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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 34501. |
Explain thales theorem |
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Answer» Basic Proportionality Theorem was first stated by Thales, a Greek mathematician. Hence it is also known as Thales Theorem.Basic Proportionality Theorem (can be abbreviated as BPT) states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.\xa0 if a line drawn parallel to any one side of the triangle jioning the other two lines then the parallel lines divides other two sides in a equal ratio.. |
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| 34502. |
For what value of k the eq kx+3y- ( k-3) =0 and 12x+ ky- k= 0 have infinite solution |
| Answer» I think you didn\'t learn properly the rules Graphical representation. It is given on page number 46.and this question is of Example no.16Don\'t you think it is a thing of shame, this shows that you are not learning properly.Open your book , go to example no. 13 of chapter 3 page no.61. It is solved already there | |
| 34503. |
If area equal to 64 π than d equal to |
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| 34504. |
All the formulas of all the characters |
| Answer» are you mad?? | |
| 34505. |
Find the roots of equation by completing the square -x^2-(√3+1)x+√3=0 |
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| 34506. |
Explain why 3×5×7+7 is a composite number. |
| Answer» ✓ 3×5×7+7= 7(3×5×1+1)= 7(15+1)= 7(16)= 112It has more than two factorsi.e.1,2,4.Therefore it is a composite number. | |
| 34507. |
3 + root 2 irrational number Prove |
| Answer» To prove 3+root 2 is irrational.Proof-Let 3+root 2 is rational number. So we can write it in the form of p/q where p and q are co-prime number. According to question 3+root2=p/qRoot 2=p-3/qHere LHS is rational so RHS should must be rational but it is contradiction the fact that 3+root 2 is rational. So, our supposition that 3+root 2 is rational is wrong. Hence, 3+root 2 is irrational. | |
| 34508. |
3+ |
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| 34509. |
Fundamental theorm of arithmetic |
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| 34510. |
If x=30 degree then,verify that ....tan 2 x = 2 tan x - 1/tan x. |
| Answer» We have,{tex} x = 30 ^ { \\circ }{/tex}{tex} \\Rightarrow 2 x = 60 ^ { \\circ }{/tex}{tex}L.H.S=\\;\\tan2x=\\tan60^\\circ=\\sqrt3{/tex}and,\xa0{tex}R.H.S=\\frac{2\\tan x}{1-\\tan^2x}{/tex}{tex} = \\frac { 2 \\tan 30 ^ { \\circ } } { 1 - \\tan ^ { 2 } 30 ^ { \\circ } }{/tex}{tex} = \\frac { 2 \\times \\frac { 1 } { \\sqrt { 3 } } } { 1 - \\left( \\frac { 1 } { \\sqrt { 3 } } \\right) ^ { 2 } }{/tex}{tex} = \\frac { 2 / \\sqrt { 3 } } { 1 - \\frac { 1 } { 3 } }{/tex}{tex} = \\frac { 2 / \\sqrt { 3 } } { 2 / 3 }{/tex}{tex} = \\frac { 2 } { \\sqrt { 3 } } \\times \\frac { 3 } { 2 } = \\sqrt { 3 }{/tex}therefore, L.H.S = R.H.S | |
| 34511. |
What is the value of (√5+√3) (√5-√3) |
| Answer» (a +b) (a -b ) = (a)2 - (b)2(√5+√3) (√5-√3) = (√5)2 - (√3)2= 5 - 3= 2 | |
| 34512. |
3x+y=11 and x-y=1 show it graphically and find the area of triangle formed |
