Match the following : Column I (A)...

MathematicsVariable Separable MethodJEE Advanced 2006Difficult

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

Visualized Solution (Hindi)

Analysis of Part (A): The Integral

Let $f (x) = (sin x)^{c o s x}$
Differentiating using $\frac{d}{d x} (u^{v}) = u^{v} [\frac{v}{u} u^{'} + v^{'} lo g u]$ :
$\frac{d}{d x} (sin x)^{c o s x} = (sin x)^{c o s x} [cos x cot x - sin x lo g (sin x)]$
Notice that $lo g (sin x)^{s i n x} = sin x lo g (sin x)$
Thus, the integrand is exactly $\frac{d}{d x} (sin x)^{c o s x}$

Evaluating Part (A)

$I = [(sin x)^{c o s x}]_{0}^{\frac{π}{2}}$
Evaluate at upper limit: $(sin \frac{π}{2})^{c o s \frac{π}{2}} = 1^{0} = 1$
Evaluate at lower limit: $(sin 0)^{c o s 0} = 0^{1} = 0$
Result: $1 - 0 = 1$
(A) matches (p)

Part (B): Intersecting Parabolas

Curves: $x = - 4 y^{2}$ and $x = 1 - 5 y^{2}$
To find intersection, set $x_{1} = x_{2}$ :
$- 4 y^{2} = 1 - 5 y^{2}$
$y^{2} = 1 \Rightarrow y = \pm 1$
Intersection points are $(- 4, 1)$ and $(- 4, - 1)$

Part (B): Area Calculation

Area $= \int_{- 1}^{1} (x_{r i g h t} - x_{l e f t}) d y$
Area $= \int_{- 1}^{1} (1 - 5 y^{2} - (- 4 y^{2})) d y$
Area $= \int_{- 1}^{1} (1 - y^{2}) d y$
Area $= [y - \frac{y ^{3}}{3}]_{- 1}^{1} = (1 - \frac{1}{3}) - (- 1 + \frac{1}{3}) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$
(B) matches (s)

Part (C): Intersection of Curves

Curves: $y_{1} = 3^{x - 1} lo g x$ and $y_{2} = x^{x} - 1$
Intersection at $x = 1$ : $y_{1} (1) = 0, y_{2} (1) = 0$
Slope $m_{1} = \frac{d y _{1}}{d x} = 3^{x - 1} \cdot \frac{1}{x} + lo g x \cdot 3^{x - 1} ln 3$
At $x = 1, m_{1} = 1$
Slope $m_{2} = \frac{d y _{2}}{d x} = x^{x} (1 + ln x)$ . At $x = 1, m_{2} = 1$
Angle $θ = 0 \Rightarrow cos θ = 1$
(C) matches (p)

Part (D): Differential Equation Substitution

Equation: $\frac{d y}{d x} = \frac{6}{x + y}$
Let $x + y = v \Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x}$
Substitute: $\frac{d v}{d x} - 1 = \frac{6}{v}$
$\frac{d v}{d x} = \frac{v + 6}{v}$

Part (D): Solving and Final Answer

Separate variables: $\int \frac{v}{v + 6} d v = \int d x$
$\int (1 - \frac{6}{v + 6}) d v = x + c$
$v - 6 ln (v + 6) = x + c$
Substitute $v = x + y$ : $y - 6 ln (x + y + 6) = c$
Using $(0, 0)$ : $0 - 6 ln 6 = c \Rightarrow c = - 6 ln 6$
When $x + y = 6$ : $y - 6 ln (12) = - 6 ln 6$
$y = 6 (ln 12 - ln 6) = 6 ln 2$
(D) matches (r)

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Conceptually Similar Problems

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2007Easy

View in:English Hindi

Match the integrals in Column I with the values in Column II.

List-I

(P)

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

(Q)

\int_{0}^{1} \frac{d x}{1 - x ^{2}}

(R)

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

(S)

\int_{1}^{2} \frac{d x}{x x ^{2} - 1}

List-II

(1)

\frac{1}{2} lo g (\frac{2}{3})

(2)

2 lo g (\frac{2}{3})

(3)

\frac{π}{3}

(4)

\frac{π}{2}

Answer:P → 4, Q → 4, R → 1, S → 3

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

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

MathematicsArea Bounded by CurvesJEE Advanced 2008Moderate

View in:English Hindi

The area of the region between the curves $y = \frac{1 + s i n x}{c o s x}$ and $y = \frac{1 - s i n x}{c o s x}$ bounded by the lines $x = 0$ and $x = π /4$ is

(A)

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

(B)

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

(C)

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

(D)

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

Answer:B

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

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

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1984Easy

View in:English Hindi

Evaluate the following $\int_{0}^{1} \frac{x s i n ^{- 1} x}{1 - x ^{2}} d x$ .

Answer:1

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

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1995Moderate

View in:English Hindi

Evaluate the definite integral : $\int_{- 1/ 3}^{1/ 3} (\frac{x ^{4}}{1 - x ^{4}}) cos^{- 1} (\frac{2 x}{1 + x ^{2}}) d x$

Answer:

\frac{π}{12} [π + 3 lo g_{e} (2 + 3) - 43]

MathematicsArea Bounded by CurvesJEE Advanced 2002Easy

View in:English Hindi

The area bounded by the curves $y = ∣ x ∣ - 1$ and $y = - ∣ x ∣ + 1$ is

(A)

(B)

(C)

22

(D)

Answer:B

Match the following :

List-I

List-II

Visualized Solution (Hindi)

Analysis of Part (A): The Integral

Evaluating Part (A)

Part (B): Intersecting Parabolas

Part (B): Area Calculation

Part (C): Intersection of Curves

Part (D): Differential Equation Substitution

Part (D): Solving and Final Answer

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