Match the statements/expressions given...

MathematicsLinear Differential EquationsJEE Advanced 2009Difficult

Match the statements/expressions given in Column-I with the values given in Column-II.

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)

Answer:P → 1, Q → 4, R → NaN, S → 3

Visualized Solution (English)

Part A: Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x e^{s i n x} - cos x = 0

Let $f (x) = x e^{s i n x} - cos x$ for $x \in (0, \frac{π}{2})$
Check boundaries: $f (0) = 0 \cdot e^{0} - cos 0 = - 1 < 0$
$f (\frac{π}{2}) = \frac{π}{2} e^{s i n (π /2)} - cos (\frac{π}{2}) = \frac{π}{2} e > 0$
By Intermediate Value Theorem, at least one root exists in $(0, \frac{π}{2})$ .

Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x)

Differentiate: $f^{'} (x) = \frac{d}{d x} (x e^{s i n x}) - \frac{d}{d x} (cos x)$
$f^{'} (x) = e^{s i n x} + x e^{s i n x} cos x + sin x$
For $x \in (0, \frac{π}{2})$ , $sin x > 0$ , $cos x > 0$ , and $e^{s i n x} > 0$
Thus, $f^{'} (x) > 0 ⟹ f (x)$ is strictly increasing.
Conclusion: Exactly one solution exists. (Matches 1)

Part B: Intersection of Planes

For planes to intersect in a line, the system must have infinite solutions.
Condition: $Δ = k 42 4 k 2 121 = 0$
Expand along the first row: $k (k - 4) - 4 (4 - 4) + 1 (8 - 2 k) = 0$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k in Part B

Simplify: $k^{2} - 4 k + 8 - 2 k = 0$
$k^{2} - 6 k + 8 = 0$
Factorize: $(k - 2) (k - 4) = 0$
Values of $k$ : $k = 2$ or $k = 4$ . (Matches 2, 4)

Part C: Absolute Value Equation

Let $f (x) = ∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣$
Minimum value occurs for $x \in [- 1, 1]$ : $f (0) = 1 + 2 + 1 + 2 = 6$
Equation: $f (x) = 4 k$ . For solutions to exist, $4 k \geq 6$ .

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

If $k = 1, 4 k = 4 < 6$ (No solution)
If $k = 2, 4 k = 8$ . $f (2) = 1 + 0 + 3 + 4 = 8$ (Integer solution $x = 2$ )
If $k = 3, 4 k = 12$ . $f (3) = 2 + 1 + 4 + 5 = 12$ (Integer solution $x = 3$ )
If $k = 4, 4 k = 16$ . $f (4) = 3 + 2 + 5 + 6 = 16$ (Integer solution $x = 4$ )
If $k = 5, 4 k = 20$ . $f (5) = 4 + 3 + 6 + 7 = 20$ (Integer solution $x = 5$ )
Matches: 2, 3, 4, 5

Part D: Differential Equation

Equation: $\frac{d y}{d x} - y = 1$ (Linear D.E.)
Integrating Factor (I.F.) $= e^{\int - 1 d x} = e^{- x}$
Solution: $y e^{- x} = \int 1 \cdot e^{- x} d x = - e^{- x} + C$
$y = - 1 + C e^{x}$

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y (ln 2)

Apply $y (0) = 1$ : $1 = - 1 + C e^{0} ⟹ C = 2$
Particular solution: $y = 2 e^{x} - 1$
Find $y (ln 2)$ : $y = 2 e^{l n 2} - 1 = 2 (2) - 1 = 3$
Matches: 3

Final Matching Summary

Final Match:
(A) $\to$ 1
(B) $\to$ 2, 4
(C) $\to$ 2, 3, 4, 5
(D) $\to$ 3

00:00 / 00:00

Conceptually Similar Problems

MathematicsVector (Cross) ProductJEE Advanced 2014Difficult

View in:English Hindi

Match List I with List II:

List-I

(P)

Let

y (x) = cos (3 cos^{- 1} x), x \in [- 1, 1], x \neq = \pm \frac{3}{2}

. Then

\frac{1}{y ( x )} {(x^{2} - 1) \frac{d ^{2} y ( x )}{d x ^{2}} + x \frac{d y ( x )}{d x}}

equals

(Q)

Let

A_{1}, A_{2}, \dots, A_{n} (n > 2)

be the vertices of a regular polygon of

n

sides with its centre at the origin. Let

a_{k}

be the position vector of the point

A_{k}, k = 1, 2, \dots, n

. If

∣ \sum_{k = 1}^{n - 1} (a_{k} \times a_{k + 1}) ∣ = ∣ \sum_{k = 1}^{n - 1} (a_{k} \cdot a_{k + 1}) ∣

, then the minimum value of

n

(R)

