If the tangent to the curve y = x^3 + ax - b | vedclass

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(A) Given the curve $y = x^3 + ax - b$. Since the point $(1, -5)$ lies on the curve,we have: $-5 = (1)^3 + a(1) - b$ $-5 = 1 + a - b$ $a - b = -6$ $\d

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$(2, -2)$

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(A) Given the curve $y = x^3 + ax - b$. Since the point $(1, -5)$ lies on the curve,we have: $-5 = (1)^3 + a(1) - b$ $-5 = 1 + a - b$ $a - b = -6$ $\dots(i)$ The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 3x^2 + a$. At the point $(1, -5)$,the slope of the tangent is $m_1 = 3(1)^2 + a = 3 + a$. The given line is $-x + y + 4 = 0$,which can be written as $y = x - 4$. The slope of this line is $m_2 = 1$. Since the tangent is perpendicular to the line,the product of their slopes must be $-1$: $m_1 	imes m_2 = -1$ $(3 + a)(1) = -1$ $3 + a = -1$ $a = -4$. Substituting $a = -4$ into equation $(i)$: $-4 - b = -6$ $-b = -2$ $b = 2$. Thus,the equation of the curve is $y = x^3 - 4x - 2$. Now,we check the given options: For $(2, -2)$: $y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2$. Since the point $(2, -2)$ satisfies the equation,it lies on the curve.

If the tangent to the curve $y = x^3 + ax - b$ at the point $(1, -5)$ is perpendicular to the line $-x + y + 4 = 0$,then which one of the following points lies on the curve?

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