JEE PYQ: Quadratic Equation Question 41
Question 41 - 2019 (09 Jan Shift 1)
If a, b and c be three distinct real numbers in G.P. and $a + b + c = xb$, then $x$ cannot be:
(1) $-2$
(2) $-3$
(3) 4
(4) 2
Show Answer
Answer: (4)
Solution
$\therefore a, b, c$ are in G.P.
$\Rightarrow b^2 = ac$
Since, $a + b + c = xb$
$\Rightarrow a + c = (x-1)b$
Take square on both sides, we get
$a^2 + c^2 + 2ac = (x-1)^2 b^2$
$\Rightarrow a^2 + c^2 = (x-1)^2 ac - 2ac$ [$\because b^2 = ac$]
$\Rightarrow a^2 + c^2 = ac[(x-1)^2 - 2]$
$\therefore a^2 + c^2$ are positive and $b^2 = ac$ which is also positive.
Then, $x^2 - 2x - 1$ would be positive but for $x = 2$, $x^2 - 2x - 1$ is negative.
Hence, $x$ cannot be taken as 2.
BETA
