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
$f(x)=x^{3}-x^{2}+100 x+1001$
$f(x)=x^{3}-x^{2}+100 x+1001$
$f^{\prime}(x)=3 x^{2}-2 x+100>0 \forall x \in R$
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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