The number of points, where the curve y = x^5 - 20x^3 + 50x + 2 c… — Solution | JEE Mathematics

NVAMediumJEE 2023Applications of Derivatives (Monotonicity, Extrema)

The number of points, where the curve $y = x^5 - 20x^3 + 50x + 2$ crosses the x-axis is :

Answer

Correct answer:5

Standard Method

Given: The function is $y = x^5 - 20x^3 + 50x + 2$ .

Find: The number of points where the curve crosses the x-axis.

To find the points where the curve crosses the x-axis, we need to solve for $y = 0$ .

The solution uses the derivative to study the turning points:

\frac{dy}{dx} = 5x^4 - 60x^2 + 50

We solve:

\frac{dy}{dx} = 0 \quad \Rightarrow \quad x^4 - 12x^2 + 10 = 0

Solving for $x^2$ , we find:

x^2 = 6 \pm \sqrt{26} \quad \Rightarrow \quad x^2 \approx 6 \pm 5.1

Thus:

x \approx \pm 3.3, \pm 0.95

Therefore, the number of points where the curve cuts the x-axis is $5$ .

Differentiating the polynomial incorrectly is a common mistake. If $\frac{d}{dx}(x^5 - 20x^3 + 50x + 2)$ is not computed carefully, the critical points will be wrong. Differentiate term by term and then solve the resulting quartic in $x^2$ .
Assuming that solving $\frac{dy}{dx} = 0$ directly gives the x-intercepts is incorrect. The derivative gives turning points, not the roots of the original equation. Use the critical points only to infer how many times the graph can cross the x-axis.
Ignoring the distinction between touching and crossing the x-axis can lead to an incorrect count. The question asks where the curve crosses the x-axis, so the graph's behavior around each root must be interpreted from the turning-point structure.

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