Question

Let ${{f}_{1}}:R\to R,{{f}_{2}}:[0,\infty )\to R,{{f}_{3}}:R\to R$ and ${{f}_{4}}:R\to [0,\infty )$ defined by {{f}_{1}}\left( x \right)=\left\\{ \begin{matrix} & \left| x \right| & \text{ if x<0} \\\ & {{e}^{x}} & \text{ if x}\ge \text{0} \\\ \end{matrix} \right.;{{f}_{2}}\left( x \right)={{x}^{2}};{{f}_{3}}\left( x \right)=\left\\{ \begin{matrix} & \sin x & \text{ if x<0} \\\ & x & \text{ if x}\ge \text{0} \\\ \end{matrix} \right.;{{f}_{4}}\left( x \right)=\left\\{ \begin{matrix} & {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right) & \text{ if x>0} \\\ & {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right) & \text{ if x}\ge \text{0} \\\ \end{matrix} \right.

List-I	List-II
P. ${{f}_{4}}$ is	1. onto but not one-one
Q. ${{f}_{3}}$ is	2. neither continuous nor one-one
R. ${{f}_{2}}o{{f}_{1}}$	3. differentiable but not one-one
S. ${{f}_{2}}$ is	4. continuous and one-one

A. $P-3,Q-1,R-4,S-2$
B. $P-1,Q-3,R-4,S-2$
C. $P-3,Q-1,R-3,S-4$
D. $P-1,Q-3,R-2,S-4$

Answer 1

Now this is a matching type question based on functions. To solve this, we will first draw rough sketches of graphs of all the functions in list-I and then check which of the options in List-II stands correct for that function. After doing that, we will check the options to see which is the most appropriate option for all the conclusions drawn from the graphs. That will give us the correct option as our answer.

Complete step by step solution:
Now, in matching type questions, we should always read the list-I first to know what the question is asking about then the list-II to know what is being asked and then the details of those in list-I from the given data.
Now, in list-I, the question asks about the continuity, differentiability, bijectivity and surjectivity of ${{f}_{4}}$ . So, we will try to draw a rough sketch of its graph based on its domain and codomain.
Now, we have been given that ${{f}_{4}}:R\to [0,\infty )$ .
Thus, its domain is R, i.e all real numbers and the co-domain is $[0,\infty )$ .
Also, the function is defined as:
{{f}_{4}}\left( x \right)=\left\\{ \begin{matrix} & {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)&\text{ if x>0} \\\ & {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)&\text{ if x}\ge \text{0} \\\ \end{matrix} \right.
Now, we also have been given that ${{f}_{1}}:R\to R$ defined as {{f}_{1}}\left( x \right)=\left\\{ \begin{matrix} & \left| x \right|&\text{ if x<0} \\\ & {{e}^{x}}&\text{ if x}\ge \text{0} \\\ \end{matrix} \right. and ${{f}_{2}}:[0,\infty )\to R$ defined as ${{f}_{2}}\left( x \right)={{x}^{2}}$.
Now, when we take a look at ${{f}_{1}}$ , we can see from its definition that it is always greater than or equal to 0, i.e. the range of ${{f}_{1}}$ is $[0,\infty )$ (because modulus function and the exponent function of a positive number always give non-negative real values). This is also the domain of ${{f}_{2}}$, thus the whole function ${{f}_{1}}$ fits into ${{f}_{2}}$ .
Now, ${{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)$ is also given by two different functions in two different domains as ${{f}_{1}}$ is given in the same way and in both the cases, the range of ${{f}_{1}}$ is a subset of the domain of ${{f}_{2}}$ .
Thus, ${{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)$ is given as:
When $x<0$:
$\begin{aligned} & {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)={{f}_{2}}\left( \left| x \right| \right) \\\ & \Rightarrow {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)={{\left| x \right|}^{2}} \\\ \end{aligned}$
Now, when $x<0$, $\left| x \right|$ is given as:

