PracticeHow it worksFeaturesPricingBlog Start practising free
Past Paper Solutions

Sets, Relations and Functions JEE 2007: Matrix-Match Solution

JEE Advanced 2007 Mathematics Sets, Relations and Functions Equations involving inverse trigonometric functions

By Founder, JEEnius - IIT Kanpur Alumni · Sep 16, 2026 · 4 min read

Hard 5 min target

Let (x,y) be such that

sin−1(ax)+cos−1(y)+cos−1(bxy)=π2

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

| Column I | Column II |
|---|---|
| (A) If a=1 and b=0, then (x,y) | (p) lies on the circle x2+y2=1 |
| (B) If a=1 and b=1, then (x,y) | (q) lies on (x2−1)(y2−1)=0 |
| (C) If a=1 and b=2, then (x,y) | (r) lies on y=x |
| (D) If a=2 and b=2, then (x,y) | (s) lies on (4x2−1)(y2−1)=0 |

Show answerAnswer

A-p, B-q, C-p, D-s

Explanation

Given equation is

sin−1(ax)+cos−1(y)+cos−1(bxy)=π2

Use the identity

cos−1(y)=π2−sin−1(y)

So the equation becomes

sin−1(ax)−sin−1(y)+cos−1(bxy)=0

Hence

cos−1(bxy)=sin−1(y)−sin−1(ax)

Taking cosine on both sides,

bxy=cos(sin−1(y)−sin−1(ax))

Using

cos(A−B)=cosAcosB+sinAsinB

we get

bxy=1−y21−a2x2+axy

Therefore

(1−y2)(1−a2x2)=(b−a)xy

Now examine each case.

For (A), a=1 and b=0.

(1−y2)(1−x2)=−xy

Squaring,

(1−y2)(1−x2)=x2y2

1−y2−x2+x2y2=x2y2

x2+y2=1

So (A) matches (p).

For (B), a=1 and b=1.

(1−y2)(1−x2)=0

Thus

(1−y2)(1−x2)=0

So

(x2−1)(y2−1)=0

Hence (B) matches (q).

For (C), a=1 and b=2.

(1−y2)(1−x2)=xy

Squaring,

(1−y2)(1−x2)=x2y2

1−y2−x2+x2y2=x2y2

x2+y2=1

So (C) matches (p).

For (D), a=2 and b=2.

(1−y2)(1−4x2)=0

Thus

(1−y2)(1−4x2)=0

This is equivalent to

(4x2−1)(y2−1)=0

So (D) matches (s).

Final matching is A-p, B-q, C-p, D-s.

Mathematics artwork for the article: Sets, Relations and Functions JEE 2007: Matrix-Match Solution

What are the correct matches for Sets, Relations and Functions JEE 2007?

The correct matches for this Sets, Relations and Functions JEE 2007 question are A–p, B–q, C–p and D–s. One square-root reduction gives all four answers, but squaring loses restrictions on the solutions.

The source is JEE 2007, Paper 2, Mathematics, a matrix-match question. The question bank files this inverse-trigonometric equation under Sets, Relations and Functions and tags it hard. That tag is not evidence of student error rates.

Real pairs satisfy:

arcsin(ax)+arccos(y)+arccos(bxy)=π2.

For each parameter choice, identify which listed relation necessarily holds:

  • A: (a,b)=(1,0)
  • B: (a,b)=(1,1)
  • C: (a,b)=(1,2)
  • D: (a,b)=(2,2)

The available relations are:

  • p: x2+y2=1
  • q:
(x2−1)(y2−1)=0
  • r: y=x
  • s:
(4x2−1)(y2−1)=0.

How do you reduce the three inverse functions once?

Begin with the real-domain conditions:

|ax|≤1,|y|≤1,|bxy|≤1.

The official method replaces the middle inverse cosine, isolates the remaining one, and takes cosine. Keep the principal-angle restriction alongside the algebra so that a necessary relation is not mistaken for the complete solution set.

Use:

arccos(y)=π2−arcsin(y).

Substitution gives:

arcsin(ax)−arcsin(y)+arccos(bxy)=0.

Rearrange, then take cosine:

arccos(bxy)=arcsin(y)−arcsin(ax),
bxy=cos(arcsin(y)−arcsin(ax)).

Expand using the cosine-difference identity:

cos(U−V)=cosUcosV+sinUsinV.

The square roots are nonnegative because cosine is nonnegative throughout the principal range of inverse sine:

arcsin(t)∈[−π2,π2],cos(arcsint)=1−t2.

Therefore:

bxy=1−y21−a2x2+axy.

This gives the common reduction:

(1−y2)(1−a2x2)=(b−a)xy.

The isolated inverse cosine is nonnegative, so the original equation also requires:

arcsin(y)−arcsin(ax)≥0⟹y≥ax.

Inverse sine is increasing, which justifies the implication. Taking cosine and later squaring are not automatically reversible: retain the domains, this angle condition and the sign of the unsquared right-hand side.

Why does A match p and B match q?

A gives the circle relation, while B gives a zero product. Both are necessary consequences of the original equation, subject to the domain and branch restrictions above.

For A, substitute:

a=1,b=0.

The reduction becomes:

(1−y2)(1−x2)=−xy.

Squaring and expanding:

(1−y2)(1−x2)=x2y2,
1−y2−x2+x2y2=x2y2.

Cancel the common product: x2+y2=1.

