How Can Sequences and Series Prove a Real Solution to the Equation x^11+2x^5=2?

[pablito21]

A={xεR:X^11+2X^5<2} let a=supA By choosing a suitable sequence of elements of belonging to A and which tends to a as n->inf, or otherwise, show that a^11+2a^5=<2.Choose another sequence this time of all real numbers not belonging to A to show that a^11+2a^5>=2 and hence show that a^11+2a^5=2,so the equation x^11+2x^5=2 has a real solution

any help would be really appreciated!how can i solve it

 

[Office_Shredder]

For the first sequence, use a-1/n. a-1/n<a for all n, so is in A (since the function is increasing, all elements less than a are in A). When you plug a-1/n into the function, you get something like:

a^11 + 2a^5 +1/n*(bunch of stuff) < 2

Similiarly for taking a+1/n for the second part gives

a^11 + 2a^5 + 1/n*(bunch of stuff) >=2

Try to work it from there