Samadhan Academy - Polynomial Worksheet

Samadhan Academy

✦ Excellence in Mathematical Education ✦

Polynomials & Quadratic Equations

Class X | Chapter: Polynomials | Zeroes & Relationships

📚 Instructions for Students

Read each problem carefully and attempt to solve it on your own first. Use rough paper for calculations. Once you've completed your solution, click the "Show Solution" button to verify your answer. Study the step-by-step solutions to understand the methodology.

Identifying Zeroes from Coordinate Points

A quadratic polynomial p(x) passes through the points:

(-6, 0), (0, -30), (4, -20), and (6, 0)

Find the zeroes of the polynomial p(x).

📊 Graph Visualization

★ Complete Solution

Understanding Zeroes:

The zeroes of a polynomial are the x-values where p(x) = 0.

These are the points where the graph intersects the x-axis (where y = 0).

Identifying from Given Points:

Looking at the given points:

Point (-6, 0): Here y = 0, so x = -6 is a zero
Point (0, -30): Here y ≠ 0, so not a zero
Point (4, -20): Here y ≠ 0, so not a zero
Point (6, 0): Here y = 0, so x = 6 is a zero

✓ Answer:

The zeroes of p(x) are -6 and 6

Finding a Polynomial with Irrational Zeroes

Find a quadratic polynomial having zeroes:

α = -52 and β = 52

★ Complete Solution

Formula:

If α and β are the zeroes, then the quadratic polynomial is:

p(x) = k[x² - (α + β)x + αβ]

Step 1: Find Sum of Zeroes

α + β = -52 + 52 = 0

Step 2: Find Product of Zeroes

αβ = (-52)(52)

= -52

Step 3: Form the Polynomial (taking k = 2)

p(x) = 2[x² - 0·x + (-52)]

= 2[x² - 52]

= 2x² - 5

✓ Answer:

The required polynomial is 2x² - 5

(or any non-zero multiple like 4x² - 10, etc.)

Polynomial from Given Sum and Product

Find a quadratic polynomial whose:

Sum of zeroes = -5
Product of zeroes = 6

★ Complete Solution

Given Information:

Sum of zeroes = α + β = -5

Product of zeroes = αβ = 6

Formula:

For a quadratic polynomial with zeroes α and β:

p(x) = k[x² - (sum of zeroes)x + (product of zeroes)]

Solution (taking k = 1):

p(x) = x² - (-5)x + 6

p(x) = x² + 5x + 6

Verification (Factorization):

x² + 5x + 6 = (x + 2)(x + 3)

Zeroes are -2 and -3

Sum = -2 + (-3) = -5 ✓

Product = (-2)(-3) = 6 ✓

✓ Answer:

The required polynomial is x² + 5x + 6

Finding Zeroes by Factorization

Find the zeroes of the polynomial:

p(x) = 3x² + 11x - 4

★ Complete Solution

Method: Factorization by Splitting the Middle Term

Given: 3x² + 11x - 4 = 0

Step 1: Find two numbers whose:

Product = 3 × (-4) = -12
Sum = 11

The numbers are 12 and -1

(Because: 12 × (-1) = -12 and 12 + (-1) = 11)

Step 2: Split the middle term

3x² + 11x - 4 = 3x² + 12x - x - 4

Step 3: Group and factor

= 3x(x + 4) - 1(x + 4)

= (3x - 1)(x + 4)

Step 4: Find zeroes

Setting p(x) = 0:

(3x - 1)(x + 4) = 0

Either 3x - 1 = 0 or x + 4 = 0

x = 13 or x = -4

✓ Answer:

The zeroes are 13 and -4

Evaluating Reciprocal Sum of Zeroes

If α and β are the zeroes of the polynomial:

p(x) = 2x² + 6x - 6

Find the value of 1α + 1β

★ Complete Solution

Given Polynomial:

p(x) = 2x² + 6x - 6

Comparing with ax² + bx + c:

a = 2, b = 6, c = -6

Step 1: Use relationships between zeroes and coefficients

Sum of zeroes: α + β = -ba = -62 = -3

Product of zeroes: αβ = ca = -62 = -3

Step 2: Simplify the required expression

1α + 1β = β + ααβ = α + βαβ

Step 3: Substitute values

1α + 1β = -3-3 = 1

✓ Answer:

