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

(1, 0)

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(1, \frac{7 π}{2})

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(- 1, \frac{7 π}{2})

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(- 1, 0)

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Solution

The correct options are
B $(- 1, \frac{7 π}{2})$
D $(1, \frac{7 π}{2})$
we get
$(x + sin (x y))^{2} + {cos}^{2} (x y) = 0$
Now we verify with the options

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Similar questions

{(x, y) : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1, 0 \leq x \leq 2}

{(x, y) : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1, 0 \leq x \leq 2}

S = x^{2} + y^{2} + z^{2} - 2 x + 3 y - 4 z - 5 = 0

A (1, 0, - 1)

B (- 1, 0, - 2)

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

(1, 0)

The area of the region

${x, y) : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1, 0 \leq x \leq 2}$

(in sq. units)

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