JEE Advanced 1998

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: Let $C_{1}$ and $C_{2}$ be the graphs of the functions $y = x^{2}$ and $y = 2 x, 0 \leq x \leq 1$ respectively. Let $C_{3}$ be the graph of a function $y = f (x), 0 \leq x \leq 1, f (0) = 0$ . For a point $P$ on $C_{1}$ , let the lines through $P$ , parallel to the axes, meet $C_{2}$ and $C_{3}$ at $Q$ and $R$ respectively (see figure.) If for every position of $P$ (on $C_{1}$ ), the areas of the shaded regions $O P Q$ and $O R P$ are equal, determine the function $f (x)$ .

Q: Why did we integrate $OPQ$ with respect to $y$ instead of $x$?

Great question! While you could integrate with respect to $x$, it would require splitting the region into two parts because the boundaries change. Integrating with respect to $y$ allows us to treat the region as a single, continuous strip from $y=0$ to $y=t^2$, making the calculation significantly cleaner and less prone to errors.

Q: Is $f(x) = x^3 - x^2$ the only possible answer?

Yes, absolutely. Because the condition must hold for all $t \in [0, 1]$, the resulting differential equation uniquely determines the function. There is no ambiguity here; the geometry of the curves forces this specific functional form.

Progressive Hints

Hint 1

Visualize the geometry. For the region

O P Q

, the boundaries are defined by

x

as a function of

y

(the parabola

x = y

and the line

x = y /2

). For

O R P

, the boundaries are defined by

y

as a function of

x

(the parabola

y = x^{2}

and the curve

y = f (x)

). Choosing the correct axis of integration for each region is the secret to avoiding messy algebra.

Hint 2

Once you have the area expressions, you will arrive at an integral equation of the form

\int_{0}^{t} f (x) d x = g (t)

. To isolate

f (t)

, you must invoke the Leibniz Integral Rule. Differentiate both sides of the equation with respect to the variable

t

Hint 3

Start by setting up the equality:

\int_{0}^{t^{2}} (y - \frac{y}{2}) d y = \int_{0}^{t} (x^{2} - f (x)) d x

. Evaluate the left integral first, then differentiate the entire equation with respect to

t

to extract

f (t)

Visualized Solution

Visualizing the Setup

Given curves:
$C_{1} : y = x^{2}$ (Blue Parabola)
$C_{2} : y = 2 x$ (Green Line)
$C_{3} : y = f (x)$ (Unknown Curve)
Let $P (t, t^{2})$ be a general point on $C_{1}$ .

Locating Points $Q$ and $R$

Horizontal line through $P$ meets $C_{2}$ at $Q$ .
$y_{Q} = y_{P} = t^{2} ⟹ 2 x_{Q} = t^{2} ⟹ x_{Q} = \frac{t ^{2}}{2}$
$Q = (\frac{t ^{2}}{2}, t^{2})$
Vertical line through $P$ meets $C_{3}$ at $R$ .
$x_{R} = x_{P} = t ⟹ y_{R} = f (t)$
$R = (t, f (t))$

Setting up Area $O P Q$

Area $O P Q$ is bounded by $x = \frac{y}{2}$ and $x = y$ .
Integrate with respect to $y$ from $0$ to $t^{2}$ :
$Area (O P Q) = \int_{0}^{t^{2}} (x_{C 1} - x_{C 2}) d y$
$Area (O P Q) = \int_{0}^{t^{2}} (y - \frac{y}{2}) d y$

Evaluating Area $O P Q$

Evaluating the integral:
$Area (O P Q) = [\frac{2}{3} y^{3/2} - \frac{y ^{2}}{4}]_{0}^{t^{2}}$
$Area (O P Q) = \frac{2}{3} (t^{2})^{3/2} - \frac{1}{4} (t^{2})^{2}$
$Area (O P Q) = \frac{2}{3} t^{3} - \frac{1}{4} t^{4}$

