13 Replies
@Apu
Note for OP
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Wait, why is degree 1?
There are 2 roots, precisely -1 and 0
After this, summation 1/f(r)=1/2(1/x-1/(x+2)); And hence this gets telescoped.
how do u get them
i understand how u related f^2(x) and f(x+1)
after that i need explanation
damn but howd you know to check at -2
The degree is the highest power of x in the expression. If the degree of f(x) is n, f^2(x) will have an order of 2n, since the highest power is the x-term gets squared. f(x+1) has the same order as f(x). By given constraints, we find 2n=n+2 or n=2. The expression is a bi-quadratic. We know that the 2 roots are 0 and -2. (From option B and plugging x=0). Now we plug in x=-1 to find f(x)=-1. Hence, the required quadratic is x(x+2).
The term $(x+2)\sqrt{1+(x+3)\sqrt{1+(x+4....)}}$ becomes zero 'cuz of the (x+2) term. So we have $-2\sqrt{1+(-2+1)}=0$
SirLancelotDuLac
Yes but like if it was any higher degree or a diff polynomial wouldn't we need to check for -3,-4,etc
But here we know the degree is 2. (Refer to the solution above) and hence we stop after finding 2 roots.
Sorted?