Chapter 1 - Real Numbers Excercise Ex. 1.1Solution 1(i) 135 and 225 Step 1:  Since 225 > 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r  such that 225 = 135q+r, 0  r<135 On dividing 225 by 135 we get quotient as 1 and remainder as 90 i.e 225 = 135 x 1 + 90 Step 2: Remainder r which is 90  0, we apply Euclid's division lemma to b =135 and r = 90 to find whole numbers q and r such that 135 = 90 x q + r, 0  r<90On dividing 135 by 90 we get quotient as 1 and remainder as 45i.e 135 = 90 x 1 + 45Step 3:  Again remainder r = 45  0 so we apply Euclid's division lemma to b =90 and r = 45 to find q and r  such that 90 = 90 x q + r,  0  r<45On dividing 90 by 45 we get quotient as 2 and remainder as 0 i.e 90 = 2 x 45 + 0Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (135, 225). Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45. (ii)   196 and 38220Step 1:  Since 38220 > 196, apply Euclid's division lemmato a =38220 and b=196 to find whole numbers q and r  such that 38220 = 196 q + r, 0  r < 196On dividing 38220 we get quotient as 195 and remainder r as 0 i.e 38220 = 196 x 195 + 0Since the remainder is zero, divisor at this stage will be HCF Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.NOTE: HCF( a,b) = a if a is a factor of b. Here, 196 is a factor of 38220 so HCF is 196. (iii)   867 and 255 Step 1: Since 867 > 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r  such that 867 = 255q + r, 0  r<255On dividing 867 by 255 we get quotient as 3 and remainder as 102i.e 867 = 255 x 3 + 102 Step 2: Since remainder 102  0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that 255 = 102q + r where 0  r<102On dividing 255 by 102 we get quotient as 2 and remainder as 51 i.e 255 = 102 x 2 + 51Step 3: Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51 to find whole numbers q and r such that 102 = 51 q + r where 0 r < 51  On dividing 102 by 51 quotient is 2 and remainder is 0i.e 102 = 51 x 2 + 0Since the remainder is zero, the divisor at this stage is the HCFSince the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51. Concept Insight: To crack such problem remember to apply the Euclid's division Lemma which states that "Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0  r < b" in the correct order. Here, a > b. Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives a number which divides a and b exactly. i.e   HCF(a,b) =HCF(b,r)Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps.