1α + 1β = 1

Constructing Polynomial with Transformed Zeroes

If α and β are zeroes of the polynomial:

x² - 3x + 2

Construct a polynomial whose zeroes are 2α + 1 and 2β + 1

★ Complete Solution

Given Polynomial:

x² - 3x + 2

Step 1: Find relationships for original zeroes

Sum of zeroes: α + β = 3

Product of zeroes: αβ = 2

Step 2: Find sum of new zeroes

(2α + 1) + (2β + 1) = 2α + 2β + 2

= 2(α + β) + 2

= 2(3) + 2 = 8

Step 3: Find product of new zeroes

(2α + 1)(2β + 1) = 4αβ + 2α + 2β + 1

= 4αβ + 2(α + β) + 1

= 4(2) + 2(3) + 1

= 8 + 6 + 1 = 15

Step 4: Construct the required polynomial

Using the formula:

p(x) = x² - (sum of zeroes)x + (product of zeroes)

p(x) = x² - 8x + 15

Verification:

Original zeroes: α = 1, β = 2 (from x² - 3x + 2 = (x-1)(x-2))

New zeroes: 2(1)+1 = 3, 2(2)+1 = 5

Check: x² - 8x + 15 = (x-3)(x-5) ✓

✓ Answer:

The required polynomial is x² - 8x + 15

Polynomial with Given Sum and Product

Find a quadratic polynomial whose sum and product of zeroes are:

Sum of zeroes = 0
Product of zeroes = -9

Also find the zeroes of the polynomial.

★ Complete Solution

Given Information:

Sum of zeroes: α + β = 0

Product of zeroes: αβ = -9

Part 1: Construct the Polynomial

Using the formula:

p(x) = k[x² - (α + β)x + αβ]

Taking k = 1:

p(x) = x² - 0·x + (-9)

p(x) = x² - 9

Part 2: Find the Zeroes

Setting p(x) = 0:

x² - 9 = 0

This is a difference of squares:

(x - 3)(x + 3) = 0

Therefore:

x = 3 or x = -3

Verification:

Sum: 3 + (-3) = 0 ✓

Product: 3 × (-3) = -9 ✓

✓ Answer:

Polynomial: x² - 9

Zeroes: 3 and -3

Samadhan Academy - Mathematics Worksheet

Samadhan Academy

✦ Excellence in Education ✦

Mathematics Problem Worksheet

Class X | Quadratic Equations & Applications

Instructions: Attempt each problem independently first. Click the "Show Solution" button to verify your answer. Diagrams are provided where applicable.

Finding Two Positive Numbers

The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number.

Find the two numbers.

★ Solution

Let the greater number be $x$ and the smaller number be $y$ .

Given: $x² - y² = 180$ ... (1)

Also: $y² = 8x$ ... (2)

Substituting (2) in (1):

$x² - 8x = 180$

$x² - 8x - 180 = 0$

$(x - 18)(x + 10) = 0$

$x = 18$ or $x = -10$ (rejected, as x is positive)

From (2): $y² = 8 \times 18 = 144$

$y = 12$

Answer: The two numbers are 18 and 12.

Quadratic Equation Coefficients

Express $x +$ 1x $= 3$ as a quadratic equation in the form $ax² + bx + c = 0$ .

Find the value of $a - b + c$ .

★ Solution

Given: $x +$ 1x $= 3$

Multiplying both sides by $x$ :

$x² + 1 = 3x$

$x² - 3x + 1 = 0$

Comparing with $ax² + bx + c = 0$ :

$a = 1, b = -3, c = 1$

Therefore: $a - b + c = 1 - (-3) + 1 = 1 + 3 + 1 = 5$

Answer: $a - b + c$ = 5

Rectangular Field Dimensions

The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side.

Find the length of the sides.

📐 Diagram: Rectangular Field

★ Solution

Let the shorter side = $x$ m

Then: Longer side = $(x + 30)$ m, Diagonal = $(x + 60)$ m

By Pythagoras theorem:

$x² + (x + 30)² = (x + 60)²$

$x² + x² + 60x + 900 = x² + 120x + 3600$

$x² - 60x - 2700 = 0$

$(x - 90)(x + 30) = 0$

$x = 90$ (rejecting negative value)

Shorter side = 90 m, Longer side = 90 + 30 = 120 m

Answer: Shorter side = 90 m, Longer side = 120 m

Fraction Problem

The denominator of a fraction is 2 more than the numerator. If 2 is added to both the numerator and denominator, the sum of the new and original fractions is 4635.

Find the original fraction.

