If y = a log |x| + b x^2 + x has its extremum | vedclass

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(B) Given $y = a \log |x| + b x^2 + x$. Since $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x 
eq 0$,we differentiate $y$ with respect to $x$: $\fra

Vedclass · Accepted Answer

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

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(B) Given $y = a \log |x| + b x^2 + x$. Since $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x 
eq 0$,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extremum values at $x = -1$ and $x = 2$,the derivative must be zero at these points. For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$ (Equation $1$). For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -\frac{1}{2}$. Substituting $b = -\frac{1}{2}$ into Equation $1$: $a + 2(-\frac{1}{2}) = 1 \implies a - 1 = 1 \implies a = 2$. Thus,$a = 2$ and $b = -\frac{1}{2}$.

If $y = a \log |x| + b x^2 + x$ has its extremum values at $x = -1$ and $x = 2$,then

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