Mixed Mathematics JEE 2015: Full Matrix Match Solution

What was the mixed mathematics JEE 2015 matrix-match question in JEE Advanced?

The correct matching for this mixed mathematics JEE 2015 question is (A)→P,R,S; (B)→P; (C)→P,Q; (D)→S,T. It presents four independent situations (A) to (D) each requiring matching of possible numerical values from Column II.

(A) uses the triangle side-angle relation $2 (a^{2} - b^{2}) = c^{2}$ and defines $λ = \frac{\sin (X - Y)}{\sin Z}$ , then asks for possible integer values of $n$ that satisfy $\cos (n π λ) = 0$ .

(B) supplies a trigonometric equation in angles $X$ and $Y$ and asks for possible ratios $a / b$ .

(C) supplies the position vectors of points $X$ , $Y$ and $Z$ with respect to origin $O$ and requires the distance of $Z$ from the acute-angle bisector of $O X$ and $O Y$ to equal $3 / \sqrt{2}$ , then asks for possible values of $| β |$ .

(D) defines $F (α)$ as the area bounded by $y^{2} = 4 x$ , $x = 0$ , $x = 2$ and the piecewise linear function $y = | α x - 1 | + | α x - 2 | + α x$ for $α = 0$ and $α = 1$ , then asks for the values taken by $F (α) + \frac{8}{3} \sqrt{2}$ .

How do you solve the 2015 mixed mathematics JEE question using the official method?

For (A), apply the sine rule so sides are proportional to sines: $a = k \sin X$ , $b = k \sin Y$ , $c = k \sin Z$ . Substitute into the given side relation:

2 k^{2} (\sin^{2} X - \sin^{2} Y) = k^{2} \sin^{2} Z

Cancel $k^{2}$ :

2 (\sin^{2} X - \sin^{2} Y) = \sin^{2} Z

Replace the difference of squares with the identity $\sin^{2} X - \sin^{2} Y = \sin (X + Y) \sin (X - Y)$ . In a triangle $X + Y = π - Z$ , so $\sin (X + Y) = \sin Z$ . This simplifies to:

2 \sin Z \cdot \sin (X - Y) = \sin^{2} Z

Divide by $\sin Z$ (non-zero):

\frac{\sin (X - Y)}{\sin Z} = \frac{1}{2}

Thus $λ = 1 / 2$ and the equation becomes $\cos (n π / 2) = 0$ . This holds only for odd integers $n$ . From the options the matches are therefore 1, 3 and 5, so (A) matches P, R, S.

For (B), start with the given equation $1 + \cos 2 X - 2 \cos 2 Y = 2 \sin X \sin Y$ . Replace $1 + \cos 2 X$ by $2 \cos^{2} X$ . Use double-angle substitutions to reach the quadratic $t^{2} + t - 2 = 0$ . The positivity argument on sines in a triangle yields $a / b = 1$ , so (B) matches P.

For (C), the position vectors are $X = (\sqrt{3}, 1)$ , $Y = (1, \sqrt{3})$ and $Z = (β, 1 - β)$ . Vector $O X$ makes $30^{\circ}$ with the positive x-axis and $O Y$ makes $60^{\circ}$ , so the acute-angle bisector is the line at $45^{\circ}$ , which is $y = x$ .

Diagram with origin O, position vector of X at (√3,1), Y at (1,√3), Z at (β,1-β), acute-angle bisector drawn as the line y=x at 45° with all labels and axes clearly marked.

The distance of point $(β, 1 - β)$ from line $y = x$ is:

\frac{| (1 - β) - β |}{\sqrt{2}} = \frac{| 1 - 2 β |}{\sqrt{2}}

Set equal to the given distance $3 / \sqrt{2}$ : $| 1 - 2 β | = 3$

Solve the two cases: $β = - 1$ or $β = 2$ . Thus $| β | = 1, 2$ and (C) matches P, Q.

For (D), first set $α = 0$ . The linear piece collapses to the constant $y = 3$ . The area between this line and the upper branch $y = 2 \sqrt{x}$ of the parabola from $x = 0$ to $x = 2$ is:

F (0) = \int_{0}^{2} (3 - 2 \sqrt{x}) d x = {[3 x - \frac{4}{3} x^{3 / 2}]}_{0}^{2} = 6 - \frac{8}{3} \sqrt{2}

Therefore $F (0) + \frac{8}{3} \sqrt{2} = 6$ , matching T.

Now set $α = 1$ . The linear expression changes at the points where the arguments of the absolute values are zero. For $0 \leq x \leq 1$ , $y = 3 - x$ . For $1 \leq x \leq 2$ , $y = x + 1$ . Split the integral:

F (1) = \int_{0}^{1} (3 - x - 2 \sqrt{x}) d x + \int_{1}^{2} (x + 1 - 2 \sqrt{x}) d x

First integral evaluates to $7 / 6$ . Second evaluates to $23 / 6 - (8 / 3) \sqrt{2}$ . Adding gives:

F (1) = 5 - \frac{8}{3} \sqrt{2}

Thus $F (1) + \frac{8}{3} \sqrt{2} = 5$ , matching S. So (D) matches S, T.

