Match List I with List II: List I P....

MathematicsVector (Cross) ProductJEE Advanced 2014Difficult

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

Visualized Solution (Hindi)

Introduction to Match List

List I contains four mathematical problems from different domains.
List II contains the possible numerical answers.
We will solve each sub-problem (P, Q, R, S) sequentially.

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

Given: $y (x) = cos (3 cos^{- 1} x)$
Let $cos^{- 1} x = θ \Rightarrow x = cos θ$
Then $y = cos 3 θ = 4 cos^{3} θ - 3 cos θ$
Substitute $x$ back: $y = 4 x^{3} - 3 x$

Part P: Differentiation

First derivative: $\frac{d y}{d x} = 12 x^{2} - 3$
Second derivative: $\frac{d ^{2} y}{d x ^{2}} = 24 x$

Part P: Final Evaluation

Expression: $\frac{1}{y} {(x^{2} - 1) (24 x) + x (12 x^{2} - 3)}$
$= \frac{1}{y} {24 x^{3} - 24 x + 12 x^{3} - 3 x}$
$= \frac{36 x ^{3} - 27 x}{4 x ^{3} - 3 x} = \frac{9 ( 4 x ^{3} - 3 x )}{4 x ^{3} - 3 x} = 9$
P matches with 4 (Value: 9)

Part Q: Regular Polygon Vectors

Angle between consecutive vectors $a_{k}$ and $a_{k + 1}$ is $θ = \frac{2 π}{n}$
$∣ a_{k} \times a_{k + 1} ∣ = R^{2} sin (\frac{2 π}{n})$
$∣ a_{k} \cdot a_{k + 1} ∣ = R^{2} cos (\frac{2 π}{n})$

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

Condition: $(n - 1) R^{2} sin (\frac{2 π}{n}) = (n - 1) R^{2} cos (\frac{2 π}{n})$
$\Rightarrow tan (\frac{2 π}{n}) = 1$
$\Rightarrow \frac{2 π}{n} = \frac{π}{4}$

Part Q: Minimum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

$n = 8$
Q matches with 3 (Value: 8)

Part R: Ellipse and Normal

Ellipse: $\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1 \Rightarrow a^{2} = 6, b^{2} = 3$
Line: $x + y = 8$ has slope $m_{1} = - 1$
Normal slope $m_{n} = \frac{- 1}{m _{1}} = 1$

Part R: Finding the Point

Normal slope: $\frac{a ^{2} y _{1}}{b ^{2} x _{1}} = \frac{6 y _{1}}{3 x _{1}} = \frac{2 y _{1}}{x _{1}}$
Set $\frac{2 y _{1}}{x _{1}} = 1 \Rightarrow x_{1} = 2 y_{1}$
Substitute into ellipse: $\frac{( 2 y _{1} ) ^{2}}{6} + \frac{y _{1}^{2}}{3} = 1 \Rightarrow y_{1}^{2} = 1 \Rightarrow y_{1} = 1, x_{1} = 2$

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

Normal at $(2, 1)$ : $\frac{6 x}{2} - \frac{3 y}{1} = 6 - 3 \Rightarrow x - y = 1$
Point $P (h, 1)$ lies on it: $h - 1 = 1 \Rightarrow h = 2$
R matches with 2 (Value: 2)

Part S: Inverse Trig Equation

Equation: $tan^{- 1} \frac{1}{2 x + 1} + tan^{- 1} \frac{1}{4 x + 1} = tan^{- 1} \frac{2}{x ^{2}}$
Using $tan^{- 1} A + tan^{- 1} B = tan^{- 1} \frac{A + B}{1 - A B}$
LHS: $tan^{- 1} \frac{\frac{1}{2 x + 1} + \frac{1}{4 x + 1}}{1 - \frac{1}{( 2 x + 1 ) ( 4 x + 1 )}} = tan^{- 1} \frac{6 x + 2}{8 x ^{2} + 6 x}$

Part S: Quadratic Equation

Equate: $\frac{6 x + 2}{8 x ^{2} + 6 x} = \frac{2}{x ^{2}} \Rightarrow \frac{3 x + 1}{4 x ^{2} + 3 x} = \frac{2}{x ^{2}}$
$3 x^{3} + x^{2} = 8 x^{2} + 6 x \Rightarrow 3 x^{2} - 7 x - 6 = 0$
$(3 x + 2) (x - 3) = 0$

Part S: Positive Solutions

Solutions: $x = 3, - \frac{2}{3}$
Only positive solution is $x = 3$
Number of positive solutions = 1
S matches with 1 (Value: 1)

Final Result

P -> 4 (9)
Q -> 3 (8)
R -> 2 (2)
S -> 1 (1)
The correct matching is (a).

00:00 / 00:00

Conceptually Similar Problems

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2010Difficult

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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}

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The values of

x

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

are

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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

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(1)

-4

(2)

(3)

(4)

(5)

Answer:P → 5, P → 3, Q → 4, S → 3

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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

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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

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, 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 following

List-I

(P)

\sum_{i = 1}^{\infty} tan^{- 1} (\frac{1}{2 i ^{2}}) = t

, then

tan t =

(Q)

Sides

a, b, c

of a triangle

A B C

are in AP and

cos θ_{1} = \frac{a}{b + c}

cos θ_{2} = \frac{b}{a + c}

cos θ_{3} = \frac{c}{a + b}

, then

tan^{2} (\frac{θ _{1}}{2}) + tan^{2} (\frac{θ _{3}}{2}) =

(R)

A line is perpendicular to

x + 2 y + 2 z = 0

and passes through

(0, 1, 0)

. The perpendicular distance of this line from the origin is

List-II

(1)

(2)

\frac{5}{3}

(3)

\frac{2}{3}

Answer:P → 1, Q → 3, R → 2

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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

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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

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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

MathematicsScalar Triple ProductJEE Advanced 2009Moderate

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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

MathematicsEquation of a PlaneJEE Advanced 2006Difficult

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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

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2014Difficult

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Following question has matching lists. Match list I with List II.

List-I

(P)

The number of polynomials

f (x)

with non-negative integer coefficients of degree

\leq 2

, satisfying

f (0) = 0

and

\int_{0}^{1} f (x) d x = 1

, is

(Q)

The number of points in the interval

[- 13, 13]

at which

f (x) = sin (x^{2}) + cos (x^{2})

attains its maximum value, is

(R)

\int_{- 2}^{2} \frac{3 x ^{2}}{1 + e ^{x}} d x

equals

(S)

\frac{\int _{- 1/2}^{1/2} c o s 2 x l o g ( \frac{1 + x}{1 - x} ) d x}{\int _{0}^{1/2} c o s 2 x l o g ( \frac{1 + x}{1 - x} ) d x}

List-II

(1)

(2)

(3)

(4)

Answer:P → 2, Q → 3, R → 1, S → 4

MathematicsAlgebraic Operations on MatricesJEE Advanced 2008Moderate

View in:English Hindi

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

Match List I with List II:

List-I

List-II

Visualized Solution (Hindi)

Introduction to Match List

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

Part P: Differentiation

Part P: Final Evaluation

Part Q: Regular Polygon Vectors

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

Part Q: Minimum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Part R: Ellipse and Normal

Part R: Finding the Point

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

Part S: Inverse Trig Equation

Part S: Quadratic Equation

Part S: Positive Solutions

Final Result

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