QJEEM20Mae3d2: JEE Main 2020 Mathematics - Class 11 JEEM20Mae3d2 Solution

Question

Verified Solution • 2025–26 Session moderate

If

\alpha

and

\beta

are the roots of the equation
x² + px + 2 = 0 and

{1 \over \alpha }

and

{1 \over \beta }

are the
roots of the equation 2x² + 2qx + 1 = 0, then

\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)

is equal to :

Accepted Answer

\alpha

and

\beta

are the roots of the

equation x² + px + 2 = 0

\therefore

\alpha + \beta = - p,\,\alpha \beta = 2

{1 \over \alpha }

and

{1 \over \beta }

are the roots of the

equation 2x² + 2qx + 1 = 0

\therefore

{1 \over \alpha } + {1 \over \beta } = - q,\,{1 \over {\alpha \beta }} = {1 \over 2}

\Rightarrow

{{\alpha + \beta } \over {\alpha \beta }} = - q \Rightarrow {{ - p} \over 2} = - q

\Rightarrow p = 2q

\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \alpha \beta + {1 \over {\alpha \beta }} + 2

= 2 + {1 \over 2} + 2 = {9 \over 2}

\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right) = \alpha \beta + {1 \over {\alpha \beta }} - {\alpha \over \beta } - {\beta \over \alpha }

= 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]

= \frac{5}{2} - [\frac{{(α + β)}^{2} - 2 α β}{α β}]

= {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]

= {{9 - {p^2}} \over 2}

\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right)

= {9 \over 4}(9 - {p^2})

Step-by-Step Verified Solution