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Dive deeper into problem-solving techniques in polynomials, We’ll explore…

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  1. Factoring Techniques
  2. Solving Quadratic Polynomials
  3. Graphical Analysis
  4. Application-based Problems

1. Factoring Techniques

Factoring is crucial for simplifying and solving polynomial equations.

Case 1: Common Factor

Find the greatest common factor (GCF) and factor it out.
Example: Factor 6×3+9x26x^3 + 9x^2. =3×2(2x+3).= 3x^2(2x + 3).

Case 2: Grouping

Group terms to create common factors.
Example: Factor x3+3×2+x+3x^3 + 3x^2 + x + 3. =(x3+3×2)+(x+3).= (x^3 + 3x^2) + (x + 3). =x2(x+3)+1(x+3).= x^2(x + 3) + 1(x + 3). =(x+3)(x2+1).= (x + 3)(x^2 + 1).

Case 3: Splitting the Middle Term

Split the middle term in quadratic polynomials.
Example: Factor x2−5x+6x^2 – 5x + 6.
Find two numbers that multiply to 66 and add to −5-5 (i.e., −2-2 and −3-3): =x2−2x−3x+6.= x^2 – 2x – 3x + 6. =x(x−2)−3(x−2).= x(x – 2) – 3(x – 2). =(x−2)(x−3).= (x – 2)(x – 3).

2. Solving Quadratic Polynomials

To solve ax2+bx+c=0ax^2 + bx + c = 0, use:

Method 1: Factoring

Solve x2−5x+6=0x^2 – 5x + 6 = 0: x2−5x+6=(x−2)(x−3)=0.x^2 – 5x + 6 = (x – 2)(x – 3) = 0. x=2, x=3.x = 2, \, x = 3.

Method 2: Completing the Square

Solve x2−4x+3=0x^2 – 4x + 3 = 0 by completing the square:

  1. Rewrite x2−4x+3x^2 – 4x + 3 as: x2−4x=−3.x^2 – 4x = -3.
  2. Add (−42)2=4\left(\frac{-4}{2}\right)^2 = 4 to both sides: x2−4x+4=−3+4.x^2 – 4x + 4 = -3 + 4.
  3. Simplify: (x−2)2=1.(x – 2)^2 = 1.
  4. Solve: x−2=±1 ⟹ x=3, x=1.x – 2 = \pm 1 \implies x = 3, \, x = 1.

Method 3: Quadratic Formula

For ax2+bx+c=0ax^2 + bx + c = 0: x=−b±b2−4ac2a.x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}.

Example: Solve 2×2−4x−6=02x^2 – 4x – 6 = 0: a=2, b=−4, c=−6.a = 2, \, b = -4, \, c = -6. x=−(−4)±(−4)2−4(2)(−6)2(2).x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)}. x=4±16+484.x = \frac{4 \pm \sqrt{16 + 48}}{4}. x=4±644.x = \frac{4 \pm \sqrt{64}}{4}. x=4±84.x = \frac{4 \pm 8}{4}. x=3, x=−1.x = 3, \, x = -1.

3. Graphical Analysis

Graphs of polynomials provide visual insights into their behavior and solutions.

Quadratic Polynomials:

  • Shape: Parabola.
  • Key points:
    • Vertex: Turning point of the parabola.
    • Zeros: Points where the parabola intersects the xx-axis.
    • Direction: Opens upwards (a>0a > 0) or downwards (a<0a < 0).

Example: y=x2−4x+3y = x^2 – 4x + 3.

  1. Factorize: (x−1)(x−3)(x – 1)(x – 3), so zeros are x=1x = 1, x=3x = 3.
  2. Vertex: Midpoint of zeros: x=1+32=2, y=22−4(2)+3=−1.x = \frac{1 + 3}{2} = 2, \, y = 2^2 – 4(2) + 3 = -1.
  3. Plot points: (1, 0), (3, 0), (2, -1).

4. Application-Based Problems

Problem 1: Cost Optimization

A farmer builds a rectangular pen with xx-meter fencing on one side. The total area is 60 m260 \, \text{m}^2. Find xx.

Area = x⋅widthx \cdot \text{width}: 60=x⋅(x−2) ⟹ x2−2x−60=0.60 = x \cdot (x – 2) \implies x^2 – 2x – 60 = 0.

