Match the Statements/Expressions in...

MathematicsAlgebraic Operations on MatricesJEE Advanced 2008Moderate

Match the Statements/Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

(P)

The minimum value of

\frac{x ^{2} + 2 x + 4}{x + 2}

(Q)

Let A and B be

3 \times 3

matrices of real numbers, where A is symmetric, B is skew-symmetric, and

(A + B) (A - B) = (A - B) (A + B)

. If

(A B)^{t} = (- 1)^{k} A B

, where

(A B)^{t}

is the transpose of the matrix AB, then the possible values of

k

are

(R)

Let

a = lo g_{3} lo g_{3} 2

. An integer

k

satisfying

1 < 2^{- k + 3^{- a}} < 2

, must be less than

(S)

sin θ = cos ϕ

, then the possible values of

\frac{1}{π} (θ \pm ϕ - \frac{π}{2})

are

List-II

(1)

(2)

(3)

(4)

Answer:P → 3, Q → 4, R → 4, P → 3

Visualized Solution (Hindi)

Introduction to the Matching Problem

Objective: Match Column I expressions with Column II values.
Part A: Minimum value of $f (x) = \frac{x ^{2} + 2 x + 4}{x + 2}$
Part B: Matrix properties involving symmetry and commutativity.
Part C: Logarithmic inequality for an integer $k$ .
Part D: General solution of a trigonometric equation.

Part A: Simplifying the Expression

Rewrite $f (x) = \frac{x ^{2} + 2 x + 4}{x + 2}$ by dividing the terms.
$f (x) = \frac{( x + 2 ) ^{2} + 4 - 2 ( x + 2 )}{x + 2}$ is not as clean as:
$f (x) = (x + 2) + \frac{4}{x + 2} - 2$
Assume $x + 2 > 0$ for a local minimum.

Part A: Applying AM-GM Inequality

Apply AM-GM on $(x + 2)$ and $\frac{4}{x + 2}$ :
$\frac{( x + 2 ) + \frac{4}{x + 2}}{2} \geq (x + 2) \cdot \frac{4}{x + 2}$
$(x + 2) + \frac{4}{x + 2} \geq 2 \cdot 4 = 4$
Minimum value $y_{min} = 4 - 2 = 2$
Match: (A) $\to$ (r)

Part B: Matrix Commutativity

Given: $A^{t} = A$ (Symmetric) and $B^{t} = - B$ (Skew-symmetric).
Expand $(A + B) (A - B) = (A - B) (A + B)$ :
$A^{2} - A B + B A - B^{2} = A^{2} + A B - B A - B^{2}$
Simplify: $2 B A = 2 A B ⟹ A B = B A$

Part B: Finding the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Calculate $(A B)^{t} = B^{t} A^{t} = (- B) A = - B A$
Since $A B = B A$ , then $(A B)^{t} = - A B$
Given $(A B)^{t} = (- 1)^{k} A B ⟹ (- 1)^{k} = - 1$
Therefore, $k$ must be an odd integer.
Possible values for $k$ : $1, 3$ . Match: (B) $\to$ (q), (s)

Part C: Logarithmic Simplification

Given $a = lo g_{3} lo g_{3} 2 ⟹ 3^{a} = lo g_{3} 2$
Calculate $3^{- a} = \frac{1}{l o g _{3} 2} = lo g_{2} 3$
Substitute into inequality: $1 < 2^{- k + l o g_{2} 3} < 2$

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

Take $lo g_{2}$ on all sides: $0 < - k + lo g_{2} 3 < 1$
Rearrange: $lo g_{2} 3 - 1 < k < lo g_{2} 3$
Since $lo g_{2} 3 \approx 1.58$ , then $0.58 < k < 1.58$
The integer satisfying this is $k = 1$ .
$k = 1$ is less than $2$ and $3$ . Match: (C) $\to$ (r), (s)

Part D: Trigonometric Equation

Given $sin θ = cos ϕ ⟹ cos (\frac{π}{2} - θ) = cos ϕ$
General solution: $\frac{π}{2} - θ = 2 nπ \pm ϕ$
Rearrange: $θ \pm ϕ - \frac{π}{2} = - 2 nπ$
Expression: $\frac{1}{π} (θ \pm ϕ - \frac{π}{2}) = - 2 n$ (Even integers)

