The use of electric vehicles will curb air pollution in the long run. The use of electric vehicles is increasing every year and the estimated number of electric vehicles in use at any time $t$ is given by the function $V$: $V(t) = \frac{1}{5}t^3 - \frac{5}{2}t^2 + 25t - 2$, where $t$ represents the time and $t = 1, 2, 3, \ldots$ corresponds to the years 2001, 2002, 2003, $\ldots$ respectively. Based on the above information, answer the following questions: (i) Can the above function be used to estimate the number of vehicles in the year 2000? Justify. (ii) Prove that the function $V(t)$ is an increasing function.

> [!TLDR] > The function $V(t)$ is a cubic polynomial defined only for $t \ge 1$ (years 2001 onward), so it cannot estimate the year 2000. Its derivative $V'(t)$ is always positive for $t \ge 1$, proving $V(t)$ is strictly increasing. --- ## The Concept: Domain and Monotonicity Two separate ideas are at play here. First, a function can only give meaningful output for inputs that lie in its **domain** — the set of $t$ values for which it is defined. The problem explicitly says $t = 1, 2, 3, \dots$ correspond to 2001, 2002, 2003, … So $t = 0$ (which would be the year 2000) is **not** in the domain. Second, to prove a function is **increasing**, we check its derivative. If $V'(t) > 0$ for all $t$ in the domain, then $V(t)$ is strictly increasing — meaning as time passes, the estimated number of electric vehicles always rises. --- ## Step-by-Step Solution ### (i) Can we estimate the number of vehicles in the year 2000? 1. **Identify the mapping.** The problem states: $t = 1$ corresponds to 2001, $t = 2$ to 2002, and so on. So the year $Y$ and the variable $t$ are related by $Y = 2000 + t$. For the year 2000, we would need $t = 0$. 2. **Check the domain.** The function $V(t)$ is given only for $t = 1, 2, 3, \ldots$ — that is, $t$ is a positive integer starting from 1. $t = 0$ is **not** included. Even if we tried to plug $t = 0$ into the formula, the result would not correspond to any valid data point in the model. 3. **Conclusion.** The function cannot be used to estimate the number of vehicles in the year 2000 because $t = 0$ lies outside the defined domain. > [!WARNING] > A common mistake is to think that because the formula is a polynomial, it works for any real $t$. But the problem explicitly restricts $t$ to $1, 2, 3, \dots$ — the domain is not all real numbers. --- ### (ii) Prove that $V(t)$ is an increasing function 1. **Find the derivative.** $$V(t) = \frac{1}{5}t^3 - \frac{5}{2}t^2 + 25t - 2$$ Differentiate te

The use of electric vehicles will curb air pollution in the long run. The use of electric vehicles is increasi

Question

t

(i) Can the above function be used to estimate the number of vehicles in the year 2000? Justify.

(ii) Prove that the function

V(t)

is an increasing function.

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The function $V(t)$ is a cubic polynomial defined only for $t \ge 1$ (years 2001 onward), so it cannot estimate the year 2000. Its derivative $V'(t)$ is always positive for $t \ge 1$ , proving $V(t)$ is strictly increasing.

The Concept: Domain and Monotonicity

Two separate ideas are at play here. First, a function can only give meaningful output for inputs that lie in its domain — the set of $t$ values for which it is defined. The problem explicitly says $t = 1, 2, 3, \dots$ correspond to 2001, 2002, 2003, … So $t = 0$ (which would be the year 2000) is not in the domain.

Second, to prove a function is increasing, we check its derivative. If $V'(t) > 0$ for all $t$ in the domain, then $V(t)$ is strictly increasing — meaning as time passes, the estimated number of electric vehicles always rises.

Step-by-Step Solution

(i) Can we estimate the number of vehicles in the year 2000?

Identify the mapping.

The problem states: $t = 1$ corresponds to 2001, $t = 2$ to 2002, and so on.

So the year $Y$ and the variable $t$ are related by $Y = 2000 + t$ .

For the year 2000, we would need $t = 0$ .
Check the domain.

The function $V(t)$ is given only for $t = 1, 2, 3, \ldots$ — that is, $t$ is a positive integer starting from 1.

$t = 0$ is not included. Even if we tried to plug $t = 0$ into the formula, the result would not correspond to any valid data point in the model.
Conclusion.

The function cannot be used to estimate the number of vehicles in the year 2000 because $t = 0$ lies outside the defined domain.

Watch out

A common mistake is to think that because the formula is a polynomial, it works for any real $t$ . But the problem explicitly restricts $t$ to $1, 2, 3, \dots$ — the domain is not all real numbers.

(ii) Prove that $V(t)$ is an increasing function

Find the derivative.

