JEE Mathematics 2023 Question with Solution

Given: The curve is $y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$ and the normal is parallel to $x + 90y + 2 = 0$.

Find: The number of points on the curve where the normal has the same slope as the given line.

The line $x + 90y + 2 = 0$ can be written as

$$y = -\frac{1}{90}x - \frac{1}{45}$$

so its slope is $-\frac{1}{90}$.

For the curve, if tangent slope is $\frac{dy}{dx}$, then normal slope is

$$m_N = -\frac{1}{\frac{dy}{dx}}$$

Since the normal is parallel to the given line,

$$-\frac{1}{\frac{dy}{dx}} = -\frac{1}{90}$$

Hence,

$$\frac{dy}{dx} = 90$$

Now differentiate the curve:

$$\frac{dy}{dx} = \frac{d}{dx}\left(54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x\right)$$ $$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$$

Set this equal to $90$:

$$270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$$ $$270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$$

From the extracted working, the roots are

$$x = 1, 2, -\frac{2}{3}, -\frac{1}{3}$$

Thus there are 4 real points on the curve where the normal is parallel to the given line.

Therefore, the correct option is C.

Given: Normal is parallel to $x + 90y + 2 = 0$.

Find: How many points satisfy the condition.

Instead of finding equations of normals, compare slopes directly. The slope of the given line is $-\frac{1}{90}$. For a curve, normal slope is $-\frac{1}{\frac{dy}{dx}}$. Therefore,

$$-\frac{1}{\frac{dy}{dx}} = -\frac{1}{90}$$

which immediately gives

$$\frac{dy}{dx} = 90$$

Now use

$$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$$

so

$$270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$$

The extracted solution gives four real roots: $x = 1, 2, -\frac{2}{3}, -\frac{1}{3}$. Hence the number of required points is 4.

Therefore, the correct option is C.