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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.
| 6901. |
Find the value of k,-4 is a zero of the polynomial x²-x-(2k+2) |
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Answer» X²-x-(2k+2)=0(-4)²-(-4)-(2k+2)=016-4-(2k+2)=0-2k-2= -12-2k= -12+2k= -10/-2=5(K =5) k=9 |
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| 6902. |
In the given figure RQ and TP perpendicular to PQ.also TS perpendicular PR prove that ST×QR=PS×PQ |
| Answer» | |
| 6903. |
5ysq + 10y |
| Answer» We need us food | |
| 6904. |
Sin45 |
| Answer» 1/√2 | |
| 6905. |
What is Thales theorem |
| Answer» it is the same as bpt | |
| 6906. |
Prove that area of a scalene triangle=sqrt.s×s-a×s-b×s-c |
| Answer» | |
| 6907. |
If two zeroes of the polynomial f(x)=x^2-4x^2-3x +12 find third term |
| Answer» Let ,\xa0{tex}\\alpha,\\beta,\\gamma {/tex}\xa0be the zeroes of the polynomial where,{tex}\\alpha = \\sqrt { 3 } , \\beta = - \\sqrt { 3 }{/tex}{tex} \\alpha + \\beta + \\gamma = - \\frac { \\text { Coeff. of } x ^ { 2 } } { \\text { Coeff. of } x ^ { 3 } } {/tex}{tex}\\alpha + \\beta + \\gamma = - \\left( \\frac { - 4 } { 1 } \\right){/tex}.{tex}\\Rightarrow \\quad \\sqrt { 3 } - \\sqrt { 3 } + \\gamma = 4{/tex}{tex}\\Rightarrow \\quad \\gamma = 4{/tex}Hence, third zero is 4. | |
| 6908. |
In an Arithmetic Progression if a+1,2a+1,4a-1are in A.P.Then find value of a. |
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| 6909. |
Show that the cube of any positive integer is of the form 4m, 4m+1, 4m+3 |
| Answer» Let a be the positive integer and b = 4.Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.{tex}(4q)^3\\;=\\;64q^3\\;=\\;4(16q^3){/tex}= 4m, where m is some integer.{tex}(4q+1)^3\\;=\\;64q^3+48q^2+12q+1=4(\\;16q^3+12q^2+3q)+1{/tex}= 4m + 1, where m is some integer.{tex}(4q+2)^3\\;=\\;64q^3+96q^2+48q+8=4(\\;16q^3+24q^2+12q+2){/tex}= 4m, where m is some integer.{tex}(4q+3)^3\\;=\\;64q^3+144q^2+108q+27{/tex}=4×(16q3+36q2+27q+6)+3= 4m + 3, where m is some integer.Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. | |
| 6910. |
Prove \'root two\' is irrational number |
| Answer» We have to prove that {tex} \\sqrt2{/tex}\xa0is an irrational number.Let\xa0{tex}\\sqrt2{/tex}\xa0be a rational number.{tex}\\therefore \\quad \\sqrt { 2 } = \\frac { p } { q }{/tex}where p and q are co-prime integers and\xa0{tex}q \\neq 0{/tex}On squaring both the sides, we get,or,\xa0{tex}2 = \\frac { p ^ { 2 } } { q ^ { 2 } }{/tex}or, p2 = 2q2{tex}\\therefore{/tex}\xa0p2 is divisible\xa0by 2.p is divisible by 2........(i)Let p = 2r for some integer ror, p2\xa0= 4r22q2= 4r2 [∵ p2 = 2q2]or, q2 = 2r2or, q2\xa0is divisible by 2.{tex}\\therefore{/tex}q is divisible by 2..........(ii)From (i) and (ii)p and q are divisible by 2, which contradicts the fact that p and q are co-primes.Hence, our assumption is wrong.{tex}\\therefore{/tex}\xa0{tex}\\sqrt2{/tex}\xa0is irrational number. | |
| 6911. |
Meaning of co-prime |
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Answer» numbers which have no common factors other than 1e.g 21 and 22\xa0the factors of 21 are 1 , 3 , 7, 21the factors of 22 are 1 , 2, 11, 22only 1 is common factor among themthus 21 and 22 are coprime\xa0But 21 and 24 are not coprime as the factors of 24 are 1 , 2, 3, 4, 6, 8 , 12 , 24the common factors are 1 AND 3 The number having factor 1 |
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| 6912. |
What is infinite series? |
| Answer» Uncountable | |
| 6913. |
find the lcm of first five prime odd number |
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Answer» first 5 prime odd numbers3 , 5 , 7, 11, 13LCM = 3x5x7x11x13= 15015 4725 4725 |
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| 6914. |
show that 8 is not a square no. |
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| 6915. |
