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← IIT-JEE 2009 · Paper 2
MathematicsMatrix match+8Question 30 of 57

IIT-JEE 2009 · Paper 2, Mathematics question 30

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

Column I

  • A
    The number of solutions of the equation xesin⁡x−cos⁡x=0xe^{\sin x}-\cos x=0 in the interval (0,π2)\left(0,\frac\pi2\right)
  • B
    Value(s) of kk for which the planes kx+4y+z=0kx+4y+z=0, 4x+ky+2z=04x+ky+2z=0 and 2x+2y+z=02x+2y+z=0 intersect in a straight line
  • C
    Value(s) of kk for which ∣x−1∣+∣x−2∣+∣x+1∣+∣x+2∣=4k|x-1|+|x-2|+|x+1|+|x+2|=4k has integer solution(s)
  • D
    If y′=y+1y'=y+1 and y(0)=1y(0)=1, then value(s) of y(ln⁡2)y(\ln2)

Column II

  • P
    1
  • Q
    2
  • R
    3
  • S
    4
  • T
    5
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

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

  • A: xesin⁡xxe^{\sin x} rises from 0 while cos⁡x\cos x falls from 1 on (0,π2)\left(0,\frac\pi2\right), so they cross exactly once. → P.
  • B: three planes through the origin share a line when the determinant vanishes: k(k−4)−4(4−4)+(8−2k)=k2−6k+8=0k(k-4)-4(4-4)+(8-2k)=k^2-6k+8=0, so k=2,4k=2,4. → Q, S.
  • C: f(x)=∣x−1∣+∣x−2∣+∣x+1∣+∣x+2∣f(x)=|x-1|+|x-2|+|x+1|+|x+2| equals 6 for ∣x∣≤1|x|\le1 and 4∣x∣4|x| for ∣x∣≥2|x|\ge2. At integers it takes the values 6, 8, 12, 16, 20, …, so 4k4k can be 8, 12, 16, 20: k=2,3,4,5k=2,3,4,5. → Q, R, S, T.
  • D: y′=y+1y'=y+1 with y(0)=1y(0)=1 gives y=2ex−1y=2e^x-1, so y(ln⁡2)=3y(\ln2)=3. → R.

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