Saturday, May 23, 2020

NCERT Class 12 maths solutions I CBSE Class 12 answers


NCERT Class 12 maths solutions, CBSE Class 12 answers

NCERT Class 12 maths solutions, CBSE Class 12 answers NCERT Class 12 maths solutions, CBSE Class 12 answers

CBSE Class 12 answers of mathematics

Now there is no need to take tension of solutions of class 12 mathematics. Here you can get all solutions of mathematics.These maths notes for class 12 have been designed in the most simple and precise format covering almost all the domains like differential calculus, algebra, trigonometry and coordinate geometry. Preparing from these notes could help the students to fetch excellent marks in their class 12th as well as competitive examinations like JEE Mains and JEE Advanced. The CBSE notes that we are offering will help the students to grasp any concept quickly and revise thoroughly before the exams. These notes have been created by subject experts and offer a huge advantage as students will be fully prepared to tackle any type of questions that may be asked in the exams.


NCERT Class 12 maths solutions
NCERT Class 12 maths solutions

NCERT Solutions for Class 12 Maths Chapter-wise PDF


For more:

Fractions (भिन्न) and Types of Fraction
NCERT Solutions for Class 6 to 10 in PDF
[BEST] Free download RS Aggarwal solutions in pdf
What are different Types of Triangles?| Geometry
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RS Aggarwal Book
RS Aggarwal Book

Tuesday, May 19, 2020

Polynomials and Types of polynomials

Polynomials and Types of polynomials

Polynomials

Polynomials are as expressions which are composed of two algebraic terms. In this lesson, all the concepts of polynomials like its definition, terms and degree, types, functions, formulas and solution are covered along with solved example problems. Below is the list of topics covered in this lesson.

Table of Contents:

  • Definition
  • Degree
  • Terms
  • Types
  • Properties and Theorems
  • Equations
  • Function
  • Solving Polynomials
  • Operations
  • Examples
Polynomials

Polynomials

What is a Polynomial?

A polynomial is defined as an expression which contains two or more algebraic terms. It is made up of two terms namely Poly (meaning “many”) and Nominal (meaning “terms.”). Polynomials are composed of:

  • Constants such as 1, 2, 3, etc.
  • Variables such as g, h, x, y, etc.
  • Exponents such as 5 in x5 etc.

Degree of a Polynomial

The degree of a polynomials is defined as the highest degree of a monomial within a polynomial. Thus, a polynomial equation having one variable which has the largest exponent is called a degree of the polynomial.

Polynomials


                       

Example: Find the degree of the polynomial 6s4+ 3x2+ 5x +19

Solution:

The degree of the polynomial is 4.

Terms of a Polynomial

The terms of polynomials are the parts of the equation which are generally separated by “+” or “-” signs. So, each part of a polynomial in an equation is a term. For example, in a polynomial, say, 2x2 + 5 +4, the number of terms will be 3. The classification of a polynomial is done based on the number of terms in it.

Types of Polynomials

Polynomials are of 3 different types and are classified based on the number of terms in it. The three types of polynomials are:

  • Monomial
  • Binomial
  • Trinomial

These polynomials can be combined using addition, subtraction, multiplication, and division but is never division by a variable. A few examples of Non Polynomials are: 1/x+2, x-3

Monomial

A monomial is an expression which contains only one term. For an expression to be a monomial, the single term should be a non-zero term. A few examples of monomials are:

  • 5x
  • 3
  • 6a4
  • -3xy

Binomial

A binomial is a polynomial expression which contains exactly two terms. A binomial can be considered as a sum or difference between two or more monomials. A few examples of binomials are:

  • – 5x+3,
  • 6a4 + 17x
  • xy2+xy

Trinomial

A trinomial is an expression which is composed of exactly three terms. A few examples of trinomial expressions are:

  • – 8a4+2x+7
  • 4x2 + 9x + 7
  • Polynomial Properties and Theorems

    Some of the important properties of polynomials along with some important polynomial theorems are as follows:

    Property 1: Division Algorithm

    If a polynomial P(x) is divided by a polynomial G(x) results in quotient Q(x) with remainder R(x), then,

    P(x) = G(x) • Q(x) + R(x)

    Property 2: Bezout’s Theorem

    Polynomial P(x) is divisible by binomial (x – a) if and only if P(a) = 0.

