Q.

If the equation  21cos1x2π21cos1xπa+12a2=0   has only one real solution then subsets of values of ‘a’ are

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a

1,

b

3,

c

3,1

d

,3

answer is B, C.

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Detailed Solution

 

let  2πcos1=t   since  0cos1xπ1π1cos1x1πcos1x2t<122π/cos1xa+122π/cos1xa2=0

t2a+12ta2=0t=a+12±a+122+4a22 From (1) 2a+12±a+122+4a224a+12±a+122+4a2±a+122+4a24a+12=72aa+122+4a272a2

a2+14+a+4a2494+a27a4a2+8a120a2+2a30(a1)(a+3)0a3 or a1

 

 

 

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