Share

Books Shortlist

# Solution for Show that F(X) = 1 1 + X 2 Decreases in the Interval [0, ∞) and Increases in the Interval (−∞, 0] ? - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

Show that f(x) = $\frac{1}{1 + x^2}$ decreases in the interval [0, ∞) and increases in the interval (−∞, 0] ?

#### Solution

$\text { Here },$

$f\left( x \right) = \frac{1}{1 + x^2}$

$\text { Case 1: Let} x_1 , x_2 \in \left( 0, \infty \right) \text { such that } x_1 < x_2 . \text { Then },$

$x_1 < x_2$

$\Rightarrow {x_1}^2 < {x_2}^2$

$\Rightarrow 1 + {x_1}^2 < 1 + {x_2}^2$

$\Rightarrow \frac{1}{1 + {x_1}^2} > \frac{1}{1 + {x_2}^2}$

$\Rightarrow f\left( x_1 \right) > f\left( x_2 \right) \forall x_1 , x_2 \in \left( 0, \infty \right)$

$\text { So, }f\left( x \right) \text { is decreasing on }\left( 0, \infty \right).$

$\text { Case } 2: Let x_1 , x_2 \in ( - \infty , 0]\text { such that } x_1 < x_2 . \text { Then, }$

$x_1 < x_2$

$\Rightarrow {x_1}^2 > {x_2}^2$

$\Rightarrow 1 + {x_1}^2 > 1 + {x_2}^2$

$\Rightarrow \frac{1}{1 + {x_1}^2} < \frac{1}{1 + {x_2}^2}$

$\Rightarrow f\left( x_1 \right) < f\left( x_2 \right)$

$\Rightarrow f\left( x_1 \right) < f\left( x_2 \right), \forall x_1 , x_2 \in ( - \infty , 0]$

$\text { So, }f\left( x \right) \text { is increasing on } ( - \infty , 0].$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [3]

Solution Show that F(X) = 1 1 + X 2 Decreases in the Interval [0, ∞) and Increases in the Interval (−∞, 0] ? Concept: Increasing and Decreasing Functions.
S