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

Concept and Intuition

Since $[x]$ (the greatest-integer / floor function) is always an integer, an equation in $[x]$ is first solved as an ordinary quadratic in the integer variable $n=[x]$; each valid integer root $n=k$ then corresponds to the whole interval $x\in[k,k+1)$, because $[x]=k$ exactly on that half-open interval.

Step-by-Step Solution

1. Let $n=[x]$ (an integer). The equation becomes $n^2+n+1=7$, i.e. $n^2+n-6=0$.
2. Factor: $(n-2)(n+3)=0 \Rightarrow n=2$ or $n=-3$. Both are integers, so both are valid values of $[x]$.
3. $[x]=2 \Leftrightarrow 2\le x<3$, i.e. $x\in[2,3)$.
4. $[x]=-3 \Leftrightarrow -3\le x<-2$, i.e. $x\in[-3,-2)$.
5. So the full solution set is $[-3,-2)\cup[2,3)$.
6. Compare with the options: $[-3,3)-[-2,2)$ means "all $x$ with $-3\le x<3$, except those with $-2\le x<2$", which leaves precisely $[-3,-2)\cup[2,3)$ — an exact match. …

Concept and Intuition

$[x]$ is the greatest-integer (floor) function, so $x - [x] = \{x\}$ is the fractional part of $x$, which by definition always lies in $[0,1)$ — it can never be negative and never reach $1$.

$g(x) = 1 + x - [x] = 1+\{x\}$ is therefore always shifted into the range $[1,2)$: it is never zero or negative, for any real $x$ (positive, negative, integer, or non-integer).

$f$ is the sign-type function: $f(t)=-1$ if $t<0$, $0$ if $t=0$, $1$ if $t>0$. Since we only ever feed $f$ a strictly positive number (namely $g(x)$), $f$ always returns $1$.

Step-by-Step Solution

1. Recall $\{x\} = x-[x]$ satisfies $0\le \{x\} < 1$ for all real $x$ (this is the definition/property of the floor function).
2. So $g(x) = 1+\{x\}$ satisfies $1 \le g(x) < 2$ for every real $x$.
3. In particular, $g(x) \ge 1 > 0$ for all $x$ — $g(x)$ is never $\le 0$. …

Concept and Intuition

The signum function is $f(x)=1$ for $x>0$, $f(x)=0$ for $x=0$, $f(x)=-1$ for $x<0$. To find which piecewise expression built from f always equals the constant 1, substitute each branch of each option against the corresponding case(s) of x.

Step-by-Step Solution

1. Recall f(x): +1 for x>0; 0 for x=0; -1 for x<0.
2. Check option (D): $g(x)=f(x)+2$ for $x<0$; $1+f(x)$ for $x=0$; $f(x)$ for $x>0$.
   • x<0: f(x)=-1, g=-1+2=1.
   • x=0: f(x)=0, g=1+0=1.
   • x>0: f(x)=1, g=1. All branches give 1, so g(x)=1 for every real x. …

Concept and Intuition

For a piecewise function, the domain is simply everywhere at least one branch applies (here, every real number falls under either x>1 or x≤1, so the domain is all of ℝ). The range is the union of the images of each piece over its own restricted interval, which can require care since a parabola or line might not cover its "naive" range on a restricted domain.

Step-by-Step Solution

1. Domain: since "x>1" and "x≤1" partition ℝ completely, A = ℝ.
2. For x ≤ 1: g(x) = x² + 1. As x ranges over (−∞, 1], x² ranges over [0, ∞) (x can be any non-positive-or-small value including large negative x, and x=0 gives x²=0). So x²+1 ranges over [1, ∞).
3. For x > 1: h(x) = 3x − 1. As x → 1⁺, h(x) → 2⁺; as x → ∞, h(x) → ∞. So this piece covers (2, ∞). …

Concept and Intuition

For a piecewise function, the overall range is the union of the ranges attained on each piece. We just need to carefully find the range each branch sweeps out, paying attention to which endpoints are included (based on strict vs non-strict inequalities in the domain).

Step-by-Step Solution

1. Piece 1 ($x>3$): $f(x)=4x-1$ is increasing; as $x\to3^+$, $f\to 4(3)-1=11$ (not attained, since $x>3$ strictly), and as $x\to\infty$, $f\to\infty$. Range: $(11,\infty)$.
2. Piece 2 ($-2\le x\le3$): $f(x)=x^2-2$. On this interval, $x^2$ attains its minimum $0$ at $x=0$ (included) and maximum $9$ at $x=3$ (included, since $|3|>|-2|$). Since the interval is connected and $f$ continuous, $x^2$ sweeps all of $[0,9]$, so $f(x)=x^2-2$ sweeps $[-2,7]$.
3. Piece 3 ($x<-2$): $f(x)=3x+4$ is increasing; as $x\to-2^-$, $f\to 3(-2)+4=-2$ (not attained, strict inequality), and as $x\to-\infty$, $f\to-\infty$. Range: $(-\infty,-2)$. …