JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

\alpha

and

\beta

are roots of the quadratic equation,

x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right)

, then

\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right) \cdot(\alpha-\beta)^{24}}

is equal to :

View answer and explanation

Correct option: C

\frac{2^{12}}{(\sin \theta+8)^{12}}

Option C (

\frac{{2^{{12}}}}{{(\sin \theta+8)^{{12}}}}

) is correct because it is obtained by simplifying the given expression using the relationships between the roots and coefficients of the quadratic equation.

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)