Thus A–p. The unsquared equation also requires: xy≤0.

For B, substitute: a=b=1.

Then:

(1−y2)(1−x2)=0,
(1−y2)(1−x2)=0⟺(x2−1)(y2−1)=0.

Both factors change sign, leaving their product unchanged. Hence B–q.

A zero product means at least one factor is zero. It does not say that the two variables are equal, so it does not justify selecting r.

Why does C match p and D match s?

C produces the same circle as A, but with the opposite product-sign restriction. D produces the listed relation s. The matches identify necessary relations, not unrestricted solution sets.

For C, substitute:

a=1,b=2.

Then:

(1−y2)(1−x2)=xy.

Square and expand:

(1−y2)(1−x2)=x2y2,
1−y2−x2+x2y2=x2y2,

x2+y2=1. Thus C–p. However, C requires a nonnegative product, whereas A requires a nonpositive one:

C: xy≥0,A: xy≤0.

For D, substitute: a=b=2.

The reduction gives:

(1−y2)(1−4x2)=0.

Consequently:

(1−y2)(1−4x2)=0⟺(4x2−1)(y2−1)=0.

Hence D–s. The supplied official answer is A–p, B–q, C–p, D–s.

“Lies on” asserts a necessary relation. It does not claim that every point of the circle, or every point on the listed lines, satisfies the original equation.

For a concrete check, take C at: (x,y)=(1,0).

This point satisfies the circle equation, but the original left side is:

arcsin(1)+arccos(0)+arccos(0)=3π2≠π2.

How can an invalid cancellation produce the wrong match r?

An unjustified zero assumption can produce B–r. Equal parameters make the square-root product vanish in the derived equation, not the remaining inverse-cosine term.

Return to B:

arccos(xy)=arcsin(y)−arcsin(x).

The invalid shortcut is:

arccos(xy)=unjustified0⟹arcsin(y)=arcsin(x)⟹y=x.

The missing condition is:

arccos(xy)=0⟺xy=1.

Equal parameters do not impose that condition on every solution. Setting the inverse cosine to zero therefore restricts the problem without justification.

Check this valid B pair: (x,y)=(−1,0).

Substitution gives:

arcsin(−1)+arccos(0)+arccos(0)=−π2+π2+π2=π2.

Yet: y≠x.

One valid counterexample rejects a relation claimed to hold for every solution. The shortcut is an unjustified cancellation or zero assumption, not an inverse-trigonometric identity.

Which two practice questions test the same domain and branch checks?

A domain question tests the input restrictions; an exact-solution-set question also tests the principal branches. Both below are original practice based on this PYQ, not additional verified past-paper questions.

Question 1. Find the real domain of:

F(x,y)=arcsin(2x)+arccos(y)+arccos(2xy).

Worked answer. Require:

|2x|≤1,|y|≤1,|2xy|≤1.

The first two imply the third: |2xy|=|2x||y|≤1.

Hence:

DomF=[−12,12]×[−1,1].

Question 2. Find the necessary quadratic relation and the part that actually satisfies:

arcsin(2x)+arccos(y)+arccos(0)=π2.

Worked answer. The same reduction gives:

(1−y2)(1−4x2)=−2xy.

Squaring and expanding:

1−y2−4x2+4x2y2=4x2y2,
4x2+y2=1.

For the exact restriction, return to the original equation:

arccos(0)=π2⟹arcsin(2x)=−arccos(y).

Principal ranges require:

−12≤x≤0,0≤y≤1.

These restrictions are sufficient on the ellipse. Write:

2x=−sinθ,y=cosθ,0≤θ≤π2.

Then:

arcsin(−sinθ)+arccos(cosθ)+π2=−θ+θ+π2=π2.

When asked for the exact set, retain the domain, sign and branch checks. Give this restricted ellipse arc, not the whole ellipse.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

For a worked example of the same idea, see Solutions JEE 2025: Osmotic Pressure and Molar Mass.

Frequently asked questions

What are the correct matches for Sets, Relations and Functions JEE 2007?

The correct matches are A–p, B–q, C–p and D–s. A and C give x² + y² = 1, B gives (x² − 1)(y² − 1) = 0, and D gives (4x² − 1)(y² − 1) = 0. These are necessary relations, not unrestricted solution sets.

How do I reduce the inverse-trigonometric equation in this JEE 2007 question?

Replace arccos(y) with π/2 − arcsin(y), isolate arccos(bxy), and take cosine using the cosine-difference identity. This gives sqrt((1 − y²)(1 − a²x²)) = (b − a)xy. Retain |ax| ≤ 1, |y| ≤ 1, |bxy| ≤ 1, the branch condition y ≥ ax, and the nonnegative sign of the right-hand side.

Why do A and C match the same circle despite having different parameters?

For A, the square-root reduction equals −xy; for C, it equals xy. Squaring either equation gives x² + y² = 1. However, A requires xy ≤ 0 while C requires xy ≥ 0, and both still need the domain and principal-branch checks.

Why is y = x not a correct match for B?

For B, equal parameters make the square-root product zero; they do not force arccos(xy) to be zero. The pair (x, y) = (−1, 0) satisfies the original equation but has y ≠ x. This counterexample rules out r as a relation that necessarily holds.

domain and rangeinverse trigonometryjee 2007matrix matchsets and relations

Practise this with JEEnius AI

25 years of PYQs, AI doubt solving, and the 2027 prediction paper.

Start practising free