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Chapter 2 : Polynomials - Ncert Solutions for Class 10 Maths CBSEChapter 2 - Polynomials Excercise Ex. 2.1Solution 1(i) The graph of P(x) does not cut the x-axis at all . So, the number of zeroes is 0.(ii) The graph of P(x) intersects the x-axis at only 1 point.So, the number of zeroes is 1.(iii) The graph of P(x) intersects the x-axis at 3 points.So, the number of zeroes is 3.(iv) The graph of P(x) intersects the x-axis at 2 points.So, the number of zeroes is 2.(v) The graph of P(x) intersects the x-axis at 4 points.So, the number of zeroes is 4.(vi) The graph of P(x) intersects the x-axis at 3 points.So, the number of zeroes is 3.Concept insight: Since the polynomial p(x) given here is a polynomial in variable x, so to find the number of zeroes, we look at the number of points where the graph intersects or touches the x-axis and not the y-axis.At all these points where the graph intersects x axis the value of the polynomial y = p(x) will be zero.Chapter 2 - Polynomials Excercise Ex. 2.2Solution 1 So, the zeroes of x² - 2x - 8 are 4 and -2.       Concept insight: The zero of a polynomial is that value of the variable which when substituted in the polynomial makes its value 0. When a quadratic polynomial is equated to 0, then the values of the variable obtained are the zeroes of that polynomial. The relationship between the zeroes of a quadratic polynomial with its coefficients is very important. Also, while verifying the above relationships, be careful about the signs of the coefficients.Chapter 2 - Polynomials Excercise Ex. 2.3Solution 1Quotient = x - 3 Remainder = 7x - 9 Quotient = x2 + x - 3Remainder = 8 Quotient = -x2 - 2Remainder = -5x + 10 Concept insight: While dividing one polynomial by another, first arrange the polynomial in descending powers of the variable. In the process of division, be careful about the signs of the coefficients of the terms of the polynomials. After performing division, one can check his/her answer obtained by the division algorithm which is as below:Dividend = Divisor x Quotient + Remainder Also, remember that the quotient obtained is a polynomial only.Solution 2(i) The polynomial 2t4 + 3t3 - 2t2 - 9t - 12 can be divided by the polynomial t2 - 3 = t2 + 0.t - 3 as follows:  Since the remainder is 0, t² - 3 is a factor of 2t4 + 3t3 - 2t2 - 9t - 12 . (ii) The polynomial 3x4 + 5x3 - 7x2 + 2x + 2 can be divided by the polynomial x2 + 3x + 1 as follows:Since the remainder is 0, x² + 3x + 1 is a factor of 3x4 + 5x3 - 7x2 + 2x + 2 (iii) The polynomial x5 - 4x3 + x2 + 3x + 1 can be divided by the polynomial x3 - 3x + 1 as follows:  Since the remainder is not equal to 0, x3 - 3x + 1 is not a factor of x5 - 4x3 + x2 + 3x + 1. Concept insight:  A polynomial g(x) is a factor of another polynomial p(x) if the remainder obtained on dividing p(x) by g(x) is zero and not just a constant. While changing the sign, make sure you do not change the sign of the terms which were not involved in the previous operation. For example in the first step of (iii), do not change the sign of 3x + 1.Solution 3Let p(x) = 3x4 + 6x3 - 2x2 - 10x -5  Now, x² + 2x + 1 = (x + 1)2So, the two zeroes of x² + 2x + 1 are -1 and -1.Concept insight: Remember that if (x - a) and (x - b) are factors of a polynomial, then (x - a)(x - b) will also be a factor of that polynomial. Also, if a is a zero of a polynomial p(x), where degree of p(x) is greater than 1, then (x - a) will be a factor of p(x), that is when p(x) is divided by (x - a), then the remainder obtained will be 0 and the quotient will be a factor of the polynomial p(x). To cross check your answer number of zeroes of the polynomial will be less than or equal to the degree of the polynomial.Solution 4Divided, p(x) = x3 - 3x2 + x + 2 Quotient = (x - 2)Remainder = (-2x + 4)Let g(x) be the divisor.According to the division algorithm, Dividend = Divisor x Quotient + Remainder Concept insight: When a polynomial is divided by any other non-zero polynomial, then it satisfies the division algorithm which is as below:Dividend = Divisor x Quotient + RemainderDivisor x Quotient = Dividend - Remainder So, from this relation, the divisor can be obtained by dividing the result of (Dividend - Remainder) by the quotient.Solution 5According to the division algorithm, if p(x) and g(x) are two polynomials with g(x)  0, then we can find polynomials q(x) and r(x) such thatp(x) = g(x) x q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).(i)   Degree of quotient will be equal to degree of dividend when divisor is constant.Let us consider the division of 18x2 + 3x + 9 by 3.Here, p(x) = 18x2 + 3x + 9 and g(x) = 3q(x) = 6x2 + x + 3 and r(x) = 0Here, degree of p(x) and q(x) is the same which is 2.Checking:p(x) = g(x) x q(x) + r(x) Thus, the division algorithm is satisfied.(ii)   Let us consider the division of 2x4 + 2x by 2x3,Here, p(x) = 2x4 + 2x and g(x) = 2x3q(x) = x and r(x) = 2xClearly, the degree of q(x) and r(x) is the same which is 1.Checking,p(x) = g(x) x q(x) + r(x)2x4 + 2x = (2x3 ) x x + 2x2x4 + 2x = 2x4 + 2xThus, the division algorithm is satisfied.(iii)   Degree of remainder will be 0 when remainder obtained on division is a constant.Let us consider the division of 10x3 + 3 by 5x2.Here, p(x) = 10x3 + 3 and g(x) = 5x2q(x) = 2x and r(x) = 3Clearly, the degree of r(x) is 0.Checking:p(x) = g(x) x q(x) + r(x)10x3 + 3 = (5x2 ) x 2x + 310x3 + 3 = 10x3 + 3Thus, the division algorithm is satisfied.Concept insight: In order to answer such type of questions, one should remember the division algorithm. Also, remember the condition on the remainder polynomial r(x). The polynomial r(x) is either 0 or its degree is strictly less than g(x). The answer may not be unique in all the cases because there can be multiple polynomials which satisfies the given conditions.

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What is light reflection and refraction?The phenomenon of sendig back of light rays after it strikes the smooth/polished surface is known as reflection. The phenomenon of bending of light rays when it changes the medium is called refraction of light.

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this is pdf which is consist a notes of class 10 th chpater 3 full this my handwritten notes you can download free of cost .so please download

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