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Question

If sinA,sinB and cosA are in GP, then the roots of x2+2xcotB+1=0 are always.


A

real

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B

imaginary

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C

greater than 1

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D

equal

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Solution

The correct option is A

real


Explanation for the correct option:

Roots of an quadratic equation:

Given that, sinA,sinB and cosA are in G.P.

Then, by definition of G.P which states that if a,b,care in G.P then b2=ac.

Here, a=sinA,b=sinB,c=cosA.

ā‡’sin2B=sinAcosA

Now using discriminant, D=b2-4ac we determine the nature of roots.

We have equation x2+2xcotB+1=0 .

ā‡’D=4cot2B-4=4cot2B-1=4cosec2B-2>0 [āˆµ1+cot2B=cosec2B] =4(1-2sin2B)sin2B =4(sin2A+cos2A-2sinAcosA)sin2B [āˆµsin2A+cos2A=1,sin2B=sinAcosA] =4(sinA-cosA)2sin2B

Since discriminant is >0 so the equation has real roots.

Hence, option A is correct.


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