& \left| x \right|=-x \\\ & \Rightarrow {{\left| x \right|}^{2}}={{\left( -x \right)}^{2}} \\\ & \Rightarrow {{\left| x \right|}^{2}}={{x}^{2}} \\\ \end{aligned}$$ Thus, our function then becomes: ${{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)={{x}^{2}}$ …..(i) When $x\ge 0$: $\begin{aligned} & {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)={{f}_{2}}\left( {{e}^{x}} \right) \\\ & \Rightarrow {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)={{\left( {{e}^{x}} \right)}^{2}} \\\ \end{aligned}$ $\Rightarrow {{f}_{2}}\left( {{f}_{1}}\left( x \right) \right)={{e}^{2x}}$ …..(ii) Thus, from equations (i) and (ii), we can write the definition of function ${{f}_{4}}$ as: ${{f}_{4}}\left( x \right)=\left\\{ \begin{matrix} & {{x}^{2}}&\text{ if x<0} \\\ & {{e}^{2x}}&\text{ if x}\ge \text{0} \\\ \end{matrix} \right.$ Thus, in the second quadrant, the graph of the function will be a parabola given by $y={{x}^{2}}$ and in the first quadrant, the graph of the function will be an exponential curve given as $y={{e}^{2x}}$. The breaking point of the domain is ‘0’. Thus, the graph for $x<0$ will terminate at x=0 and the graph for $x\ge 0$ will start from x=0. At x=0: ${{f}_{4}}\left( 0 \right)=1$ as 0 is included in this case only. Thus, for $x<0$ , the point at which x=0, i.e. the terminating point will be an open point. Now, from all this information about the function, we can draw the rough sketch of the graph for the required function. Thus, the graph of the function is given as:  The first observation we get from this graph is that the function is breaking at x=0. Thus, we can say that the function is discontinuous at x=0 and we know that if there exists a point of discontinuity, then that point is also the point of non-differentiability. Thus, we can say that function is discontinuous and non-differentiable. Now, we can also see from the graph that the range of the function is $[0,\infty )$ which is also the co-domain of the function. We know that if the range and codomain of a function are equal, then the function is onto. Thus the function is also an onto function. Now, we will check if the function is one-one or not. To check that, we do a test called vertical and horizontal line test. In this test, if we draw two lines, each one parallel to one of the axes, then if both the lines intersect the curve at maximum of one point then the function is one-one. In all the other cases, it is many one. Now, if we do the above mentioned test on this function, we will get: From this, we can see that the function does not satisfy the horizontal line test. Thus, the function is many-one. From all these observations, we can say that ${{f}_{4}}$ is discontinuous, non-differentiable, many-one and onto. Thus, (P) from list-I satisfies (1) and (2) from list-II. Now, we will check for (Q) in list-II. (Q) is ${{f}_{3}}$ which is from $R\to R$ defined as: ${{f}_{3}}\left( x \right)=\left\\{ \begin{matrix} & \sin x\text{ if x<0} \\\ & x\text{ if x}\ge \text{0} \\\ \end{matrix} \right.$ Thus, from the definition of the function, we can see that the function is a periodic function given by $y=\sin x$ when $x<0$ and a straight line given by $y=x$ when $x\ge 0$. Thus, the graph of the periodic function will be in the second and the third quadrant and the graph of the straight line will be in the first. So, from all this information about the function, we can draw its graph. Hence, the graph of the function is given as: Now, the first observation we have from the graph is that it is continuous everywhere. The graph is also differentiable as the graph is changing only at x=0 but it is a smooth turn not a sharp turn which tells us that the graph is differentiable. Thus, the function is continuous and differentiable in R. Now, we will check its range. When $x\ge 0$ the function is given by $y=x$. Thus in ${{R}^{+}}$, the function will obtain all the real