IIT-JEE 2009 · Paper 2 · Mathematics Q29

MathematicsMatrix match+8Question 29 of 57

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

Column I

A
Root(s) of the equation $2\sin^2\theta+\sin^22\theta=2$
B
Points of discontinuity of the function $f(x)=\left[\dfrac{6x}\pi\right]\cos\left[\dfrac{3x}\pi\right]$ , where $[y]$ denotes the largest integer less than or equal to $y$
C
Volume of the parallelopiped with its edges represented by the vectors $\hat i+\hat j$ , $\hat i+2\hat j$ and $\hat i+\hat j+\pi\hat k$
D
Angle between vectors $\vec a$ and $\vec b$ where $\vec a$ , $\vec b$ and $\vec c$ are unit vectors satisfying $\vec a+\vec b+\sqrt3\,\vec c=\vec0$

Column II

Answer and worked solution

Official answer: A → Q, S; B → P, R, S, T; C → T; D → R

A: with $s=\sin^2\theta$ , $2s+4s(1-s)=2$ gives $2s^2-3s+1=0$ , so $s=1$ or $\frac12$ : $\theta=\frac\pi2,\frac\pi4$ . → Q, S.
B: $\left[\frac{6x}\pi\right]$ jumps at every multiple of $\frac\pi6$ . At $\frac\pi6,\frac\pi3,\frac\pi2,\pi$ the product changes value (e.g. at $\frac\pi6$ it jumps from 0 to $\cos0=1$ ; at π from $5\cos2$ to $6\cos3$ ). $\frac\pi4$ is not such a point. → P, R, S, T.
C: $\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi\end{vmatrix}=\pi(2-1)=\pi$ . → T.
D: $|\vec a+\vec b|^2=3|\vec c|^2=3$ , so $2+2\cos\theta=3$ and $\theta=\frac\pi3$ . → R.

Answer: A → Q, S; B → P, R, S, T; C → T; D → R

IIT-JEE 2009 · Paper 2, Mathematics question 29