    Property 3: Remainder Theorem

    If P(x) is divided by (x – a) with remainder r, then P(a) = r.

    Property 4: Factor Theorem

    A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x).

    Property 5: Intermediate Value Theorem

    If P(x) is a polynomial, and P(x) ≠ P(y) for (x < y), then P(x) takes every value from P(x) to P(y) in the closed interval [x, y].

    Property 6:

    The addition, subtraction and multiplication of polynomials P and Q result in a polynomial where,

    Degree(P ± Q) ≤ Degree(P or Q)

    Degree(P × Q) = Degree(P) + Degree(Q)

    Property 7:

    If a polynomial P is divisible by a polynomial Q, then every zero of Q is also a zero of P.

    Property 8:

    If a polynomial P is divisible by two coprime polynomials Q and R, then it is divisible by (Q • R).

    Property 9:

    If P(x) = a0 + a1x + a2x2 + …… + anxn is a polynomial such that deg(P) = n ≥ 0 then, P has at most “n” distinct roots.

    Property 10: Descartes’ Rule of Sign

    The number of positive real zeroes in a polynomial function P(x) is the same or less than by an even number as the number of changes in the sign of the coefficients. So, if there are “K” sign changes, the number of roots will be “k” or “(k – a)”, where “a” is some even number.

    Property 11: Fundamental Theorem of Algebra

    Every non-constant single-variable polynomial with complex coefficients has at least one complex root.

    Property 12:

    If P(x) is a polynomial with real coefficients and has one complex zero (x = a – bi), then x = a + bi will also be a zero of P(x). Also, x2 – 2ax + a2 + b2 will be a factor of P(x).

    Polynomial Equations

    The polynomial equations are those expressions which are made up of multiple constants and variables. The standard form of writing a polynomial equation is to put the highest degree first then, at the last, the constant term. An example of a polynomial equation is:

    b = a4 +3a3 -2a2 +a +1

    Polynomial Functions

    A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n can be written as:

    f(x) = a0xn + a1xn-1 + a2xn-2 + ….. + an-2x+ an-1x + an

    Solving Polynomials

    Any polynomial can be easily solved using basic algebra and factorization concepts. To solve a polynomial equation, the first step is to set the right-hand side as 0. The explanation of a polynomial solution are explained for two types:

    • Solving Linear Polynomials
    • Solving Quadratic Polynomials

    Solving Linear Polynomials

    Getting the solution of linear polynomials is easy and simple. First, isolate the variable term and make the equation as equal to zero. Then solve as basic algebra operation. An example of finding the solution of a linear equation is given below:

    Example: Solve 3x – 9

    Solution:

    First, make the equation as 0. So,

    3x – 9 = 0

    ⇒ 3x = 9

    ⇒ x = 9/3

    Or, x = 3.

    Thus, the solution of 3x-9 is x = 3.

    Solving Quadratic Polynomials

    To solve a quadratic polynomial, first, rewrite the expression in the ascending order of degree. Then, equate the equation and perform polynomial factorization to get the solution of the equation. An example to find the solution of a quadratic polynomial is given below for better understanding.

    Example: Solve 3x2 – 6x + x3 – 18

    Solution:

    First, arrange the polynomial in the ascending order of degree and equate to zero.

