Q.The domain and range of the real function $f$ defined by $f(x) = \dfrac{4 - x}{x - 4}$ is given by
(A) Domain $= \mathbf{R}$, Range $= \{-1, 1\}$
(B) Domain $= \mathbf{R} - \{1\}$, Range $= \mathbf{R}$
(C) Domain $= \mathbf{R} - \{4\}$, Range $= \{-1\}$
(D) Domain $= \mathbf{R} - \{-4\}$, Range $= \{-1, 1\}$

Concept and Intuition

For $f(x)=\sqrt{\dfrac{[x]-1}{[x]^2-[x]-6}}$ to be real, the fraction inside the square root must be $\ge 0$, and the denominator must be non-zero. Since the expression depends only on $[x]$ (an integer), first solve the inequality treating $n=[x]$ as an integer variable, then translate back to $x$ using $[x]=n \iff n\le x<n+1$.

Step-by-Step Solution

1. Factor the denominator: $n^2-n-6=(n-3)(n+2)$.
2. Need $\dfrac{n-1}{(n-3)(n+2)}\ge 0$, $n\neq 3,-2$.
3. Sign chart with critical points $-2,1,3$: for $n<-2$: numerator negative, denominator (neg)(neg)=positive -> ratio negative. For $-2<n<1$: numerator negative, denominator (neg)(pos)=negative -> ratio positive. For $1<n<3$: numerator positive, denominator (neg)(pos)=negative -> ratio negative. For $n>3$: all positive -> ratio positive. At $n=1$: ratio $=0$ (allowed, included).
4. So real-valued solution set for $n$: $(-2,1]\cup(3,\infty)$.
5. Integers in this set: $n=-1,0,1$ (inside $(-2,1]$) and $n=4,5,6,\dots$ (inside $(3,\infty)$, note $n=3$ excluded). …

Concept and Intuition

For a square root to be a real, well-defined function, the expression under it must be $\ge 0$, AND if that expression is a fraction, the denominator must additionally be non-zero (division by zero is never allowed, even though $0/\text{anything nonzero}=0\ge 0$ would otherwise be fine). Working with $|x|$ makes the problem symmetric, so it helps to substitute $t=|x|\ge 0$ and solve the resulting rational inequality in $t$ first, then translate back to $x$.

Step-by-Step Solution

1. Domain condition: $\dfrac{|x|-2}{|x|-3}\ge 0$ and $|x|\ne 3$.
2. Let $t=|x|\ge 0$. Solve $\dfrac{t-2}{t-3}\ge 0$. The critical points are $t=2$ (numerator zero, included) and $t=3$ (denominator zero, excluded).
3. Sign analysis: for $t<2$, both $t-2<0$ and $t-3<0$, so the quotient is positive. At $t=2$, quotient $=0$ (included, since $\ge 0$). For $2<t<3$, $t-2>0,\,t-3<0$: quotient negative (excluded). For $t>3$, both positive: quotient positive (included). So the solution in $t$ is $t\in[0,2]\cup(3,\infty)$ (recall $t\ge0$ always).
4. Convert back: $t\le 2 \iff |x|\le 2 \iff x\in[-2,2]$. And $t>3 \iff |x|>3 \iff x\in(-\infty,-3)\cup(3,\infty)$.
5. Combined domain: $x\in(-\infty,-3)\cup[-2,2]\cup(3,\infty)$. …

Concept and Intuition

For the given expression to be a real number, two separate conditions must both hold:

1. The quantity under the square root in the numerator must be non-negative.
2. The argument of the logarithm in the denominator must be strictly positive (log undefined for zero/negative arguments), and the denominator itself must not vanish.

The domain is the intersection of all such conditions.

Step-by-Step Solution

1. Numerator condition: $|x|^2 - 2|x| - 8 \ge 0$. Let $u = |x| \ge 0$. Then $u^2-2u-8\ge0 \Rightarrow (u-4)(u+2)\ge0 \Rightarrow u\le -2 \text{ or } u\ge 4$. Since $u\ge0$ always, only $u\ge4$ survives, giving $|x|\ge4$, i.e. $x\le-4$ or $x\ge4$.
2. Denominator condition: need $2-x-x^2 > 0 \Rightarrow x^2+x-2<0 \Rightarrow (x+2)(x-1)<0 \Rightarrow -2<x<1$. …

Concept and Intuition

$f(x)=\sin\!\big(\log(g(x))\big)$ is defined wherever $g(x)=\dfrac{\sqrt{4-x^2}}{1-x}$ is a well-defined positive real number (since $\log$ needs a strictly positive argument; $\sin$ itself is defined for all reals, so it imposes no further restriction).

Step-by-Step Solution

1. Square root real: $4-x^2\ge0 \Rightarrow -2\le x\le2$.
2. Log argument positive: $g(x) = \dfrac{\sqrt{4-x^2}}{1-x}>0$. The numerator $\sqrt{4-x^2}\ge0$, being exactly zero only at $x=\pm2$. For $g(x)$ to be strictly positive (log of 0 is undefined), we need $x\ne\pm2$ so the numerator is strictly positive, and then the sign of $g(x)$ matches the sign of the denominator $1-x$.
3. So we need $1-x>0 \Rightarrow x<1$. …

Concept and Intuition

For a real-valued function built from square roots, every expression under a radical must be $\ge 0$ for the square root to be a real number, and any radical sitting in the denominator must additionally be strictly positive (since division by zero is undefined, and a square root can equal zero when its argument is zero). The domain is the intersection of all these individual conditions.