Part D: Final Matches

Possible values are even integers: $..., - 2, 0, 2, 4, ...$
From Column II, the values are $0$ and $2$ .
Match: (D) $\to$ (p), (r)
Final Summary: A-r; B-q,s; C-r,s; D-p,r

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Conceptually Similar Problems

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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

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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

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Match the following:

List-I

(P)

R^{2}

, if the magnitude of the projection vector of the vector

α \hat{i} + β \hat{j}

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

3

and if

α = 2 + 3 β

, then possible value of

∣ α ∣

is/are

(Q)

Let

a

and

b

be real numbers such that the function

f (x) = {- 3 a x^{2} - 2, b x + a^{2}, x < 1 x \geq 1

if differentiable for all

x \in R

. Then possible value of

a

is (are)

(R)

Let

ω \neq = 1

be a complex cube root of unity. If

(3 - 3 ω + 2 ω^{2})^{4 n + 3} + (2 + 3 ω - 3 ω^{2})^{4 n + 3} + (- 3 + 2 ω + 3 ω^{2})^{4 n + 3} = 0

, then possible value (s) of

n

is (are)

(S)

Let the harmonic mean of two positive real numbers

a

and

b

be 4. If

q

is a positive real number such that

a, 5, q, b

is an arithmetic progression, then the value(s) of

∣ q - a ∣

is (are)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → 2, P → 2, R → NaN, Q → 5

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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

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

List-I

(P)

a = \hat{j} + 3 \hat{k}

b = - \hat{j} + 3 \hat{k}

and

c = 23 \hat{k}

form a triangle, then the internal angle of the triangle between

a

and

b

(Q)

\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}

, then the value of

f (\frac{π}{6})

(R)

The value of

\frac{π ^{2}}{l n 3} \int_{7/6}^{5/6} sec (π x) d x

(S)

The maximum value of

Arg (\frac{1}{1 - z})

for

∣ z ∣ = 1, z \neq = 1

is given by

List-II

(1)

\frac{π}{6}

(2)

\frac{2 π}{3}

(3)

\frac{π}{3}

(4)

π

(5)

\frac{π}{2}

Answer:P → 2, Q → 1, R → 4, S → 5

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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

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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

MathematicsLinear Differential EquationsJEE Advanced 2009Difficult

View in:English Hindi

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

MathematicsProperties of DeterminantsJEE Advanced 1988Moderate

View in:English Hindi

The values of $θ$ lying between $θ = 0$ and $θ = π /2$ and satisfying the equation $1 + sin^{2} θ sin^{2} θ sin^{2} θ cos^{2} θ 1 + cos^{2} θ cos^{2} θ 4 sin 4 θ 4 sin 4 θ 1 + 4 sin 4 θ = 0$ are

* Multiple Correct Options

(A)

7 π /24

(B)

5 π /24

(C)

11 π /24

(D)

π /24

Answer:A, C

MathematicsProperties of DeterminantsJEE Advanced 2000Moderate

View in:English Hindi

Prove that for all values of $θ$ , $sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) sin 2 θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) = 0$ .

Answer:Proof by operation

R_{2} \to R_{2} + R_{3}

List-I

List-II

Visualized Solution (Hindi)

Introduction to the Matching Problem

Part A: Simplifying the Expression

Part A: Applying AM-GM Inequality

Part B: Matrix Commutativity

Part B: Finding the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Part C: Logarithmic Simplification

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

Part D: Trigonometric Equation

Part D: Final Matches

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Part B: Finding the value of $k$

Part C: Solving for $k$

The values of $θ$ lying between $θ = 0$ and $θ = π /2$ and satisfying the equation $1 + sin^{2} θ sin^{2} θ sin^{2} θ cos^{2} θ 1 + cos^{2} θ cos^{2} θ 4 sin 4 θ 4 sin 4 θ 1 + 4 sin 4 θ = 0$ are

Prove that for all values of $θ$ , $sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) sin 2 θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) = 0$ .

The values of $θ$ lying between $θ = 0$ and $θ = π /2$ and satisfying the equation $1 + sin^{2} θ sin^{2} θ sin^{2} θ cos^{2} θ 1 + cos^{2} θ cos^{2} θ 4 sin 4 θ 4 sin 4 θ 1 + 4 sin 4 θ = 0$ are

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