Solve using the quadratic formula: x=−(−2)±(−2)2−4(1)(−60)2(1).x = \frac{-(-2) \pm \sqrt{(-2)^2 – 4(1)(-60)}}{2(1)}. x=2±4+2402.x = \frac{2 \pm \sqrt{4 + 240}}{2}. x=2±2442.x = \frac{2 \pm \sqrt{244}}{2}. x=2±2612.x = \frac{2 \pm 2\sqrt{61}}{2}. x=1±61.x = 1 \pm \sqrt{61}.

Problem 2: Physics – Motion

The height hh of a ball is modeled by h(t)=−5t2+20t+25h(t) = -5t^2 + 20t + 25. Find the time when the ball hits the ground.
Set h(t)=0h(t) = 0: −5t2+20t+25=0.-5t^2 + 20t + 25 = 0.

Divide through by −5-5: t2−4t−5=0.t^2 – 4t – 5 = 0.

Factorize: (t−5)(t+1)=0.(t – 5)(t + 1) = 0. t=5, t=−1.t = 5, \, t = -1.

Since tt cannot be negative, t=5 secondst = 5 \, \text{seconds}.

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Polynomials, a critical evaluation and detailed explanation of its key concepts, with examples and problem-solving techniques

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Key Concepts in Polynomials

  1. Definition of a Polynomial: A polynomial is an algebraic expression consisting of variables, coefficients, and exponents, where the exponents are whole numbers. Examples:
    • Polynomial: 2×3−3×2+5x−72x^3 – 3x^2 + 5x – 7
    • Not a Polynomial: 2x−1+x2x^{-1} + \sqrt{x} (exponent is not a whole number).
  2. Degrees of a Polynomial:
    • The degree is the highest power of the variable in the polynomial.
    • Examples:
      • 5×3−2×2+45x^3 – 2x^2 + 4: Degree = 3.
      • 7y5+3y4+y7y^5 + 3y^4 + y: Degree = 5.
  3. Types of Polynomials Based on Degree:
    • Linear Polynomial: Degree = 1 (e.g., 3x+23x + 2).
    • Quadratic Polynomial: Degree = 2 (e.g., x2−4x+3x^2 – 4x + 3).
    • Cubic Polynomial: Degree = 3 (e.g., 2×3−x2+3x−52x^3 – x^2 + 3x – 5).
  4. Types of Polynomials Based on Terms:
    • Monomial: One term (e.g., 3x3x).
    • Binomial: Two terms (e.g., x2+2xx^2 + 2x).
    • Trinomial: Three terms (e.g., x3+x2+2x^3 + x^2 + 2).

Key Operations and Concepts

1. Addition, Subtraction, and Multiplication

  • Combine like terms while performing operations.

Example: Simplify (2×2+3x+4)+(x2−5x+6)(2x^2 + 3x + 4) + (x^2 – 5x + 6): =(2×2+x2)+(3x−5x)+(4+6).= (2x^2 + x^2) + (3x – 5x) + (4 + 6). =3×2−2x+10.= 3x^2 – 2x + 10.

2. Division of Polynomials

Division is performed using long division.

Example: Divide 2×3+3×2−x+52x^3 + 3x^2 – x + 5 by x+1x + 1.

  1. Divide 2x32x^3 by xx: Quotient = 2x22x^2.
  2. Multiply 2x22x^2 by x+1x + 1: 2×3+2x22x^3 + 2x^2.
  3. Subtract: (2×3+3×2−x+5)−(2×3+2×2)=x2−x+5(2x^3 + 3x^2 – x + 5) – (2x^3 + 2x^2) = x^2 – x + 5.
  4. Repeat until the degree of the remainder is less than the degree of the divisor.

The Remainder Theorem

If p(x)p(x) is divided by x−ax – a, the remainder is p(a)p(a).

Example: Find the remainder when p(x)=x3−3×2+4x−2p(x) = x^3 – 3x^2 + 4x – 2 is divided by x−2x – 2.

  1. Substitute x=2x = 2 into p(x)p(x): p(2)=23−3(22)+4(2)−2.p(2) = 2^3 – 3(2^2) + 4(2) – 2. =8−12+8−2=2.= 8 – 12 + 8 – 2 = 2.
  2. The remainder is 2.

The Factor Theorem

A polynomial p(x)p(x) has x−ax – a as a factor if p(a)=0p(a) = 0.

Example: Check if x−1x – 1 is a factor of p(x)=x3−4×2+3xp(x) = x^3 – 4x^2 + 3x.

  1. Substitute x=1x = 1 into p(x)p(x): p(1)=13−4(12)+3(1)=1−4+3=0.p(1) = 1^3 – 4(1^2) + 3(1) = 1 – 4 + 3 = 0.
  2. Since p(1)=0p(1) = 0, x−1x – 1 is a factor.

