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(C) Given $f(x) = \ln|x| + bx^2 + ax$. The derivative is $f'(x) = \frac{1}{x} + 2bx + a$.Since $f(x)$ has extreme values at $x = -1$ and $x = 2$,we mu

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Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$.

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(C) Given $f(x) = \ln|x| + bx^2 + ax$. The derivative is $f'(x) = \frac{1}{x} + 2bx + a$.Since $f(x)$ has extreme values at $x = -1$ and $x = 2$,we must have $f'(-1) = 0$ and $f'(2) = 0$.For $x = -1$: $-1 - 2b + a = 0 \implies a - 2b = 1$.For $x = 2$: $\frac{1}{2} + 4b + a = 0 \implies a + 4b = -\frac{1}{2}$.Subtracting the equations: $(a + 4b) - (a - 2b) = -\frac{1}{2} - 1 \implies 6b = -\frac{3}{2} \implies b = -\frac{1}{4}$.Substituting $b = -\frac{1}{4}$ into $a - 2b = 1$: $a - 2(-\frac{1}{4}) = 1 \implies a + \frac{1}{2} = 1 \implies a = \frac{1}{2}$.Thus,Statement-$2$ is true.Now,$f''(x) = -\frac{1}{x^2} + 2b = -\frac{1}{x^2} - \frac{1}{2}$.At $x = -1$,$f''(-1) = -1 - \frac{1}{2} = -\frac{3}{2} At $x = 2$,$f''(2) = -\frac{1}{4} - \frac{1}{2} = -\frac{3}{4} Thus,Statement-$1$ is true and Statement-$2$ is a correct explanation for Statement-$1$.

Let $a, b \in R$ be such that the function $f(x) = \ln|x| + bx^2 + ax, x \neq 0$ has extreme values at $x = -1$ and $x = 2$.
Statement-$1$: $f$ has a local maximum at $x = -1$ and $x = 2$.
Statement-$2$: $a = \frac{1}{2}$ and $b = -\frac{1}{4}$.

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Let $a, b \in R$ be such that the function $f(x) = \ln|x| + bx^2 + ax, x \neq 0$ has extreme values at $x = -1$ and $x = 2$.Statement-$1$: $f$ has a local maximum at $x = -1$ and $x = 2$.Statement-$2$: $a = \frac{1}{2}$ and $b = -\frac{1}{4}$.