If $f(x)=\left\{\begin{array}{ll}2 x^2+\frac{2}{x^2}, & 0 2\end{array}\right.$, then

$x=1,-1$ are the points of global minima

$x=0$ is the points of local minima

JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Analyzing +4 | -1

f(x)=\left\{\begin{array}{ll}2 x^2+\frac{2}{x^2}, & 0<|x| \leq 2 \\ 3, & x>2\end{array}\right.

, then

View answer and explanation

Correct option: B

x=1,-1

are the points of local minima

Option B (

x=1,-1

are the points of local minima) is correct because the second derivative is positive at these points.

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p

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