Zeros of a Polynomial

The zeros of a polynomial are the values of xx for which p(x)=0p(x) = 0.

Key Results:

  1. A polynomial of degree nn has at most nn zeros.
  2. Zeros can be real or complex numbers.

Example: Find the zeros of p(x)=x2−5x+6p(x) = x^2 – 5x + 6.

  1. Factorize: p(x)=(x−2)(x−3).p(x) = (x – 2)(x – 3).
  2. Zeros: x=2,x=3x = 2, x = 3.

Graphical Representation

The graph of a polynomial depends on its degree and coefficients:

  1. Linear Polynomials: Straight line.
  2. Quadratic Polynomials: Parabola.
  3. Cubic Polynomials: S-shaped curve.

Application Problems

Problem 1: Sum and Product of Zeros

If p(x)=ax2+bx+cp(x) = ax^2 + bx + c, the sum and product of zeros are: Sum of zeros=−ba,Product of zeros=ca.\text{Sum of zeros} = -\frac{b}{a}, \quad \text{Product of zeros} = \frac{c}{a}.

Example: For p(x)=2×2−5x+3p(x) = 2x^2 – 5x + 3:

  1. Sum of zeros: =−−52=52.= -\frac{-5}{2} = \frac{5}{2}.
  2. Product of zeros: =32.= \frac{3}{2}.

Problem 2: Verify Factor Theorem

Verify that x−2x – 2 is a factor of p(x)=x3−3×2+4x−8p(x) = x^3 – 3x^2 + 4x – 8.

  1. Substitute x=2x = 2 into p(x)p(x): p(2)=23−3(22)+4(2)−8=8−12+8−8=0.p(2) = 2^3 – 3(2^2) + 4(2) – 8 = 8 – 12 + 8 – 8 = 0.
  2. Since p(2)=0p(2) = 0, x−2x – 2 is a factor.

Critical Observations

  1. Importance of Remainder and Factor Theorem:
    • Simplifies finding roots and divisors of polynomials.
    • Basis for synthetic division.
  2. Role of Degree:
    • The degree dictates the behavior of polynomials (number of zeros, shape of graph).
  3. Graphical Interpretation:
    • Understanding graphs builds intuition about polynomial behavior.
  4. Applications:
    • Used in physics (trajectory of objects), economics (profit functions), and data fitting.

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Number System basics

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1. Sets and Classification of Numbers

Sets provide a foundation to understand the hierarchy and relationships between types of numbers.SE

Hierarchy of Numbers:

  • Natural Numbers (N\mathbb{N}): {1,2,3,… }\{1, 2, 3, \dots\}
  • Whole Numbers (W\mathbb{W}): {0,1,2,3,… }\{0, 1, 2, 3, \dots\}
  • Integers (Z\mathbb{Z}): {…,−3,−2,−1,0,1,2,3,… }\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}
  • Rational Numbers (Q\mathbb{Q}): Numbers of the form pq\frac{p}{q}, where p,qp, q are integers, q≠0q \neq 0.
  • Irrational Numbers (I\mathbb{I}): Non-terminating, non-repeating decimals like 2,π\sqrt{2}, \pi.
  • Real Numbers (R\mathbb{R}): R=Q∪I\mathbb{R} = \mathbb{Q} \cup \mathbb{I}.

2. Surds

Surds are irrational numbers expressed in root form.

Basic Rules:

  1. a⋅b=ab\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}
  2. ab=ab\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}
  3. (a)2=a(\sqrt{a})^2 = a

Simplification Examples:

  1. Simplify 50\sqrt{50}: 50=25⋅2=25⋅2=52.\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}.
  2. Simplify 182\frac{\sqrt{18}}{\sqrt{2}}: 182=182=9=3.\frac{\sqrt{18}}{\sqrt{2}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3.

3. Complex Numbers (Introduction)

Complex numbers are an extension of real numbers, forming C\mathbb{C}, and include the imaginary unit ii, where i2=−1i^2 = -1.

Example:

  1. Solve x2+1=0x^2 + 1 = 0: x2=−1 ⟹ x=±i.x^2 = -1 \implies x = \pm i.

Usage:

  • Solving quadratic equations with negative discriminants.
  • Representing numbers on the Argand plane.

4. Prime Numbers and Factorization

Prime numbers are natural numbers greater than 1 with only two factors: 1 and itself.

