JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The minimum value of

a

for which

a^{2} x^{2}+(2 a-1) x+1

is non-negative for any real

x

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (

\frac{{1}}{{4}}

) is correct because the expression being non-negative implies that the discriminant is non-positive.

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{x^2} - px + q = 0

{\alpha _1},{\beta _1}

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{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

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9{x^2} - 18\left| x \right| + 5 = 0

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