Match the statements / expressions...

MathematicsScalar Triple ProductJEE Advanced 2009Moderate

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

Visualized Solution (Hindi)

Introduction to the Problem

Match Column-I (Expressions) with Column-II (Values).
Part A: Trigonometric roots of $2 sin^{2} θ + sin^{2} 2 θ = 2$ .
Part B: Discontinuity of $f (x) = [\frac{6 x}{π}] cos [\frac{3 x}{π}]$ .
Part C: Volume of a parallelepiped with given vector edges.
Part D: Angle between unit vectors $a$ and $b$ given $a + b + 3 c = 0$ .

Part A: Simplifying the Equation

Equation: $2 sin^{2} θ + sin^{2} 2 θ = 2$
Apply identity: $sin 2 θ = 2 sin θ cos θ$
Substitute: $2 sin^{2} θ + (2 sin θ cos θ)^{2} = 2$
Simplify: $2 sin^{2} θ + 4 sin^{2} θ cos^{2} θ = 2$
Divide by $2$ : $sin^{2} θ + 2 sin^{2} θ cos^{2} θ = 1$

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

Substitute $cos^{2} θ = 1 - sin^{2} θ$ :
$sin^{2} θ + 2 sin^{2} θ (1 - sin^{2} θ) = 1$
$2 sin^{4} θ - 3 sin^{2} θ + 1 = 0$
Factorize: $(2 sin^{2} θ - 1) (sin^{2} θ - 1) = 0$
Roots: $sin^{2} θ = \frac{1}{2} ⟹ θ = π /4$ (Matches q)
Roots: $sin^{2} θ = 1 ⟹ θ = π /2$ (Matches s)

Part B: Analyzing Discontinuity

Function: $f (x) = [\frac{6 x}{π}] cos [\frac{3 x}{π}]$
Discontinuity occurs when $\frac{6 x}{π} \in Z$ or $\frac{3 x}{π} \in Z$ .
Check values from Column-II: $π /6, π /4, π /3, π /2, π$ .

Part B: Testing Values

At $x = π /6$ : $\frac{6 x}{π} = 1$ (Integer) $⟹$ Discontinuous (p)
At $x = π /4$ : $\frac{6 x}{π} = 1.5, \frac{3 x}{π} = 0.75$ $⟹$ Continuous
At $x = π /3$ : $\frac{6 x}{π} = 2, \frac{3 x}{π} = 1$ (Integers) $⟹$ Discontinuous (r)
At $x = π /2$ : $\frac{6 x}{π} = 3$ (Integer) $⟹$ Discontinuous (s)
At $x = π$ : $\frac{6 x}{π} = 6, \frac{3 x}{π} = 3$ (Integers) $⟹$ Discontinuous (t)

Part C: Volume of Parallelepiped

Edges: $u = \hat{i} + \hat{j}$ , $v = \hat{i} + 2 \hat{j}$ , $w = \hat{i} + \hat{j} + π \hat{k}$
Volume $V = ∣ [u v w] ∣ = 111121 00 π$
Expand along $C_{3}$ : $V = ∣ π (2 - 1) ∣ = π$
Matches (t).

Part D: Angle Between Vectors

Given: $a + b + 3 c = 0 ⟹ a + b = - 3 c$
Square both sides: $∣ a + b ∣^{2} = ∣ - 3 c ∣^{2}$
Expand: $∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b = 3∣ c ∣^{2}$
Since unit vectors: $1 + 1 + 2 cos θ = 3 (1)$
$2 cos θ = 1 ⟹ cos θ = \frac{1}{2} ⟹ θ = π /3$
Matches (r).

Final Summary

Final Matches:
(A) $\to$ (q), (s)
(B) $\to$ (p), (r), (s), (t)
(C) $\to$ (t)
(D) $\to$ (r)
Key Takeaway: Multi-concept problems require systematic step-by-step analysis of each part independently.

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