Q.Find the domain of the function $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

Step 1 — Set up the condition

For $\sqrt{\dfrac{x-7}{9-x}}$ to be real, we need

Step 2 — Sign analysis

The critical points are $x = 7$ (numerator zero) and $x = 9$ (denominator zero).

• For $x < 7$: numerator $(x-7) < 0$, denominator $(9-x) > 0$ $\Rightarrow$ quotient $< 0$. Rejected.
• At $x = 7$: quotient $= 0 \geq 0$. Included.
• For $7 < x < 9$: numerator $> 0$, denominator $> 0$ $\Rightarrow$ quotient $> 0$. Included.
• At $x = 9$: denominator $=0$, undefined. Excluded.
• For $x > 9$: numerator $> 0$, denominator $< 0$ $\Rightarrow$ quotient $< 0$. Rejected.

Step 3 — Combine

The domain is all $x$ with $7 \leq x < 9$, i.e. $[7, 9)$.

Step 1 — Evaluate the greatest-integer constants.

Why $-10$ and not $-9$: $[y]$ is the greatest integer not exceeding $y$. Since $-10 \le -9.87 < -9$, the greatest integer below $-9.87$ is $-10$. (Students routinely write $-9$ here — that is the whole trap of this problem.)

Step 2 — Simplify $f$.

Since $\sin$ is odd, $\sin(-10x) = -\sin(10x)$:

Step 3 — Test each option.

(A) $f(0)$:

(B) $f(\pi/2)$:

(C) $f(\pi/4)$:

(D) $f(\pi)$:

Both arguments are integer multiples of $\pi$, and $\sin(n\pi)=0$ for every integer $n$:

Step 4 — Answer the question as asked.

The question asks which statement is not true; three statements check out and only $f(\pi)=-1$ fails (the true value is $0$).

ANSWER: D

The core idea here is the meaning of pre‑image (also called inverse image). For a function $f: \mathbb{R} \to \mathbb{R}$, the pre‑image of a number $y$ is the set of all $x$ in the domain that map to $y$:

It is not about finding an inverse function — it is about solving the equation $f(x)=y$ and collecting all solutions.

Now $f(x)=x^2+1$ is a parabola opening upward, with minimum value $1$ at $x=0$. So $f(x)$ can never be less than $1$. That immediately tells us something about the pre‑image of $-3$.

Let’s work through each case.

1. Pre‑image of $17$ Set $f(x)=17$:

Both $4$ and $-4$ are real numbers, so the pre‑image is $\{4, -4\}$.

2. Pre‑image of $-3$ Set $f(x) = -3$:

No real number squared gives $-4$. Hence there is no $x \in \mathbb{R}$ satisfying this equation. The pre‑image is the empty set $\phi$.

Thus the pre‑images are $\{4,-4\}$ for $17$ and $\phi$ for $-3$.

Concept & Intuition

A rational function $f(x) = \frac{P(x)}{Q(x)}$ is defined everywhere except where $Q(x)=0$. At such an $x$, the function may have a vertical asymptote (non-removable discontinuity) or a hole (removable discontinuity) if the same factor also appears in $P(x)$. The question asks for the number of points of discontinuity — that includes both holes and asymptotes, because at those $x$-values the function is not defined. So we must find all real zeros of the denominator, then check for cancellation.

Step-by-step solution

1. Factor numerator and denominator Numerator: $x^2 - 3x + 2 = (x-1)(x-2)$. Denominator: $4x - x^3 = -x(x^2 - 4) = -x(x-2)(x+2)$. So

2. Identify all zeros of the denominator

   Set denominator = 0: $-x(x-2)(x+2)=0$ gives $x = 0,\; x = 2,\; x = -2$.

   These are the only candidates for discontinuity.

3. Check for cancellation (removable vs. non-removable)

   The factor $(x-2)$ appears in both numerator and denominator. Cancel it:

At $x=2$, the original function is undefined, but the limit exists (it’s a hole). So $x=2$ is a discontinuity (removable).

The factors $x$ and $(x+2)$ remain in the denominator only, so at $x=0$ and $x=-2$ the function blows up (vertical asymptotes). These are also discontinuities.

