Prove that the function f x = log a x is increasing on 0 , ∞ if a > 1 and decreasing on 0 , ∞ , if 0 < a < 1 .

f ( x ) = log a x , x ∈ ( 0 , ∞ ) = log x log a ⇒ f ' ( x ) = 1 x log a Case (I): a > 1 ⇒ log a > 0 ⇒ 1 x log a > 0 ⇒ f ' ( x ) > 0 ⇒ f ( x ) is increasing function. Case (II): 0 < a < 1 ⇒ log a < 0 ⇒ 1 x log a < 0 ⇒ f ' ( x ) < 0 ⇒ f ( x ) is decreasing function.

Show that f x = 1 x is a decreasing function on 0 , ∞ .

Here, f x = 1 x Let x 1 , x 2 ∈ 0 , ∞ such that x 1 < x 2 . Then, x 1 < x 2 ⇒ 1 x 1 > 1 x 2 ⇒ f x 1 > f x 2 ∴ x 1 < x 2 ⇒ f x 1 > f x 2 , ∀ x 1 , x 2 ∈ 0 , ∞ So, f x is decreasing function on 0 , ∞ .

Show that f x = 1 1 + x 2 decreases in the interval [ 0 , ∞ ) and increases in the interval ( − ∞ , 0 ] .

Here, f x = 1 1 + x 2 Case 1 : Let x 1 , x 2 ∈ 0 , ∞ such that x 1 < x 2 . Then, x 1 < x 2 ⇒ x 1 2 < x 2 2 ⇒ 1 + x 1 2 < 1 + x 2 2 ⇒ 1 1 + x 1 2 > 1 1 + x 2 2 ⇒ f x 1 > f x 2 ∀ x 1 , x 2 ∈ 0 , ∞ So, f x is decreasing on 0 , ∞ . Case 2 : Let x 1 , x 2 ∈ ( - ∞ , 0 ] such that x 1 < x 2 Then, x 1 < x 2 ⇒ x 1 2 > x 2 2 ⇒ 1 + x 1 2 > 1 + x 2 2 ⇒ 1 1 + x 1 2 < 1 1 + x 2 2 ⇒ f x 1 < f x 2 ⇒ f x 1 < f x 2 , ∀ x 1 , x 2 ∈ ( - ∞ , 0 ] So, f x is increasing on ( - ∞ , 0 ] .

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MATHEMATICS CLASS XII VOLUME-1>Chapter 17 - Increasing and Decreasing Functions>EXERCISE

R D Sharma Solutions for Chapter: Increasing and Decreasing Functions, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Increasing and Decreasing Functions, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 17: Increasing and Decreasing Functions, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.