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Q.

The value of a in (−π,0) satisfying  sin⁡α+∫α2α cos⁡2xdx=0 is

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a

−π/2

b

−π

c

−π/3

d

−π/4

answer is C.

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Detailed Solution

sin⁡α+∫a2a cos⁡2xdx=0⇒sin⁡α+12(sin⁡4α−sin⁡2α)=0⇒sin⁡α[1+cos⁡(3α)]=0As −π<α<0,sin⁡α≠0 thereforecos⁡(3α)=−1⇒α=−π/3.
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