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← JEE Advanced 2015 · Paper 1
MathematicsMatrix match+8 / -4Question 60 of 60

JEE Advanced 2015 · Paper 1, Mathematics question 60

Match the entries in Column I with the entries in Column II.

Column I

  • A
    In a triangle ΔXYZ\Delta XYZ, let aa, bb and cc be the lengths of the sides opposite to the angles XX, YY and ZZ, respectively. If 2(a2−b2)=c22(a^2-b^2)=c^2 and λ=sin⁡(X−Y)sin⁡Z\lambda=\dfrac{\sin(X-Y)}{\sin Z}, then possible values of nn for which cos⁡(nπλ)=0\cos(n\pi\lambda)=0 is (are)
  • B
    In a triangle ΔXYZ\Delta XYZ, let aa, bb and cc be the lengths of the sides opposite to the angles XX, YY and ZZ, respectively. If 1+cos⁡2X−2cos⁡2Y=2sin⁡Xsin⁡Y1+\cos2X-2\cos2Y=2\sin X\sin Y, then possible value(s) of ab\dfrac ab is (are)
  • C
    In R2\mathbb R^2, let 3i^+j^\sqrt3\hat i+\hat j, i^+3j^\hat i+\sqrt3\hat j and βi^+(1−β)j^\beta\hat i+(1-\beta)\hat j be the position vectors of XX, YY and ZZ with respect to the origin OO, respectively. If the distance of ZZ from the bisector of the acute angle of OX→\overrightarrow{OX} with OY→\overrightarrow{OY} is 32\dfrac3{\sqrt2}, then possible value(s) of ∣β∣|\beta| is (are)
  • D
    Suppose that F(α)F(\alpha) denotes the area of the region bounded by x=0x=0, x=2x=2, y2=4xy^2=4x and y=∣αx−1∣+∣αx−2∣+αxy=|\alpha x-1|+|\alpha x-2|+\alpha x, where α∈{0,1}\alpha\in\{0,1\}. Then the value(s) of F(α)+832F(\alpha)+\dfrac83\sqrt2, when α=0\alpha=0 and α=1\alpha=1, is (are)

Column II

  • P
    1
  • Q
    2
  • R
    3
  • S
    5
  • T
    6
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

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

(A) In any triangle, sin⁡(X−Y)sin⁡Z=a2−b2c2\dfrac{\sin(X-Y)}{\sin Z}=\dfrac{a^2-b^2}{c^2}, so λ=12\lambda=\frac12. Then cos⁡nπ2=0\cos\dfrac{n\pi}2=0 for odd nn: 1, 3, 5, i.e. P, R, S.

(B) 1+cos⁡2X−2cos⁡2Y=4sin⁡2Y−2sin⁡2X1+\cos2X-2\cos2Y=4\sin^2Y-2\sin^2X. Setting this equal to 2sin⁡Xsin⁡Y2\sin X\sin Y:
(sin⁡X+2sin⁡Y)(sin⁡X−sin⁡Y)=0,(\sin X+2\sin Y)(\sin X-\sin Y)=0,
so a=ba=b and a/b=1a/b=1: P.

(C) ∣OX∣=∣OY∣=2|OX|=|OY|=2, so the acute bisector is y=xy=x. The distance of Z=(β,1−β)Z=(\beta,1-\beta) from it is
∣2β−1∣2=32,\frac{|2\beta-1|}{\sqrt2}=\frac3{\sqrt2},
so β=2\beta=2 or −1-1 and ∣β∣=2,1|\beta|=2,1: P, Q.

(D) The curve lies above y=2xy=2\sqrt x on [0,2][0,2].

  • α=0\alpha=0: y=3y=3, so F=6−832F=6-\frac83\sqrt2.
  • α=1\alpha=1: y=3−xy=3-x on [0,1][0,1] and x+1x+1 on [1,2][1,2], so F=52+52−832F=\frac52+\frac52-\frac83\sqrt2.

Adding 832\frac83\sqrt2 gives 6 and 5: S, T.

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