V(t) = \frac{1}{5}t^3 - \frac{5}{2}t^2 + 25t - 2

Differentiate term by term:

V'(t) = \frac{3}{5}t^2 - 5t + 25

Check the sign of $V'(t)$ for $t \ge 1$ . This is a quadratic in $t$ . To see if it is always positive, compute its discriminant: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 24 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$5$
›Reveal solutionSolution
We need to find the rate of change of the sphere's volume with respect to its diameter, which is $\frac{dV}{dD}$ . Using the chain rule, $\frac{dV}{dD} = \frac{dV}{dr} \cdot \frac{dr}{dD}$ , we find this rate to be $2\pi r^2$ . For a radius of $5$ cm, the rate is $\boxed{50\pi \text{ cm}^3/\text{cm}}$ .

When we talk about the "rate of change of the volume of a sphere with respect to its diameter," we are essentially asking for the derivative of the volume ( $V$ ) with respect to the diameter ( $D$ ). In mathematical terms, this is $\frac{dV}{dD}$ .

The volume of a sphere is typically expressed in terms of its radius, $r$ . The diameter, $D$ , is related to the radius by $D = 2r$ . To find $\frac{dV}{dD}$ , we can either express $V$ entirely in terms of $D$ and then differentiate, or we can use the chain rule. The chain rule is often more intuitive for problems like this, as it breaks down the problem into smaller, more manageable derivatives. It states that if $V$ depends on $r$ , and $r$ depends on $D$ , then $\frac{dV}{dD} = \frac{dV}{dr} \cdot \frac{dr}{dD}$ .

Let's work through the problem step-by-step.

Identify the relevant formulas and relationships. The volume of a sphere is given by:

$V = \frac{4}{3}\pi r^3$
The relationship between the radius ($r$) and the diameter ($D$) is:
$D = 2r$
From this, we can express $r$ in terms of $D$:
$r = \frac{D}{2}$
We are given that the radius $r = 5$ cm. We need to find $\frac{dV}{dD}$ at this specific radius.

2. Find the rate of change of volume with respect to radius ( $\frac{dV}{dr}$ ).

We differentiate the volume formula $V = \frac{4}{3}\pi r^3$ with respect to $r$ :
$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$ $\frac{dV}{dr} = \frac{4}{3}\pi \cdot 3r^2$ $\frac{dV}{dr} = 4\pi r^2$
This tells us how quickly the volume changes as the radius changes.

3. Find the rate of change of radius with respect to diameter ( $\frac{dr}{dD}$ ).

We use the relationship $r = \frac{D}{2}$ and differentiate it with respect to $D$ :
$\frac{dr}{dD} = \frac{d}{dD}\left(\frac{D}{2}\right)$ $\frac{dr}{dD} = \frac{1}{2}$
This makes sense: for every unit increase in diameter, the radius increases by half a unit.

4. Apply the Chain Rule to find $\frac{dV}{dD}$ . …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The rate of change of the area of a circle with respect to its radius r at r = 5 cm is$
(a) $12\pi$
(b) $8\pi$
(c) $5\pi$
(d) $10\pi$
›Reveal solutionSolution

The area of a circle is $A=\pi r^2$ ; its rate of change with $r$ is $dA/dr=2\pi r$ .

$A=\pi r^2 \Rightarrow \dfrac{dA}{dr}=2\pi r$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The rate of change of area of circle w.r.t. its radius 'r' when r = 4 cm is$
(a) 8π cm²/cm
(b) 6π cm²/cm
(c) 4π cm²/cm
(d) 2π cm²/cm
›Reveal solutionSolution

The rate of change of area with radius is $dA/dr = 2\pi r$ ; substitute $r=4$ .

Area of a circle: $A = \pi r^2$ .

$\frac{dA}{dr} = 2\pi r$

At $r = 4\ \text{cm}$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$r$
(a) $10\pi$ cm
(b) $12\pi$ cm
(c) $8\pi$ cm
(d) $11\pi$ cm
›Reveal solutionSolution

Differentiate the area formula $A=\pi r^2$ with respect to $r$ and evaluate at $r=6$ .

Area of a circle: $A = \pi r^2$

$\frac{dA}{dr} = 2\pi r$

At $r=6$ cm: …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The rate of change of the area of a circle with respect to its radius r at r=6 cm is:$
(a)
(i) 8\pi
(b)
(ii) 10\pi
(c)
(iii) 11\pi
(d)
(iv) 12\pi
›Reveal solutionSolution
$\dfrac{dA}{dr}=2\pi r = 12\pi$ at $r=6$ cm — option (iv).

Concept. The rate of change of a quantity with respect to a variable is its derivative with respect to that variable.

Steps.

Area of a circle: $A=\pi r^2$ .

Differentiate with respect to $r$ : $\dfrac{dA}{dr}=2\pi r$ . …
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CBSE 2025Set A1 mark
$r = 6$
›Reveal solutionSolution

Area of a circle is $A = \pi r^2$ ; differentiate with respect to $r$ and substitute $r=6$ .

