JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

\tan \alpha

and

\tan \beta

are roots of the equation

{x^2} + px + q = 0

with

p \ne 0,

then

View answer and explanation

Correct option: A

{\sin ^2}\left( {\alpha + \beta } \right) + p\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right) + q{\cos ^2}\left( {\alpha + \beta } \right) = q

Option A (

{\sin ^2}\left( {\alpha + \beta } \right) + p\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right) + q{\cos ^2}\left( {\alpha + \beta } \right) = q

) is correct because it is derived by substituting the sum and product of roots into the trigonometric identity.

More Questions from QUADRATIC EQUATIONS

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

\alpha ,\beta

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

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

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

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

{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\;

\;a \ne 0,

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

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