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← IIT-JEE 2008 · Paper 2
MathematicsMatrix match+6Question 21 of 66

IIT-JEE 2008 · Paper 2, Mathematics question 21

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

Column I

  • A
    The minimum value of x2+2x+4x+2\dfrac{x^2+2x+4}{x+2} is
  • B
    Let AA and BB be 3×33\times3 matrices of real numbers, where AA is symmetric, BB is skew-symmetric, and (A+B)(A−B)=(A−B)(A+B)(A+B)(A-B)=(A-B)(A+B). If (AB)t=(−1)kAB(AB)^t=(-1)^kAB, where (AB)t(AB)^t is the transpose of the matrix ABAB, then the possible values of kk are
  • C
    Let a=log⁡3log⁡32a=\log_3\log_32. An integer kk satisfying 1<2(−k+3−a)<21<2^{(-k+3^{-a})}<2, must be less than
  • D
    If sin⁡θ=cos⁡φ\sin\theta=\cos\varphi, then the possible values of 1π(θ±φ−π2)\frac1\pi\left(\theta\pm\varphi-\frac\pi2\right) are

Column II

  • P
    0
  • Q
    1
  • R
    2
  • S
    3
Mark every match for each entry
PQRS
A
B
C
D
Answer and worked solution

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

  • A. x2+2x+4x+2=(x+2)+4x+2−2≥4−2=2\dfrac{x^2+2x+4}{x+2}=(x+2)+\dfrac4{x+2}-2\ge4-2=2 for x>−2x>-2 (AM-GM), equality at x=0x=0. The minimum (local) value is 2. → R.
  • B. (A+B)(A−B)=(A−B)(A+B)(A+B)(A-B)=(A-B)(A+B) gives AB=BAAB=BA. Then (AB)t=BtAt=−BA=−AB(AB)^t=B^tA^t=-BA=-AB, so (−1)k=−1(-1)^k=-1: kk odd, e.g. 1 or 3. → Q, S.
  • C. 3−a=1log⁡32=log⁡23≈1.5853^{-a}=\dfrac1{\log_32}=\log_23\approx1.585. The condition 0<−k+log⁡23<10<-k+\log_23<1 gives 0.585<k<1.5850.585<k<1.585, so k=1k=1, which is less than 2 and less than 3. → R, S.
  • D. sin⁡θ=cos⁡φ\sin\theta=\cos\varphi gives θ=2nπ+π2∓φ\theta=2n\pi+\frac\pi2\mp\varphi, so one of 1π(θ±φ−π2)\frac1\pi\left(\theta\pm\varphi-\frac\pi2\right) equals 2n2n: even integers, e.g. 0 or 2. → P, R.

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