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The minimum value of the function $f(x)=2 x^2-\ln |x|$, when $x \geq 1$ is
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Verified Answer
The correct answer is:
$2$
Wc have,
$f(x)=2 x^2-\ln |x|$
$\therefore \quad f^{\prime}(x)=4 x-\frac{1}{x}$
For minima
$f^{\prime}(x)=0$
$\begin{aligned} & \Rightarrow \quad 4 x-\frac{1}{x}=0 \Rightarrow 4 x^2-1=0 \\ & \Rightarrow \quad x= \pm \frac{1}{2} \\ & \text { But }\end{aligned}$
$\therefore \quad f_{\min }=f(1)=2-\ln \mathrm{l}=2$
$f(x)=2 x^2-\ln |x|$
$\therefore \quad f^{\prime}(x)=4 x-\frac{1}{x}$
For minima
$f^{\prime}(x)=0$
$\begin{aligned} & \Rightarrow \quad 4 x-\frac{1}{x}=0 \Rightarrow 4 x^2-1=0 \\ & \Rightarrow \quad x= \pm \frac{1}{2} \\ & \text { But }\end{aligned}$
$\therefore \quad f_{\min }=f(1)=2-\ln \mathrm{l}=2$
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