Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class12 Chapter Functions exercise 2.1  question 13 

Answers (1)

Given:

            f:R_0^+\rightarrow R given by f(x)\log _ax:a >0.

To prove:

            f(x)\log _ax:a  is bijection.

Hint:

For any function to be bijective,  the function should be one-one and onto.

Solution:

Let x,y \in R_0^+  such that

f(x)=f(y)

\begin{array}{ll} \Rightarrow & \log _{a} x=\log _{a} y \\ \Rightarrow & \log _{a} \frac{x}{y}=0 \\ \Rightarrow & \frac{x}{y}=1 \\ \Rightarrow & x=y \end{array}

\therefore f is one-one.

Now, let y\in R be arbitrary then

    f(x)=y

\begin{aligned} &\Rightarrow \quad \log _{a} x=y \\ &\Rightarrow \quad x=a^{y} \in R_{0}^{+} \\ &{\left[\because a>0 \Rightarrow a^{y}>0\right]} \end{aligned}                                                                                             

Thus for all y\in R, there exist x=a^y  such  that f(x)=y.

\therefore f is onto

\therefore f  is one-one and onto.

Hence f  is bijective.

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads