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Question

If cos(x+π/3)cosx=a has real solutions, then

A
number of integral value of a are 3
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B
sum of number of integral value of a is 0
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C
when a=1 number of solutions for xϵ[0,2π] are 3
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D
when a=1, number of solutions for xϵ[0,2π] are 2
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Solution

The correct option is B sum of number of integral value of a is 0
cos(x+π3)cosx=a

(cosx.cosπ3sinx.sinπ3)cosx=a ........ Using cos(a+b)=cosacosbsinasinb

(cosx3sinx).cosx=2a

cos2x3sinx.cosx=2a .

1+cos2x232sin2x=2a ......... [cos2x=1+cos2x2]

cos2x3sin2x=4a1

Thus equation is only solvable if,

24a12 ......... From concept of extrema of trigonametric function

a[14,34]

Number of integral solution is 1 which is 0.

sum of number of integral value of a is 0.

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