JEE Main 2022 — Application Of Derivatives Question with Solution

Given curve,

$$y=x^3+3x^2+5$$

Slope of tangent,

$$\frac{dy}{dx}=3x^2+6x$$

Slope of line joined by (x₁, y₁) and (0, 0) is $$=\frac{y_0-0}{x_1-0}$$

$$\therefore$$ $$\frac{y_1}{x_1}=3x_1^2+6x_1$$

$$\Rightarrow y_1=3x_1^3+6x_1^2$$ ...... (1)

Point (x₁, y₁) is on the line,

$$\therefore$$ $$y_1=x_1^3+3x_1^2+5$$ ..... (2)

$$\therefore$$ $$3x_1^3+6x_1^2=x_1^3+3x_1^2+6$$

$$\Rightarrow 2x_1^3+3x_1^2-5=0$$

x = 1 satisfy the equation

Now, $$y_1=x_1^3+3x_1^2+5$$

$$=1+3+5$$

$$=9$$

$$\therefore$$ (x₁, y₁) = (1, 9)

Option D does not satisfy point (1, 9).