If y = a ln x + bx^2 + x has its extremum val | vedclass

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(D) Given the function $y = a \ln x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extremum values

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$ (-2/3, -1/6) $

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(D) Given the function $y = a \ln x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extremum values at $x = 1$ and $x = 2$,the derivative must be zero at these points. At $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0 \implies a + 2b = -1$ (Equation $1$). At $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b = -1 \implies a + 8b = -2$ (Equation $2$). Subtract Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - (-1) \implies 6b = -1 \implies b = -1/6$. Substitute $b = -1/6$ into Equation $1$: $a + 2(-1/6) = -1 \implies a - 1/3 = -1 \implies a = -1 + 1/3 = -2/3$. Thus,$(a, b) = (-2/3, -1/6)$.

If $y = a \ln x + bx^2 + x$ has its extremum value at $x = 1$ and $x = 2,$ then $(a, b) =$

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