4I think ans is c
bx + c = I - x^2 (If I is an integer)
bx + c ε Z
As Nishant Sir said in the below post c ε Z
So , bx ε Z
if x = 1 , b ε Z
So , both b & c must be integers
ANS c
f(x)=x^2+bx+c.
If f(x) assumes only interger values for all integer x then
:
a)b must be integer but c may not be.
b)c must be integer but b may not be.
c)b+c must be integer.
d)None of these.
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8 Answers
4
62THis is not very tough..
if this is integer for all integer values of x, then it is also an integer for x=0
hence f(0) is integer.. so c is an integer..
now can you try the whole question and get the right answer?
62No uttara there is a small mistake
Hint: See what happens if x=1 is substituted..
11Sir,i had already done it by putting f(0).
So, c is an integer.
Now f(x is an int;
X^2 is an int;
So,f(x)-x^2-c is an integer i.e bx is an int
X is an int.
Now,b=int /Some int which may not be an integer.(eg.2/3)
So, the correct ans should be b.
But ans given in the book is c(Where am i going wrong?!!)
62if it c .. the book is correct
f(0)=c is an integer
f(1) = 1+b+c is an integer
hence b is an integer
so b+c is an integer... hence the answer is c....
@Sagnik.. you are not wrong.. only that you are not using all the information availlable..
1@Nishant sir..as u said when x=1,then
f(1)=1+b+c
since
f(1) is an integer..then b+c is an integer....i think i did not get ur point!!!
621+b+c is == f(1) which is an integer..
so b+c = f(1) -1 has to be an integer as well..
I thought this was obvious!