| Answer» On a graph paper, draw a horizontal line XOX\' and a vertical line YOY\' as the x-axis and they-axis respectively.{tex}3x + y - 11\xa0= 0{/tex}{tex}\\Rightarrow{/tex}{tex}y = (11 -3x){/tex}. ...(i)Putting {tex}x = 2{/tex} in (i), we get {tex}y = 5{/tex}.Putting {tex}x = 3{/tex} in (i), we get {tex}y = 2{/tex}.Putting {tex}x = 5{/tex} in (i), we get {tex}y = -4{/tex}.\tx235y52-4\tOn the graph paper, plot the points {tex}A (2, 5), B(3, 2)\\ and\\ C(5, -4).{/tex}Join AB and BC to get the graph line ABC.Thus, the line ABC is the graph of the equation {tex}3x + y - 11\xa0= 0{/tex}.{tex}x - y - 1\xa0= 0{/tex}{tex}\\Rightarrow{/tex}{tex}y = (x - 1){/tex} . ...(ii)Putting {tex}x = -3{/tex} in (ii), we get {tex}y = -4{/tex}.Putting {tex}x = 0{/tex} in (ii), we get {tex}y = -1{/tex}.Putting {tex}x = 3{/tex} in (ii), we get {tex}y = 2{/tex}.\tx-303y-4-12\tOn the same graph paper as above, plot the points {tex}P(-3, -4)\\ and\\ Q(0, -1){/tex}. The third point {tex}B(3,2){/tex} is already plotted.Join PQ and QB to get the graph line PQB.Thus, line PQB is the graph of the equation {tex}x - y - 1 = 0{/tex}.The two graph lines intersect at the point {tex}B(3,2){/tex}. {tex}x = 3, y = 2{/tex} is the solution of the given system of equations.The region bounded by these lines and the y-axis has been shaded.On extending the graph lines on both sides, we find that these graph lines intersect the y-axis at the points Q(0, -1) and R(0,11).Area of triangle =\xa0{tex}\\frac 12 \\times 12 \\times 3 = 18{/tex}\xa0sq units. | |
| 34513. |
If -3 is one of the zeroes of the polynomial (k-1)x^2+kx+1, find the value of k . |
| Answer» p(x) = (k-1)x2\xa0+ kx + 1Since, -3 is a zero of the polynomial.p(-3) = (k - 1)(-3)2\xa0+ k(-3) + 1 = 09(k - 1) - 3k + 1 = 09k - 9 - 3k + 1 = 06k - 8 = 06k = 8k = 8/6k = 4/3 | |
| 34514. |
√2x^2 - 3x - 2√2 = 0 |
| Answer» Proved | |
| 34515. |
If sec theta + tan theta =p. Then find the value of cosec theta |
| Answer» Here is the solutino of this question . I hope that you will understand thishttps://www.teachoo.com/9007/2862/Question-30/category/CBSE-Class-10-Sample-Paper-for-2019-Boards/ | |
| 34516. |
What is Euclid division lemma..?Kindly tell .. |
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Answer» Thanks mates☺☺☺ For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such thata=bq+r , where 0≤r Also r is less than b r | |
| 34517. |
Cos2A+1/1+cot2A=1 |
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| 34518. |
Honey paii us question ke option kya hai ?? Vector vale question ka |
| Answer» Honey batao ? | |
| 34519. |
Solve for x and y: 3(?+?)?+?-2(?−?)?−?= 18(?+?)?+?+ 7(?−?)?−?= 1 |
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| 34520. |
Find hcf of272 and 420 using euclids division algorithms |
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Answer» 420 = 272×1 + 148272 = 148×1+124148 = 124×1 + 24124 = 24×5 + 424 = 4×6 + 0.Therefore,4 is the HCF of 272 and 420. 4 is right answer |
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| 34521. |
Show that 12n cannot end with the digit 0 or 5 for any natural number n |
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Answer» 12 ka factor2×2×3ha or jo digit 0 sa end hota ha 2 or 5 come complesari ha 12n can be written as (3×4)n and 3 and 4 are the factors of 12 and 5 is not the factor so 11n can nvr end with 0 |
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| 34522. |
