JEE Challenger
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MathMedAdv 2009 • Paper 2

Match Trigonometric Function Properties Discontinuity Vectors to Values

Match the statements/expressions given in Column I with the values given in Column II.

Column IColumn II(A) Root(s) of the equation 2sin⁡2θ+sin⁡22θ=2(p) π6(B) Points of discontinuity of the function f(x)=[6xπ]cos⁡[3xπ],(q) π4where [y] denotes the largest integer less than or equal to y(C) Volume of the parallelopiped with its edges represented by the vectors(r) π3i^+j^,  i^+2j^ and i^+j^+πk^(D) Angle between vectors a⃗ and b⃗ where a⃗,b⃗ and c⃗ are unit vectors satisfying(s) π2a⃗+b⃗+3 c⃗=0⃗(t) π\begin{array}{ll} \textbf{Column I} & \textbf{Column II} \\ \text{(A) Root(s) of the equation } 2\sin^2\theta + \sin^2 2\theta = 2 & \text{(p) } \frac{\pi}{6} \\[6pt] \text{(B) Points of discontinuity of the function } f(x) = \left[\frac{6x}{\pi}\right]\cos\left[\frac{3x}{\pi}\right], & \text{(q) } \frac{\pi}{4} \\[4pt] \quad \text{where } [y] \text{ denotes the largest integer less than or equal to } y & \\[6pt] \text{(C) Volume of the parallelopiped with its edges represented by the vectors} & \text{(r) } \frac{\pi}{3} \\[4pt] \quad \hat{i}+\hat{j}, \; \hat{i}+2\hat{j} \text{ and } \hat{i}+\hat{j}+\pi\hat{k} & \\[6pt] \text{(D) Angle between vectors } \vec{a} \text{ and } \vec{b} \text{ where } \vec{a}, \vec{b} \text{ and } \vec{c} \text{ are unit vectors satisfying} & \text{(s) } \frac{\pi}{2} \\[4pt] \quad \vec{a}+\vec{b}+\sqrt{3}\,\vec{c} = \vec{0} & \text{(t) } \pi \end{array}

Options

A

(A)→(Q,S),  (B)→(P,R,S,T),  (C)→(T),  (D)→(R)(\text{A}) \rightarrow (\text{Q}, \text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{R}, \text{S}, \text{T}),\; (\text{C}) \rightarrow (\text{T}),\; (\text{D}) \rightarrow (\text{R})

Correct
B

(A)→(Q),  (B)→(P,R,S,T),  (C)→(T),  (D)→(R)(\text{A}) \rightarrow (\text{Q}),\; (\text{B}) \rightarrow (\text{P}, \text{R}, \text{S}, \text{T}),\; (\text{C}) \rightarrow (\text{T}),\; (\text{D}) \rightarrow (\text{R})

C

(A)→(Q,S),  (B)→(P,S,T),  (C)→(R),  (D)→(R)(\text{A}) \rightarrow (\text{Q}, \text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{S}, \text{T}),\; (\text{C}) \rightarrow (\text{R}),\; (\text{D}) \rightarrow (\text{R})

D

(A)→(S),  (B)→(P,R,S),  (C)→(T),  (D)→(P,R)(\text{A}) \rightarrow (\text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{T}),\; (\text{D}) \rightarrow (\text{P}, \text{R})

Topics & Concepts

Vector AlgebraScalar Triple ProductDot and Cross ProductTrigonometric EquationsContinuity

Step-by-Step Solution

To solve each part:

(A) Root(s) of the equation 2sin⁡2θ+sin⁡22θ=22\sin^2\theta + \sin^2 2\theta = 2

Using the double-angle identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta:
2sin⁡2θ+4sin⁡2θcos⁡2θ=22\sin^2\theta + 4\sin^2\theta\cos^2\theta = 2
Dividing by 22:
sin⁡2θ+2sin⁡2θ(1−sin⁡2θ)=1\sin^2\theta + 2\sin^2\theta(1 - \sin^2\theta) = 1
sin⁡2θ+2sin⁡2θ−2sin⁡4θ=1\sin^2\theta + 2\sin^2\theta - 2\sin^4\theta = 1
2sin⁡4θ−3sin⁡2θ+1=02\sin^4\theta - 3\sin^2\theta + 1 = 0
Factoring the quadratic equation in terms of sin⁡2θ\sin^2\theta:
(2sin⁡2θ−1)(sin⁡2θ−1)=0(2\sin^2\theta - 1)(\sin^2\theta - 1) = 0
This gives:

