We identify which of the functions are continuous / differentiable / one-one / onto (the usual matching-type interpretation for such questions), and then match them with the labels $P,Q,R,S$.

1. Analyze $f_1$

Given

For $x<0$, we have $|x|=-x$, so

Continuity at $x=0$

Left limit:

Right limit and value:

Since $0\ne 1$, $f_1$ is discontinuous at $0$.

One-one / onto

• For $x<0$, range is $(0,\infty)$.
• For $x\ge 0$, range is $[1,\infty)$. So total range is $(0,\infty)$.

Not one-one, because for example many positive values can come from both branches. Also not onto $\mathbb R$, since negative values and $0$ are not attained.

So $f_1$ is the function with the characteristic “discontinuous”.

2. Analyze $f_2$

Given

Since domain is $[0,\infty)$, $x^2$ is strictly increasing there. Hence $f_2$ is one-one.

Its range is also $[0,\infty)$, so it is not onto $\mathbb R$. It is continuous and differentiable on $[0,\infty)$.

Thus $f_2$ is the function with the characteristic “one-one”.

3. Analyze $f_3$

Given

Continuity at $x=0$

Left limit:

Right limit and value:

So $f_3$ is continuous at $0$.

Differentiability at $x=0$

Left derivative:

Right derivative:

Hence $f_3$ is differentiable at $0$.

But it is not one-one on $\mathbb R$ because $\sin x$ on $(-\infty,0)$ is not one-one. Also not onto $\mathbb R$? Actually for $x\ge 0$, $f_3(x)=x$ gives all $[0,\infty)$, and for $x<0$, $\sin x$ gives values in $[-1,1]$, so negative values less than $-1$ are not attained. Hence not onto $\mathbb R$.

Thus $f_3$ is the function with the characteristic “differentiable”.

4. Analyze $f_4$

Given

Now compute explicitly.

For $x<0$:

• $f_1(x)=|x|=-x$
• $f_2(f_1(x))=(|x|)^2=x^2$

So for $x<0$,

For $x\ge 0$:

• $f_1(x)=e^x$
• $f_2(f_1(x))=(e^x)^2=e^{2x}$

So for $x\ge 0$,

Hence

Continuity at $x=0$

Left limit:

Right limit and value:

So $f_4$ is continuous at $0$.

Differentiability at $x=0$

Left derivative:

Right derivative:

Since $0\ne 2$, $f_4$ is not differentiable at $0$.

Onto check

Codomain is $[0,\infty)$.

• For $x<0$, $x^2\in(0,\infty)$
• For $x\ge 0$, $e^{2x}-1\in[0,\infty)$ So every value in $[0,\infty)$ is attained. Hence $f_4$ is onto $[0,\infty)$.

Thus $f_4$ is the function with the characteristic “onto but not differentiable”; in the matching this corresponds to the remaining label.

5. Matching

From the above:

• $f_1 \to$ discontinuous
• $f_3 \to$ differentiable
• $f_2 \to$ one-one
• $f_4 \to$ onto

Thus the matching is

which is Option D.

6. Comparison with stored answer

Stored correct answer: D

Derived answer: D

So they agree.