Applications:

  1. Prime Factorization: Decompose a number into its prime factors.
    • Example: 72=23⋅3272 = 2^3 \cdot 3^2.
  2. Finding HCF and LCM: Use prime factorization.
    • HCF: Product of the smallest powers of common factors.
    • LCM: Product of the highest powers of all factors.
    • Example:
      • 12=22⋅312 = 2^2 \cdot 3, 18=2⋅3218 = 2 \cdot 3^2.
      • HCF = 2⋅3=62 \cdot 3 = 6, LCM = 22⋅32=362^2 \cdot 3^2 = 36.

5. Roots and Their Properties

Roots generalize square roots to higher powers.

Examples:

  1. Simplify 273\sqrt[3]{27}: 273=3 (since 33=27).\sqrt[3]{27} = 3 \text{ (since \( 3^3 = 27 \))}.
  2. Simplify 814\sqrt[4]{81}: 814=81=9=3.\sqrt[4]{81} = \sqrt{\sqrt{81}} = \sqrt{9} = 3.

6. Exponentiation with Rational Numbers

Rational exponents provide a way to express roots and powers together.

Examples:

  1. am/n=amna^{m/n} = \sqrt[n]{a^m}:
    • 82/3=(83)2=22=48^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4.
  2. 27−1/3=1273=1327^{-1/3} = \frac{1}{\sqrt[3]{27}} = \frac{1}{3}.

7. Infinite Series (Optional Advanced Topic)

An introduction to infinite series helps understand repeating decimals.

Example: Convert 0.3‾0.\overline{3} into a fraction.

  1. Let x=0.3‾x = 0.\overline{3}.
  2. Multiply by 10: 10x=3.3‾.10x = 3.\overline{3}.
  3. Subtract xx: 10x−x=3.3‾−0.3‾.10x – x = 3.\overline{3} – 0.\overline{3}. 9x=3 ⟹ x=39=13.9x = 3 \implies x = \frac{3}{9} = \frac{1}{3}.

8. Approximation Techniques

Irrational numbers like 2\sqrt{2} or π\pi are approximated using sequences.

Example: Approximate 2\sqrt{2}.

  • Using the Babylonian method:
    • Start with x0=1x_0 = 1.
    • Use xn+1=xn+2xn2x_{n+1} = \frac{x_n + \frac{2}{x_n}}{2}.
    • Iterations:
      • x1=1.5x_1 = 1.5, x2=1.4167x_2 = 1.4167, x3=1.4142x_3 = 1.4142.

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The first chapter of CBSE Class 9 Maths is “Number Systems”

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1. Introduction to Number Systems

  • The chapter revisits the concept of numbers, starting from natural numbers, whole numbers, and integers, and introduces students to new types of numbers such as rational and irrational numbers.
  • Importance: Understanding the classification of numbers is foundational for higher mathematics. It forms the basis of algebra, geometry, and advanced calculus.

2. Real Numbers

  • Definition: Real numbers include all rational and irrational numbers.
  • Decimal Representation:
    • Rational numbers have either terminating or recurring decimals.
    • Irrational numbers have non-terminating, non-recurring decimals.
  • Examples:
    • Rational: 12=0.5\frac{1}{2} = 0.5, 13=0.333…\frac{1}{3} = 0.333…
    • Irrational: 2,π\sqrt{2}, \pi

3. Irrational Numbers

  • Concept: Numbers that cannot be expressed as pq\frac{p}{q} where pp and qq are integers and q≠0q \neq 0.
  • Examples: 2,3,π\sqrt{2}, \sqrt{3}, \pi.
  • Properties:
    • Adding, subtracting, multiplying, or dividing a rational number with an irrational number generally results in an irrational number.

4. Laws of Exponents for Real Numbers

  • Laws:
    • am⋅an=am+na^m \cdot a^n = a^{m+n}
    • aman=am−n\frac{a^m}{a^n} = a^{m-n}
    • (am)n=amn(a^m)^n = a^{mn}
    • (ab)m=am⋅bm(ab)^m = a^m \cdot b^m
    • ambm=(ab)m\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m
  • Importance: These laws simplify calculations and are essential for algebraic expressions and equations.

5. Representation of Real Numbers on the Number Line

  • Concept: Real numbers can be represented on the number line using successive magnification.
  • Visualization: Techniques like zooming in on the number line help locate irrational numbers like 2\sqrt{2}.

6. Rationalization

  • Definition: The process of converting an expression with an irrational denominator into a rational form by multiplying numerator and denominator with a suitable conjugate.
  • Example:
    • 12×22=22\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}.