Setting up Area $O R P$

Area $O R P$ is bounded by $y = x^{2}$ and $y = f (x)$ .
Integrate with respect to $x$ from $0$ to $t$ :
$Area (O R P) = \int_{0}^{t} (y_{C 1} - y_{C 3}) d x$
$Area (O R P) = \int_{0}^{t} (x^{2} - f (x)) d x$

Expanding Area $O R P$

Splitting the integral:
$Area (O R P) = \int_{0}^{t} x^{2} d x - \int_{0}^{t} f (x) d x$
$Area (O R P) = [\frac{x ^{3}}{3}]_{0}^{t} - \int_{0}^{t} f (x) d x$
$Area (O R P) = \frac{t ^{3}}{3} - \int_{0}^{t} f (x) d x$

Equating the Areas

Given condition: $Area (O P Q) = Area (O R P)$
Equating the expressions:
$\frac{2}{3} t^{3} - \frac{1}{4} t^{4} = \frac{t ^{3}}{3} - \int_{0}^{t} f (x) d x$

Isolating the Integral

Rearranging to isolate the integral:
$\int_{0}^{t} f (x) d x = \frac{t ^{3}}{3} - \frac{2}{3} t^{3} + \frac{1}{4} t^{4}$
$\int_{0}^{t} f (x) d x = \frac{1}{4} t^{4} - \frac{1}{3} t^{3}$

Differentiating using Leibniz Rule

Differentiate both sides w.r.t $t$ using Leibniz Rule:
$\frac{d}{d t} \int_{0}^{t} f (x) d x = \frac{d}{d t} (\frac{1}{4} t^{4} - \frac{1}{3} t^{3})$
$f (t) = \frac{1}{4} (4 t^{3}) - \frac{1}{3} (3 t^{2})$

The Final Function

Simplifying the expression:
$f (t) = t^{3} - t^{2}$
Replacing $t$ with $x$ :
$f (x) = x^{3} - x^{2}$

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

Welcome, future engineer! Today, we are not just solving a problem; we are embarking on a journey to uncover a hidden function. We have a blue parabola,

C_{1}

, a green line,

C_{2}

, and a mysterious red curve,

C_{3}

Our goal is to find the equation of

C_{3}

, defined by

y = f (x)

. We assign a point

P

on the parabola

C_{1}

with coordinates

(t, t^{2})

. This point

P

serves as our anchor for the geometric construction.

The Geometry of $O P Q$

First, let us focus on the region

O P Q

. This area is trapped between the green line

C_{2}

and the blue curve

C_{1}

. To calculate this area, we integrate with respect to

y

to create a simple, continuous strip.

The curve on the right is

C_{1}

, which gives us

x = y

. The curve on the left is

C_{2}

, which gives us

x = y /2

. Our limits for

y

run from

0

t^{2}

. Thus, the area is given by the integral:

A r e a (O P Q) = \int_{0}^{t^{2}} (y - \frac{y}{2}) d y

Evaluating this integral, we find the anti-derivative of

y

\frac{2}{3} y^{3/2}

and the anti-derivative of

y /2

y^{2} /4

. Applying the limits, we get:

A r e a (O P Q) = \frac{2}{3} (t^{2})^{3/2} - \frac{1}{4} (t^{2})^{2} = \frac{2}{3} t^{3} - \frac{1}{4} t^{4}

The Mystery of $O R P$

Now, let us turn our attention to the region

O R P

. This area is bounded by the upper blue curve

C_{1}

and the lower red curve

C_{3}

. Integrating along the

x

-axis is the most natural path here.

