Let (x, y) be such that sin⁻¹(ax) +...

MathematicsSolving Inverse Trigonometric EquationsJEE Advanced 2007Moderate

Let $(x, y)$ be such that $sin^{- 1} (a x) + cos^{- 1} (y) + cos^{- 1} (b x y) = \frac{π}{2}$ . Match the statements in Column I with statements 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)

a = 1

and

b = 0

, then

(x, y)

(Q)

a = 1

and

b = 1

, then

(x, y)

(R)

a = 1

and

b = 2

, then

(x, y)

(S)

a = 2

and

b = 2

, then

(x, y)

List-II

(1)

lies on the circle

x^{2} + y^{2} = 1

(2)

lies on

(x^{2} - 1) (y^{2} - 1) = 0

(3)

lies on

y = x

(4)

lies on

(4 x^{2} - 1) (y^{2} - 1) = 0

Answer:P → 1, Q → 2, R → 1, S → 4

Visualized Solution (Hindi)

Analyze the Equation

Given equation: $sin^{- 1} (a x) + cos^{- 1} (y) + cos^{- 1} (b x y) = \frac{π}{2}$
Rearrange to isolate the cosine terms: $cos^{- 1} (y) + cos^{- 1} (b x y) = \frac{π}{2} - sin^{- 1} (a x)$

Apply Complementary Identity

Using identity: $\frac{π}{2} - sin^{- 1} (θ) = cos^{- 1} (θ)$
Equation becomes: $cos^{- 1} (y) + cos^{- 1} (b x y) = cos^{- 1} (a x)$

Trigonometric Substitution

Let $α = cos^{- 1} y$ , $β = cos^{- 1} (b x y)$ , and $γ = cos^{- 1} (a x)$
Then $α + β = γ ⟹ β = γ - α$
Taking cosine on both sides: $cos β = cos (γ - α)$

Expand and Substitute Back

Expansion: $cos β = cos γ cos α + sin γ sin α$
Substitute back: $b x y = (a x) (y) + 1 - a^{2} x^{2} 1 - y^{2}$

Isolate the Radical and Square

Isolate root: $(b - a) x y = 1 - a^{2} x^{2} 1 - y^{2}$
Square both sides: $(b - a)^{2} x^{2} y^{2} = (1 - a^{2} x^{2}) (1 - y^{2})$

Case A: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a = 1, b = 0

Substitute $a = 1, b = 0$ : $(0 - 1)^{2} x^{2} y^{2} = (1 - x^{2}) (1 - y^{2})$
$x^{2} y^{2} = 1 - x^{2} - y^{2} + x^{2} y^{2}$
Result: $x^{2} + y^{2} = 1$ (Matches statement p)

Case B: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a = 1, b = 1

Substitute $a = 1, b = 1$ : $(1 - 1)^{2} x^{2} y^{2} = (1 - x^{2}) (1 - y^{2})$
$0 = (1 - x^{2}) (1 - y^{2})$
Result: $(x^{2} - 1) (y^{2} - 1) = 0$ (Matches statement q)

Case C: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a = 1, b = 2

Substitute $a = 1, b = 2$ : $(2 - 1)^{2} x^{2} y^{2} = (1 - x^{2}) (1 - y^{2})$
$x^{2} y^{2} = 1 - x^{2} - y^{2} + x^{2} y^{2}$
Result: $x^{2} + y^{2} = 1$ (Matches statement p)

Case D: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a = 2, b = 2

Substitute $a = 2, b = 2$ : $(2 - 2)^{2} x^{2} y^{2} = (1 - 4 x^{2}) (1 - y^{2})$
$0 = (1 - 4 x^{2}) (1 - y^{2})$
Result: $(4 x^{2} - 1) (y^{2} - 1) = 0$ (Matches statement s)

Final Summary

Key Takeaways:
Identity $sin^{- 1} θ + cos^{- 1} θ = \frac{π}{2}$ is powerful for simplification.
Squaring is necessary to remove radicals but can introduce extraneous solutions; always check constraints.
Final Match:
A $\to$ p; B $\to$ q; C $\to$ p; D $\to$ s

00:00 / 00:00

Conceptually Similar Problems

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 2001Moderate

View in:English Hindi

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

(A)

1/2

(B)

(C)

-1/2

(D)

-1

Answer:B

MathematicsStandard Equations of Parabola, Ellipse, and HyperbolaJEE Advanced 2007Moderate

View in:English Hindi

Match the statements in Column I with the properties 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)

Two intersecting circles

(Q)

Two mutually external circles

(R)

Two circles, one strictly inside the other

(S)

