If f(x) = x^3-x^2+100 ,x , +1001 ,; then - Vidyadip

MCQ

If $f(x) = x^3-x^2+100\,x \, +1001\,;$ then

A
$f(2010) > f(2011)$
B
$f(3x -5) > f(3x)$
C
$f(x + 1) < f(x -1)$
✓
$f\left( {\frac{1}{{1999}}} \right) > f\left( {\frac{1}{{2000}}} \right)$

✓

Answer

Correct option: D.

$f\left( {\frac{1}{{1999}}} \right) > f\left( {\frac{1}{{2000}}} \right)$

d
$f(x)=x^{3}-x^{2}+100 x+1001$

$f^{\prime}(x)=3 x^{2}-2 x+100>0 \forall x \in R$

Therefore, $f(x)$ is increasing (strictly).

Therefore,

$f\left(\frac{1}{1999}\right)>f\left(\frac{1}{2000}\right)$

$\Rightarrow f(x+1)>f(x-1)$

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