Question

If f(x)=x^2+lambda x +u be an integral function of the integral variable x then
1)lambda is an integer and u is a rational fraction
2)lambda and u are integers
3)u is an integer and lambda is a rational funcion
4)lambda and u are rational functions

Answer 1

@yrelhan4 @shubhamsrg @electrokid

Answer 2

\[\Large f(x)=x^2+\lambda x+u\]

Answer 3

Since X is an integer variable,lets fix it as 2
now I don't think lambda has to be necessarily integer,
lets take lambda as 2.5 so it becomes
\[2^2+2 \times 2.5 +u\]
still the condition is satisfied,the overall function is yielding integral value.
while u has to be integer,I'm not quite sure about my views though ^_^

Answer 4

It can't be 1
It can't be 3
It can't be 4
It has to be (2) but If I had an option like lambda CAN be a decimal but u will always be integer I would prefer it or not? However it only implies for rare values,it won't apply for 1.1,1.2..etc..

Answer 5

@shubhamsrg THATS THE BEST I CAN DO. \m/

Answer 6

:|

Answer 7

how about approaching it this way
assume "x" and f(x) are rationals and cane be differentiated
the \[f(x_0+\Delta x)=f(x_0)+\Delta xf'(x_0)+\frac{\Delta x^2}{2}f''(x_0)\\
f(x_0+\Delta x)=f(x_0)+(2x_0+\lambda)\Delta x+(\Delta x)^2\\\;\\
n+1=n+(2+\lambda)+1\implies \lambda=-2\\\;\\
n-1=n+(2+\lambda)+1\implies\lambda=-4
\]
hence, we may conclude that both lambda and u must be integers

Answer 8

but if you have had
\[f(n)=an^2+bn+c\]
where n and f(n) are integers,
then all of them must be rational functions.