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Let $f(x)=x^3+3x^2+6x+2009$ and $g(x)=\large\frac{1}{x-f(1)}+\frac{2}{x-f(2)}+\frac{3}{x-f(3)}$.The no of real solutions of $g(x)$=0 is :

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;2&(D)\;\text{infinite}\end{array} $

1 Answer

$f'(x)=3x^2+6x+6$
$\Rightarrow 3(x+1)^2+3 > 0\forall x$
$\Rightarrow f(x)$ increases on R
Let $f(1)=a,f(2)=b,f(3)=c$ then $a < b< c$ and
$g(x)=(x-b)(x-c)+2(x-a)(x-c)+3(x-a)(x-c)$
$g(a) >0,g(b)<0$ and $g(c) > 0,g(x)=0$ has exactly two real solution.
Hence (C) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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