    ⇒ x+ 3x2 -6x – 18 = 0

    Now, take the common terms.

    x2(x+3) – 6(x+3)

    ⇒ (x2-6)(x+3)

    So, the solutions will be x =-3 and

    x2 = 6

    Or, x = √6

  • Polynomial Operations

    There are four main polynomial operations which are:

    • Addition of Polynomials
    • Subtraction of Polynomials
    • Multiplication of Polynomials
    • Division of Polynomials

    Each of the operations on polynomials are explained below using solved examples.

    Addition of Polynomials

    To add polynomials, always add the like terms i.e. the terms having the same variable and power. The addition of polynomials always results in a polynomial of the same degree. For example,

    Example: Find the sum of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5

    Solution:

    First, combine the like terms while leaving the unlike terms as they are. Hence,

    (5x3+3x2y+4xy−6y2)+(3x2+7x2y−2xy+4xy2−5)

    = 5x3+3x2+(3+7)x2y+(4−2)xy+4xy2−6y2−5

    = 5x3+3x2+10x2y+2xy+4xy2−6y2−5

    Subtraction of Polynomials

    Subtracting polynomials is similar to addition, the only difference being the type of operation. So, subtract the like terms to obtain the solution. It should be noted that subtraction of polynomials also result in a polynomial of the same degree.

    Example: Find the difference of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5

    Solution:

    First, combine the like terms while leaving the unlike terms as they are. Hence,

    (5x3+3x2y+4xy−6y2)-(3x2+7x2y−2xy+4xy2−5)

    = 5x3-3x2+(3-7)x2y+(4+2)xy-4xy2−6y2+5

    = 5x3-3x2-4x2y+6xy-4xy2−6y2+5

    Multiplication of Polynomials

    Two or more polynomial when multiplied always result in a polynomial of higher degree (unless one of them is a constant polynomial). An example of multiplying polynomials is given below:

    Example: Solve (6x−3y)×(2x+5y)

    Solution:

    ⇒ 6x ×(2x+5y)–3y × (2x+5y) ———- Using distributive law of multiplication

    ⇒ (12x2+30xy) – (6yx+15y2) ———- Using distributive law of multiplication

    ⇒12x2+30xy–6xy–15y—————– as xy = yx

    Thus, (6x−3y)×(2x+5y)=12x2+24xy−15y2

    Division of Polynomials

    Division of two polynomial may or may not result in a polynomial. Let us study below the division of polynomials in details. To divide polynomials, follow the given steps:

    Polynomial Division Steps:

    If a polynomial has more than one term, we use long division method for the same. Following are the steps for it.

    1. Write the polynomial in descending order.
    2. Check the highest power and divide the terms by the same.
    3. Use the answer in step 2 as the division symbol.
    4. Now subtract it and carry down the next term.
    5. Repeat step 2 to 4 until you have no more terms to carry down.
    6. Note the final answer including remainder in the fraction form (last subtract term).

    Polynomial Examples

    Example:

    Given two polynomial 7s3+2s2+3s+9 and 5s2+2s+1.

    Solve these using mathematical operation.

    Solution:

    Given polynomial:

    7s3+2s2+3s+9 and 5s2+2s+1

    Polynomial Addition: (7s3+2s2+3s+9) + (5s2+2s+1)

    = 7s3+(2s2+5s2)+(3s+2s)+(9+1)

    = 7s3+7s2+5s+10

    Hence, addition result in a polynomial.

    Polynomial Subtraction: (7s3+2s2+3s+9) – (5s2+2s+1)

    = 7s3+(2s2-5s2)+(3s-2s)+(9-1)

    = 7s3-3s2+s+8

    Hence addition result in a polynomial.