Identify root 49/147 as rational or irrational. |
| Answer» {tex}\\huge\\implies \\sqrt{\\frac {49}{147}}\\\\\\huge\\implies \\sqrt\\frac {7\\times7}{21\\times7}\\\\\\huge\\implies \\sqrt\\frac {1}{3}\\\\\\hugewhich\\space is\\space not\\space a\\space rational\\space number.{/tex} | |
| 34523. |
Solve the quadratic equation by factorization x+1/x-1-x-1/x+1=5/6 |
| Answer» We have,{tex}\\frac{{x + 1}}{{x - 1}} - \\frac{{x - 1}}{{x + 1}} = \\frac{5}{6}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}= \\frac56{/tex}{tex}\\Rightarrow{/tex} (x2 + 1 + 2x) - (x2 + 1 - 2x) = {tex}\\frac{5}{6}{/tex}(x2 - 12){tex}\\Rightarrow{/tex} x2 + 1 + 2x - x2 - 1 + 2x = {tex}\\frac{5}{6}{/tex}(x2 - 1){tex}\\Rightarrow{/tex} 4x = {tex}\\frac{5}{6}{/tex}(x2 - 1){tex}\\Rightarrow{/tex} 24x = 5(x2 - 1){tex}\\Rightarrow{/tex}\xa024x = 5x2 - 5{tex}\\Rightarrow{/tex}\xa05x2 - 24x - 5 = 0{tex}\\Rightarrow{/tex} 5x2 - 25x + 1x - 5 = 0{tex}\\Rightarrow{/tex} 5x(x - 5) + 1(x - 5) = 0{tex}\\Rightarrow{/tex} (x - 5)(5x + 1) = 0{tex}\\Rightarrow{/tex} x - 5 = 0 or 5x + 1 = 0{tex}\\Rightarrow{/tex} x = 5 or {tex}x = - \\frac { 1 } { 5 }{/tex} | |
| 34524. |
if two zeroes of polynomial f( x) = x4_6x3 _ 26x2+138x _35 are 2_3 , 2+3 , find other zeroes |
| Answer» Try to seek help from example 9 of chapter 2 page no.35 | |
| 34525. |
In a graph of y=p (x), find a no. Of zeros of p(x) ? |
| Answer» Very poor you are, This question is same as of exercise 3.1 | |
| 34526. |
Explain the meaning of polynomials |
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Answer» Polynomial is\xa0an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s). \'Poly\' means many. \'Nomial\' means terms. Together it means many terms. Let x be a variable, n be a positive integer and as a real number us called polynomials |
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| 34527. |
write one rational and irrational number lying between 0.25and0.32 |
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Answer» Rational number between 0.25 and\xa00.32 is 0.28Irrational number between 0.25 and\xa00.32 is 0.290290029000......... 32096877551 |
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| 34528. |
find thethe largest number whichof divides 129 and 545 leaving remainders 3 and 5 respectively |
| Answer» To find the largest number which divides 129 and 545 leaving remainder 3 and 5 i.e. HCF.Consider HCF be x.In order to make 129 and 545 completely divisible by x, we need to deduct the remainder 3 and 5\xa0from the cases.126 =2× 3 x 3 x 7 ,540= 2×2×3×3 x 3 x 5⇒ x = 2×3×3 = 18∴\xa0largest number which divides 126 and 540 leaving remainder 3 and 5 in case is 18. | |
| 34529. |
Prove that root 3 is irrational number |
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Answer» Let us assume that √3 is a rational number.then, as we know a rational number should be in the form of p/q\xa0where p and q are co- prime number.So,√3 = p/q { where p and q are co- prime}√3q = pNow, by squaring both the side\xa0we get,(√3q)² = p²3q² = p² ........ ( i )So,if 3 is the factor of p²then, 3 is also a factor of p ..... ( ii )=> Let p = 3m { where m is any integer }squaring both sidesp² = (3m)²p² = 9m²\xa0putting the value of p² in equation ( i )3q² = p²\xa03q² = 9m²q² = 3m²So,if 3 is factor of q²then, 3 is also factor of qSince3 is factor of p & q bothSo, our assumption that p & q are co- prime is wrong Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. |
