Algebra> functions...

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)

mohammed saquib iqbal , 15 Years ago

Grade 12

2 Answers

dvm srikant

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

Last Activity: 15 Years ago

jagdish singh singh

$$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 $2000>1999\Leftrightarrow \frac{1}{2000}<\frac{1}{1999}\Leftrightarrow f(\frac{1}{2000})<f(\frac{1}{1999})$\\\\ Similarly $x+1>x-1\Leftrightarrow f(x+1)>f(x-1)$\\\\ Similarly $3x>3x-5\Leftrightarrow f(3x)>f(3x-5)$\\\\ So (ii) and (iii) option is Right.$