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Question

If the inequality sin2x+acosx+a2>1+cosx holds for any xR, then the largest negative integral value of a is

A
4
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B
3
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C
2
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D
1
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Solution

The correct option is B 3
sin2x+acosx+a2>1+cosx
cos2x+(1a)cosxa2<0
Now let cosx=t
t[1,1], xR
So, t2+(1a)ta2<0(i)
For the given inequality to be true, [1,1] should lie in between the roots of equation (i)

Hence required condition are
(1) f(1)<0aa2<0a(a1)>0a<0 or a>1(2) f(1)<02aa2<0a2+a2>0(a+2)(a1)>0a<2 or a>1

So from the above condtitions, we get
a(,2)(1,)

Hence, the largest negative integral value of a is 3

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