Q.Redefine the function $f(x) = |x - 2| + |2 + x|$, $-3 \le x \le 3$

We are asked whether Rolle’s theorem and/or Lagrange’s mean value theorem (LMVT) apply to the function

on the interval $\left[\frac13, 3\right]$.

The core idea

Both Rolle’s theorem and Lagrange’s mean value theorem require the function to be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Rolle’s theorem additionally requires $f(a) = f(b)$.

If the function has a break (discontinuity) inside the interval, neither theorem can be applied — no matter how nice the pieces are individually.

Step-by-step reasoning

1. Identify the interval and the potential trouble point

   The interval is $\left[\frac13, 3\right]$. The function’s definition changes at $x = 2$. Since $2$ lies strictly between $\frac13$ and $3$, the point $x=2$ is inside our interval. This is where we must check continuity.

2. Check continuity at $x = 2$

   • Left-hand limit: as $x \to 2^-$, we use the first piece:

• Right-hand limit: as $x \to 2^+$, we use the second piece:

• Function value at $x=2$: since $x=2$ belongs to the first piece ($0 \leq x \leq 2$),

So the left-hand limit, right-hand limit, and function value all equal $5$.

Thus $f$ is continuous at $x=2$.

3. Check differentiability at $x=2$
   • Left-hand derivative: derivative of $3x-1$ is $3$, so

• Right-hand derivative: derivative of $\sqrt{25(x-1)} = 5\sqrt{x-1}$ is

 so at $x=2$:  

Since $3 \neq 2.5$, the left and right derivatives are different.

Therefore $f$ is not differentiable at $x=2$. …

We are given

and told its range is exactly $\left[\frac13, 3\right]$. That means for every $y$ in that interval, there is some real $x$ with $f(x)=y$, and no $y$ outside the interval can be attained.

1. Set up the equation $f(x) = y$

Write

Cross-multiply (the denominator is never zero for the $x$ we care about; we'll check later):

Bring all terms to one side:

which simplifies to

2. For a given $y$, when does a real $x$ exist?

This is a quadratic in $x$. For real $x$ to exist, either:

• The coefficient of $x^2$ is zero and the resulting linear equation gives a real $x$, or
• The coefficient is nonzero and the discriminant is $\ge 0$.

Case 1: $1 - y = 0$, i.e. $y = 1$. Then the equation becomes

which is real. So $y=1$ is always attainable (provided $x=0$ doesn't make denominator zero; $f(0)=1$ works for any $k$). So $1$ is in the range for all $k$.

Case 2: $y \neq 1$. Then we need the discriminant $\Delta \ge 0$:

So the condition for $y$ to be in the range is

3. The range is exactly $[1/3, 3]$ — what does that tell us?

The endpoints $y = 1/3$ and $y = 3$ should be the boundary where $\Delta = 0$ (since inside the interval $\Delta > 0$, outside $\Delta < 0$). Also note that $y=1$ is inside $[1/3, 3]$, consistent.

Set $\Delta = 0$ at $y = 1/3$:

Multiply by $9$: $16 - 16k = 0 \implies k = 1$.

Check at $y = 3$:

$(1+3)^2 - 4k(1-3)^2 = 16 - 4k \cdot 4 = 16 - 16k.$ …

We need the set of all real $x$ for which $f(x)$ is defined.

$f(x) = \dfrac{\sqrt{\log_{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2 - 5[x] + 6}}$

Two square roots appear: one in the numerator, one in the denominator.

• The numerator’s square root requires its inside to be $\ge 0$.
• The denominator’s square root requires its inside to be $> 0$ (strictly positive, because denominator cannot be zero).

We handle each condition separately, then intersect them.

1. Numerator condition: $\log_{10}\left(\frac{x}{x-2}\right) \ge 0$

Since $\log_{10} t \ge 0$ iff $t \ge 1$, we need

Solve:

The fraction $\frac{2}{x-2} \ge 0$ means $x-2 > 0$, i.e. $x > 2$.

Also, the argument of the log must be positive: $\frac{x}{x-2} > 0$. For $x > 2$, both numerator and denominator are positive, so this holds automatically.