    Polynomial Multiplication:(7s3+2s2+3s+9) × (5s2+2s+1)

    = 7s3 (5s2+2s+1)+2s2 (5s2+2s+1)+3s (5s2+2s+1)+9 (5s2+2s+1))

    = (35s5+14s4+7s3)+ (10s4+4s3+2s2)+ (15s3+6s2+3s)+(45s2+18s+9)

    = 35s5+(14s4+10s4)+(7s3+4s3+15s3)+ (2s2+6s2+45s2)+ (3s+18s)+9

    = 35s5+24s4+26s3+ 53s2+ 21s +9

    Polynomial Division: (7s3+2s2+3s+9) ÷ (5s2+2s+1)

    (7s3+2s2+3s+9)/(5s2+2s+1)

    This cannot be simplified. Therefore division of these polynomial do not result in a Polynomial.


  • For more:

    Fractions (भिन्न) and Types of Fraction

    यहां क्लिक करें

    NCERT Solutions for Class 6 to 10 in PDF

    यहां क्लिक करें 

    [BEST] Free download RS Aggarwal solutions in pdf

    यहां क्लिक करें 

    What are different Types of Triangles?| Geometry

    यहां क्लिक करें 


    RS Aggarwal Book
    RS Aggarwal Book

Sunday, May 17, 2020

आयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindi

आयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindiआयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindi

Age Related Question of Maths in Hindi सभी प्रकार की प्रतियोगिता परीक्षाओं में विभिन्न तरह के गणित के सवाल पूछे जाते है जिनमे से आयु संबंधित प्रश्न (Math Age Trick) अवश्य पूछे जाते है ! परीक्षाओं में जिस तरह के प्र्श्न पूछे जाते है  उन प्रश्नो पर आधारित कुछ प्रश्न नीचे दिए गए है जिससे कि आपको ये पता चल सके कि प्रतियोगिता परीक्षाओं में आयु संबंधित प्रश्नो का प्रकार क्या होता है , किस तरह के सवाल पूछे जाते है उनका कठिनाई स्तर क्या होता है ! नीचे दिए गए प्रश्नो का अध्ययन करने से आपको आयु संबंधित सवालों को परीक्षा में हल करने में आसानी होगी ! उदाहरण के रूप में नीचे कुछ प्रश्न हल सहित दिए गए है I

Age Related Question of Maths in Hindi
आयु सम्बन्धी प्रश्न और हल  – Age Related Question of Maths in Hindi

Age Related Question of Maths in Hindi

01  समीना तथा सुहाना की आयु का अनुपात 7:3 है 6 वर्ष बाद इनकी आयु का अनुपात 5:3 होगा इनकी आयु मे कितना अंतर है ?

 हल –  माना समीना की वर्तमान आयु = 7x वर्ष तथा सुहाना की वर्तमान आयु = 3x वर्ष

 तब 7x+6 / 3x+6 = 5 / 3 3 (7x + 6) = 5 (3x =6)

  21x + 18 = 15x + 30 6x + 12 x = 2

 इनकी आयु मे अंतर = (7x – 3x) वर्ष = 4x वर्ष = (4*2) वर्ष = 8 वर्ष


02   रमेश की आयु अपने पुत्र राहुल आयु से 4 गुना है 5 वर्ष पूर्व रमेश की आयु राहुल की आयु से 9 गुना थी रमेश की वर्तमान आयु कितनी होगी ?

 हल –   माना राहुल की वर्तमान आयु = x वर्ष तब रमेश की वर्तमान आयु = 4x वर्ष (4x – 5) = 9 (x-5) 4x – 5 = 9x – 45

  5x = 40

  x =8

 रमेश की वर्षमान आयु = (4*8) वर्ष = 32 वर्ष


03.  8 वर्ष पूर्व विशाल की आयु अपने पुत्र शेखर की आयु से 4 गुना थी 8 वर्ष बाद विशाल की आयु शेखर की आयु से दुगुनी होगी प्रत्येक की वर्तमान आयु ज्ञात कीजिये ?

