The curve y = 2x^3 + ax^2 + bx + c passes thr | vedclass

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(B) Since the curve passes through the origin $(0, 0)$,we have $0 = 2(0)^3 + a(0)^2 + b(0) + c$,which implies $c = 0$.The derivative of the curve is $

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$-3, -12, 0$

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(B) Since the curve passes through the origin $(0, 0)$,we have $0 = 2(0)^3 + a(0)^2 + b(0) + c$,which implies $c = 0$.The derivative of the curve is $\frac{dy}{dx} = 6x^2 + 2ax + b$.Given that the tangents at $x = -1$ and $x = 2$ are parallel to the $X$-axis,their slopes must be zero.At $x = -1$: $6(-1)^2 + 2a(-1) + b = 0 \implies 6 - 2a + b = 0 \implies 2a - b = 6 \quad (1)$.At $x = 2$: $6(2)^2 + 2a(2) + b = 0 \implies 24 + 4a + b = 0 \implies 4a + b = -24 \quad (2)$.Adding equations $(1)$ and $(2)$:$(2a - b) + (4a + b) = 6 - 24$$6a = -18 \implies a = -3$.Substituting $a = -3$ into equation $(1)$:$2(-3) - b = 6 \implies -6 - b = 6 \implies b = -12$.Thus,the values are $a = -3, b = -12, c = 0$.

The curve $y = 2x^3 + ax^2 + bx + c$ passes through the origin,and the tangents at $x = -1$ and $x = 2$ are parallel to the $X$-axis. Then the values of $a, b,$ and $c$ are respectively:

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