QJEEM19Mab85b: JEE Main 2019 Mathematics - Class 11 JEEM19Mab85b Solution: If , and are three consecutive terms of a non-constant G.P.…

Let the common ratio of G.P is r

Then the equation of $\alpha {x^2} + 2\beta x + \gamma = 0$

$\Rightarrow$ $\alpha {x^2} + 2\alpha rx + \alpha {r^2} = 0$

$\Rightarrow$ x² + 2rx + r² = 0 ........(i)
Equation (i) and x² + x - 1 = 0 ......... (ii) has a common root.

Now (i) - (ii) $\Rightarrow$ (2r-1)x + (r²+1) = 0

$\Rightarrow$ x = ${{ - \left( {{r^2} + 1} \right)} \over {2r - 1}}$ ....... (iii)

Now putting (iii) in equation (ii)

$\Rightarrow$ (r²+1)² - (r²+1) (2r - 1) - (2r - 1)² = 0

$\Rightarrow$ r⁴ - 2r³ - r² + 2r + 1 = 0 .......... (iv)

dividing equation (iv) by r²

$\Rightarrow$ r² - 2r - 1 + ${2 \over r} + {1 \over {{r^2}}} = 0$

$\Rightarrow$ ${\left( {r - {1 \over r}} \right)^2} - 2\left( {r - {1 \over r}} \right) + 1 = 0$

$\Rightarrow$ ${\left( {r - {1 \over r} - r} \right)^2} = 0$

$\Rightarrow$ ${{{r^2} - 1} \over r} = 1$ $\Rightarrow$ r² = 1 + r

Now $\alpha$ ( $\beta$ + $\gamma$ ) $\Rightarrow$ $\alpha$ ( $\alpha$ r + $\alpha$ r²)

$\Rightarrow$ $\alpha ^2 r (1 + r)$   [ $\because$ r² = 1 + r]

$\Rightarrow$ $\alpha ^2 r . r^2$ $\Rightarrow$ $( \alpha r )\,( \alpha r^2 )$ $\Rightarrow$ $\beta$ $\gamma$