So numerator condition gives: $x > 2$.

2. Denominator condition: $[x]^2 - 5[x] + 6 > 0$

Let $n = [x]$, an integer. Then we need

Factor: $(n-2)(n-3) > 0$.

This quadratic in $n$ is positive when $n < 2$ or $n > 3$. …

The core idea is that a function has an inverse only if it is bijective — both one‑to‑one (injective) and onto (surjective). When we say $f$ and $g$ are inverses, we mean $g(f(x)) = x$ for all $x$ in $A$ and $f(g(y)) = y$ for all $y$ in $B$. That forces $f$ to be a bijection from $A$ onto $B$, and $g$ to be its reverse.

The square function $f(x)=x^2$ is not injective on all real numbers because $(-2)^2 = 2^2$. The square‑root function $g(x)=x^{1/2}$ is defined only for non‑negative inputs and gives only non‑negative outputs. So the natural question is: on what sets do these two become perfect mirrors of each other?

Let’s examine each option.

1. Option (A): $A = B = \mathbb{R}$

   $f(x)=x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is not injective (since $f(-1)=f(1)$) and not onto (negative numbers have no pre‑image). So $f$ has no inverse. $g(x)=x^{1/2}$ is not even defined for negative $x$ in $\mathbb{R}$. These cannot be inverses.

2. Option (B): $A = \mathbb{R} \setminus \mathbb{R}^-$ (all non‑negative reals), $B = \mathbb{R} \setminus \mathbb{R}^+$ (all non‑positive reals)

   Here $A = [0,\infty)$ and $B = (-\infty,0]$. For $x \in A$, $f(x)=x^2$ gives non‑negative outputs, but $B$ contains only non‑positive numbers. So $f$ does not map $A$ into $B$ at all — the composition $g(f(x))$ is not defined for most $x$. Not possible.

3. Option (C): $A = \mathbb{R}$, $B = \mathbb{R} \setminus \mathbb{R}^-$ (non‑negative reals)

   $f$ from $\mathbb{R}$ to $[0,\infty)$ is onto but not injective (still $f(-1)=f(1)$). So $f$ is not bijective. $g$ from $[0,\infty)$ to $\mathbb{R}$ is injective but not onto (it never gives negative outputs). The compositions fail: $g(f(-1)) = g(1)=1 \neq -1$, so $g(f(x)) \neq x$ for negative $x$. …

The question gives you a mapping rule $y(x)$ and asks for the set $A$ — the domain of this function. The key idea: a function’s domain is the set of all inputs for which the output is defined and real. So we must find every real number $x$ that the rule can handle.

The rule has two pieces: a rational expression for $x \neq -1$, and a constant value for $x = -1$. Let’s examine each.

1. The rational part: $\frac{5x}{(x-3)(x+3)}$ is defined only when the denominator is non-zero. The denominator $(x-3)(x+3)$ equals zero when $x = 3$ or $x = -3$. So $x = 3$ and $x = -3$ are not allowed.

2. The $x = -1$ case: The rule explicitly says: if $x = -1$, then $y = -1$. That’s perfectly fine — no division, no problem. So $x = -1$ is allowed.

3. What about $x = 0$? Plug $x = 0$ into the rational part: $\frac{5(0)}{(0-3)(0+3)} = \frac{0}{(-3)(3)} = \frac{0}{-9} = 0$. That’s defined. So $x = 0$ seems allowed — but wait. Look closely at the denominator: $(x-3)(x+3)$. At $x = 0$, it’s $(-3)(3) = -9 \neq 0$, so no issue. So $x = 0$ is allowed.

4. Check the options: …

Continuity at the join $x=4$.

Both branches are polynomials and they agree at $x=4$, so $f$ is continuous on all of $[2,10]$. (This rules out option B.)

Endpoint values.

So $f(2)=f(10)$ (this rules out option A).

Differentiability at $x=4$.

$f'(x)=x-6\ \text{ for }x<4\ \Rightarrow\ f'(4^-)=-2,\qquad f'(x)=1\ \text{ for }x>4\ \Rightarrow\ f'(4^+)=1.$ …