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  • please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise 16.1 , question 4 maths textbook solution

please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise 16.1 , question 4 maths textbook solution

Answers (1)

               f(x)=ax+b is a decreasing function on R.

Hint:

               A function f(x)is said to be a decreasing function on (a,b), if x_{1}<x_{2}\Rightarrow f(x_{1})>f(x_{2}) for all x_{1},x_{2}\epsilon (a,b).

Given:

               f(x)=ax+b, where a,b are constant and a<0

To prove:

               f(x)=ax+b is a decreasing function on R.

Solution:

Let x_{1},x_{2}\epsilon R Such that x_{1}< x_{2}

Then,
             \begin{aligned} & x_{1}<x_{2} \\ \Rightarrow \quad & a x_{1}>a x_{2} \end{aligned} \quad[\because a<0]

\Rightarrow \quad a x_{1}+b>a x_{2}+b     [Given that b is constant]

\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right)

Thus, x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right) for all x_{1},x_{2}\epsilon R

So, f(x)=ax+b is a decreasing function on R.

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