23no one??? [12]????????
let P[x] = x3 - rx + r + 11 be a polynomial , where r is a +ve integer . Let r vary
[Q1] the no . positive integral solutions of p[x] = 0 are
[a] 6 [b) 3 [c] nil [d] infinitely many
[Q2] the sum of all +ve integral roots of p[x] = 0
is
[a] 34 [b) 21 [c] -12 [d] infinity
[PS : i got ans of Q1 as [a]...i.e 6+ ve roots and also got their sum as 34 ....but not sure ...coz option [d] can also b correct...... pls confirm ]
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11 Answers
23yaar koi to reply karo ..... [7]
admins ye bhi post dekho pls !! [2]
@ kaymant sir ,nishant sir ...and oder expert members pls help
39i have a doubt::
let us suppose the roots are a,b,c
then we have a+b+c=0
ab+bc+ca=-r
abc=-(r+11)
abc is negative implies all a,b,c should be negative or only one of them is negative.
all negative is not possible as their sum is 0.
so one of them is negative.. thas what i am getting.
but there is no such option for the first question?????
23yes u r correct ..but that is true for only 1 value of r ...
for example ..... for r= 19,
we get x = 2,3 as +ve integers ...
for other values of r ..u wil get other values of x ....
it is given that r can vary ..
39dude, how can u vary `r`?
its given r` is a positive integer
its not given `r` is any postive integer
23its given in the question dude .......
they hav written dat " r can vary " thats y i m varying it ..LOL
39sorry , i din see that before.
then for (1) the asnwer has to be `d`
and even for (@) it has to be `d` then.
u have the asnwers?
23its like consider the general form of a circle .....
x2 + y2 = a2
where we kno dat r is constant
for a particular circle suppose a = 5
but this doesnt mean dat all the circles in the world hav radius as 5 ....
that means a varies
23i don hav d answers ..... but i also think dat it shud be [d] in both cases
but i hav done it dis way
x3 - rx + r + 11 = 0
so x3 + 11 = r
(x - 1)
i tried putting x = 1,2,3,4......
upto 13 i got 6 values of x and their sum is also 34 ....
and after 13 i m not getting any suitable value of x ...
i tried upto next 10 to 15 numbers ...but i don think thats how dis problem is to be solved .........hence i m confused since option 34 is der in d next ans
my method seems a bit rash.....
it shud hav some formal method ...with wich we can solve the ques .....
it is a question frm the major tests of iitian's pace ..a famous coaching institute in bombay
66Qwerty, I think what you have done could be formalized as follows:
We require values of x for which
x3+11x-1
is an integer. The above fraction is same as
x3-1+12x-1=x2+x+1 + 12x-1
Obviously, x≠1. For any other integer x, the epression x2+x+1 is always an integer, so we require all positive integers x such that x-1 should divide 12. This is obviously true only for x = 2, 3, 4, 5, 7, 13
So these must be the required values. So for Q1, the correct option is (a) while for Q2, its again (a).