★ Solution

Let the numerator = $x$

Then denominator = $x + 2$

Original fraction = xx + 2

New fraction = x + 2x + 4

Given: xx + 2 + x + 2x + 4 = 4635

Solving: $35[x(x+4) + (x+2)²] = 46(x+2)(x+4)$

$35[x² + 4x + x² + 4x + 4] = 46[x² + 6x + 8]$

$70x² + 280x + 140 = 46x² + 276x + 368$

$24x² + 4x - 228 = 0$

$6x² + x - 57 = 0$

$(6x + 19)(x - 3) = 0$

$x = 3$ (rejecting negative value)

Answer: The original fraction is 35

Age Problem

Sourav's age is 3 years more than the square of Ravi's age. When Ravi reaches Sourav's current age, Sourav will be 6 years less than 13 times Ravi's current age.

Find their present ages.

★ Solution

Let Ravi's present age = $x$ years

Then Sourav's present age = $x² + 3$ years

Years until Ravi reaches Sourav's current age = $(x² + 3) - x = x² - x + 3$

Sourav's age at that time = $(x² + 3) + (x² - x + 3) = 2x² - x + 6$

Given: $2x² - x + 6 = 13x - 6$

$2x² - 14x + 12 = 0$

$x² - 7x + 6 = 0$

$(x - 6)(x - 1) = 0$

$x = 6$ or $x = 1$

If x = 1, Sourav = 4, but Ravi reaching age 4 in 3 years doesn't fit context.

So x = 6: Ravi = 6 years, Sourav = 36 + 3 = 39 years

Answer: Ravi is 6 years old and Sourav is 39 years old.

Weaving Sweaters (LCM Problem)

Ranjita, Neha, and Salma take 15, 18, and 20 days respectively to weave a sweater. They all start on the same day.

When will they all start a new sweater at the same time, and how many sweaters will they have completed in total?

★ Solution

To find when they all start a new sweater together, we need LCM of 15, 18, and 20.

Prime factorization:

15 = 3 × 5

18 = 2 × 3²

20 = 2² × 5

LCM = 2² × 3² × 5 = 4 × 9 × 5 = $180$ days

Sweaters completed:

Ranjita: 180 ÷ 15 = 12 sweaters

Neha: 180 ÷ 18 = 10 sweaters

Salma: 180 ÷ 20 = 9 sweaters

Total = 12 + 10 + 9 = 31 sweaters

Answer: After 180 days, total sweaters completed = 31

2-Digit Number Problem

A 2-digit number (where tens digit > unit digit) is obtained by:

Multiplying the sum of digits by 7 and adding 3, OR
Multiplying the difference of digits by 19 and subtracting 1

Find the 2-digit number.

★ Solution

Let tens digit = $x$ , units digit = $y$ (where x > y)

The number = $10x + y$

From condition 1: $10x + y = 7(x + y) + 3$

$10x + y = 7x + 7y + 3$

$3x - 6y = 3$

$x - 2y = 1$ ... (1)

From condition 2: $10x + y = 19(x - y) - 1$

$10x + y = 19x - 19y - 1$

$-9x + 20y = -1$

$9x - 20y = 1$ ... (2)

From (1): $x = 1 + 2y$

Substituting in (2): $9(1 + 2y) - 20y = 1$

$9 + 18y - 20y = 1$

$-2y = -8$

$y = 4$

Therefore: $x = 1 + 2(4) = 9$

Answer: The 2-digit number is 94

Ratio of Areas: Square and Circle

If the perimeter of a square is equal to the circumference of a circle.

Find the ratio of their areas.

📐 Diagram: Square and Circle Comparison

★ Solution

Let side of square = $a$ , radius of circle = $r$

Given: Perimeter of square = Circumference of circle

$4a = 2πr$

$a =$ πr2

Area of square = $a² =$ π²r²4

Area of circle = $πr²$

Ratio = Area of SquareArea of Circle = π²r²/4πr² = π4

Answer: Ratio of areas (Square : Circle) = π : 4

Circle and Semicircle Radii

The area of a circle is equal to the perimeter of a semicircular disc of equal radius.

Find the radius.

📐 Diagram: Circle and Semicircular Disc

★ Solution

Let the common radius = $r$

Area of circle = $πr²$

Perimeter of semicircular disc = Half circumference + Diameter

= $πr + 2r = r(π + 2)$

Given: Area of circle = Perimeter of semicircular disc

$πr² = r(π + 2)$

$πr = π + 2$ (dividing by r, assuming r ≠ 0)

$r =$ π + 2π $= 1 +$ 2π

≈ 1 + 0.637 ≈ 1.637 units

Answer: $r =$ π + 2π ≈ 1.64 units

Math Worksheet - Samadhan Academy

Samadhan Academy

Mathematics Excellence Series: Step-by-Step Solutions

1. Solve the Rational Equation

Solve for $x$: ; where $ x \neq -4, 7 $

Step 1: Find a common denominator:
$ \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} $

Step 2: Simplify the numerator:
$ \frac{-11}{x^2 - 3x - 28} = \frac{11}{30} $

Step 3: Cross-multiply and solve the quadratic:
$ x^2 - 3x + 2 = 0 $

Answer: $ x = 1 $ or $ x = 2 $

2. Linear Equations: Infinitely Many Solutions

Find $k$ for: $ kx + y = k^2 $ and $ x + ky = 1 $

Condition: $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{k}{1} = \frac{1}{k} = \frac{k^2}{1} $

From $ k^2 = 1 $, we get $ k = \pm 1 $. Verifying with the third ratio, only $ k = 1 $ satisfies all parts.

Answer: $ k = 1 $

3. Geometry: Surface Area of a Cone

Assertion (A): CSA of a cone (r=3.5, l=4) is 44 sq cm.
Reason (R): CSA formula is $ \pi rl $.

Calculation: $ CSA = \frac{22}{7} \times 3.5 \times 4 = 11 \times 4 = 44 $.

Conclusion: Both (A) and (R) are true; (R) is the correct explanation.

4. Y-axis Intersection

Equation: $ 2y - x = 4 $

Set $ x = 0 $: $ 2y - 0 = 4 \implies y = 2 $.

Answer: (C) (0, 2)

5. Graphical Representation

Lines: $ 8x - 4y + 12 = 0 $ and $ 2x - y + 5 = 0 $

Compare ratios: $ \frac{8}{2} = 4 $, $ \frac{-4}{-1} = 4 $, $ \frac{12}{5} = 2.4 $.

Since $ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $, the lines never meet.

Answer: (C) Parallel

The admission details for B.Sc. programs

https://drive.google.com/file/d/13tBfLcb5kKsquvwheELhIoR8yzutcY5F/view?usp=sharing

Event Announcement:
Samadhan Academy CBSE Classes Inauguration

**Event Announcement: Samadhan Academy CBSE Classes Inauguration**

**Date:** July 25, 2024

**Time:** 10:00 AM

**Location:** University Gate 02, Chhatarpur, M.P. 471001

Samadhan Academy is pleased to announce the inauguration of its new CBSE classes. The event will take place on July 25, 2024, starting at 10:00 AM, at University Gate 02 in Chhatarpur, Madhya Pradesh. This initiative aims to provide high-quality education and academic support to students in the region.

The event will include an introduction to the academy's curriculum, a tour of the facilities, and opportunities to meet the faculty. Prospective students and their parents are invited to attend and learn more about the academy's offerings.

For more information, please contact Samadhan Academy at University Gate 02, Chhatarpur. Call: 9754624944

We look forward to seeing you there!

**Anoop Wari**

**CEO, Samadhan Academy**

University Gate 02

Chhatarpur, M.P. 471001

Phone: 9754624944

### IES Academy : Pure Mathematics Workshop

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#### Welcome to the IES Academy 2024!

We are thrilled to announce an exciting new addition to our educational offerings – the **Pure Mathematics Workshop**. This workshop is designed for students, educators, and enthusiasts who are passionate about the elegance and depth of pure mathematics. Below, you'll find all the details about this enriching experience.

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#### Workshop Overview

**Date:** August 1 - August 5, 2024

**Time:** 9:00 AM - 4:00 PM daily

**Venue:** IES Academy, Main Campus, Room 205

This intensive five-day workshop will cover fundamental and advanced topics in pure mathematics, offering participants the opportunity to deepen their understanding and explore new areas of interest.

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#### Key Topics Covered

- **Number Theory:** Delve into the properties and relationships of numbers, including prime numbers, divisibility rules, and modular arithmetic.

- **Abstract Algebra:** Explore the structures of algebraic systems such as groups, rings, and fields, and their applications.

- **Real Analysis:** Understand the rigorous underpinnings of calculus, including sequences, series, and continuity.

- **Topology:** Learn about the properties of space that are preserved under continuous transformations.

---

#### Featured Speakers

We are honored to host a panel of esteemed mathematicians and educators who will lead the sessions:

- **Anupam Sir**, Professor of Mathematics, IES Academy,

- **Anoop Sir**, Research Scientist, IES Institute of Mathematical Sciences

---

#### **Who Should Attend?**

- **Students:** High school and college students with a keen interest in mathematics.

- **Educators:** Teachers and professors looking to enhance their curriculum and teaching methods.

- **Math Enthusiasts:** Anyone with a passion for mathematics and a desire to learn more about its theoretical aspects.

