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Q.

If ∫π/2x 3−2sin2⁡zdz+∫0y cos⁡tdt=0 then  dydx at (π/2,π) is

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a

1

b

2

c

3

d

0

answer is A.

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Detailed Solution

Differentiating w ·r· t· x, we get3−2sin2⁡x+dydxcos⁡y=0Putting x=π/2 and y=π we getdydx(π/2,π)=1.
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