Area of a circle: $A = \pi r^2$ .

$\dfrac{dA}{dr} = 2\pi r$

At $r = 6$ cm: …

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CBSE 2025Set ANNUAL1 mark
$r$
›Reveal solutionSolution

The rate of change of the area of a circle with respect to its radius is $\dfrac{dA}{dr}$ .

$A=\pi r^2 \;\Rightarrow\; \dfrac{dA}{dr}=2\pi r$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The rate of change of area of a circle w.r.t. its radius 'r' when r = 5 cm is :$
(a) 8π cm²/cm
(b) 10π cm²/cm
(c) 9π cm²/cm
(d) 16π cm²/cm
›Reveal solutionSolution

Area of a circle $A=\pi r^2$ ; rate of change $\dfrac{dA}{dr}=2\pi r$ , which at $r=5$ gives $10\pi$ .

Given $A=\pi r^2$ , differentiate with respect to $r$ :

$\frac{dA}{dr} = 2\pi r.$

At $r=5\ \text{cm}$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$r=6$
(a) $6\pi\ cm^2/cm$
(b) $6\pi\ cm/cm^2$
(c) $12\pi\ cm^2/cm$
(d) $12\pi\ cm/cm^2$
›Reveal solutionSolution

Differentiate the area formula $A=\pi r^2$ w.r.t. $r$ and substitute $r=6$ .

Area of circle: $A = \pi r^2$ .

$\dfrac{dA}{dr} = 2\pi r$

At $r=6$ cm: …

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CBSE 2024Set 65/3/11 markMCQ
$1.5$
›Reveal solutionSolution
The perimeter of a square is directly proportional to its side length. If the side decreases at $1.5$ cm/s, the perimeter decreases at $4$ times that rate, which is $6$ cm/s. The correct option is (B).

This problem asks us to find the rate at which the perimeter of a square is decreasing, given the rate at which its sides are decreasing. This is a classic "related rates" problem in calculus. The core idea is to establish a relationship between the quantities involved (side length and perimeter), and then differentiate that relationship with respect to time to find how their rates of change are related.

When we talk about a "rate of change," we are essentially talking about a derivative with respect to time. If a quantity is decreasing, its rate of change will be negative.

Here's how we approach it:

Identify the variables and given rates.

Let $s$ be the side length of the square at any given time $t$ .

Let $P$ be the perimeter of the square at any given time $t$ .

We are given that the sides of the square are decreasing at the rate of $1.5$ cm/s. In calculus terms, this means the derivative of the side length with respect to time, $\frac{ds}{dt}$ , is $-1.5$ cm/s. The negative sign indicates a decrease.

Our goal is to find the rate of decrease of the perimeter, which means we need to find $\frac{dP}{dt}$ .

Establish a relationship between the variables.

The formula for the perimeter of a square with side length $s$ is:

$P = 4s$

Differentiate the relationship with respect to time. To find how the rates of change are related, we differentiate both sides of the equation $P = 4s$ with respect to time $t$ . We use the chain rule here.

$\frac{dP}{dt} = \frac{d}{dt}(4s)$
Since $4$ is a constant, we can pull it out of the differentiation:
$\frac{dP}{dt} = 4 \frac{ds}{dt}$
This equation tells us that the rate of change of the perimeter is $4$ times the rate of change of the side. This makes intuitive sense: if each of the four sides shrinks by a certain amount, the total perimeter shrinks by four times that amount.

4. Substitute the given rate and calculate. …
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CBSE 2024Set D1 markMCQ
$r$
(a) $10\pi$
(b) $12\pi$
(c) $8\pi$
(d) $11\pi$
›Reveal solutionSolution

$\frac{dA}{dr}=2\pi r$ , which at $r=6$ equals $12\pi$ .

Area of a circle: $A = \pi r^2$ .

Rate of change of area with respect to radius: $\dfrac{dA}{dr} = 2\pi r$ .

…

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CBSE 2024Set ANNUAL1 mark
$r$
›Reveal solutionSolution

Differentiate the area formula $A=\pi r^2$ w.r.t. $r$ and evaluate at $r=3$ .

$A = \pi r^2 \Rightarrow \dfrac{dA}{dr} = 2\pi r$

…

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Concept understanding — Rate Of Change

Rate of Change

Average versus instantaneous rate

Rates with respect to time

Reading the sign and size

The Concept: Domain and Monotonicity

Step-by-Step Solution

(i) Can we estimate the number of vehicles in the year 2000?

(ii) Prove that $V(t)$ is an increasing function

The Concept: Domain and Monotonicity

Step-by-Step Solution

(i) Can we estimate the number of vehicles in the year 2000?

(ii) Prove that V(t)V(t)V(t) is an increasing function

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(ii) Prove that $V(t)$ is an increasing function

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