Column I Column II A In a triangle ∆ X Y Z let a , b and c be the length of sides opposite to the…

 Column I  Column II  
  A            

 In a triangle XYZ let a,b and c be the length of sides opposite to the angles X,Y and Z, respectively. If 2a2b2=c2 and λ=sinXYsinZthen possible values of n for which cosnπλ=0 is are

  P         1       
  B               

 In a triangle ΔXYZ, let a,b and c be the lengths of the sides opposite to the angles X,Y and Z, respectively. If 1+cos2X-2cos2Y=2sinXsinY, then possible value(s) of ab is (are)

  Q   2
  C                   In R2, let 3 i^+j^, i^+3 j^ and βi^+1-β j^ be the position vectors of X,Y and Z with respect to the origin O, respectively. If the distance of Z from the bisector of the acute angle of OX with OY is 32, then possible value (s) of β is (are)  R  3
  D                

 Suppose that F(α) denotes the area of the region bounded by x=0, x=2, y2=4x andy=ax-1+α x-2+αx,  where α0, 1.  Then the value (s) of Fα+83 2,  when α=0 and α=1, is (are)

 S  5
     T  6

 

  1. a-p,r,s;b-p;c-p,q;d-s,t;
  2. a-q,r,s,t;b-q,r,s,t;c-r,s,t;d-q,r,s;
  3. a-q;b-r;c-r,s;d-r,s;
  4. a-q,r,s,t;b-t;c-r,s,t;d-q,r,s,t;

Solution

A λ=sinX-YsinZ
sinXcosY-cosXsinYsinZ
acosY-bcosXc
aa2+c2-b22ac-bb2+c2-a22bcc=λ
a2-b2c2=λ
As 2a2-b2=c2
λ=12
cosnπ2=0
n=1, 3, 5
B 1+cos2X-2cos2Y=2sinXsinY
Solving
4sin2Y-2sin2X=2sinXsinY
4b2-2a2=2ab
4-2ab2=2ab


ab=X

X2+X-2=0

X=1

ab=1

x=-2

ab=-2 (rejected)

C OYy=3x

OXy=13x

Equation of bisector of OX, OY

y=x

x-y2=32

β-1-β2=±32

2β-1=±3

β=2

β=-1

β=1, 2

(D)

Area =6-022x dx

=6-823

Required value =6-823+823=6


Area= Shaded region - area under parabola.

=5-823

Required value =5

Asked in: JEE Advanced 2015 (Paper 1)

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