 हल –   माना 8 वर्ष पूर्व शेखर की आयु = x वर्ष तथा विशाल की आयु 4x वर्ष है

 शेखर की वर्तमान आयु = (x+8) वर्ष तथा विशाल की वर्तमान आयु = (4x + 8) वर्ष

 (4x+8) + 8= 2 [ (x+8) + 8]

  4x + 16 = 2x + 32 2x = 16 x = 8

 अतः शेखर की वर्तमान आयु = ( 8+8 ) वर्ष = 16 वर्ष

 विशाल की वर्तमान आयु = (4 *8 + 8) वर्ष = 40 वर्ष


04.   A तथा B की आयु का अंतर 16 वर्ष है 6 वर्ष पहले A की आयु B की आयु से तिगुनी थी प्रत्येक की वर्तमान आयु ज्ञात कीजिये ?

हल –    माना 6 वर्ष पहले B की आयु = x वर्ष तथा A की आयु = 3x वर्ष

 B की वर्तमान आयु = ( x+6) वर्ष तथा A की वर्तमान आयु = (3x + 6)

 क्योकि (3x + 6) – (x+6) = 16 2x = 16 x = 8

 अतः B की वर्तमान आयु = (8+6) वर्ष = 14 वर्ष

 A की वर्तमान आयु = (3 * 8 + 6) वर्ष = 30 वर्ष

 

05.   एक व्यक्ति की आयु अपने पुत्र की आयु के तिगुनी से 3 वर्ष अधिक है यदि 3 वर्ष बाद उस व्यक्ति की आयु  अपने पुत्र की आयु के दुगने से 10 वर्ष अधिक हो तो पुत्र की वर्तमान आयु ज्ञात कीजिये ?

 हल –   माना पुत्र की आयु = x वर्ष तब व्यक्ति की आयु = (3x + 3) वर्ष

 (3x + 3) + 3 = 2 (x + 3) + 10 3x + 6 = 2x + 16 x = 10

 अतः पुत्र की वर्तमान आयु = 10 वर्ष

 

06.   रीना तथा उसकी माता की आयु का योग 49 वर्ष है 7 वर्ष पूर्व माता की आयु रीना की आयु से 4 गुना थी रीना की माता की वर्तमान आयु ज्ञात कीजिये ?

 हल –    माना रीना की आयु = x वर्ष अतः उसकी माता की आयु = (49 – x) वर्ष

 7 वर्ष पूर्व रीना की आयु = (x – 7) वर्ष

7 वर्ष पूर्व माता की आयु = (49 – x – 7) वर्ष = (42 -x) वर्ष

 क्योकि 42 – x = 4 (x – 7) 42 – x = 4x – 28

  5x = 70 x =14

 अतः रीना की माता की वर्तमान आयु = (49 – 14) वर्ष = 35 वर्ष

इस तरह के प्रश्न हमेशा प्रतियोगिता परीक्षा में पूछे जाते है अगर आपने एक बार अच्छे से इन सवलो का अध्ययन कर लिया तो आप परीक्षा में इस तरह के सवाल आसानी से हल कर पायेंगे ! इस तरह के प्रश्नो में अक्सर एक व्यक्ति की आयु दी जाती है एवं दूसरे व्यक्ति की आयु के बारे में पूछा जाता है , इन सवलो को बीजगणित की सहायता से आसानी से हल किया जा सकता है ! जिसकी आयु पूछी जाती है उसे x माना जाता है फिर प्रश्न के अनुसार हल करके उत्तर निकाल  लिया जाता है.


For more:

Fractions (भिन्न) and Types of Fraction

यहां क्लिक करें

NCERT Solutions for Class 6 to 10 in PDF

यहां क्लिक करें 

[BEST] Free download RS Aggarwal solutions in pdf

यहां क्लिक करें 

What are different Types of Triangles?| Geometry

यहां क्लिक करें 

Math Notes PDF In Hindi

पब्लिकेशन का नाम                      S. Chand

लेखंक का नाम                        R.S. Aggarwal

विषय                                                 गणित

यह किताब किस भाषा में है ?               हिंदी

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आयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindi PDF – Square And Square Roots

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