JEE Advanced 2009

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Match the statements/expressions given in Column-I with the values given in Column-II.

Q: How do we find the minimum of the absolute value sum in Part C?

The function $f(x) = |x - 1| + |x - 2| + |x + 1| + |x + 2|$ is a sum of distances. The minimum occurs between the two middle critical points, which are $-1$ and $1$. By testing any value in the interval $[-1, 1]$, we find the minimum value is $6$.

List-I

(P)

The number of solutions of the equation

x e^{s i n x} - cos x = 0

in the interval

(0, \frac{π}{2})

(Q)

Value(s) of

k

for which the planes

k x + 4 y + z = 0

4 x + k y + 2 z = 0

and

2 x + 2 y + z = 0

intersect in a straight line

(R)

Value(s) of

k

for which

∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣ = 4 k

has integer solution(s)

(S)

y^{'} = y + 1

and

y (0) = 1

, then value(s) of

y (ln 2)

List-II

(1)

(2)

(3)

(4)

(5)

Select Matching Pairs:

* Multiple Allowed

PMatches

QMatches

RMatches

SMatches

Visualized Solution

Introduction to the Matching Challenge

We need to match the mathematical statements in Column-I with their correct values in Column-II.
The problems cover diverse topics: transcendental equations, systems of linear equations, absolute value functions, and differential equations.
Let's break down each part systematically to find the correct matches.

Part A: Analyzing $x e^{s i n x} - cos x = 0$

Let's define the function: $f (x) = x e^{s i n x} - cos x$ for the interval $x \in (0, \frac{π}{2})$
We want to find the number of solutions in this open interval.
Let's first check the values of the function at the boundaries: $x = 0$ and $x = \frac{π}{2}$ .

Part A: Applying the Intermediate Value Theorem

At the lower boundary: $f (0) = 0 \cdot e^{s i n 0} - cos 0 = 0 - 1 = - 1 < 0$
At the upper boundary: $f (\frac{π}{2}) = \frac{π}{2} e^{s i n (\frac{π}{2})} - cos (\frac{π}{2}) = \frac{π}{2} e^{1} - 0 = \frac{π}{2} e > 0$
Since $f (0) < 0$ and $f (\frac{π}{2}) > 0$ , by the Intermediate Value Theorem, there is at least one root in $(0, \frac{π}{2})$ .

Part A: Proving Uniqueness via Monotonicity

Let's differentiate $f (x)$ to check if it is monotonic:
$f^{'} (x) = \frac{d}{d x} (x e^{s i n x}) - \frac{d}{d x} (cos x)$
$f^{'} (x) = e^{s i n x} + x e^{s i n x} cos x + sin x$
For $x \in (0, \frac{π}{2})$ , we have $sin x > 0$ , $cos x > 0$ , and $e^{s i n x} > 0$ .
Thus, $f^{'} (x) > 0$ , meaning $f (x)$ is strictly increasing.
Conclusion: Exactly one solution exists. This matches 1 in Column-II.

Part B: Condition for Planes Intersecting in a Line

We are given three planes:
1) $k x + 4 y + z = 0$
2) $4 x + k y + 2 z = 0$
3) $2 x + 2 y + z = 0$
For these planes to intersect in a straight line, the system of equations must have infinitely many solutions.
This requires the determinant of the coefficient matrix to be zero: $Δ = 0$ .

Part B: Solving the Determinant Equation

Set up the determinant: $Δ = k 42 4 k 2 121 = 0$
Expand along the first row:
$k (k \cdot 1 - 2 \cdot 2) - 4 (4 \cdot 1 - 2 \cdot 2) + 1 (4 \cdot 2 - k \cdot 2) = 0$
$k (k - 4) - 4 (0) + (8 - 2 k) = 0$
Simplify: $k^{2} - 6 k + 8 = 0$
Factorize: $(k - 2) (k - 4) = 0 ⟹ k = 2$ or $k = 4$ .
This matches 2 and 4 in Column-II.

Part C: Analyzing the Absolute Value Function

Let $f (x) = ∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣$
We want to find the values of $k$ for which $f (x) = 4 k$ has integer solutions.
Let's find the minimum value of $f (x)$ . Since the critical points are $- 2, - 1, 1, 2$ , the minimum occurs in the central interval $[- 1, 1]$ .
For $x \in [- 1, 1]$ : $f (x) = (1 - x) + (2 - x) + (x + 1) + (x + 2) = 6$ .
Thus, the minimum value of $f (x)$ is $6$ . For solutions to exist, we must have $4 k \geq 6 ⟹ k \geq 1.5$ .

