Discussion Forum

Let f₁:R $\rightarrow$R, f₂:[0,$\infty$] $\rightarrow$ R, f_a : R $\rightarrow$ R

 and f₄:R$\rightarrow$ [0,$\infty$] be defined by

   $f_{1}(x)=\begin{cases}|x |& if x < 0\\e^{x} &if  x \geq 0\end{cases}$

        $f_{2}(x)=x^{2},$

$f_{3}(x)=\begin{cases}sin x &if  x < 0\\x &if  x \geq 0\end{cases}$

$f_{4}(x)=\begin{cases}f_{2}(f_{1}(x)) &if  x < 0\\f_{2}(f_{1}(x)) -1 &if  x \geq 0\end{cases}$

Answer:

Explanation:

(P) Plan

 (i)    For such  questions , we need to properly define the functions and then we draw their graphs.

  (ii)  From the graphs,we can examine the function for continuity differentiability , one-one and onto 

   $f_{1}(x)=\begin{cases}-x & x < 0\\e^{x} & x \geq 0\end{cases}$

        $f_{2}(x)=x^{2},x\geq0$

$f_{3}(x)=\begin{cases}sin x & x < 0\\x & x \geq 0\end{cases}$

$f_{4}(x)=\begin{cases}f_{2}(f_{1}(x)) & x < 0\\f_{2}(f_{1}(x)) -1 & x \geq 0\end{cases}$

 Now,   $f_{2}(f_{1}(x)=\begin{cases} x^{2},& x < 0\\e^{2x}, & x \geq 0\end{cases}$

$f_{4}=\begin{cases} x^{2},& x < 0\\e^{2x}-1, & x \geq 0\end{cases}$

 As f₄(x) is contiinuous

                $f'_{4}=\begin{cases} 2x,& x < 0\\2e^{2x}, & x > 0\end{cases}$

 $f'_{4}(0)$  is not defined

 Its range is [0, $\infty$]

 Thus, range= codomain =[0,$\infty$]  thus f₄  is onto.

 Also, horizontal line (drawn parallel to x-axis) meets  the curve more than once function is not one-one 

  (Q)  Plan  $f_{3}(x)$

   differentiable at x=0  and not one-one

  As evident,  from the graph it is  continuous and no sharp turn at X=0 , thus f(x)  is  differentiable  at x=0

 Also, a horizontal line intersects the graph more than once.

 $\therefore$    It is not one-one

 (R)   Plan $f_{2}(f_{1}(x))$

 It is neither continuous nor one-one.

 From the graph , it can be observed that the function is not continuous at x=0

  Also , the horizontal line intersect,the curve at more than one point So,  $f_{2}(f_{1}(x))$ , is not one-one

(S) f₂(x)

 It is continuous  and one-one

 As evident from graphs , the function is continuous 

 also, the function is one-one , as any horizontal line  will meet the graph  only one 

$P\rightarrow(i), Q\rightarrow(iii),R\rightarrow(ii),S\rightarrow(iv)$