---

#### Registration Information

**Early Bird Registration (until July 15, 2024):** 200

**Regular Registration (after July 15, 2024):** 250

How to Register:

Visit our [registration page](#) to sign up. Spaces are limited, so be sure to secure your spot early!

---

#### **Contact Us**

For more information, please contact us at:

- **Email:** workshopsiesacademy@gmail.com

- **Phone:** 9754624944

- **Address:** IES Academy, Bajrang nagar , chhatarpur, M..P., 471001

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We look forward to welcoming you to this transformative experience. Join us at the IES Academy Pure Mathematics Workshop and immerse yourself in the beauty and power of pure mathematics.

---

**Stay Connected with IES Academy:**

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Best regards,

**IES Academy Team**

Excellence in Education and Beyond

90 Days Summer Camps at Samadhan CBSE Academy

Dear Parents and Students,

Get ready for an exciting learning adventure at Samadhan CBSE Academy! We are thrilled to announce our exclusive 90-day summer camps designed especially for CBSE Class X students. With a focus on academic excellence and comprehensive preparation, our program aims to complete the Class X Math and Science syllabus in just 90 days!

Program Highlights:

- **Subject Coverage:** Our summer camps are tailored to cover the entire Class X Math and Science syllabus, ensuring that students are fully prepared for their upcoming examinations.

- **Expert Faculty:** Benefit from the guidance of experienced and dedicated faculty members who will provide comprehensive explanations, solve queries, and offer valuable insights to enhance learning.

- **Interactive Learning:** Engage in interactive sessions, group discussions, and practical exercises that make learning enjoyable and effective.

- **Regular Assessments:** Track your progress with regular assessments and mock tests designed to simulate exam conditions and boost confidence.

- **Personalized Attention:** Receive personalized attention and support from our educators to address individual learning needs and challenges.

- **Fun Activities:** Beyond academics, participate in fun and engaging activities to unwind, relax, and rejuvenate while making new friends.

Registration Details:

- Contact Number: For registration and inquiries, please call us at 9754624944.

- Address: Visit us at Bajrang Nagar, Panna Road, Chhatarpur, to register for the summer camp.

- Eligibility: This summer camp is exclusively for CBSE Class X students.

Don't miss this opportunity to embark on a journey towards academic success with Samadhan CBSE Academy. Registrations are now open! Secure your spot today and make this summer a memorable and productive one.

Warm regards,

ANOOP SIR

[ HOD OF ]

Samadhan CBSE Academy

UNIT I: NUMBER SYSTEMS

1. REAL NUMBER (15) Periods

Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality of

UNIT II: ALGEBRA

1. POLYNOMIALS (8) Periods

Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials.

2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods

Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situational problems.

3. QUADRATIC EQUATIONS (15) Periods

Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities to be incorporated.

4. ARITHMETIC PROGRESSIONS (10) Periods

Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.

UNIT III: COORDINATE GEOMETRY Coordinate Geometry (15) Periods Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division).

UNIT IV: GEOMETRY

1. TRIANGLES (15) Periods

Definitions, examples, counter examples of similar triangles. 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. 3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar. 4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar. 5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.

2. CIRCLES (10) Periods

Tangent to a circle at, point of contact 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.

UNIT V: TRIGONOMETRY

1. INTRODUCTION TO TRIGONOMETRY (10) Periods

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios whichever are defined at 0o and 90o. Values of the trigonometric ratios of 300 , 450 and 600 . Relationships between the ratios.

2. TRIGONOMETRIC IDENTITIES (15) Periods

Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be given.

3. HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression. (10)Periods

Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, and 60°.

UNIT VI: MENSURATION

1. AREAS RELATED TO CIRCLES (12) Periods

Area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60°, 90° and 120° only. 2. SURFACE AREAS AND VOLUMES (12) Periods

Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.

UNIT VII: STATISTICS AND PROBABILITY

1. STATISTICS (18) Periods

Mean, median and mode of grouped data (bimodal situation to be avoided).

2. PROBABILITY (10) Periods

Classical definition of probability. Simple problems on finding the probability of an event.

Mathematics - Textbook for class X - NCERT Publication

Guidelines for Mathematics Laboratory in Schools, class X - CBSE Publication

Laboratory Manual - Mathematics, secondary stage - NCERT Publication

Mathematics exemplar problems for class X, NCERT publication.

