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A solution $(x, y) of x^{2} + 2 x \sin x y + 1 = 0 is$

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(1, 0)

(1, 7π/2)

(– 1, –7π/2)

(– 1, 0)

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detailed solution

Correct option is B

When x=1, 2+2sin⁡ y=0⇒sin⁡ y=−1and 7π2=sin⁡ 3π+π2=−1So (1, 7π/2) is a solution.For x = – 1, sin y = 1 which is not satisfied by y = 7π/2 or 0.

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A solution (x, y) of x2+2xsin⁡xy+1=0 is

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A solution $(x, y) of x^{2} + 2 x \sin x y + 1 = 0 is$