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JEE Advanced · Mathematics · 21. ITF

Match the following.
 
  List – I   List – II
(A) Let yx=cos3cos-1x, x ϵ -1, 1, x±32. Then 1yx x2-1d2yxdx2+xdyxdx equals (P) 1
(B) Let A1, A2, , Ann>2 be the vertices of a regular polygon of n sides with its centre at the origin. Let ak be the position vector of the point Ak, k=1, 2, n. If k=1n-1 ak× ak+1=k=1n-1 ak  ak+1 , then the minimum value of n is (Q) 2
(C) If the normal from the point Ph, 1 on the ellipse x26+y23=1 is perpendicular to the line x+y=8, then the value of h is (R) 8
(D) Number of positive solutions satisfying the equation tan-112x+1+tan-114x+1=tan-12x2 is (S) 9

  1. A a-r;b-q;c-s;d-p;
  2. B a-s;b-r;c-q;d-p;
  3. C a-r;b-q;c-s;d-p;
  4. D a-q;b-p;c-r;d-s;
Verified Solution

Answer & Solution

Correct Answer

(B) a-s;b-r;c-q;d-p;

Step-by-step Solution

Detailed explanation

P   y=cos3cos-1x
y=3sin3cos-1x1-x2
1-x2 y=3sin3cos-1x
   -x1-x2 y+1-x2  y"=3 cos (3 cos-1x) . -31-x2
  -xy+1-x2 y"=-9y
  1y[x2-1 y"+xy] =9
Q  ak×ak+1=r2sin2πn
ak . ak+1=r2cos2πn
   k=1n-1ak× ak+1=k=1n-1ak .  ak+1
   r2n-1sin2πn=r2n-1cos2πn
tan2πn=1    n=84k+1 
   n=8
R h26+123=1, h=±2
Tangent at (2, 1) is 2x6+y3=1  x+y=3
S  tan-112x+1+tan-114x+1=tan-12x2
tan-13x+14x2+3x=tan-12x2
   3x2-7x-6=0
x=-23, 3
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