मराठी

Show that F(X) = 1 1 + X 2 is Neither Increasing Nor Decreasing on R ?

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प्रश्न

Show that f(x) = \[\frac{1}{1 + x^2}\] is neither increasing nor decreasing on R ?

बेरीज
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उत्तर

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

\[\text { R can be divided into two intervals }\left( 0, \infty \right)\text { and }( - \infty , 0] . \]

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

\[ x_1 < x_2 \]

\[ \Rightarrow {x_1}^2 < {x_2}^2 \]

\[ \Rightarrow 1 + {x_1}^2 < 1 + {x_2}^2 \]

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

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

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

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

\[ x_1 < x_2 \]

\[ \Rightarrow {x_1}^2 > {x_2}^2 \]

\[ \Rightarrow 1 + {x_1}^2 > 1 + {x_2}^2 \]

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

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

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

\[\text { Here }, f\left( x \right)\text { is decreasing on}\left( 0, \infty \right)\text { and increasing on }( - \infty , 0].\]

\[\text { Thus },f\left( x \right) \text { is neither increasing nor decreasing on R } . \]

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पाठ 16: Increasing and Decreasing Functions - Exercise 17.1 [पृष्ठ १०]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 16 Increasing and Decreasing Functions
Exercise 17.1 | Q 7 | पृष्ठ १०

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