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Let $a, b \in R$ be such that the function $f$ given by $f(x)=\ln |x|+b x^2+a x, x \neq 0$ has extreme values at $x=-1$ and $x=2$.
Statement 1: $f$ has local maximum at $x=-1$ and at $x=2$.
Statement 2: $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{-1}{4}$
Options:
Statement 1: $f$ has local maximum at $x=-1$ and at $x=2$.
Statement 2: $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{-1}{4}$
Solution:
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Verified Answer
The correct answer is:
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$
$f^{\prime}(x)=\frac{1}{x}+2 b x+a$
$f$ has extremevalues and differentiable
$\Rightarrow f^{\prime}(-1)=0 \quad \Rightarrow a-2 b=1$
$f^{\prime}(2)=0 \quad \Rightarrow a+4 b=-\frac{1}{2} \quad \Rightarrow a=\frac{1}{2}, b=-\frac{1}{4}$
$f^{\prime \prime}(-1), f^{\prime \prime}(2)$ are negative. $f$ has local maxima at $-1,2$
$f$ has extremevalues and differentiable
$\Rightarrow f^{\prime}(-1)=0 \quad \Rightarrow a-2 b=1$
$f^{\prime}(2)=0 \quad \Rightarrow a+4 b=-\frac{1}{2} \quad \Rightarrow a=\frac{1}{2}, b=-\frac{1}{4}$
$f^{\prime \prime}(-1), f^{\prime \prime}(2)$ are negative. $f$ has local maxima at $-1,2$
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