  1. sin⁡2θ=12  ⟹  sin⁡θ=±12  ⟹  θ=π4\sin^2\theta = \frac{1}{2} \implies \sin\theta = \pm \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} (which matches (q)).
  2. sin⁡2θ=1  ⟹  sin⁡θ=±1  ⟹  θ=π2\sin^2\theta = 1 \implies \sin\theta = \pm 1 \implies \theta = \frac{\pi}{2} (which matches (s)).

Thus, (A) →\to (q), (s).

(B) Points of discontinuity of f(x)=[6xπ]cos⁡[3xπ]f(x) = \left[\frac{6x}{\pi}\right]\cos\left[\frac{3x}{\pi}\right]

We test each of the given values from Column II:

  1. For x=π6x = \frac{\pi}{6} (p):

    • Left-hand limit: As x→(π6)−x \to \left(\frac{\pi}{6}\right)^-, 6xπ→1−\frac{6x}{\pi} \to 1^- and 3xπ→(12)−\frac{3x}{\pi} \to \left(\frac{1}{2}\right)^-.
      lim⁡x→(π/6)−f(x)=[1−]cos⁡[0.5−]=0⋅cos⁡(0)=0\lim_{x \to (\pi/6)^-} f(x) = [1^-] \cos[0.5^-] = 0 \cdot \cos(0) = 0
    • Right-hand limit: As x→(π6)+x \to \left(\frac{\pi}{6}\right)^+, 6xπ→1+\frac{6x}{\pi} \to 1^+ and 3xπ→(12)+\frac{3x}{\pi} \to \left(\frac{1}{2}\right)^+.
      lim⁡x→(π/6)+f(x)=[1+]cos⁡[0.5+]=1⋅cos⁡(0)=1\lim_{x \to (\pi/6)^+} f(x) = [1^+] \cos[0.5^+] = 1 \cdot \cos(0) = 1
      Since the limits are not equal, f(x)f(x) is discontinuous at x=π6x = \frac{\pi}{6}.

  2. For x=π4x = \frac{\pi}{4} (q):

    • At x=π4x = \frac{\pi}{4}, 6xπ=32\frac{6x}{\pi} = \frac{3}{2} and 3xπ=34\frac{3x}{\pi} = \frac{3}{4}. Both arguments are non-integers, so both floor functions are locally constant and hence f(x)f(x) is continuous at x=π4x = \frac{\pi}{4}.

  3. For x=π3x = \frac{\pi}{3} (r):

    • As x→(π3)−x \to \left(\frac{\pi}{3}\right)^-, 6xπ→2−\frac{6x}{\pi} \to 2^- and 3xπ→1−\frac{3x}{\pi} \to 1^-.
      lim⁡x→(π/3)−f(x)=1⋅cos⁡(0)=1\lim_{x \to (\pi/3)^-} f(x) = 1 \cdot \cos(0) = 1
    • As x→(π3)+x \to \left(\frac{\pi}{3}\right)^+, 6xπ→2+\frac{6x}{\pi} \to 2^+ and 3xπ→1+\frac{3x}{\pi} \to 1^+.
      lim⁡x→(π/3)+f(x)=2⋅cos⁡(1)≠1\lim_{x \to (\pi/3)^+} f(x) = 2 \cdot \cos(1) \neq 1
      Thus, f(x)f(x) is discontinuous at x=π3x = \frac{\pi}{3}.

