# Q :if f(x) = x3 – x2 + 100x + 1001, then(a) f(2000) > f(2001)(b) f(1/1999)  > f (1/2000)(c) f(x+1) > f(x–1)(d) f(3x – 5) > f(3x)why....??  (reason)

dvm srikant
39 Points
13 years ago

friend you directly substitute the options in question for 3 and 4 and take any x value and verify and for 1 and d also substitute and approximate the value                                                                                                                                                                                          pls approve

jagdish singh singh
173 Points
13 years ago

$Here f(x)=x^3-x^2+100x+1001\\\\ Now Diff. both side w.r.to x, We Get\\\\ f^{'}(x)=3x^2-2x+100>0\forall x\in\mathbb{R}\\\\ So f(x) is Strictly increasing function.\\\\ i.e for \boxed{x_{1}>x_{2}\Leftrightarrow f(x_{1})>f(x_{2})}\\\\ Here 2001>2000\Leftrightarrow f(2001)>f(2000)\\\\ Similarly 2000>1999\Leftrightarrow \frac{1}{2000}<\frac{1}{1999}\Leftrightarrow f(\frac{1}{2000})x-1\Leftrightarrow f(x+1)>f(x-1)\\\\ Similarly 3x>3x-5\Leftrightarrow f(3x)>f(3x-5)\\\\ So (ii) and (iii) option is Right.$