Step-by-Step Solution

1. Numerator term $\sqrt{2-x}$ requires $2-x\ge0 \Rightarrow x\le 2$.
2. Numerator term $\sqrt{1+x}$ requires $1+x\ge0 \Rightarrow x\ge -1$.
3. Denominator term $\sqrt{x+3}$ requires $x+3>0$ (strict, since it's a denominator) $\Rightarrow x>-3$.
4. Intersecting all three: $x\ge-1$, $x\le2$, and $x>-3$ (automatically satisfied whenever $x\ge-1$). The combined domain is $-1\le x\le 2$, i.e. $[-1,2]$.

Common Mistakes …

Concept and Intuition

For a square root to be real, the radicand must be $\ge 0$, and the denominator must never vanish. Substituting $t=|x|$ turns this into an ordinary rational-inequality problem.

Step-by-Step Solution

1. Need $\dfrac{2-|x|}{3-|x|}\ge 0$ and $|x|\ne 3$.
2. Let $t=|x|\ge0$. Critical points are $t=2$ and $t=3$.
3. For $0\le t<2$: numerator $2-t>0$, denominator $3-t>0$ $\Rightarrow$ ratio $>0$. Valid.
4. At $t=2$: ratio $=0$. Valid (square root of $0$ is defined).
5. For $2<t<3$: numerator negative, denominator positive $\Rightarrow$ ratio $<0$. Invalid.
6. At $t=3$: denominator zero — excluded.
7. For $t>3$: numerator negative, denominator negative $\Rightarrow$ ratio $>0$. Valid.
8. So valid $t$: $t\in[0,2]\cup(3,\infty)$. …

Concept and Intuition

The domain of a sum of functions is the intersection of each piece's individual domain — every term must be simultaneously defined.

Step-by-Step Solution

1. For $\dfrac{-5}{4x^2+1}$: the denominator $4x^2+1 \ge 1$ for all real $x$ (since $x^2\ge0$), so this term is defined for every real $x$ — domain $=\mathbb{R}$.
2. For $\sqrt{x^2-4}$: need $x^2-4\ge0 \Rightarrow x^2\ge4 \Rightarrow x\le-2$ or $x\ge2$ — domain $=(-\infty,-2]\cup[2,\infty)$.
3. Intersecting the two domains: since the first is all of $\mathbb{R}$, the overall domain is exactly the second piece's domain: $(-\infty,-2]\cup[2,\infty)$.

Common Mistakes …

Concept and Intuition

$|x| - x$ measures "how far $x$ is below zero, doubled": for non-negative $x$ it's identically

zero (since $|x|=x$), and for negative $x$ it becomes $2|x|$, always positive. Since this

quantity sits under a square root that is itself in a denominator, we need it to be strictly

greater than zero — zero would make the denominator zero (undefined), and any negative value

would make the square root undefined.

Step-by-Step Solution

1. Require $|x| - x > 0$ (strict inequality, both for the square root to be real and non-zero, and to avoid division by zero).
2. Case $x \ge 0$: $|x| = x \Rightarrow |x| - x = 0$, which fails the strict inequality for every such $x$. …

Concept and Intuition

For a square root to be real, the expression inside must be $\ge 0$, and since this is a ratio, we also need the denominator to be nonzero (division by zero is undefined). Substituting $t=|x|\ge 0$ turns this into a one-variable rational-inequality problem, which is solved by sign analysis across the critical points where numerator or denominator vanish.

Step-by-Step Solution

1. Let $t=|x|\ge 0$. We require $\dfrac{1-t}{2-t}\ge 0$ and $2-t\ne 0$ (i.e. $t\ne2$).
2. Critical points of the rational expression: $t=1$ (numerator zero) and $t=2$ (denominator zero).
3. For $t\in[0,1)$: $1-t>0$, $2-t>0$ → ratio positive. At $t=1$: ratio $=0$, allowed since $\ge0$ is satisfied (square root of 0 is defined). So $[0,1]$ is valid.
4. For $t\in(1,2)$: $1-t<0$, $2-t>0$ → ratio negative. Invalid (would need square root of a negative number).
5. At $t=2$: denominator is zero — undefined, excluded.
6. For $t>2$: $1-t<0$, $2-t<0$ → ratio $=\dfrac{\text{neg}}{\text{neg}}>0$. Valid, so $(2,\infty)$ works. …

Concept and Intuition

For a rational function of the form $\dfrac{ax+b}{cx+d}$, the range excludes exactly the value of $y$ for which solving for $x$ produces a zero denominator — this corresponds to the horizontal asymptote value $a/c$.

Step-by-Step Solution

1. Let $y = \dfrac{x-2}{x+3}$.
2. Cross-multiply: $y(x+3) = x-2 \Rightarrow yx + 3y = x - 2 \Rightarrow x(y-1) = -2-3y$.
3. $x = \dfrac{-2-3y}{y-1}$, defined for every $y$ except $y=1$.
4. So every real number except $1$ is in the range. …