Home
School Syllabus
CBSE
Class XI
Mathematics
अध्याय 5: रैखिक असमिकाएं (Linear Inequalities)

माध्यम चुनें / Select Medium:

CBSE • कक्षा XI • Mathematics • अध्याय 5

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

रैखिक असमिकाएं (Linear Inequalities)

In Class 11 Mathematics, "Linear Inequalities" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

⚖️ Have You Ever Wondered?

In real-world engineering and business, you rarely have exact equalities like 'Cost = $1,000'; instead, you have constraints like 'Budget $\le$ $10,00...

In real-world engineering and business, you rarely have exact equalities like 'Cost = $1,000'; instead, you have constraints like 'Budget $\le$ $10,000$' and 'Load capacity $\ge$ 500 tons'. Linear inequalities define feasible regions of optimization.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Mathematics, "Linear Inequalities" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Linear equations in one and two variables.
Number line representations.
Coordinate geometry.

इस अध्याय के लक्ष्य

Solve linear inequalities in one variable algebraically and represent solutions on the real number line.
Understand reversal of inequality signs when multiplying or dividing by negative numbers.
Solve systems of linear inequalities in one variable and find intersection solution intervals.
Graph linear inequalities in two variables using boundary lines and half-planes.
Determine the feasible solution region for a system of simultaneous linear inequalities graphically.

अध्याय रूपरेखा एवं प्रगति

1 1. Algebraic Solutions in One Varia...

2 2. Graphical Solutions in Two Varia...

3 3. Systems of Inequalities

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Algebraic Solutions in One Variable

Two real numbers or algebraic expressions related by $<, >, \le,$ or $\ge$ form an Inequality.
• Cardinal Golden Rule: If both sides are multiplied or divided by a negative number, the inequality sign MUST be reversed! $$\mathbf{-2x < 6 \implies x > -3}$$ Solutions are expressed as intervals: $(a, b)$ open, $[a, b]$ closed.

2. Graphical Solutions in Two Variables

The equation $ax + by = c$ divides the Cartesian plane into two half-planes. To determine which half-plane satisfies $ax + by < c$, test the origin $(0, 0)$: if $(0, 0)$ satisfies the inequality, shade the half-plane containing the origin; otherwise, shade the opposite side! (Dashed boundary line for strict $<, >$; solid line for $\le, \ge$).

3. Systems of Inequalities

The solution region of a system of simultaneous inequalities is the common intersecting shaded region satisfying all given constraints simultaneously.

चित्रात्मक व्याख्या एवं मॉडल

Linear Inequalities Master Matrix

Conceptual framework, core mechanisms, and analytical relationships

Academic Architecture

1. Algebraic Solutions in One Variable • 2. Graphical Solutions in Two Variables

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Golden Rule: Multiplying or dividing by a negative number inverts the inequality sign.

मुख्य बिंदु 2

Interval Notation: Brackets $[a, b]$ indicate inclusion; parentheses $(a, b)$ indicate exclusion.

मुख्य बिंदु 3

Half-Plane: Region of Cartesian plane bounded by a line satisfying an inequality.

मुख्य बिंदु 4

Origin Test: Plugging $(0, 0)$ into an inequality to identify the correct shaded half-plane.

मुख्य बिंदु 5

Feasible Region: Intersection of multiple half-planes satisfying all constraints.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

Solve $30x < 200$ when: (i) $x$ is a natural number, (ii) $x$ is an integer.

उत्तर एवं व्याख्या देखें

उत्तर: $x < \frac{200}{30} \implies x < 6.67$. (i) If $x \in \mathbb{N}$, solution set is $\{1, 2, 3, 4, 5, 6\}$. (ii) If $x \in \mathbb{Z}$, solution set is $\{\dots, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$.
(i) {1,2,3,4,5,6}; (ii) {..., 0, 1, ..., 6}.

Solve the inequality: $\frac{3(x-2)}{5} \le \frac{5(2-x)}{3}$.

उत्तर एवं व्याख्या देखें

उत्तर: Multiply both sides by 15: $9(x - 2) \le 25(2 - x) \implies 9x - 18 \le 50 - 25x \implies 34x \le 68 \implies x \le 2$. Solution interval is $(-\infty, 2]$.
x ∈ (-∞, 2].

Solve $-12x > 30$ when $x$ is a real number.

उत्तर एवं व्याख्या देखें

उत्तर: Divide both sides by $-12$ and reverse the inequality sign: $x < -\frac{30}{12} \implies x < -\frac{5}{2}$. Solution interval is $(-\infty, -2.5)$.
x ∈ (-∞, -2.5).

How do you determine whether a boundary line should be drawn solid or dashed when graphing an inequality in two variables?

उत्तर एवं व्याख्या देखें

उत्तर: If the inequality includes equality ($\le$ or $\ge$), the points on the boundary line are included in the solution set and the line is drawn solid; if strict ($<$ or $>$), the line is dashed.
Solid for ≤ and ≥; dashed for < and >.

Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.

उत्तर एवं व्याख्या देखें

उत्तर: Let consecutive odd integers be $x$ and $x + 2$. Constraints: $x < 10$, $x+2 < 10 \implies x < 8$. Sum: $x + (x+2) > 11 \implies 2x + 2 > 11 \implies 2x > 9 \implies x > 4.5$. Odd integers for $x$ are 5 and 7. The pairs are $(5, 7)$ and $(7, 9)$.
Pairs are (5, 7) and (7, 9).

अध्याय का अध्ययन पूर्ण हुआ?

अभ्यास के लिए तैयार?

ऑनलाइन CBT टेस्ट देकर तैयारी का मूल्यांकन करें

झारखण्ड बोर्ड परीक्षा पैटर्न पर आधारित बहुविकल्पीय प्रश्नों का ऑनलाइन टेस्ट दें। तुरंत परिणाम, समय विश्लेषण और प्रत्येक प्रश्न का विस्तृत हल प्राप्त करें।

फाउंडेशन टेस्ट

15 MCQs • मूल अवधारणाएँ

एडवांस्ड टेस्ट

15 MCQs • HOTS और ट्रैप्स