Calculus

Application of Derivatives

Medium

core

A solid sphere of gold of radius 5 cm, is being melted in a furnace such that the radius is decreasing uniformly. When the radius is 1cm, the rate at which surface area (A) is decreasing w.r.t. time ( $t$ ) is equal to

( $\frac{dv}{dt}$ = rate at which volume is decreasing w.r.t. $t$ )

$\frac{dv}{dt}$

$\frac{2dv}{dt}$

$\frac{1}{2}\left(\frac{dv}{dt}\right)$

$3\left(\frac{dv}{dt}\right)$

✅ Correct Option: 2

$V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$ , and $A = 4\pi r^2 \Rightarrow \frac{dA}{dt} = 8\pi r\frac{dr}{dt}$ .

Dividing, $\frac{dA/dt}{dV/dt} = \frac{2}{r}$ . At $r=1$ this equals 2, so $\frac{dA}{dt} = 2\frac{dV}{dt}$ .

2026: 25 May Shift 1

Let $C(x)$ and $R(x)$ be the cost function and the revenue function for $x$ units of a commodity such that marginal cost of producing 2 units is equal to marginal revenue of selling 1 unit. If $C(x) = ax^3 + bx^2 - 12x + 6$ and $R(x) = 100 + 6ax - bx^2$ , then which one of the following is TRUE?

2026: 25 May Shift 1

The rate of change of volume of a sphere with respect to its surface area when radius of the sphere is 4 cm, is

2025: 29 May Shift 2

Match List-I with List-II

List-I	List-II
Function f(x)	Interval for increasing/decreasing of function f(x)
---	---
(A) $f(x) = x\vert x\vert$	(I) Decreases on $(0, \infty)$
(B) $f(x) = x^2 + 2x - 5$	(II) Increases on $(3, \infty)$
(C) $f(x) = x^2 - 6x + 9$	(III) Decreases on $(-\infty, -1)$
(D) $f(x) = -x^2$	(IV) Increases on $(-\infty, \infty)$

Choose the correct answer from the options given below:

2025: 21 May Shift 1

The interval in which the function $f(x) = 2x^3 + 3x^2 - 12x + 1$ is strictly increasing, is

2026: 30 May Shift 1

The function $f(x)=17-9x+6x^2-x^3$ is strictly increasing in the interval, is:

2026: 21 May Shift 1

The maximum and the minimum value of the function $f(x) = -||x+3|-5| + 2$ are :

A solid sphere of gold of radius 5 cm, is being melted in a furnace such that the radius is decreasing uniformly. When the radius is 1cm, the rate at which surface area (A) is decreasing w.r.t. time (ttt) is equal to (dvdt\frac{dv}{dt}dtdv​ = rate at which volume is decreasing w.r.t. ttt)

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

A solid sphere of gold of radius 5 cm, is being melted in a furnace such that the radius is decreasing uniformly. When the radius is 1cm, the rate at which surface area (A) is decreasing w.r.t. time ( $t$ ) is equal to

( $\frac{dv}{dt}$ = rate at which volume is decreasing w.r.t. $t$ )