JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

If the larger root of equation

x^2+\left(2-a^2\right) x+\left(1-a^2\right)=0

is less than both the roots of the equation

x^2-\left(a^2+4 a+1\right) x+a^2+4 a=0

, then the range of

a,

View answer and explanation

Correct option: B

\left(-\frac{1}{4}, \sqrt{2}\right)

Option A (

\sin ^{-1}\left(\tan \frac{\alpha}{2} \cdot|\cot \alpha|\right)

) is correct because it is derived by finding the angle between the vector product of two unit vectors and the plane containing the other two unit vectors.

More Questions from QUADRATIC EQUATIONS

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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)