4. Count the points of discontinuity All three $x$-values $0, 2, -2$ are points where $f$ is not defined. Therefore there are 3 points of discontinuity.

Concept & Intuition

A function’s domain is the set of inputs for which every piece of its expression is defined. Here we have two pieces: a square root and a fraction whose denominator contains a logarithm. Each imposes its own restrictions, and the domain is the intersection of all those conditions. The trick is to handle the logarithm carefully: the argument $3-x$ must be positive, the base $10$ is fine (positive and not 1), but the denominator $\log_{10}(3-x)$ cannot be zero — that happens when $3-x = 1$.

Step-by-step reasoning

1. Square root condition $\sqrt{x+7}$ requires the radicand to be non‑negative:

2. Logarithm argument condition $\log_{10}(3-x)$ is defined only when its argument is positive:

3. Denominator non‑zero condition The denominator is $\log_{10}(3-x)$. A fraction is undefined when its denominator is zero, so we need

Since $\log_{10}(u)=0$ exactly when $u=1$, this gives

4. Combine all conditions From steps 1–3 we have:

In interval notation, $x \ge -7$ and $x < 3$ together give $[-7,3)$. Excluding $x=2$ yields $[-7,3) - \{2\}$.

Step 1 — The concept: what a modulus inequality means.

For any real $a$ and any $b > 0$,

The reason: $|a|$ is the distance of $a$ from $0$ on the number line, so $|a|\le b$ says "$a$ lies within $b$ units of the origin", i.e. $a$ lies in the closed interval $[-b,\,b]$.

(Contrast with $|a| \ge b$, which splits into $a \le -b$ or $a \ge b$ — that one gives two separate rays, not a single interval. Here the sign is $\le$, so we get one interval.)

Step 2 — Apply it with $a = 3x-5$ and $b = 2$.

Step 3 — Add 5 to all three parts.

Adding the same number to every part of a double inequality preserves it:

Step 4 — Divide throughout by 3.

Dividing by a positive number preserves the direction of the inequalities:

Step 5 — Sanity-check the endpoints and the options.

• At $x=1$: $|3(1)-5| = |-2| = 2 \le 2$ ✓ (boundary satisfied, so $\le$ is right).
• At $x=\tfrac{7}{3}$: $\left|3\cdot\tfrac{7}{3}-5\right| = |7-5| = 2 \le 2$ ✓.
• At $x=0$ (in option (A)'s range but not ours): $|0-5| = 5 \not\le 2$ ✗ — so (A), with its lower bound $-1$, is too wide.
• Options (C) and (D) carry the upper bound $\tfrac{9}{3} = 3$; at $x = 3$, $|9-5| = 4 \not\le 2$ ✗.

Only (B) matches exactly.

ANSWER: B

The core idea here is that the absolute value function $|x|$ behaves differently depending on whether $x$ is positive or negative. When $x$ is positive, $|x| = x$; when $x$ is negative, $|x| = -x$. This means the expression $\frac{|x|}{x}$ simplifies to two different constants on either side of zero. For a limit to exist at a point, the function must approach the same value from both sides. Here, it doesn't — so the limit does not exist.

1. Understand the function piecewise.

   For $x > 0$, $|x| = x$, so $\frac{|x|}{x} = \frac{x}{x} = 1$.

   For $x < 0$, $|x| = -x$, so $\frac{|x|}{x} = \frac{-x}{x} = -1$.

   The function is not defined at $x = 0$ itself (division by zero), but that is irrelevant for a limit — we only care about values near zero.

2. Compute the right-hand limit (as $x \to 0^+$).

   When $x$ approaches 0 from the positive side, $x$ is always positive, so $\frac{|x|}{x} = 1$.

   Hence, $\lim_{x \to 0^+} \frac{|x|}{x} = 1$.

3. Compute the left-hand limit (as $x \to 0^-$).

   When $x$ approaches 0 from the negative side, $x$ is always negative, so $\frac{|x|}{x} = -1$.

   Hence, $\lim_{x \to 0^-} \frac{|x|}{x} = -1$.

4. Compare the two one-sided limits.

   The right-hand limit is $1$, and the left-hand limit is $-1$. Since $1 \neq -1$, the two-sided limit $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.