If $\dfrac{|x+1|}{x+1} > 0$, $x \in \mathbb{R}$, then : (a) $x \in [-1, \infty)$ (b) $x \in (-1, \infty)$ (c) $x \in (-\infty, -1)$ (d) $x \in (-\infty, -1]$

> [!TLDR] > $\dfrac{|x+1|}{x+1}>0$ forces $x+1>0$, i.e. $x\in(-1,\infty)$. > [!FORMULA] > For $t\neq 0$: $\dfrac{|t|}{t}=+1$ if $t>0$ and $-1$ if $t 0$ holds only when $t>0$. 2. So $x+1>0 \Rightarrow x>-1$. …

If (|x+1|)/(x+1) > 0, x ∈ mathbbR, then : (a) x ∈ [-1, — CBSE Class 12 Applied Mathematics 2023 PYQ

Question

\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]

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$\dfrac{|x+1|}{x+1}>0$ 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 $x+1>0 \Rightarrow x>-1$ . …

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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 $\dfrac{|t|}{t} = -1 < 0$ . ✗ …
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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. …
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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 $z<0$ . …
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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 $x+1>0 \Rightarrow x>-1$ . …
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Q.If $\dfrac{|x+1|}{x+1} > 0$ , $x \in \mathbb{R}$ , then :

Concept understanding — Linear and Modulus Inequalities

Q.If ∣x+1∣x+1>0\dfrac{|x+1|}{x+1} > 0x+1∣x+1∣​>0, x∈Rx \in \mathbb{R}x∈R, then :

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Q.If $\dfrac{|x+1|}{x+1} > 0$ , $x \in \mathbb{R}$ , then :

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