If the normal from the point

P (h, 1)

on the ellipse

\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

is perpendicular to the line

x + y = 8

, then the value of

h

(S)

Number of positive solutions satisfying the equation

tan^{- 1} (\frac{1}{2 x + 1}) + tan^{- 1} (\frac{1}{4 x + 1}) = tan^{- 1} (\frac{2}{x ^{2}})

List-II

(1)

(2)

(3)

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2010Difficult

View in:English Hindi

Match the statement in Column-I with the values in Column-II

List-I

(P)

A line from the origin meets the lines

\frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1}

and

\frac{x - \frac{8}{3}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1}

P

and

Q

respectively. If length

P Q = d

, then

d^{2}

(Q)

The values of

x

satisfying

tan^{- 1} (x + 3) - tan^{- 1} (x - 3) = sin^{- 1} (\frac{3}{5})

are

(R)

Non-zero vectors

a, b

and

c

satisfy

a \cdot b = 0

(b - a) \cdot (b + c) = 0

and

2∣ b + c ∣ = ∣ b - a ∣

. If

a = μ b + 4 c

, then the possible values of

μ

are

(S)

Let

f

be the function on

[- π, π]

given by

f (0) = 9

and

f (x) = \frac{s i n ( \frac{9 x}{2} )}{s i n ( \frac{x}{2} )}

for

x \neq = 0

. The value of

\frac{2}{π} \int_{- π}^{π} f (x) d x

List-II

(1)

-4

(2)

(3)

(4)

(5)

Answer:P → 5, P → 3, Q → 4, S → 3

MathematicsScalar Triple ProductJEE Advanced 2009Moderate

View in:English Hindi

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

(P)

Root(s) of the equation

2 sin^{2} θ + sin^{2} 2 θ = 2

(Q)

Points of discontinuity of the function

f (x) = [\frac{6 x}{π}] cos [\frac{3 x}{π}]

, where

[y]

denotes the largest integer less than or equal to

y

(R)

Volume of the parallelopiped with its edges represented by the vectors

\hat{i} + \hat{j}, \hat{i} + 2 \hat{j}

and

\hat{i} + \hat{j} + π \hat{k}

(S)

Angle between vector

a

and

b

where

a, b

and

c

are unit vectors satisfying

a + b + 3 c = 0

List-II

(1)

π /6

(2)

π /4

(3)

π /3

(4)

π /2

(5)

π

Answer:Q → 4, Q → NaN, R → 5, S → 3

MathematicsEquation of a PlaneJEE Advanced 2006Difficult

View in:English Hindi

Match the following :

List-I

(P)

Two rays

x + y = ∣ a ∣

and

a x - y = 1

intersects each other in the first quadrant in the interval

a \in (a_{0}, \infty)

, the value of

a_{0}

(Q)

Point

(α, β, γ)

lies on the plane

x + y + z = 2

. Let

a = α \hat{i} + β \hat{j} + γ \hat{k}

\hat{k} \times (\hat{k} \times a) = 0

, then

γ =

(R)

\int_{0}^{1} (1 - y^{2}) d y + \int_{1}^{0} (y^{2} - 1) d y

(S)

sin A sin B sin C + cos A cos B = 1

, then the value of

sin C =

List-II

(1)

(2)

4/3

(3)

\int_{0}^{1} 1 - x d x + \int_{- 1}^{0} 1 + x d x

(4)

Answer:P → 4, Q → 1, Q → 3, S → 4

MathematicsComponents of a VectorJEE Advanced 2015Difficult

View in:English Hindi

Match the following:

List-I

(P)

In a triangle

Δ X Y Z

, let

a, b

, and

c

be the lengths of the sides opposite to the angles

X, Y

and

Z

, respectively. If

2 (a^{2} - b^{2}) = c^{2}

and

λ = \frac{s i n ( X - Y )}{s i n Z}

, then possible values of

n

for which

cos (nπ λ) = 0

is (are)

(Q)

In a triangle

Δ X Y Z

, let

a, b

and

c

be the lengths of the sides opposite to the angles

X, Y

, and

Z

respectively. If

1 + cos 2 X - 2 cos 2 Y = 2 sin X sin Y

, then possible value(s) of

\frac{a}{b}

is (are)

(R)

R^{2}

, let

3 \hat{i} + \hat{j}, \hat{i} + 3 \hat{j}

and

β \hat{i} + (1 - β) \hat{j}

be the position vectors of

X, Y

and

Z

with respect to the origin

O

, respectively. If the distance of

Z

from the bisector of the acute angle of

O X

with

O Y

\frac{3}{2}

, then possible value(s) of

∣ β ∣

is (are)

(S)

