me book mathematics

Question

Which of following function is strictly bounded?

Comments

0 <= 1/(e^-(x^2)) <= e 

 How you know the range ?? in which case you get 0 as a lower bound can I know ???

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x square doesn't have an upper bound

I already say it's a bounded below function

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For any function to be bounded the range should be finite on both upper and lower bounds, so $x^2$ cannot be bounded

Answer 1

Ans (D) $e^{-x^{2}}$

$1/x^{2}$ when x tends to infinity becomes 0 and when x tends to 0 becomes infinity, therefore it has no bounds range [0,$\infty$]

Similarly for $x^{2}$ when x tends to infinity becomes infinity and when x tends to 0 becomes 0 , Again [0,$\infty$]

For $e^{x}$ when x tends to infinity and zero it becomes infinity and 1 respectively (Anything to power 0 is 1) Again [0,$\infty$]

But for $e^{-x^{2}} = 1/e^{x^{2}}$ becomes 0 and 1 when x tends to infinity and 0 respectively. range [0,1]

As x→∞⇒e−∞=1/e∞=1∞→0 =0

For strictly bounded to be true, range should be [o, Finite]