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| 34530. |
Using prime FACTRISATION find the Hcf and lcm of: 144,198 |
| Answer» HCF OF 144 & 198 is 18 and LCM is 1584. | |
| 34531. |
Solve : If Sin 2A=SinA*CosA , then find the value of A..... |
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| 34532. |
Find the positive values of m for which the distance between the points A(3,1)B(13,m)is 10 unit |
| Answer» Distance formula=\xa0√ (x2\xa0- x1)2 + ( y2\xa0- y1)2Distance = 10 units√ (13-3)2\xa0+ (m-1)2 = 10√ (10)2\xa0\xa0+ m2\xa0+ (1)2\xa0- 2(m)(1) = 10 Squaring both sides(10)2\xa0\xa0+ m2\xa0+ (1)2\xa0- 2(m)(1) = (10)2 100 + m2\xa0+ 1 - 2m = 100 m2\xa0-2m + 1 = 100 - 100 m2\xa0-2m+1 = 0 m2\xa0- 1m - 1m +1 =0 m (m -1) -1( m-1) =0 ( m-1) (m-1) =0 m-1=0 Therefore, m= 1 | |
| 34533. |
If an acute angle |
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| 34534. |
Show that p square will leave a remainder 1 when divided by 8 if P is an odd positive integer |
| Answer» P = 2n + 1 where n = 0, 1, 2, 3, 4, ...=> p² = 4n² + 4n + 1= 4(n² + n) + 1= 4n(n+1) + 1but n(n+1) is even because if n is odd, n+1 is even - otherwise n is eventherefore 4n(n+1) + 1 = 8m + 1 where m is an integer.=> p² = 1 | |
| 34535. |
State whether 1.456bar on 4 is rational |
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Answer» Let x=1.456456......... eq-110x=14.56456100x=145.64561000x=1456.456...............eq-2eq-2_eq-11000x-x=1456.456-1.456456999x=1455x=1455/999 {tex}\\begin{array}{l}\\mathrm{let}\\;\\mathrm x=1.\\overline{456}\\\\\\mathrm{then}\\;1000\\mathrm x=1456.\\overline{456}\\\\1000000=1456456.\\overline{456}\\\\\\mathrm{subtracting}\\;\\mathrm{we}\\;\\mathrm{get}\\\\999000\\mathrm x=1455000\\\\\\mathrm x=\\frac{1455000}{999000}=\\frac{1455}{999}=\\frac{485}{333}\\\\\\end{array}{/tex}Hence\xa0the number is in p/q ,so it is a rational |
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| 34536. |
1.1÷1 |
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Answer» 1 1.1 |
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| 34537. |
Draw the graph of quadratic polynomial F(x)=3-2x-x^2 |
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| 34538. |
We need to find two consecutive odd integers, the sum of whose square is 202 |
| Answer» let the consecutive odd integers are x and x+2then x2+(x+2)2=202x2+x2+4x+4-202=02x2+4x-198=0x2+2x-99=0x2+11x-9x-99=0x(x+11)-9(x+11)=0(x+11)(x-9)=0x=-11,9So other no are -11+2 ad 9+2 or -9,11\xa0So the numbers are9,11 or -9,-11\xa0 | |
| 34539. |
Can you please also add RD solutions ....for class 10 maths |
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Answer» But it is not available for all the chapters. They are available in each chapter\'s category. |
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| 34540. |
Rd shrma questions ex 2.1 |
| Answer» Get RD Sharma Solutions of each chapter :\xa0https://mycbseguide.com/course/cbse-class-10-mathematics/1202/ | |
| 34541. |
Hcf of 180 ,252 and 324 using Euclid\'s division lemma |
| Answer» 324 = 252 x 1 + 72252 = 72 x 3+ 3672 = 36 x 2 + 0HCF(324, 25) = 36180 = 36 x 5 + 0HCF (36, 180) = 36:. HCF of 180, 252 and 324 is 36. | |