non negative values. Hence, when $x\ge 0$, the range of the function will be $[0,\infty )$. But when $x<0$, the function is given by $y=\sin x$. Now, we know that the range of a sin function is given by $\left[ -1,1 \right]$ for all real values of ‘x’. Hence, when $x<0$, the range of the function will be $\left[ -1,1 \right]$. Hence, the range of ${{f}_{3}}$ is: $\left[-1,1\right]\cup\left[ 0,\infty\right)$ $\Rightarrow \left[-1,\infty\right)$ But, here the co-domain of ${{f}_{3}}$ is given as R. Hence, the function is not onto. Now, we will check if it is one-one or not. Now, we can see that this graph satisfies the vertical line test but it does not satisfy the horizontal line test. Hence, the function ${{f}_{3}}$ is not a one-one function. From all these observations, we can say that ${{f}_{3}}$ is continuous, differentiable, many-one and into. Thus, (Q) from list-I satisfies only (3) from list-II. Now, we will check for (R) from list-I. We have been given (R) as ${{f}_{2}}o{{f}_{1}}$. We can see in the question that ${{f}_{4}}$ given to us is ${{f}_{2}}o{{f}_{1}}$ only but it differs in domain and co-domain. ${{f}_{2}}o{{f}_{1}}$ has the co-domain of ${{f}_{2}}$ as it is still the principal function and ${{f}_{1}}$ acts as an argument in it and as already established while solving for ${{f}_{4}}$, that the domain of ${{f}_{2}}o{{f}_{1}}$ is the same as that of ${{f}_{2}}$. Hence, the function ${{f}_{2}}o{{f}_{1}}$ and ${{f}_{4}}$ will have the same graphs. The rough sketch of graph of ${{f}_{2}}o{{f}_{1}}$ is given as follows:  Now, as we established above, this function is discontinuous and non-differentiable at x=0 and it is many one and its range is given by $[0,\infty )$. But we already mentioned above that the co-domain of ${{f}_{2}}o{{f}_{1}}$ will be the same as that of ${{f}_{2}}$ which is given to us as R. Hence, the function ${{f}_{2}}o{{f}_{1}}$ is not onto. From all these observations, we can say that the function ${{f}_{2}}o{{f}_{1}}$ is discontinuous, non-differentiable, many-one and into. Thus, (R) from list-I satisfies only (2) from list-II. Now, we will check for (S) from list-I. (S) is given to us as ${{f}_{2}}$ which is from $[0,\infty )\to R$ defined as ${{f}_{2}}\left( x \right)={{x}^{2}}$ Now, $y={{x}^{2}}$ is a very common function whose graph is known by everyone. It is given by an upward parabola with the origin as its vertices. But here, we have been given that the domain of the function is $[0,\infty )$. Thus, we will only draw the graph of $y={{x}^{2}}$in the first quadrant. The graph of the function ${{f}_{2}}$ is shown as:  Now, we can see that this graph is continuously differentiable everywhere in the domain of ${{f}_{2}}$. We can see that the range of the function is $[0,\infty )$ but here, the co-domain of ${{f}_{2}}$ is given as R. hence, the function ${{f}_{2}}$ is not an onto function. Now, we will check if it is one-one or not. Now, we can see that the graph satisfies both the horizontal line and vertical line test. Hence, the function ${{f}_{2}}$ is one-one. From all these observations, we can say that the function ${{f}_{2}}$ is continuous, differentiable, many-one and into. Thus, (S) from list-I satisfies only (4) from list-II. Thus, we have obtained the following: (P): (1) and (2) (Q): (3) (R): (2) (S): (4) From looking at all these observations, we can see that only option (D) satisfies them **Hence, option (D) is the correct option.** **Note:** Here, we have checked all 4 components of list-I. But in competitive examinations, it may take a lot of time. To save time, we can only check for two components from the list looking at the options such that those 2 components form different combinations in all 4 options. Here, if we look carefully, (Q) and (R) always form different combinations in the given options. So we can just check those two and we will still get the correct answer in half the time.