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Explain solution RD Sharma class 12 chapter Functions exercise 2.1 question 12 maths

Answers (1)

Given:

f:R \rightarrow R, given byf(x)=e^x.

To prove:

f  is one-one but not onto.

Hint:

If different element in A have distinct images in B then the function is one-one and  if every element in B has at least one pre-image in A then the function is onto.

Solution:

First we will check whether the function is one-one or not.

Here, let x,y \in R such that

f(x)=f(y)

\begin{array}{ll} \Rightarrow & e^{x}=e^{y} \\ \Rightarrow & e^{x-y}=1=e^{0} \\ \Rightarrow & x-y=0 \\ \Rightarrow & x=y \end{array}                                                                                              

\therefore f is one-one

Now, we check onto

f(x)=e^x

Let f(x)=y such that y \in R

y=e^x

clearly the range ofe^{x} \text { is }(0, \infty)=R^{+} but the given range isR

co-domain≠ Range

\therefore f is not onto.

Note:

When co-domain is replaced byR_0^+, i.e. (0, \infty) then f become onto function.    

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infoexpert24

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