| 34542. |
For what value of k-4 is a zero of polynomial x2-x-(2k+2). |
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Answer» Given that, -4 is a zero of the polynomial f(x) =\xa0x2 - x - (2k + 2), so, we havef(-4) = 0{tex}\\Rightarrow{/tex}\xa0(-4)2\xa0- (-4) - 2k - 2 = 0{tex}\\Rightarrow{/tex}\xa016 + 4 - 2k - 2 = 0{tex}\\Rightarrow{/tex}\xa018 - 2k = 0{tex}\\Rightarrow{/tex}\xa02k = 18{tex}\\Rightarrow{/tex} k = 9. Please fast |
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| 34543. |
if one zero of cubic polynomial x^3-ax^2+x+3 is a then find value of a |
| Answer» Let p(x) = x3 + ax2 + bx + cAs -1 is one of the zeroes of p(x), p(-1) = 0⇒ (-1)3 + a(-1)2 + b(-1) + c = 0⇒\xa0- 1 + a – b + c = 0⇒ c = b –a + 1 …. (i)Let {tex}\\alpha{/tex}, β be two other zeroes of p(x),then product of zeroes =-1×\xa0{tex}\\alpha{/tex} × β ={tex}\\;\\;-\\frac{\\;\\;Cons\\tan t\\;term\\;}{coefficient\\;of\\;x^3}{/tex}⇒ (-1) (α β) = {tex}\\;\\;-\\frac{\\;\\;c}1{/tex}⇒ -\xa0{tex}\\alpha{/tex} β\xa0=\xa0- c⇒\xa0{tex}\\alpha{/tex} β\xa0= c⇒\xa0{tex}\\alpha{/tex} β = b –a + 1 [using (i)]\xa0Hence, the product of other two zeroes of the given cubic polynomial is b – a + 1. | |
| 34544. |
For what values of k are the roots of the quadratic equation 3xsq +2kx+27=0 real and equal |
| Answer» We have the following equation,3x2\xa0- 2kx + 27\xa0= 0a = 3, b = -2k\xa0and c = 27{tex}\\therefore{/tex}\xa0D = b2\xa0- 4ac= (-2k)2\xa0- 4(3)(27)= 4k2 - 324Roots are real and equal if D = 0{tex}\\Rightarrow{/tex}\xa04k2\xa0- 324 = 0k\u200b\u200b\u200b\u200b\u200b\u200b2=\xa0{tex}\\frac { 324 } { 4 }{/tex} = 81{tex}\\therefore k = \\pm 9{/tex} | |
| 34545. |
Kx+2y=53x+y=1This is question of pair of linear equation in two variable.....plz solve it |
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| 34546. |
Find the quardinate of the midpoint of tha line segment joining the point A (-5 , 4 ) , B (7,-8) |
| Answer» (-5+7/2,4-8/2)(2/2,-4/2)=(1,-2) | |
| 34547. |
If n is an odd integer then show that n^2-1 is divisible by 8. |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 34548. |
Using Euclid Division Lemma, prove that for any positive integer n, n^3-n, is divisible by 6. |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0\xa0 | |
| 34549. |
Show that one and only one out of q, q+2, q+4 is divisible by 3, where q is a positive integer. |
| Answer» On dividing n by 3, let q be the quotient and r be the remainder.Then, {tex}n = 3q + r{/tex}, where {tex}0 \\leq r < 3{/tex}{tex}\\Rightarrow\\;n = 3q + r{/tex} , where r = 0,1 or 2{tex}\\Rightarrow{/tex}{tex}n = 3q \\;or \\;n = (3q + 1) \\;or\\; n = (3q + 2){/tex}.Case I If n = 3q then n is clearly divisible by 3.Case II If\xa0{tex}\\;n = (3q + 1)\\; {/tex} then {tex} (n + 2)= (3q + 1 + 2) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.In this case, {tex}(n + 2){/tex} is divisible by 3.Case III If n = {tex}(3q + 2){/tex} then {tex}(n + 1) = (3q + 2 + 1) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.In this case,{tex} (n + 1){/tex} is divisible by 3.Hence, one and only one out of {tex}n, (n + 1){/tex} and {tex}(n + 2){/tex} is divisible by 3. | |
| 34550. |
If Alpha and bita be the zeros of the polynomial 2xsquare + 3x-6 |
| Answer» | |