If y = a log x + bx^2 + x has extreme values | vedclass

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(D) Given $y = f(x) = a \log x + bx^2 + x$. First,find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values

Vedclass · Accepted Answer

$a = 2, b = -\frac{1}{2}$

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(D) Given $y = f(x) = a \log x + bx^2 + x$. First,find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = -1$ and $x = 2$,the derivative must be zero at these points. For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$. For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$. Subtracting the first equation from the second: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -\frac{1}{2}$. Substituting $b = -\frac{1}{2}$ into $a + 2b = 1$: $a + 2(-\frac{1}{2}) = 1 \implies a - 1 = 1 \implies a = 2$. Thus,$a = 2$ and $b = -\frac{1}{2}$.

If $y = a \log x + bx^2 + x$ has extreme values at $x = -1$ and $x = 2$,then:

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