Find the equation of the tangent and the normal to the curve $y = \dfrac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis.

Find the equation of the tangent and the normal to the curve y = (x-7)/((x-2)(x-3)) at the point where it cuts

Question

y = \dfrac{x-7}{(x-2)(x-3)}

CBSECBSE Class XII Board 2019Subjective· 4mImportance★★★★★

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The curve cuts the x‑axis at $(7,0)$ . The tangent at that point is $y = \frac{1}{20}(x-7)$ and the normal is $y = -20(x-7)$ .

Concept and intuition

When a curve cuts the x‑axis, its y‑coordinate is zero. So the first job is to find that point. Once we have the point, the slope of the tangent is simply the derivative $dy/dx$ evaluated there. The normal is perpendicular to the tangent, so its slope is the negative reciprocal.

The curve is a rational function. Differentiating it directly with the quotient rule is fine, but we can simplify first: factor the numerator? No — the numerator is $x-7$ , and the denominator is $(x-2)(x-3)$ . The point where $y=0$ is when $x-7=0$ , i.e. $x=7$ . That’s the only x‑intercept.

Now the derivative. Rather than expanding the denominator, we can use logarithmic differentiation or the quotient rule directly. I’ll use the quotient rule — it’s clean enough.

Step‑by‑step solution

1. Find the point where the curve cuts the x‑axis.

Set $y = 0$ :

\frac{x-7}{(x-2)(x-3)} = 0 \quad\Rightarrow\quad x-7 = 0 \quad\Rightarrow\quad x = 7.

So the point is $(7,0)$ .

2. Differentiate $y$ with respect to $x$ .

Let $u = x-7$ and $v = (x-2)(x-3)$ . Then $u' = 1$ and $v' = (x-2)+(x-3) = 2x-5$ (by the product rule).

Quotient rule:

\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{1\cdot (x-2)(x-3) - (x-7)(2x-5)}{(x-2)^2(x-3)^2}.

3. Simplify the numerator.

First expand $(x-2)(x-3) = x^2 - 5x + 6$ .

Now expand $(x-7)(2x-5) = 2x^2 -5x -14x +35 = 2x^2 -19x +35$ .

So numerator:

(x^2 - 5x + 6) - (2x^2 - 19x + 35) = x^2 - 5x + 6 - 2x^2 + 19x - 35 = -x^2 + 14x - 29.

Thus

\frac{dy}{dx} = \frac{-x^2 + 14x - 29}{(x-2)^2(x-3)^2}.

4. Evaluate the slope at $x=7$ .

Substitute $x=7$ :

Numerator: $-49 + 98 - 29 = 20$ .

Denominator: $(5)^2 \cdot (4)^2 = 25 \cdot 16 = 400$ .

So …

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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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Q.Find the equation of the tangent and the normal to the curve $y = \dfrac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis.

Concept understanding — Rate Of Change

Rate of Change

Average versus instantaneous rate

Rates with respect to time

Reading the sign and size

Concept and intuition

Step‑by‑step solution

Q.Find the equation of the tangent and the normal to the curve y=x−7(x−2)(x−3)y = \dfrac{x-7}{(x-2)(x-3)}y=(x−2)(x−3)x−7​ at the point where it cuts the x-axis.

Concept and intuition

Step‑by‑step solution

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Q.Find the equation of the tangent and the normal to the curve $y = \dfrac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis.

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