Part C: Finding Integer Solutions for $k$

Since $k \geq 1.5$ , let's test the integer values of $k$ from Column-II:
If $k = 2 ⟹ 4 k = 8$ . Let's test $x = 2$ : $f (2) = ∣1∣ + ∣0∣ + ∣3∣ + ∣4∣ = 8$ . (Integer solution $x = 2$ exists)
If $k = 3 ⟹ 4 k = 12$ . Let's test $x = 3$ : $f (3) = ∣2∣ + ∣1∣ + ∣4∣ + ∣5∣ = 12$ . (Integer solution $x = 3$ exists)
If $k = 4 ⟹ 4 k = 16$ . Let's test $x = 4$ : $f (4) = ∣3∣ + ∣2∣ + ∣5∣ + ∣6∣ = 16$ . (Integer solution $x = 4$ exists)
If $k = 5 ⟹ 4 k = 20$ . Let's test $x = 5$ : $f (5) = ∣4∣ + ∣3∣ + ∣6∣ + ∣7∣ = 20$ . (Integer solution $x = 5$ exists)
Thus, $k \in {2, 3, 4, 5}$ all yield integer solutions. This matches 2, 3, 4, 5 in Column-II.

Part D: Solving the Differential Equation

We are given: $y^{'} = y + 1$ with initial condition $y (0) = 1$ .
Rewrite as a linear differential equation: $\frac{d y}{d x} - y = 1$
The Integrating Factor is: $I.F. = e^{\int - 1 d x} = e^{- x}$
Multiply both sides by the integrating factor: $y e^{- x} = \int 1 \cdot e^{- x} d x = - e^{- x} + C$

Part D: Finding the Particular Solution and $y (ln 2)$

The general solution is: $y = - 1 + C e^{x}$
Apply the initial condition $y (0) = 1$ :
$1 = - 1 + C e^{0} ⟹ C = 2$
So, the particular solution is: $y = 2 e^{x} - 1$
Now, find $y (ln 2)$ :
$y (ln 2) = 2 e^{l n 2} - 1 = 2 (2) - 1 = 3$
This matches 3 in Column-II.

Final Matching Summary

Let's summarize the final matches:
Part A $\to$ 1
Part B $\to$ 2, 4
Part C $\to$ 2, 3, 4, 5
Part D $\to$ 3

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Transcendental Equation

We begin with the function

f (x) = x e^{s i n x} - cos x

in the interval

(0, \frac{π}{2})

At the lower bound, we find

f (0) = 0 \cdot e^{0} - cos (0) = - 1

. At the upper bound, we have:

f (\frac{π}{2}) = \frac{π}{2} e^{s i n (π /2)} - cos (\frac{π}{2}) = \frac{π}{2} e - 0 > 0

Since the function transitions from negative to positive, the Intermediate Value Theorem guarantees at least one root.

To check for uniqueness, we calculate the derivative:

f^{'} (x) = e^{s i n x} + x e^{s i n x} cos x + sin x

For

x \in (0, \frac{π}{2})

, every term in

f^{'} (x)

is strictly positive. Thus, the function is strictly increasing, confirming that it crosses the x-axis exactly once.

Geometry of Intersecting Planes

We consider the system of planes: 1.

k x + 4 y + z = 0

4 x + k y + 2 z = 0

2 x + 2 y + z = 0

For these planes to intersect in a straight line, the determinant of the coefficient matrix must be zero:

Δ = k 42 4 k 2 121 = 0

Expanding the determinant, we obtain:

k (k - 4) - 4 (4 - 4) + 1 (8 - 2 k) = 0

Simplifying this expression leads to the quadratic equation

k^{2} - 6 k + 8 = 0

. Factoring gives

(k - 2) (k - 4) = 0

, resulting in the values $k = 2$ and $k = 4$ .

Absolute Value Functions

We analyze the equation

f (x) = ∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣ = 4 k

. This function represents the sum of distances from

x

to the points

1, 2, - 1, - 2

The minimum value of this sum occurs in the interval between the two central points,

[- 1, 1]

. Evaluating

f (x)

in this range:

f (x) = (1 - x) + (2 - x) + (x + 1) + (x + 2) = 6

For the equation to have solutions, we require

4 k \geq 6

, or

k \geq 1.5

. Consequently, for integer values of

k \geq 2

, the equation yields valid solutions.

