If $\alpha, \beta$ are the roots of $x^2-6 x-2=0.$ If $a_n=\alpha^n-\beta^n$ for $n \geq 1.$ Then the value of $\frac{a_{12}-2 a_{10}}{3 a_{11}}$ is

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

\alpha, \beta

are the roots of

x^2-6 x-2=0.

a_n=\alpha^n-\beta^n

for

n \geq 1.

Then the value of

\frac{a_{12}-2 a_{10}}{3 a_{11}}

View answer and explanation

Correct option: A

Option A (2) is correct because by using the given recurrence relation and the properties of roots, we can simplify the expression to obtain the value 2.

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The solution of the equation

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is If

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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

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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)