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(D) For the quadratic equation $(a + b)x^2 + 2bx + 1 = 0$ to have real and distinct roots,the discriminant $D$ must be greater than $0$.$D = (2b)^2 -

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lies outside the parabola $y = x^2 - x$

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(D) For the quadratic equation $(a + b)x^2 + 2bx + 1 = 0$ to have real and distinct roots,the discriminant $D$ must be greater than $0$.$D = (2b)^2 - 4(a + b)(1) > 0$$4b^2 - 4a - 4b > 0$$b^2 - b > a$Let $f(x, y) = y - (x^2 - x)$. For the point $(b, a)$,we evaluate $f(b, a) = a - (b^2 - b) = a - b^2 + b$.Since $b^2 - b > a$,it follows that $a - b^2 + b This implies that the point $(b, a)$ lies inside the parabola $y = x^2 - x$ (or below the curve $y = x^2 - x$).However,checking the options provided,the condition $a < b^2 - b$ corresponds to the region inside the parabola $y = x^2 - x$.

If the quadratic equation $(a + b)x^2 + 2bx + 1 = 0$ has real and distinct roots,then a point having coordinates $(b, a)$:

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