JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 5)
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 :
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 :
$${9 \over 4}\left( {9 - {q^2}} \right)$$
$${9 \over 4}\left( {9 + {q^2}} \right)$$
$${9 \over 4}\left( {9 - {p^2}} \right)$$
$${9 \over 4}\left( {9 + {p^2}} \right)$$
Explanation
$$\alpha $$ 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
$$ \therefore $$ $${1 \over \alpha } + {1 \over \beta } = - q,\,{1 \over {\alpha \beta }} = {1 \over 2}$$
$$ \Rightarrow $$ $${{\alpha + \beta } \over {\alpha \beta }} = - q \Rightarrow {{ - p} \over 2} = - q$$
$$ \Rightarrow p = 2q$$
$$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \alpha \beta + {1 \over {\alpha \beta }} + 2$$ $$ = 2 + {1 \over 2} + 2 = {9 \over 2}$$
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right) = \alpha \beta + {1 \over {\alpha \beta }} - {\alpha \over \beta } - {\beta \over \alpha }$$
$$ = 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]$$$$
= {5 \over 2} - \left[ {{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \over {\alpha \beta }}} \right]$$
$$ = {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]$$
$$ = {{9 - {p^2}} \over 2}$$
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right)$$
$$ = {9 \over 4}(9 - {p^2})$$
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
$$ \therefore $$ $${1 \over \alpha } + {1 \over \beta } = - q,\,{1 \over {\alpha \beta }} = {1 \over 2}$$
$$ \Rightarrow $$ $${{\alpha + \beta } \over {\alpha \beta }} = - q \Rightarrow {{ - p} \over 2} = - q$$
$$ \Rightarrow p = 2q$$
$$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \alpha \beta + {1 \over {\alpha \beta }} + 2$$ $$ = 2 + {1 \over 2} + 2 = {9 \over 2}$$
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right) = \alpha \beta + {1 \over {\alpha \beta }} - {\alpha \over \beta } - {\beta \over \alpha }$$
$$ = 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]$$$$
= {5 \over 2} - \left[ {{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \over {\alpha \beta }}} \right]$$
$$ = {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]$$
$$ = {{9 - {p^2}} \over 2}$$
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right)$$
$$ = {9 \over 4}(9 - {p^2})$$
Comments (0)
