A polyomial problem to show that Q (x)/$x^k $ is strictly positive

Question

I am extremely sorry I couldnot type as it requires much time to do MathJax.

This problem Question 8 was asked in an entrance examination at the 10+2 level. My approqch for part (a) was to use differentiation and find Q'(x) but it is not so ....helping me to arrive at the result. Any hints or help is appreciable.

Thank you

Notice that $Q(x)/x^k$ has value $1$ at $x=0$. Also notice that $|x|<1$ so higher powers of $x$ are smaller in absolute value. The triangle inequality is your friend. — Jyrki Lahtonen, Dec 03 '20 at 05:23
Dhanvi...can you tell me how to check if a problem is already existing or has been asked...this helps me to check before posting — , Dec 03 '20 at 05:27

score 1 · Answer 1 · answered Dec 03 '20 at 05:18

1

(Fill in the gaps as needed. If you're stuck, show what you've considered and explain why you're stuck.)

Hint: Show that $ |x^k | > |\sum a_i x^{k+i}|$.

answered Dec 03 '20 at 05:18

Calvin Lin

68,864

Can you tell me how to proceed from here... I am just a bit slow at learning. .so please @Calvin Lin – Maths lover Dec 03 '20 at 06:08
You may want to specify the domain of the inequality – tryst with freedom Jul 06 '21 at 22:55
@Buraian The domain that we're interested in is stated in the question, so I'm not sure why I should specify that again. – Calvin Lin Jul 07 '21 at 01:08

Gabriel Burzacchini · Answer 2 · 2020-12-03T05:32:21.243

I don't know if it could be the correct method, but I show you what I was thinking of. So, imagine to plug arbitrary numbers inside the inequality. You'd notice that the sum (with sigma symbol) is always positive (because it starts with 1 and approaches to positive infinity). Additionally, it has an absolute value, so it is surely always positive. the fraction is always positive too, because it has "1+[always positive stuff]". It approaches to 0 from the right (to prove that, imagine to plug every positive or negative number and see what would happen). Therefore, it's always positive. zero isn't included because of the sign of the inequality "i.e <, which means "is less than and not equal to [something]".

Moreover, x is inside the absolute value too, so you could surely say Q(X) is always positive.

EDIT (because I forgot the main part of this exercise): imagine to plug arbitrary values inside "k" variable (because k is a variable), you'd notice that it doesn't matter what you plug into it, because you'd get a positive value. secondly, if x is negative, it's not a problem because in the inequality you have the absolute values.

A polyomial problem to show that Q (x)/$x^k $ is strictly positive

2 Answers2