We analyze each function on the interval $(-1,1)$ for the properties:

• $(P)$ continuous in $(-1,1)$
• $(Q)$ differentiable in $(-1,1)$
• $(R)$ strictly increasing in $(-1,1)$
• $(S)$ not differentiable at least at one point in $(-1,1)$

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1. Function (A): $f(x)=x|x|$

Write it piecewise:

$$x|x|= \begin{cases} -x^2, & x<0 \\ x^2, & x\ge 0 \end{cases}$$

Continuity

Both pieces are polynomial, and at $x=0$,

$$\lim_{x\to 0^-}(-x^2)=0,\qquad \lim_{x\to 0^+}(x^2)=0=f(0)$$

So it is continuous in $(-1,1)$. Hence $(P)$ is true.

Differentiability

Differentiate piecewise:

$$f'(x)= \begin{cases} -2x, & x<0 \\ 2x, & x>0 \end{cases}$$

At $x=0$,

$$f'(0)=\lim_{h\to 0}\frac{h|h|-0}{h}=\lim_{h\to 0}|h|=0$$

So it is differentiable at $0$ also. Hence $(Q)$ is true.

Strictly increasing?

For $x<0$, $f(x)=-x^2$, and

$$f'(x)=-2x>0$$

since $x<0$. For $x>0$,

$$f'(x)=2x>0$$

So $f$ is increasing on both sides, and across $0$ also values increase. Thus it is strictly increasing on $(-1,1)$. Hence $(R)$ is true.

So for (A): $(P),(Q),(R)$.

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2. Function (B): $f(x)=\sqrt{|x|}$

Piecewise,

$$\sqrt{|x|}= \begin{cases} \sqrt{-x}, & x<0 \\ \sqrt{x}, & x\ge 0 \end{cases}$$

Continuity

Since $|x|$ and square root are continuous where defined, and $|x|\ge 0$, this is continuous for all $x$, hence on $(-1,1)$. So $(P)$ is true.

Differentiability

For $x>0$,

$$f'(x)=\frac{1}{2\sqrt{x}}$$

For $x<0$,

$$f'(x)=\frac{-1}{2\sqrt{-x}}$$

At $x=0$, derivative does not exist because the slopes blow up. Therefore it is not differentiable at $x=0$. So $(S)$ is true, and $(Q)$ is false.

Strictly increasing?

Take $x_1=-\frac14$ and $x_2=0$ with $x_1<x_2$, then

$$f(x_1)=\sqrt{1/4}=\frac12>0=f(0)$$

So it is not increasing on all of $(-1,1)$. Thus $(R)$ is false.

So for (B): $(P),(S)$.

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3. Function (C): $f(x)=x+[x]$

On $(-1,1)$,

$$[x]= \begin{cases} -1, & -1\le x<0 \\ 0, & 0\le x<1 \end{cases}$$

Hence

$$f(x)=x+[x]= \begin{cases} x-1, & -1<x<0 \\ x, & 0\le x<1 \end{cases}$$

Continuity / differentiability

At $x=0$,

$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(x-1)=-1,$$

while

$$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}x=0=f(0)$$

So there is a jump discontinuity at $x=0$. Therefore it is not continuous and hence not differentiable at $x=0$. Thus $(S)$ is true, $(P)$ and $(Q)$ are false.

Strictly increasing?

On each side, slope is $1$, and for any $x_1<0<x_2$,

$$f(x_1)=x_1-1<0\le x_2=f(x_2)$$

So whenever $x_1<x_2$, we get $f(x_1)<f(x_2)$. Therefore it is strictly increasing on $(-1,1)$. Hence $(R)$ is true.

So for (C): $(R),(S)$.

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4. Function (D): $f(x)=|x-1|+|x+1|$

For $x\in(-1,1)$,

• $x-1<0$, so $|x-1|=1-x$
• $x+1>0$, so $|x+1|=x+1$

Thus,

$$f(x)=(1-x)+(x+1)=2$$

for all $x\in(-1,1)$.

Properties

A constant function is continuous and differentiable everywhere in the interval. So $(P)$ and $(Q)$ are true.

It is not strictly increasing, so $(R)$ is false. It is differentiable everywhere, so $(S)$ is false.

So for (D): $(P),(Q)$.

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5. Final matching

Thus the correct matching is:

• $A \to (P),(Q),(R)$
• $B \to (P),(S)$
• $C \to (R),(S)$
• $D \to (P),(Q)$

This corresponds to Option A.

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6. Comparison with stored answer

Stored correct answer: A

My derived answer: A

So the derived answer agrees with the stored answer.