Theme: Materials

Unit I: Chemical Substances - Nature and Behaviour Chemical reactions: Chemical equation, Balanced chemical equation, implications of a balanced chemical equation, types of chemical reactions: combination, decomposition, displacement, double displacement, precipitation, endothermic exothermic reactions, oxidation and reduction. Acids, bases and salts: Their definitions in terms of furnishing of H+ and OH– ions, General properties, examples and uses, neutralization, concept of pH scale (Definition relating to logarithm not required), importance of pH in everyday life; preparation and uses of Sodium Hydroxide,Bleaching powder, Baking soda, Washing soda and Plaster of Paris. Metals and nonmetals: Properties of metals and non-metals; Reactivity series; Formation and properties of ionic compounds; Basic metallurgical processes; Corrosion and its prevention. Carbon compounds: Covalent bonding in carbon compounds. Versatile nature of carbon. Homologous series. Nomenclature of carbon compounds containing functional groups (halogens, alcohol, ketones, aldehydes, alkanes and alkynes), difference between saturated hydro carbons and unsaturated hydrocarbons. Chemical properties of carbon compounds (combustion, oxidation, addition and substitution reaction). Ethanol and Ethanoic acid (only properties and uses), soaps and detergents.

Theme: The World of the Living

Unit II: World of Living Life processes: ‘Living Being’. Basic concept of nutrition, respiration, transport and excretion in plants and animals.

Control and co-ordination in animals and plants: Tropic movements in plants; Introduction of plant hormones; Control and co-ordination in animals: Nervous system; Voluntary, involuntary and reflex action; Chemical co-ordination: animal hormones. Reproduction: Reproduction in animals and plants (asexual and sexual) reproductive health - need and methods of family planning. Safe sex vs HIV/AIDS. Child bearing and women’s health. Heredity and Evolution: Heredity; Mendel’s contribution- Laws for inheritance of traits: Sex determination: brief introduction: (topics excluded - evolution; evolution and classification and evolution should not be equated with progress).

Theme: Natural Phenomena

Unit III: Natural Phenomena Reflection of light by curved surfaces; Images formed by spherical mirrors, centre of curvature, principal axis, principal focus, focal length, mirror formula (Derivation not required),magnification. Refraction; Laws of refraction, refractive index. Refraction of light by spherical lens; Image formed by spherical lenses; Lens formula(Derivation not required); Magnification. Power of a lens. Functioning of a lens in human eye, defects of vision and their corrections, applications of spherical mirrors and lenses. Refraction of light through a prism, dispersion of light, scattering of light, applications in dailylife (excluding colour of the sun at sunrise and sunset).

Theme: How Things Work

Unit IV: Effects of Current Electric current, potential difference and electric current. Ohm’s law; Resistance, Resistivity, Factors on which the resistance of a conductor depends. Series combination of resistors, parallel combination of resistors and its applications in daily life. Heating effect of electric current and its applications in daily life. Electric power, Interrelation between P, V, I and R. Magnetic effects of current : Magnetic field, field lines, field due to a current carryingconductor, field due to current carrying coil or solenoid; Force on current carrying conductor, Fleming’s Left Hand Rule, Direct current. Alternating current: frequency of AC. Advantage of AC over DC. Domestic electric circuits. Theme: Natural Resources Unit V: Natural Resources Our environment: Eco-system, Environmental problems, Ozone depletion, waste production and their solutions. Biodegradable and non-biodegradable substances.

Note for the Teachers:

1. The chapter Management of Natural Resources (NCERT Chapter 16) will not be assessed in the year-end examination. However, learners may be assigned to read this chapter and encouraged to prepare a brief write up to any concept of this chapter in their Portfolio. This

2. The NCERT text books present information in boxes across the book. These help students to get conceptual clarity. However, the information in these boxes would not be assessed in the year-end examination.

Theme: Materials

Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics - shape, volume, density; change of statemelting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture). Particle nature and their basic units: Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses. Structure of atoms: Electrons, protons and neutrons, Valency, Atomic Number and Mass Number, Isotopes and Isobars.

Theme: The World of the Living

Unit II: Organization in the Living World Cell - Basic Unit of life : Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes - basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants)

Theme: Moving Things, People and Ideas

Unit III: Motion, Force and Work Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion. Force and Newton’s laws : Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, Energy and Power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy). Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food

Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

Note for the Teachers: 1. The chapter Natural Resources (NCERT Chapter 14) will not be assessed in the year-end examination. However, learners may be assigned to read this chapter and encouraged to prepare a brief write up on any concept of this chapter in their Portfolio. This may be for Internal Assessment and credit may be given for Periodic Assessment/Portfolio. 2. The NCERT text books present information in boxes across the book. These help students to get conceptual clarity. However, the information in these boxes would not be assessed in the year-end examination.

