1my method is also same as shubodip's
i considered
P(x)=f(x)-1
since -a,-b,-c are roots of this quadratic ,means its an identity
so ,it is true for all x
hence f(x) is a constant function
$Q:$\Rightarrow$ Determine The range of the function\\\\ $f(x)=\frac{(x+a)^2}{(a-b).(a-c)}+\frac{(x+b)^2}{(b-c).(b-a)}+\frac{(x+c)^2}{(c-a).(c-b)}$\\\\ Where $a,b$ and $c$ are Different Real no.
-
UP 0 DOWN 0 0 6
6 Answers
21f(a) = f(b) = f(c) = 1
By Rolle's theorem f'(x) = 0 ,somewhere between a and b , and b and c. i.e f'(x) has two zeros.
but f is a polynomial function of degree two.
so it can't have two roots.
f'(x) must be zero for all real x
so f(x) must be constant
f(x) = f(a) = 1
1
1or we can simply find coefficients of x^2, and x, by taking l.c.m, they'll come out to be 0..
and the constant term will come out to be 1..
1@ seoni i seriously dont mean to offend you , but the question setter doesnt want you do that way