Solving the Differential Equation

We are given the differential equation

y^{'} = y + 1

with the initial condition

y (0) = 1

. This is a linear first-order equation, which we rewrite as:

\frac{d y}{d x} - y = 1

The integrating factor is

e^{\int - 1 d x} = e^{- x}

. Multiplying both sides by

e^{- x}

and integrating:

\int \frac{d}{d x} (y e^{- x}) d x = \int e^{- x} d x

y e^{- x} = - e^{- x} + C \Rightarrow y = C e^{x} - 1

Applying the initial condition

y (0) = 1

, we find

C = 2

. The specific solution is

y = 2 e^{x} - 1

. Evaluating at

x = ln 2

y (ln 2) = 2 (e^{l n 2}) - 1 = 2 (2) - 1 = 3

Common Traps & FAQs

Wait, why do we check the derivative in Part A? Can't we just use the Intermediate Value Theorem?

The Intermediate Value Theorem only guarantees the existence of at least one root. To be absolutely certain there is exactly one root, we must prove the function is strictly monotonic. By showing

f^{'} (x) > 0

, we confirm the function never turns back, ensuring the root is unique.

Why does the determinant have to be zero in Part B?

For three planes to intersect in a single straight line, the system of equations must have infinitely many solutions. In a homogeneous system, this occurs if and only if the determinant of the coefficient matrix is zero, indicating that the equations are linearly dependent.

How do we find the minimum of the absolute value sum in Part C?

The function

f (x) = ∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣

is a sum of distances. The minimum occurs between the two middle critical points, which are

- 1

and

1

. By testing any value in the interval

[- 1, 1]

, we find the minimum value is

6

Animated Solution for Mathematics - Differential Equations: Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to the Matching Challenge

Part A: Analyzing $x e^{s i n x} - cos x = 0$

Part A: Applying the Intermediate Value Theorem

Part A: Proving Uniqueness via Monotonicity

Part B: Condition for Planes Intersecting in a Line

Part B: Solving the Determinant Equation

Part C: Analyzing the Absolute Value Function

Part C: Finding Integer Solutions for $k$

Part D: Solving the Differential Equation

Part D: Finding the Particular Solution and $y (ln 2)$

Final Matching Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Transcendental Equation

Geometry of Intersecting Planes

Absolute Value Functions

Solving the Differential Equation

Similar Questions

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to

Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then, the number of points of intersection of $y = y_{1} (x)$ and $y = y_{2} (x)$ is

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

Let $y = y (x)$ satisfies the equation $\frac{d y}{d x} - ∣ A ∣ = 0$ , for all $x > 0$ , where $A = y 02 sin x - 1 0 11 \frac{1}{x}$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is :

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is :

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is:

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Animated Solution for Mathematics - Differential Equations: Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to the Matching Challenge

Part A: Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xesinx−cosx=0

Part A: Applying the Intermediate Value Theorem

Part A: Proving Uniqueness via Monotonicity

Part B: Condition for Planes Intersecting in a Line

Part B: Solving the Determinant Equation

Part C: Analyzing the Absolute Value Function

Part C: Finding Integer Solutions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Part D: Solving the Differential Equation

Part D: Finding the Particular Solution and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(ln2)

Final Matching Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Transcendental Equation

Geometry of Intersecting Planes

Absolute Value Functions

Solving the Differential Equation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be solution of the following differential equation eydxdy​−2eysinx+sinxcos2x=0,y(2π​)=0. If y(0)=loge​(α+βe−2), then 4(α+β) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the equation esinycosydxdy​+esinycosx=cosx,y(0)=0; then 1+y(2π​)+2π​​y′(2π​)+2​1​y′′(2π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) satisfies the equation dxdy​−∣A∣=0, for all x>0, where A=​y02​sinx−10​11x1​​​. If y(π)=π+2, then the value of y(2π​) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution curve of the differential equation, (y2−x)dxdy​=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A solution curve of the differential equation (x2+xy+4x+2y+4)dxdy​−y2=0,x>0, passes through the point (1, 3). Then the solution curve

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution curve of the differential equation, (y2−x)dxdy​=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the differential equation, y2dx+(x−y1​)dy=0. If value of y is 1 when x = 1, then the value of x for which y = 2, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation 2 cos xdxdy​=sin 2x−4y sin x,x∈(0,2π​). If y(3π​)=0 then y′(4π​)+y(4π​) is equal to

Part A: Analyzing $x e^{s i n x} - cos x = 0$

Part C: Finding Integer Solutions for $k$

Part D: Finding the Particular Solution and $y (ln 2)$

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to

Let $y = y (x)$ satisfies the equation $\frac{d y}{d x} - ∣ A ∣ = 0$ , for all $x > 0$ , where $A = y 02 sin x - 1 0 11 \frac{1}{x}$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is :

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is :

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is:

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to