एमसीक्यू परीक्षकों को अपेक्षाकृत कम अवधि के भीतर ज्ञान और समझ की एक विस्तृत श्रृंखला का आकलन करने की अनुमति देते हैं। यह विशेष रूप से बड़े पैमाने की परीक्षाओं में उपयोगी है जहां समय की कमी चिंता का विषय है। MCQs provide an objective method of evaluation. grading is straightforward, reducing the potential for subjectivity in assessment.

can ensure that a broad spectrum of the curriculum is tested,

एमसीक्यू में अक्सर ध्यान भटकाने वाले (गलत उत्तर विकल्प) शामिल होते हैं जो, This helps to minimize the effectiveness of guesswork, requiring students to demonstrate genuine understanding to select the correct answer.

MCQs can be adapted to test different cognitive levels, from basic recall of facts to higher-order thinking skills such as analysis and synthesis.यह बहुमुखी प्रतिभा परीक्षकों को कई प्रकार के कौशल और क्षमताओं का आकलन करने की अनुमति देती है।(This versatility allows examiners to assess a range of skills and abilities.)

MCQs facilitate the standardization of exams, ensuring that all students are assessed on the same content and under the same conditions.

इससे evaluation. में fairness और consistency को बढ़ावा मिलता है।

MCQs may provide immediate feedback to students.

जिससे उन्हें to identify areas of weakness और focus their study efforts accordingly.

Overall, while MCQs may have limitations in assessing certain types of knowledge and skills,

Overall, the curriculum(पाठ्यक्रम) aims to equip students with a solid foundation in mathematics while nurturing their ability to apply mathematical concepts in various contexts.

Summary: The use of Multiple-Choice Questions (MCQs) in examinations allows examiners to evaluate students' knowledge and understanding efficiently within a limited time frame, making them particularly valuable in large-scale assessments. MCQs provide an objective method of evaluation, reducing the potential for subjectivity in grading. They cover a broad spectrum of the curriculum, minimizing the effectiveness of guesswork and requiring genuine understanding from students to select the correct answer. Moreover, MCQs can be adapted to test different cognitive levels, ensuring that a range of skills and abilities are assessed. Standardization of exams is facilitated by MCQs, promoting fairness and consistency in evaluation. Immediate feedback provided by MCQs helps students identify areas of weakness and focus their study efforts accordingly. Despite their limitations, MCQs play a crucial role in assessing students' knowledge and skills, aligning with the goal of equipping students with a solid foundation in mathematics and the ability to apply mathematical concepts effectively in various contexts.

शीर्षक: कॉर्पोरेट प्रभाव के आरोपों के बीच राजनीतिक फंडिंग पर भाजपा का दबदबा

पिछले एक दशक में राजनीतिक फंडिंग के हालिया विश्लेषण में यह बात सामने आई है कि भारत में सत्तारूढ़ दल भारतीय जनता पार्टी (बीजेपी) ने वित्तीय सहायता के मामले में अपने समकक्षों को काफी पीछे छोड़ दिया है। रिपोर्टों के अनुसार, भाजपा ने पिछले एक दशक में विभिन्न चुनावी ट्रस्टों (ईटी) के माध्यम से 1,893 करोड़ रुपये की आश्चर्यजनक राशि अर्जित की है, जिससे राजनीतिक परिदृश्य में उसका वित्तीय प्रभुत्व मजबूत हुआ है।

सबसे उल्लेखनीय चैनलों में से एक जिसके माध्यम से भाजपा ने धन हासिल किया है वह चुनावी बांड के माध्यम से है, जहां वह अग्रणी प्राप्तकर्ता के रूप में उभरी है। आंकड़ों से पता चलता है कि पार्टी को चुनावी ट्रस्टों के माध्यम से योगदान का बड़ा हिस्सा प्राप्त हुआ, 2018 और 2022 के बीच इसकी लगभग 61 प्रतिशत फंडिंग अकेले चुनावी बांड से हुई।

हालाँकि, यह वित्तीय वर्चस्व विवाद से रहित नहीं है। राजनीतिक दलों की कॉरपोरेट फंडिंग लंबे समय से विश्व स्तर पर एक विवादास्पद मुद्दा रही है और भारत भी इसका अपवाद नहीं है। बारीकी से जांच करने पर पता चलता है कि भाजपा को कॉर्पोरेट चंदा और उसके बाद केंद्र सरकार की एजेंसियों की कार्रवाइयों के बीच एक चिंताजनक संबंध है।

