Q38 / 60 · JEE Advanced 2014 (Paper 2) · Mathematics

Match the following.
 
  List – I   List – II
(A) Let yx=cos⁡3cos-1⁡x, 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-1⁡12x+1+tan-1⁡14x+1=tan-1⁡2x2 is (S) 9

  • A
    a-r;b-q;c-s;d-p;
  • Correct answer
    B
    a-s;b-r;c-q;d-p;
  • C
    a-r;b-q;c-s;d-p;
  • D
    a-q;b-p;c-r;d-s;

Solution

P   y=cos⁡3cos-1⁡x
y′=3sin⁡3cos-1⁡x1-x2
1-x2 y′=3sin⁡3cos-1⁡x
⇒   -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=r2sin⁡2πn
ak . ak+1=r2cos⁡2πn
⇒   ∑k=1n-1ak→× ak+1→=∑k=1n-1ak .  ak+1
⇒   r2n-1sin⁡2πn=r2n-1cos⁡2πn
tan⁡2πn=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-1⁡12x+1+tan-1⁡14x+1=tan-1⁡2x2
tan-1⁡3x+14x2+3x=tan-1⁡2x2
⇒   3x2-7x-6=0
x=-23, 3

Hence, the correct answer is option: 'B'.

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