JEE 2023 — Mathematics PYQ

1. Analyze the G.P. condition:

   Since $a,b,c$ are in Geometric Progression (G.P.), we have the relationship:

   $$b^2=ac$$

2. Find the intersection point:

   The parabolas intersect at $y=1$. Substituting $y=1$ into the first equation:

   $$a{x}^2+2bx+c(1)=0⟹a{x}^2+2bx+c=0$$

   The discriminant of this quadratic is $D=(2b{)}^2-4ac=4b^2-4ac$. Since $b^2=ac$, $D=0$.

   This means the equation has equal roots:

   $$x=\frac{-2b}{2a}=-\frac{b}{a}$$

3. Substitute into the second equation:

   Since they intersect at the same point, $x=-b/a$ and $y=1$ must satisfy the second parabola:

   $$d{\left(-\frac{b}{a}\right)}^2+2e\left(-\frac{b}{a}\right)+f(1)=0$$
   $$d\frac{b^2}{a^2}-\frac{2eb}{a}+f=0$$

4. Simplify using $b^2=ac$:

   Substitute $b^2=ac$ into the equation:

   $$d\frac{ac}{a^2}-\frac{2eb}{a}+f=0⟹\frac{dc}{a}-\frac{2eb}{a}+f=0$$

   Multiply the entire equation by $\frac{1}{c}$:

   $$\frac{d}{a}-\frac{2eb}{ac}+\frac{f}{c}=0$$

   Since $ac=b^2$:

   $$\frac{d}{a}-\frac{2eb}{b^2}+\frac{f}{c}=0⟹\frac{d}{a}-\frac{2e}{b}+\frac{f}{c}=0$$
   $$\frac{d}{a}+\frac{f}{c}=2\left(\frac{e}{b}\right)$$

This is the condition for an Arithmetic Progression (A.P.). Therefore, $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in A.P.

Correct Option: (2)