Our limits for

x

run from

0

t

. The upper boundary is

y = x^{2}

, and the lower boundary is

y = f (x)

. The area is:

A r e a (O R P) = \int_{0}^{t} (x^{2} - f (x)) d x = \int_{0}^{t} x^{2} d x - \int_{0}^{t} f (x) d x

The first part evaluates to

\frac{t ^{3}}{3}

. Thus, our expression for the area of

O R P

is:

A r e a (O R P) = \frac{t ^{3}}{3} - \int_{0}^{t} f (x) d x

The Bridge of Equality

We are told that for every position of

P

, the areas of

O P Q

and

O R P

are equal. We equate our two expressions:

\frac{2}{3} t^{3} - \frac{1}{4} t^{4} = \frac{t ^{3}}{3} - \int_{0}^{t} f (x) d x

Rearranging the terms to isolate the integral, we obtain:

\int_{0}^{t} f (x) d x = \frac{t ^{3}}{3} - \frac{2}{3} t^{3} + \frac{1}{4} t^{4}

Combining the

t^{3}

terms, we find:

\int_{0}^{t} f (x) d x = \frac{1}{4} t^{4} - \frac{1}{3} t^{3}

The Final Revelation

We now use the Newton-Leibniz rule to differentiate both sides of the equation with respect to

t

. On the left, the derivative of the integral

\int_{0}^{t} f (x) d x

is simply

f (t)

On the right, the derivative of

\frac{1}{4} t^{4} - \frac{1}{3} t^{3}

is:

f (t) = \frac{d}{d t} (\frac{1}{4} t^{4} - \frac{1}{3} t^{3}) = t^{3} - t^{2}

Replacing

t

with

x

, we arrive at our final answer:

$f (x) = x^{3} - x^{2}$

Common Traps & FAQs

Why did we integrate

O P Q

with respect to

y

instead of

x

Great question! While you could integrate with respect to

x

, it would require splitting the region into two parts because the boundaries change. Integrating with respect to

y

allows us to treat the region as a single, continuous strip from

y = 0

y = t^{2}

, making the calculation significantly cleaner and less prone to errors.

Wait, why do we differentiate with respect to

t

We are given that the areas are equal for every position of

P

. This means the equality holds for all

t

. Since we have an integral involving

f (x)

on one side, differentiating with respect to the upper limit

t

is the standard, powerful tool to 'peel away' the integral sign and reveal the function

f (t)

hidden inside.

f (x) = x^{3} - x^{2}

the only possible answer?

Yes, absolutely. Because the condition must hold for all

t \in [0, 1]

, the resulting differential equation uniquely determines the function. There is no ambiguity here; the geometry of the curves forces this specific functional form.

Visualized Solution

Visualizing the Setup

Locating Points $Q$ and $R$

Setting up Area $O P Q$

Evaluating Area $O P Q$

Setting up Area $O R P$

Expanding Area $O R P$

Equating the Areas

Isolating the Integral

Differentiating using Leibniz Rule

The Final Function

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Geometry of $O P Q$

The Mystery of $O R P$

The Bridge of Equality

The Final Revelation

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Comprehension Passage

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Visualized Solution

Visualizing the Setup

Locating Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Setting up Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPQ

Evaluating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPQ

Setting up Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ORP

Expanding Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ORP

Equating the Areas

Isolating the Integral

Differentiating using Leibniz Rule

The Final Function

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Geometry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPQ

The Mystery of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ORP

The Bridge of Equality

The Final Revelation

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Locating Points $Q$ and $R$

Setting up Area $O P Q$

Evaluating Area $O P Q$

Setting up Area $O R P$

Expanding Area $O R P$

The Geometry of $O P Q$

The Mystery of $O R P$

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The area of the region bounded by the curve $y = f (x)$ , the x-axis, and the lines $x = a$ and $x = b$ , where $- \infty < a < b < - 2$ , is

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Let $f (x) =$ Maximum ${x^{2}, (1 - x)^{2}, 2 x (1 - x)}$ , where $0 \leq x \leq 1$ . Determine the area of the region bounded by the curves $y = f (x), x$ -axis, $x = 0$ and $x = 1$ .