JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

a \le 0,

then the real values of

x

satisfying

{x^2} - 2a\left| {x - a} \right| - 3{a^2} = 0

are

View answer and explanation

Correct option: A

a\left( {1 - \sqrt 2 } \right),a\left( { - 1 + \sqrt 6 } \right)

Option A (

a\left( {{{{1 - \sqrt 2 }}}} \right),a\left( {{{{ - 1 + \sqrt 6 }}}} \right)

) is correct because these are the real values of x that satisfy the given equation when considering both cases for the absolute value and the condition a <= 0.

More Questions from QUADRATIC EQUATIONS

Practice All Chapter MCQs →

The solution of the equation

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

is If

\alpha ,\beta

are the roots of

{x^2} - 3x + a = 0,a \in R

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

be the roots of the equation

{x^2} - qx + p = 0,

then the equation whose roots are

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

is If

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

for all values of

x

, then

a

lies in the interval If

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

and

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

and

\;a \ne 0,

then Root(s) of the equation

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

belonging to the domain of definition of the function

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)

is (are)