If ax² + bx + c [AP EAPCET 21 May 2025 S2]

Correct Answer is : Maximum value =3b4= 3b/4=43b

Correct Answer is : Maximum value =3b4= 3b/4=43b

Test Index

Quadratic Equations

Section: Mathematics

Question : 26 of 51

Marks: +1, -0

ax^2 + bx + c < 0 \forall x \in \mathbb{R}

cx^2 + ax + b

ax^2 + bx + c

x

cx^2 + ax + b

Solution:

We have,

a x^2 + b x + c < 0

a < 0

, then

\Delta < 0

a < 0

and

b^2 - 4ac < 0

\therefore b^2 < 4ac

And

c x^2 + a x + b

and

a x^2 + b x + c

have same extreme value

\because \;\; -\frac{a}{2c} = -\frac{b}{2a} \Rightarrow a^2 = bc

Now, the minimum value of

c x^2 + a x + b

-\frac{\Delta}{4a} = -\frac{a^2 - 4bc}{4c} = -\frac{bc - 4bc}{4c} = -\left( -\frac{3bc}{4c} \right) = \frac{3b}{4}

Because

b^2 =

positive and

a < 0 \Rightarrow c < 0

\therefore \frac{3b}{4}

is the maximum value of

c x^2 + a x + b

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