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Question

If (sina)x2+(sina)x+(1-cosa)=0, then the value of a for which roots of this equation are real and distinct, is


A

0,2tan-114

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B

0,2π3

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C

(0,π)

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D

(0,2π)

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Solution

The correct option is A

0,2tan-114


Explanation for the correct option.

Step 1: Find the determinant

Given that, (sina)x2+(sina)x+(1-cosa)=0.

We know that in equation ax2+bx+c=0 roots are real & distinct when D>0.

And D=b2-4ac

sin2a-4sina(1-cosa)>0(1-cosa)(1+cosa-4sina)>0(1+cosa-4sina)>02cos2a2-4×2sina2cosa2>0sinθ=2sinθ2cosθ2andcosθ=2cos2θ212cos2a21-4tana2>0

Step 2: Find the value of a

As cos2a2>0 so we only need to check another factor.

1-4tana2>04tana2<10<a<2tan-114

Hence, option A is correct.


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