चौंकाने वाले खुलासे से संकेत मिलता है कि कम से कम 30 कंपनियों ने सामूहिक रूप से भाजपा को 335 करोड़ रुपये का चंदा दिया, उन्होंने खुद को जांच के दायरे में पाया और उसी अवधि के दौरान सरकारी एजेंसियों की कार्रवाई का सामना करना पड़ा। यह कॉर्पोरेट हितों और राजनीतिक फंडिंग के बीच सांठगांठ के बारे में प्रासंगिक सवाल उठाता है, जो संभावित रूप से लोकतंत्र और कॉर्पोरेट प्रभाव के बीच की रेखाओं को धुंधला कर देता है।

आलोचकों का तर्क है कि कॉर्पोरेट दानदाताओं और राजनीतिक दलों के बीच इतना घनिष्ठ संबंध शासन में पारदर्शिता और जवाबदेही के सिद्धांतों को कमजोर करता है। इन खुलासों के मद्देनजर राजनीतिक निर्णय लेने पर कॉर्पोरेट संस्थाओं के अनुचित प्रभाव को रोकने के लिए और अधिक कड़े नियमों और निरीक्षण तंत्र की मांग जोर पकड़ रही है।

जैसे-जैसे राजनीतिक फंडिंग को लेकर बहस तेज होती जा रही है, भारत की लोकतांत्रिक प्रक्रियाओं की अखंडता सुनिश्चित करने के लिए अधिक पारदर्शिता और जवाबदेही की तत्काल आवश्यकता है। भाजपा की वित्तीय सर्वोच्चता और व्यापक राजनीतिक परिदृश्य पर इसके निहितार्थ के बारे में खुलासे एक निष्पक्ष और न्यायसंगत लोकतंत्र के आदर्शों को बनाए रखने के लिए इन चिंताओं को संबोधित करने की तात्कालिकता को रेखांकित करते हैं।

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WELCOME LINE

📚 Instructions for Students

Identifying Zeroes from Coordinate Points

★ Complete Solution

Finding a Polynomial with Irrational Zeroes

★ Complete Solution

Polynomial from Given Sum and Product

★ Complete Solution

Finding Zeroes by Factorization

★ Complete Solution

Evaluating Reciprocal Sum of Zeroes

★ Complete Solution

Constructing Polynomial with Transformed Zeroes

★ Complete Solution

Polynomial with Given Sum and Product

★ Complete Solution

Finding Two Positive Numbers

★ Solution

Quadratic Equation Coefficients

★ Solution

Rectangular Field Dimensions

★ Solution

Fraction Problem

★ Solution

Age Problem

★ Solution

Weaving Sweaters (LCM Problem)

★ Solution

2-Digit Number Problem

★ Solution

Ratio of Areas: Square and Circle

★ Solution

Circle and Semicircle Radii

★ Solution

1. Solve the Rational Equation

2. Linear Equations: Infinitely Many Solutions

3. Geometry: Surface Area of a Cone

4. Y-axis Intersection

5. Graphical Representation

The admission details for B.Sc. programs

### IES Academy : Pure Mathematics Workshop

#### **Welcome to the IES Academy 2024!**

#### **Workshop Overview**

#### **Key Topics Covered**

#### **Featured Speakers**

#### **Registration Information**

**How to Register:**

**90 Days Summer Camps at Samadhan CBSE Academy**

Dear Parents and Students,

**Program Highlights:**

**Registration Details:**

- **Contact Number:** For registration and inquiries, please call us at 9754624944.

- **Address:** Visit us at Bajrang Nagar, Panna Road, Chhatarpur, to register for the summer camp.

- **Eligibility:** This summer camp is exclusively for CBSE Class X students.

UNIT I: NUMBER SYSTEMS

1. REAL NUMBER (15) Periods

Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality of

UNIT II: ALGEBRA

1. POLYNOMIALS (8) Periods

Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials.

2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods

Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situational problems.

3. QUADRATIC EQUATIONS (15) Periods

4. ARITHMETIC PROGRESSIONS (10) Periods

Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.

UNIT III: COORDINATE GEOMETRY Coordinate Geometry (15) Periods Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division).

UNIT IV: GEOMETRY

1. TRIANGLES (15) Periods

2. CIRCLES (10) Periods

Tangent to a circle at, point of contact 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.

#### Welcome to the IES Academy 2024!

#### Workshop Overview

#### Key Topics Covered

#### Featured Speakers

#### Registration Information

How to Register:

90 Days Summer Camps at Samadhan CBSE Academy

Program Highlights:

Registration Details:

- Contact Number: For registration and inquiries, please call us at 9754624944.

- Address: Visit us at Bajrang Nagar, Panna Road, Chhatarpur, to register for the summer camp.

- Eligibility: This summer camp is exclusively for CBSE Class X students.