7. Applications

  • These concepts are used in solving algebraic equations, scientific calculations, and advanced topics like trigonometry, calculus, and statistics.

Critical Understanding:

  • Students should focus on differentiating between rational and irrational numbers.
  • Practice representing numbers on the number line for a clear geometric understanding.
  • Emphasize mastering laws of exponents as they recur in multiple mathematical topics.

Would you like detailed explanations of specific problems or examples from this chapter?

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CBSE Class 10, 12 Sample Papers 2025 released at cbseacademic.nic.in, download links here

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Sep 06, 2024 08:10 AM IST

CBSE Class 10, 12 Sample Papers 2025 has been released. The download links is given here.

Central Board of Secondary Education has released CBSE Class 10, 12 Sample Papers 2025. The sample question papers for classes 10 and 12 for the current academic session 2024-25 is available to candidates on the official website of CBSE Academic at cbseacademic.nic.in.

CBSE Class 10, 12 Sample Papers 2025 released, download links here
CBSE Class 10, 12 Sample Papers 2025 released, download links here

As per the official notice, the Board issues Sample Question Papers (SQPs) and Marking Schemes (MS) for classes X and XII to provide a broad template to serve as a guide for ensuring uniformity and proper coverage of the curricula.

CBSE Class 10, 12 sample question papers give a broad understanding about the question paper design which should be used for classroom teaching and learning activities with an overall focus on promoting the application of concepts in real-life.

CBSE Class 10, 12 Sample Papers 2025: How to download

To download the sample papers for Class 10, 12 candidates can follow these simple steps given below.

  • Visit the official website of CBSE Academic at cbseacademic.nic.in.
  • Click on CBSE Class 10, 12 Sample Papers 2025 link available on the home page.
  • A new PDF file will open where the Class 10 and Class 12 sample papers links are attached.
  • Click on those links and the sample paper will be displayed on the screen.
  • Now click on the subject you want to check the sample paper for.
  • Once done, download the page and keep a hard copy of the same for further need.

Meanwhile, CBSE has started the submission of list of candidates for Class 10, 12 examination 2025. The submission of data started on September 5, 2024 on Pariksha Sangam website. Only those students will be allowed to appear for Class 10, 12 board examination in 2025, whose names will be submitted through online process of submission of LOC. For more related details candidates can check the official website of CBSE.

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UP NEET UG 2024 Counselling: Round 2 schedule out, registration begins Sept 9

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Sep 07, 2024 09:06 AM IST

UP NEET UG 2024 Counselling Round 2 schedule released. The registration will begin on September 9, 2024.

Medical Education and Training, Uttar Pradesh has released UP NEET UG 2024 Counselling Round 2 schedule. The complete schedule is available on the official website of UP NEET at upneet.gov.in.

UP NEET UG 2024 Counselling: Round 2 schedule out, registration begins Sept 9(Representational image)
UP NEET UG 2024 Counselling: Round 2 schedule out, registration begins Sept 9(Representational image)

The registration process begins on September 9 and will end on September 13, 2024. The last date for payment of the application fee is September 13, at 2 pm. The merit list will be displayed on September 14, 2024.

AP EdCET Counselling 2024: Revised Phase 1 schedule out at edcet-sche.aptonline.in, check seat allotment date

The choice filling will begin on September 14 and will conclude on September 18, 2024. The Round 2 seat allotment result will be announced on September 19, and candidates can download the allotment letter from September 20 to September 25, 2024.

UP NEET UG 2024 Counselling: How to apply for Round 2

To apply for the Round 2 counselling process, candidates can follow the steps given below.

  • Visit the official website of UP NEET at upneet.gov.in.
  • Click on UP NEET UG 2024 Counselling Round 2 registration link available on the home page.
  • Enter the registration details and click on submit.
  • Once done, fill the application form and make the payment of application fee.
  • Click on submit and download the confirmation page.
  • Keep a hard copy of the same for further need.

New candidates who want to participate in Round 2 counselling will have to pay ₹30000 as security money for government seats and ₹2 lakh for private medical college seats. For private sector dental colleges, it is mandatory to deposit Rs. 1,00,000/-.

NEET PG 2024 Counselling Live Updates

Un-allotted/admitted/not-reported candidates from the first round will have to participate in the second round of counselling. For those candidates participating in the second round of counselling who had not registered in the first round of counselling, it will be mandatory for such candidates to register online on the website by depositing the registration fee of Rs. 2000/-. For more related details candidates can check the official website of UP NEET.

