If $\dfrac{|x+1|}{x+1} > 0, x \in R$, then (A) $x \in [-1, \infty)$ (B) $x \in (-1, \infty)$ (C) $x \in (-\infty, -1)$ (D) $x \in (-\infty, -1]$

> [!TLDR] > $\dfrac{|x+1|}{x+1}>0$ holds precisely where $x+1>0$, i.e. $x>-1$, excluding $x=-1$ where it is undefined. > [!FORMULA] > For $t \ne 0$: $\dfrac{|t|}{t} = \begin{cases} +1, & t>0 \\ -1, & t 0$, $|t| = t$, so $\dfrac{|t|}{t} = 1 > 0$. ✓ 3. If $t 0 \Rightarrow x+1>0 \Rightarrow x>-1$, giving $x \in (-1,\infty)$. > [!ANSWER] > **(B) $x \in (-1, \infty)$**

If (|x+1|)/(x+1) > 0, x ∈ R, then (A) x ∈ [-1, ∞) (B) x — CBSE Class 12 Applied Mathematics 2025 PYQ

Question

\dfrac{|x+1|}{x+1} > 0, x \in R

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★est

✓ Free question

$\dfrac{|x+1|}{x+1}>0$ holds precisely where $x+1>0$ , i.e. $x>-1$ , excluding $x=-1$ where it is undefined.

t \ne 0

Let $t = x+1$ ; the expression is $\dfrac{|t|}{t}$ .
If $t>0$ , $|t| = t$ , so $\dfrac{|t|}{t} = 1 > 0$ . ✓
If $t<0$ , $|t| = -t$ , so $\dfrac{|t|}{t} = -1 < 0$ . ✗
At $t=0$ the fraction is undefined, so $x=-1$ is excluded.
Thus the inequality needs $t>0 \Rightarrow x+1>0 \Rightarrow x>-1$ , giving $x \in (-1,\infty)$ .

✓Final answer

x \in (-1, \infty)

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/S/WXYZ/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0, x \in R$
›Reveal solutionSolution
$\dfrac{|x+1|}{x+1}>0$ holds precisely where $x+1>0$ , i.e. $x>-1$ , excluding $x=-1$ where it is undefined.

$t \ne 0$

Let $t = x+1$ ; the expression is $\dfrac{|t|}{t}$ .

If $t>0$ , $|t| = t$ , so $\dfrac{|t|}{t} = 1 > 0$ . ✓

If $t<0$ , $|t| = -t$ , so $\dfrac{|t|}{t} = -1 < 0$ . ✗

At $t=0$ the fraction is undefined, so $x=-1$ is excluded.

Thus the inequality needs $t>0 \Rightarrow x+1>0 \Rightarrow x>-1$ , giving $x \in (-1,\infty)$ .

✓Final answer
$x \in (-1, \infty)$
CBSE 2025Set 465/S/WXYZ/41 markMCQ
$|3x - 2| \leq \dfrac{1}{2}$
›Reveal solutionSolution
The Assertion's solution set $[\tfrac12,\tfrac56]$ is correct, but the Reason misstates the modulus rule (it describes $\ge$ , not $\le$ ), so (C).

$|x-a|\le r \iff a-r \le x \le a+r$

Solve the Assertion: $|3x-2|\le \tfrac12 \iff -\tfrac12 \le 3x-2 \le \tfrac12$ .

Add 2 throughout: $\tfrac32 \le 3x \le \tfrac52$ .

Divide by 3: $\tfrac12 \le x \le \tfrac56$ , i.e. solution set $\left[\tfrac12,\tfrac56\right]$ — Assertion is true.

Examine the Reason: it claims $|x-a|\le r \Leftrightarrow x\le a-r \text{ or } x\ge a+r$ . That "or" form is the rule for $|x-a|\ge r$ , not $\le r$ . So the Reason is false.

Assertion true, Reason false $\Rightarrow$ option (C).

✓Final answer
$(C) Assertion (A) is true, but Reason (R) is false.$
CBSE 2024Set 465/RQPS/41 markMCQ
$x > y$
›Reveal solutionSolution
Dividing both sides of $x>y$ by the negative number $z$ reverses the inequality, so $\dfrac{x}{z}<\dfrac{y}{z}$ .

$a>b$

Given $x>y$ and $z<0$ .

Multiplying by $z<0$ reverses the sign: $xz<yz$ — so options (A) and (B) are wrong.

Dividing by $z<0$ also reverses the sign: $\dfrac{x}{z}<\dfrac{y}{z}$ — so (C) is wrong and (D) is correct.

✓Final answer
$\dfrac{x}{z}<\dfrac{y}{z}$
CBSE 2023Set 465/EF1GH/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0$
(a) $x \in [-1, \infty)$
(b) $x \in (-1, \infty)$
(c) $x \in (-\infty, -1)$
(d) $x \in (-\infty, -1]$
›Reveal solutionSolution
$\dfrac{|x+1|}{x+1}>0$ forces $x+1>0$ , i.e. $x\in(-1,\infty)$ .

$t\neq 0$

Let $t=x+1$ . The expression $\dfrac{|t|}{t}>0$ holds only when $t>0$ .

So $x+1>0 \Rightarrow x>-1$ .

At $x=-1$ the expression is $\tfrac{0}{0}$ , undefined, so $-1$ is excluded — the endpoint is open.

Hence the solution set is the open interval $(-1,\infty)$ ; option (a) wrongly includes $-1$ .

✓Final answer
$x\in(-1,\infty)$

Q.If $\dfrac{|x+1|}{x+1} > 0, x \in R$ , then (A) $x \in [-1, \infty)$ (B) $x \in (-1, \infty)$ (C) $x \in (-\infty, -1)$ (D) $x \in (-\infty, -1]$

Concept understanding — Linear and Modulus Inequalities

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