Suppose that

F (α)

denotes the area of the region bounded by

x = 0, x = 2, y^{2} = 4 x

and

y = ∣ α x - 1∣ + ∣ α x - 2∣ + α x

, where

α \in {0, 1}

. Then the value(s) of

F (α) + \frac{8}{3} 2

, when

α = 0

and

α = 1

, is (are)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → NaN, Q → 1, P → 2, S → 5

MathematicsVariable Separable MethodJEE Advanced 2006Difficult

View in:English Hindi

Match the following :

List-I

(P)

\int_{0}^{π /2} (sin x)^{c o s x} (cos x cot x - lo g (sin x)^{s i n x}) d x

(Q)

Area bounded by

- 4 y^{2} = x

and

x - 1 = - 5 y^{2}

(R)

Cosine of the angle of intersection of curves

y = 3^{x - 1} lo g x

and

y = x^{x} - 1

(S)

Let

\frac{d y}{d x} = \frac{6}{x + y}

where

y (0) = 0

then value of

y

when

x + y = 6

List-II

(1)

(2)

(3)

6 ln 2

(4)

4/3

Answer:P → 1, Q → 4, R → 1, S → 3

MathematicsTrigonometric Ratios and IdentitiesJEE Advanced 2002Easy

View in:English Hindi

The number of integral values of $k$ for which the equation $7 cos x + 5 sin x = 2 k + 1$ has a solution is

(A)

4

(B)

8

(C)

10

(D)

12

Answer:B

MathematicsSolving Inverse Trigonometric EquationsJEE Advanced 2013Moderate

View in:English Hindi

Match List I with List II:

List-I

(P)

[\frac{1}{y ^{2}} (\frac{c o s ( t a n ^{- 1} y ) + y s i n ( t a n ^{- 1} y )}{c o t ( s i n ^{- 1} y ) + t a n ( s i n ^{- 1} y )})^{2} + y^{4}]^{1/2}

takes value

(Q)

cos x + cos y + cos z = 0 = sin x + sin y + sin z

then possible value of

cos \frac{x - y}{2}

(R)

cos (\frac{π}{4} - x) cos 2 x + sin x sin 2 x sec x = cos x sin 2 x sec x + cos (\frac{π}{4} + x) cos 2 x

then possible value of

sec x

(S)

cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6))

x \neq = 0

, then possible value of

x

List-II

(1)

\frac{1}{2} \frac{5}{3}

(2)

2

(3)

\frac{1}{2}

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Main 2011Easy

View in:English Hindi

The number of values of $k$ for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

(A)

2

(B)

1

(C)

zero

(D)

3

Answer:A

MathematicsCube Roots and nth Roots of UnityJEE Advanced 2014Moderate

View in:English Hindi

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

List-I

(P)

For each

z_{k}

there exists a

z_{j}

such that

z_{k} \cdot z_{j} = 1

(Q)

There exists a

k \in {1, 2, \dots, 9}

such that

z_{1} \cdot z = z_{k}

has no solution

z

in the set of complex numbers

(R)

\frac{∣1 - z _{1} ∣∣1 - z _{2} ∣ \dots ∣1 - z _{9} ∣}{10}

equals

(S)

1 - \sum_{k = 1}^{9} cos (\frac{2 k π}{10})

equals

List-II

(1)

True

(2)

False

(3)

(4)

Answer:P → 1, Q → 2, R → 3, S → 4

Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

List-II

Visualized Solution (English)

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

Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Part B: Intersection of Planes

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k in Part B

Part C: Absolute Value Equation

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

Part D: Differential Equation

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

Final Matching Summary

Conceptually Similar Problems

Match List I with List II:

List-I

List-II

Match the statement in Column-I with the values in Column-II

List-I

List-II

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

List-II

Match the following :

List-I

List-II

Match the following:

List-I

List-II

Match the following :

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is

Match List I with List II:

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of values of k for which the linear equations 4x+ky+2z=0, kx+4y+z=0 and 2x+2y+z=0 possess a non-zero solution is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…,9. Match the statements in List-I with those in List-II.

List-I

List-II

Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

List-II

Match List I with List II:

List-I

List-II

Match the statement in Column-I with the values in Column-II

List-I

List-II

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

List-II

Match the following :

List-I

List-II

Match the following:

List-I

List-II

Match the following :

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is

Match List I with List II:

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of values of k for which the linear equations 4x+ky+2z=0, kx+4y+z=0 and 2x+2y+z=0 possess a non-zero solution is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…,9. Match the statements in List-I with those in List-II.

List-I

List-II

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

Monotonicity of $f (x)$

Solving for $k$ in Part B

Checking Integer Solutions for $k$

Finding $y (ln 2)$

The number of integral values of $k$ for which the equation $7 cos x + 5 sin x = 2 k + 1$ has a solution is

The number of values of $k$ for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

The number of integral values of $k$ for which the equation $7 cos x + 5 sin x = 2 k + 1$ has a solution is

The number of values of $k$ for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.