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  • please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise 16.1 , question 2 maths textbook solution

please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise 16.1 , question 2 maths textbook solution

Answers (1)

Answer:

             f(x)  is decreasing on(0,\infty )if 0<a<1

Hint:

  1. f(x)is increasing on(a,b), if the values of f(x) increase with the increase in the values of x.
  2. f(x)is decreasing on(a,b), if the values of f(x) decrease with the increase in the values of x.

Given:

Here given that,

               f(x)=\log _{a} x

To prove:

               Function f(x)=\log _{a} x is increasing on (0,\infty ), if a>1 and decreasing on (0,\infty ) , if 0<a<1

Solution:

Here we have two cases:

Case I: when a>1

Let us consider x_{1},x_{2}\epsilon (0,\infty )

Such that x_{1}<x_{2}
\begin{array}{ll} \Rightarrow & \log _{a} x_{1}<\log _{a} x_{2} \\ \Rightarrow & f\left(x_{1}\right)<f\left(x_{2}\right) \end{array}

Thus, f(x) is increasing in(0,\infty )if a>1

Case II: when 0<a<1

We have f(x)=\log _{a} x

We know that \log _{a} x=\frac{\log x}{\log a}

Therefore, f(x)=\frac{\log x}{\log a}

When a<1, then a<0

Let x_{1}<x_{2}

Taking log on both sides,

\begin{aligned} &\Rightarrow \quad & \log x_{1}<\log x_{2} \\ &\Rightarrow & \frac{\log x_{1}}{\log a}>\frac{\log x_{2}}{\log a} \\ &\Rightarrow & f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}

 

Thus, f(x) is decreasing on \left ( 0,\infty \right ) if 0<a<1

 

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