Functions Ques 5
- Match the conditions/expressions in Column I with statement in Column II.
$ \begin{aligned} & \text { Let } f_1: R \rightarrow R, f_2:[0, \infty] \rightarrow R, f_3: R \rightarrow R \text { and } \\ & f_4: R \rightarrow[0, \infty) \text { be defined by } \\ & f_1(x)=\left\{\begin{array}{rl} |x|, & \text { if } x<0 \\ e^x, & \text { if } x \geq 0 \end{array} ; f_2(x)=x^2 ; f_3(x)=\left\{\begin{aligned} \sin x, & \text { if } x<0 \\ x, & \text { if } x \geq 0 \end{aligned}\right.\right. \\ & \text { and } \quad f_4(x)=\left\{\begin{array}{cc} f_2\left[f_1(x)\right], & \text { if } x<0 \\ f_2\left[f_1(x)\right]-1, & \text { if } x \geq 0 \end{array}\right. \\ \end{aligned} $
| Column I | Column II | ||
|---|---|---|---|
| A. | $t_4$ is | p. | onto but not one-one |
| B. | $f_3$ is | q. | neither continuous nor one-one |
| C. | $t_2 o f_1$ is | r. | differentiable but not one-one |
| D. | $t_2$ is | s. | continuous and one-one |
Codes
| A | B | C | D | |
|---|---|---|---|---|
| (a) | r | p | s | q |
| (b) | p | r | s | q |
| (c) | r | p | q | s |
| (d) | p | r | q | s |
Show Answer
Answer:
Correct Answer: 5.(d)
Solution: (d) 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.
$ \begin{aligned} & f_1(x)= \begin{cases}-x, & x<0 \\ e^x, & x \geq 0\end{cases} \\ & f_2(x)=x^2, x \geq 0 \\ & f_3(x)= \begin{cases}\sin x, & x<0 \\ x, & x \geq 0\end{cases} \\ & f_4(x)= \begin{cases}f_2\left(f_1(x)\right), & x<0 \\ f_2\left(f_1(x)\right)-1, & x \geq 0\end{cases} \\ \end{aligned} $
Now, $\quad f_2\left(f_1(x)\right)= \begin{cases}x^2, & x<0 \\ e^{2 x}, & x \geq 0\end{cases}$
$\Rightarrow \quad f_4= \begin{cases}x^2, & x<0 \\ e^{2 x}-1, & x \geq 0\end{cases}$
As $f_4(x)$ is continuous, $f_4^{\prime}(x)= \begin{cases}2 x, & x<0 \\ 2 e^{2 x}, & x>0\end{cases}$
$f_4^{\prime}(0)$ is not defined. Its range is $[0, \infty)$.
Thus, range $=$ codomain $=[0, \infty)$, thus $f_4$ is onto.
Also, horizontal line (drawn parallel to $X$-axis) meets the curve more than once, thus function is not one-one.
BETA
