JEE 2023 — Mathematics PYQ

Let $t=e^{2x}$. Since $e^{2x}$ is always positive for any real $x$, we must have t > 0.

The equation becomes:

2. Simplify the Polynomial:

Since $t\ne 0$, we can divide the entire equation by $t^2$:

Rearrange the terms:

3. Second Substitution:

Let $y=t+\frac{1}{t}$.

Squaring both sides gives $y^2=t^2+\frac{1}{t^2}+2$, so $t^2+\frac{1}{t^2}=y^2-2$.

Substitute these into the equation:

4. Solve for $y$:

Using the quadratic formula:

Since t > 0, by the AM-GM inequality, $t+\frac{1}{t}\ge 2$. Therefore, we must have $y\ge 2$.

• $y_1=\frac{1+\sqrt{21}}{2}\approx \frac{1+4.58}{2}\approx 2.79$ (Valid, as 2.79 > 2)

• $y_2=\frac{1-\sqrt{21}}{2}\approx -1.79$ (Invalid, as it is less than 2)

5. Determine the number of $x$ values:

For the valid value $y=\frac{1+\sqrt{21}}{2}$, we solve for $t$:

The discriminant of this quadratic in $t$ is $D=y^2-4$.

Since $y\approx 2.79$, then y^2 > 4, meaning D > 0. This gives two distinct positive values for $t$.

Since $t=e^{2x}$, each positive value of $t$ corresponds to exactly one real value of $x$ ($x=\frac{1}{2}\ln t$).

Final Answer:

The number of points where the curve cuts the $x$-axis is 2.