Two branches of a hyperbola

List-II

(1)

have a common tangent

(2)

have a common normal

(3)

do not have a common tangent

(4)

do not have a common normal

Answer:P → 2, P → 2, Q → 3, Q → 3

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2016Moderate

View in:English Hindi

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

* Multiple Correct Options

(A)

the circle with radius

1/2 a

and centre

(1/2 a, 0)

for

a > 0, b \neq = 0

(B)

the circle with radius

- 1/2 a

and centre

(- 1/2 a, 0)

for

a < 0, b \neq = 0

(C)

the x-axis for

a \neq = 0, b = 0

(D)

the y-axis for

a = 0, b \neq = 0

Answer:A, B, C, D

MathematicsProperties of Inverse Trigonometric FunctionsJEE Main 2005Easy

View in:English Hindi

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

(A)

2 \sin 2\alpha

(B)

(C)

4 \sin^2 \alpha

(D)

- 4 \sin^2 \alpha

Answer:C

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

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

MathematicsDirector Circle and Family of CirclesJEE Main 2005Moderate

View in:English Hindi

If the circles $x^{2} + y^{2} + 2 a x + cy + a = 0$ and $x^{2} + y^{2} - 3 a x + d y - 1 = 0$ intersect in two distinct points $P$ and $Q$ then the line $5 x + b y - a = 0$ passes through $P$ and $Q$ for

(A)

exactly one value of

a

(B)

no value of

a

(C)

infinitely many values of

a

(D)

exactly two values of

a

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

MathematicsStandard Equations of Parabola, Ellipse, and HyperbolaJEE Advanced 2009Moderate

View in:English Hindi

Match the conics in Column I with the statements/expressions in Column II.

List-I

(P)

Circle

(Q)

Parabola

(R)

Ellipse

(S)

Hyperbola

List-II

(1)

The locus of the point

(h, k)

for which the line

h x + k y = 1

touches the circle

x^{2} + y^{2} = 4

(2)

Points

z

in the complex plane satisfying

∣ z + 2∣ - ∣ z - 2∣ = \pm 3

(3)

Points of the conic have parametric representation

x = 3 (\frac{1 - t ^{2}}{1 + t ^{2}}), y = \frac{2 t}{1 + t ^{2}}

(4)

The eccentricity of the conic lies in the interval

1 \leq x < \infty

(5)

Points

z

in the complex plane satisfying

R e (z + 1)^{2} = ∣ z ∣^{2} + 1

Answer:P → 1, S → 5, R → 3, Q → 4

MathematicsEquation of Tangent and NormalJEE Advanced 2016Moderate

View in:English Hindi

Let $R S$ be the diameter of the circle $x^{2} + y^{2} = 1$ , where $S$ is the point $(1, 0)$ . Let $P$ be a variable point (other than $R$ and $S$ ) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$ . The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $R S$ at point $E$ . Then the locus of $E$ passes through the point(s)

* Multiple Correct Options

(A)

(\frac{1}{3}, \frac{1}{3})

(B)

(\frac{1}{4}, \frac{1}{2})

(C)

(\frac{1}{3}, - \frac{1}{3})

(D)

(\frac{1}{4}, - \frac{1}{2})

Answer:A, C

List-I

List-II

Visualized Solution (Hindi)

Analyze the Equation

Apply Complementary Identity

Trigonometric Substitution

Expand and Substitute Back

Isolate the Radical and Square

Case A: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=1,b=0

Case B: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=1,b=1

Case C: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=1,b=2

Case D: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=2,b=2

Final Summary

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin−1(x−2x2​+4x3​−…)+cos−1(x2−2x4​+4x6​−…)=2π​ for 0<∣x∣<2​, then x equals

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List-I

List-II

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List-I

List-II

Match the following:

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List-I

List-II

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List-I

List-II

Match the following:

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List-II

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List-I

List-II

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List-I

List-II

Case A: $a = 1, b = 0$

Case B: $a = 1, b = 1$

Case C: $a = 1, b = 2$

Case D: $a = 2, b = 2$

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

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

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

If the circles $x^{2} + y^{2} + 2 a x + cy + a = 0$ and $x^{2} + y^{2} - 3 a x + d y - 1 = 0$ intersect in two distinct points $P$ and $Q$ then the line $5 x + b y - a = 0$ passes through $P$ and $Q$ for

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

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

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

If the circles $x^{2} + y^{2} + 2 a x + cy + a = 0$ and $x^{2} + y^{2} - 3 a x + d y - 1 = 0$ intersect in two distinct points $P$ and $Q$ then the line $5 x + b y - a = 0$ passes through $P$ and $Q$ for