

Assertion (A): The maximum value of quadratic function $-x^2+3x+1$ is $\frac{11}{4}$.
Reason (R): The maximum value of $a{x}^2+bx+cis\frac{4ac-b^2}{4a}$ if $a<0$.

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not the correct explanation of A
• C. A is true and R is false
• D. A is false and R is true

Answer 1

Answer: (D)

A is false and R is true

$f^{\prime }\left(x\right)=-2x+3$
$f^{\prime }\left(x\right)=0$ at $x=\frac{3}{2}$
So $f_{max}\left(x\right)$ exist at $x=\frac{3}{2}$
$f\left(\frac{3}{2}\right)=\frac{-9}{4}+\frac{9}{2}+1$
$f\left(\frac{3}{2}\right)=\frac{13}{4}$
$f^{\prime }\left(x\right)=2ax+b$
$f^{\prime }\left(x\right)=0$ $x=\frac{-b}{2a}$
$f_{max}\left(\frac{-b}{2a}\right)=a\left(\frac{b^2}{4a^2}\right)-\frac{b^2}{2a}+c$
$=\frac{4ac-b^2}{4a}$