CUET UG 2025 — Mathematics Calculus

The car is moving along the curve $y = x^3 + 12$y=x3+12.

The rate of change of the y-coordinate is $\frac{dy}{dt}$dtdy​ and the rate of change of the x-coordinate is $\frac{dx}{dt}$dtdx​.

The condition states: $\frac{dy}{dt} = 3 \times \frac{dx}{dt}$dtdy​=3×dtdx​

---

Given: $y = x^3 + 12$y=x3+12

Using the chain rule:

$$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$$dtdy​=dxdy​×dtdx​

---

Differentiating $y = x^3 + 12$y=x3+12 with respect to $x$x:

$$\frac{dy}{dx} = 3x^2$$dxdy​=3x2

Therefore:

$$\frac{dy}{dt} = 3x^2 \times \frac{dx}{dt}$$dtdy​=3x2×dtdx​

---

Applying the given condition $\frac{dy}{dt} = 3 \times \frac{dx}{dt}$dtdy​=3×dtdx​:

$$3x^2 \times \frac{dx}{dt} = 3 \times \frac{dx}{dt}$$3x2×dtdx​=3×dtdx​

Since the car is moving, $\frac{dx}{dt} \neq 0$dtdx​=0:

$$3x^2 = 3$$3x2=3

$$x^2 = 1$$x2=1

$x = 1$x=1 or $x = -1$x=−1

---

Using $y = x^3 + 12$y=x3+12 to find the corresponding y-coordinates:

When $x = 1$x=1:

$$y = (1)^3 + 12$$y=(1)3+12

$$y = 1 + 12$$y=1+12

$$y = 13$$y=13

Point: $(1, 13)$(1,13)

---

When $x = -1$x=−1:

$$y = (-1)^3 + 12$$y=(−1)3+12

$$y = -1 + 12$$y=−1+12

$$y = 11$$y=11

Point: $(-1, 11)$(−1,11)

---

Therefore, the points on the curve are $(1, 13)$(1,13) and $(-1, 11)$(−1,11).