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The absolute maximum & minimum values of functions can be found by their monotonic & asymptotic behavior provided they exist. We may agree that finite limiting values may be regarded as absolute maximum or minimum. For example, the absolute maximum value of $$\displaystyle \frac{1}{1+x^{2m}}\left ( m\epsilon N \right )$$ is 1. When $$x=0$$, on the other side absolute minimum value of the some function is 0, which is limiting value of the function when $$x\rightarrow -\infty $$ or $$x\rightarrow +\infty $$. Sometime $${f}'\left ( x \right )=0$$ & $${f}''\left ( x \right )=0$$ for $$x=a$$ but $$f^{'"}\left ( x \right )\neq 0$$ for $$x=a$$, then f(x) is neither absolute maximum nor absolute minimum at $$x=a$$, then $$x=a$$ is called point of inflexion.
On the basis of above information answer the following questions.

The function $$x^{4}-4x+1$$ will have


A
Absolute maximum value
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B
Absolute minimum value
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C
Both absolute maximum & minimum values
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D
None of these
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Solution

The correct option is B Absolute minimum value
Given $$f\left ( x \right )=x^{4}-4x+1$$
$$\Rightarrow $$   $${f}'\left ( x \right )=4\left ( x^{3}-1 \right )$$, so $${f}''\left ( x \right )=12x^{2}> 0\left ( \forall x\epsilon R \right )$$
$$\therefore $$   $${f}'\left ( x \right )=0$$
$$\Rightarrow $$ $$x=1$$
& $$x^{2}+x+1=0$$ which have no real value of $$x$$ as $$D=b^{2}-4ac< 0$$
$$\therefore $$ $$x=1$$ is only one extreme point at which f(x) attains absolute minimum value.
Ans: B

Mathematics

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