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Equation `1+x^2+2x"sin"(cos^(-1)y)=0`is satisfied byexactly one value of `x`exactly two value of `x`exactly one value of `y`exactly two value of `y`A. exactly one value of xB. exactly two values of xC. exactly one value of yD. exactly two values of y |
Answer» Correct Answer - A::C Given equation is `x^(2) + 2x sin(cos^(-1) y) + 1 =0` Since x is real, `D ge 0`. Therefore, `4(sin(cos^(-1)y))^(2) -4 ge 0` or `(sin(cos^(-1)y))^(2) ge 1` or `sin(cos^(-1)y) = +- 1` or `cos^(-1) y = (pi)/(2) rArr y = 0` Putting value of y in the original equation, we have `x^(2) + 2x + 1 = 0 rArr x -1` Hence, the equation has only one solution |
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