(A)→P,R,S; (B)→P; (C)→P,Q; (D)→S,T.

What single algebraic slip ruins the matching in part A?

Students who reach the line $2 \sin Z \cdot \sin (X - Y) = \sin^{2} Z$ sometimes drop the factor of 2 when dividing. They write $\sin (X - Y) / \sin Z = 1$ instead of $1 / 2$ , so $λ = 1$ .

The equation then becomes $\cos (n π \cdot 1) = (- 1)^{n} = 0$ . This equation has no integer solution. The student therefore matches none of the odd values 1, 3 or 5 and loses P, R and S completely.

The mistake is purely procedural. The student knows the sin-squared identity and the supplementary-angle substitution but fails to carry the coefficient through the simplification. One coefficient error collapses the entire column.

Which practice questions should I solve right now to lock in these methods?

Triangle trigonometry: In $Δ A B C$ with sides $a, b, c$ opposite angles $A, B, C$ , suppose $a^{2} + b^{2} = 2 c^{2}$ . Convert sides to sines first, apply the companion identity for $\sin^{2} A + \sin^{2} B$ , and find the exact value of $\cos (A - B)$ . Hint: the official 2015 replacement $X + Y = π - Z$ works identically here.

Vectors in plane: Position vectors of points $P$ and $Q$ are $2 𝐢 + 𝐣$ and $𝐢 + 2 𝐣$ . A point $R$ has position vector $(μ, 3 - μ)$ . Find possible $| μ |$ if the distance from $R$ to the acute-angle bisector of $O P$ and $O Q$ equals $\sqrt{2}$ . Hint: locate the angles, confirm the bisector is still $y = x$ , then apply the absolute-value distance formula exactly as in the 2015 vector part.

Area under curves: Compute the area bounded by $y^{2} = 4 x$ , the x-axis from $x = 0$ to $x = 4$ , and the graph of $y = | x - 1 | + | x - 3 | + x / 2$ . Split the integral at the points where the expression inside each modulus is zero. Hint: treat the upper parabola branch $y = 2 \sqrt{x}$ as the reference curve and evaluate the piecewise linear pieces separately before adding, following the identical interval-splitting steps used for $α = 1$ in the original question.

Search every JEE Advanced paper from 2007 in the past-paper archive by chapter to find more matrix matches that mix the same three topics.

Which core techniques does this mixed mathematics JEE 2015 question reinforce?

Conversion of sides to sines via sine rule before substituting any side relation.
Standard identity $\sin^{2} A - \sin^{2} B = \sin (A + B) \sin (A - B)$ and its companion with $X + Y = π - Z$ .
Coordinate geometry route for angle bisector: locate angles of given vectors then average for the 45° line $y = x$ .
Strict interval splitting at points where expression inside modulus equals zero before integrating against parabola $y = 2 \sqrt{x}$ .

These four steps appear in many later problems. Master the sequence once and the combined questions lose their terror.

Why do mixed-concept matrix matches reward method over memory?

Mixed-concept matrix matches reward method over memory because each part tests one clean concept but the matrix format forces simultaneous accuracy across trigonometry, vectors and calculus. The 900-second expected time indicates deliberate design to punish procedural slips such as the one shown above.

The same year also tested mixed concepts in an ellipse common-tangent question. Work through it next at 2015 coordinate-geometry ellipse tangent question for additional mixed practice.

If a similar mixed match blocks you, photograph any similar doubt using the 20-free doubt service for instant step-by-step clarification.

Frequently asked questions

What is the correct matching for mixed mathematics JEE 2015?

The correct matching is (A)→P,R,S; (B)→P; (C)→P,Q; (D)→S,T. Part A yields lambda = 1/2 so only odd n satisfy cos(n pi lambda)=0. Part B simplifies to a/b=1. Part C gives |beta|=1 and 2. Part D evaluates to 6 and 5 after adding (8/3)sqrt(2).

What single algebraic slip ruins the matching in part A of mixed mathematics JEE 2015?

After reaching 2 sin Z sin(X-Y) = sin^2 Z, students sometimes forget the factor of 2 and obtain lambda=1 instead of 1/2. Then cos(n pi * 1) = (-1)^n = 0 has no integer solutions. This causes them to miss matching with P, R and S entirely.

How do you solve the trigonometry part of mixed mathematics JEE 2015?

Apply the sine rule to substitute sides, simplify 2(sin^2 X - sin^2 Y) = sin^2 Z using the identity sin^2 X - sin^2 Y = sin(X+Y)sin(X-Y) and X+Y=pi-Z. This reduces to lambda = sin(X-Y)/sin Z = 1/2. Thus cos(n pi /2)=0 holds for odd integers n matching P, R, S.

How to solve the vector bisector part in mixed mathematics JEE 2015?

What practice questions should I solve for mixed mathematics JEE 2015?

Practice the triangle with a^2 + b^2 = 2c^2 to find cos(A-B). Solve the similar vector problem with points (2,1), (1,2) and R(mu,3-mu) at distance sqrt(2) from y=x. Compute area bounded by y^2=4x and y=|x-1|+|x-3|+x/2 from x=0 to 4 by splitting at modulus points.

Mixed Mathematics JEE 2015: Official Matrix Match Solution