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Efforts being made to recognise teachers using innovative methods: PM Modi | Education

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Prime Minister Narendra Modi has said efforts were being made to recognise bright teachers practising innovative methods as he took an apparent swipe at past governments, saying he would not talk about what used to happen earlier.

Prime Minister Narendra Modi said on Friday that the efforts were being by the government to recognise bright teachers practising innovative methods. (Photo Credit: ANI)
Prime Minister Narendra Modi said on Friday that the efforts were being by the government to recognise bright teachers practising innovative methods. (Photo Credit: ANI)

In an interaction with teachers who were conferred National Teachers’ Awards, he said they could prepare young students for ‘Viksit Bharat’.

The interaction took place on Friday while its video was shared on Saturday.

Also read: UGC NET June 2024 Answer Key awaited: Where, how to check provisional key

Modi suggested that the teachers could involve the students in selecting the top 100 tourist destinations in the country by encouraging them to learn about different places they might visit as part of education tours.

The teachers awarded were selected after a long process as their efforts can be useful in the new National Education Policy (NEP), Modi said while noting that there would be many others doing excellent work.

He told the teachers to take their students to nearby universities and watch sporting events as the experience could fire up their dreams.

Also read: Calicut University Result 2024 for regular, supplementary exams released, here’s how to check at results.uoc.ac.in

As many as 82 teachers from around the country were selected for the award. They included 50 teachers by the Department of School Education & Literacy, 16 by the Department of Higher Education and 16 by the Union Ministry of Skill Development and Entrepreneurship, an official statement said.

Modi suggested that the awardee teachers connect with each other on social media and share their best practices so that everyone could learn, adapt and benefit.

Also read: IIT Kanpur’s SIIC-incubated startups showcase innovative defence technologies at Armed Forces Festival 2024

He highlighted the impact of the NEP and spoke about the significance of attaining education in the mother tongue.

The prime minister suggested that the teachers teach local folklore to their students in different languages so that they learn multiple languages and also get exposure to India’s vibrant culture.

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IIT Kanpur’s SIIC displays cutting-edge defence tech at Armed Forces Fest 2024 | Education

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Indian Institute of Technology (IIT) Kanpur’s Startup Incubation and Innovation Centre (SIIC) took part in the three-day Armed Forces Festival ‘Sashakt aur Surakshit Bharat: Transforming the Armed Forces’ inaugurated by Chief Minister Yogi Adityanath in Lucknow on Friday, September 6.

Uttar Pradesh CM Yogi Adityanath checking defence technology displayed by a SIIC-incubated startup during the Armed Forces Festival 2024.
Uttar Pradesh CM Yogi Adityanath checking defence technology displayed by a SIIC-incubated startup during the Armed Forces Festival 2024.

The event witnessed an impressive display of military might and innovation, showcasing modern technology-based military weapon systems, including tanks, infantry weapons, and artillery guns, a press release informed.

Demonstrations during the program included slithering, para jumps, and insertion of special forces via helicopters.

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Additionally, there were other highlights such as paramotors, microlight flying, fighter aircraft flypasts, and horse, dog, and motorcycle displays.

As part of the program, Animesh Mishra, the Manager of SIIC, IIT Kanpur, led a group of startups incubated under SIIC, all making significant contributions to the defence sector.

Some of the technological innovations highlighted were the Alakh, Sabal 10/20, and Vibhram, a suite of surveillance and logistics drones developed by EndureAir Systems Pvt. Ltd which have been designed to enhance operational capabilities.

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The event also featured the Maraal Solar UAV by Maraal Aerospace Pvt. Ltd., an indigenously designed and developed solar-powered UAV with extended endurance and cyber-attack resilient features.

Likewise, Dream Aerospace Pvt. Ltd. showcased its Green Monopropellant Thruster named the Atom thruster, which utilizes a HAN-based non-toxic green fuel available in various thrust levels starting from 1N.

Other startups that featured their technological innovations included VU-Dynamics Pvt. Ltd., Space Philic Pvt Ltd., and Cyethack Solutions Pvt Ltd.

It is worth mentioning here that the products of these startups have been developed under the SIIC and aim to significantly contribute to the defence sector.

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Prof. Deepu Philip, Professor-in-Charge of SIIC at IIT Kanpur, expressed his pleasure at the technological innovations brought by the startups to the defence sector.

He said that these young companies have showcased technologies that can significantly enhance the capabilities of armed forces.

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