  4. For x=π2x = \frac{\pi}{2} (s):

    • As x→(π2)−x \to \left(\frac{\pi}{2}\right)^-, 6xπ→3−\frac{6x}{\pi} \to 3^- and 3xπ→(32)−\frac{3x}{\pi} \to \left(\frac{3}{2}\right)^-.
      lim⁡x→(π/2)−f(x)=2⋅cos⁡(1)\lim_{x \to (\pi/2)^-} f(x) = 2 \cdot \cos(1)
    • As x→(π2)+x \to \left(\frac{\pi}{2}\right)^+, 6xπ→3+\frac{6x}{\pi} \to 3^+ and 3xπ→(32)+\frac{3x}{\pi} \to \left(\frac{3}{2}\right)^+.
      lim⁡x→(π/2)+f(x)=3⋅cos⁡(1)\lim_{x \to (\pi/2)^+} f(x) = 3 \cdot \cos(1)
      Since cos⁡(1)≠0\cos(1) \neq 0, the limits are not equal, so f(x)f(x) is discontinuous at x=π2x = \frac{\pi}{2}.

  5. For x=πx = \pi (t):

    • As x→π−x \to \pi^-, 6xπ→6−\frac{6x}{\pi} \to 6^- and 3xπ→3−\frac{3x}{\pi} \to 3^-.
      lim⁡x→π−f(x)=5⋅cos⁡(2)\lim_{x \to \pi^-} f(x) = 5 \cdot \cos(2)
    • As x→π+x \to \pi^+, 6xπ→6+\frac{6x}{\pi} \to 6^+ and 3xπ→3+\frac{3x}{\pi} \to 3^+.
      lim⁡x→π+f(x)=6⋅cos⁡(3)≠5cos⁡(2)\lim_{x \to \pi^+} f(x) = 6 \cdot \cos(3) \neq 5 \cos(2)
      Thus, f(x)f(x) is discontinuous at x=πx = \pi.

Hence, the points of discontinuity are π6,π3,π2,π\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi.

Thus, (B) →\to (p), (r), (s), (t).

(C) Volume of the parallelopiped

The volume VV of a parallelopiped formed by edges u⃗=i^+j^\vec{u} = \hat{i}+\hat{j}, v⃗=i^+2j^\vec{v} = \hat{i}+2\hat{j}, and w⃗=i^+j^+πk^\vec{w} = \hat{i}+\hat{j}+\pi\hat{k} is given by the scalar triple product:

V=∣[u⃗    v⃗    w⃗]∣=∣det⁡(11012011π)∣V = |[\vec{u} \;\; \vec{v} \;\; \vec{w}]| = \left| \det \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{pmatrix} \right|

Expanding along the third column:
V=∣π⋅(1⋅2−1⋅1)∣=π⋅∣1∣=πV = \left| \pi \cdot (1 \cdot 2 - 1 \cdot 1) \right| = \pi \cdot |1| = \pi
This matches (t).

Thus, (C) →\to (t).

(D) Angle between vectors a⃗\vec{a} and b⃗\vec{b}

We are given that a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} are unit vectors, so:
∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec{a}| = |\vec{b}| = |\vec{c}| = 1
The given vector relation is:
a⃗+b⃗+3c⃗=0⃗  ⟹  a⃗+b⃗=−3c⃗\vec{a} + \vec{b} + \sqrt{3}\vec{c} = \vec{0} \implies \vec{a} + \vec{b} = -\sqrt{3}\vec{c}
Taking the square of the magnitude on both sides:
∣a⃗+b⃗∣2=∣−3c⃗∣2|\vec{a} + \vec{b}|^2 = |-\sqrt{3}\vec{c}|^2
∣a⃗∣2+∣b⃗∣2+2(a⃗⋅b⃗)=3∣c⃗∣2|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 3|\vec{c}|^2
Substitute the magnitudes:
1+1+2cos⁡θ=3(1)1 + 1 + 2\cos\theta = 3(1)
2+2cos⁡θ=3  ⟹  2cos⁡θ=1  ⟹  cos⁡θ=122 + 2\cos\theta = 3 \implies 2\cos\theta = 1 \implies \cos\theta = \frac{1}{2}
Since θ∈[0,π]\theta \in [0, \pi]:
θ=π3\theta = \frac{\pi}{3}
This matches (r).

Thus, (D) →\to (r).

Final Match:

  • (A) →\to (q), (s)
  • (B) →\to (p), (r), (s), (t)
  • (C) →\to (t)
  • (D) →\to (r)
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