Find the roots of the equation 5x^2 - 6x -2=0 by the method of completing the square |
| Answer» 5x2 - 6x - 2 = 0Multiplying the above equation by 1/5{tex} \\Rightarrow {x^2} - \\frac{6}{5}x - \\frac{2}{5} = 0{/tex}{tex}\\Rightarrow x ^ { 2 } - \\frac { 6 } { 5 } x + \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\frac { 2 } { 5 } = 0{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 } { 25 } + \\frac { 2 } { 5 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 + 10 } { 25 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 19 } { 25 }{/tex}{tex}\\Rightarrow x - \\frac { 3 } { 5 } = \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 } { 5 } \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 + \\sqrt { 19 } } { 5 } \\text { or } x = \\frac { 3 - \\sqrt { 19 } } { 5 }{/tex} | |
| 6916. |
Find the root of 1÷x-1÷x-2=3 |
| Answer» x=1/2 | |
| 6917. |
What are the identies of trigonometry |
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Answer» Sin square theta + cos square theta =11 + tan square theta =sec square theta1 + cot square theta =cosec square theta PBP/HHB |
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| 6918. |
Sin 60 cos 30 +sin 30 cos 60 |
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Answer» Sin(A+B)=Sin(60+30)=Sin90=1 1 answer sin60cos(90-30)+sin(90-30)cos60sin60sin60+cos60cos60sin²60+cos²601Ans. |
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| 6919. |
What is sec thetha? |
| Answer» the ratio of hypotenuse and perpendicular is called sec theta | |
| 6920. |
Prove that √3+√5 is irrational |
| Answer» Let √3+√5 is rational√3+√5= p/q ( where p and q are some integer)√5=p-√3/q ( then, we take a l.c.m.)√5=p-√3q/q√5 is a rational no.but,p-q√3 is a irrational no.Irrational is not equal to rational no.So, our supposition is wrong√3+√5 is irrational no. | |
| 6921. |
1/1+tan^2+1/1+cot^2=1/sin^2-sin^4 |
| Answer» LHS\xa0{tex}= \\left( 1 + \\frac { 1 } { \\tan ^ { 2 } \\theta } \\right) \\left( 1 + \\frac { 1 } { \\cot ^ { 2 } \\theta } \\right){/tex}\xa0{tex}= (1 + cot^2\\theta) (1 + tan^2\\theta){/tex}{tex}= \\left( 1 + \\frac { \\cos ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta } \\right) \\left( 1 + \\frac { \\sin ^ { 2 } \\theta } { \\cos ^ { 2 } \\theta } \\right){/tex}{tex}= \\left( \\frac { \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta } \\right) \\left( \\frac { \\cos ^ { 2 } \\theta + \\sin ^ { 2 } \\theta } { \\cos ^ { 2 } \\theta } \\right){/tex}{tex}= \\frac { 1 } { \\sin ^ { 2 } \\theta } \\cdot \\frac { 1 } { \\cos ^ { 2 } \\theta }{/tex}{tex}= \\frac { 1 } { \\sin ^ { 2 } \\theta \\left( 1 - \\sin ^ { 2 } \\theta \\right) }{/tex}{tex}= \\frac { 1 } { \\sin ^ { 2 } \\theta - \\sin ^ { 4 } \\theta }{/tex}\xa0= RHS | |
| 6922. |
Math half yearly chapter s |
| Answer» Hii,well according to NCERT th pe half yearly chapters are\xa0\t✅ \t✅ \t✅ \t\t\xa0\t\xa0✅\xa0\t\xa0\t\xa0✅\xa0\t\xa0\t\xa0\t\xa0\t\xa0\t\xa0\t\xa0✅\xa0\tThey are confirm\xa0firm otherwise this year schools hav authority to change the syllabus according to their convenience. | |
| 6923. |
2+2+5+1 |
| Answer» 10 | |
| 6924. |
When will exam be held? |
| Answer» On my birthday 15 february | |
| 6925. |
Sum of no between 50 and 500 dividisible by 7 |
| Answer» first no div by 7 = 56 = a1last no div by 7 = 497=aL497=56+(n-1)7497-49=7nn=448/7=64S64=64/2(2a+63d) = 32 x (112+441) = 32 x 553 = 17696 | |
| 6926. |
7√5+3√5 prove it irrational |
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Answer» Let 7√5+3√5 is rational number Then, 7√5+3√5=p÷q. (p and q are the integer which has no other factor expect 1 and q is not equal0)Or,. √5(7+3)=p÷qOr,. √5=(p÷q)÷(7+3) Here, (p÷q)÷(7+3) is a rational no. which is equal to a irrational no. So,our contraction was wrong . HENCE,7√5+3√5 is irrational no. 10√5=a/bThen.√5=a/10bSo √5 is a rational as a,b,10 are rational no.,But it is a contradiction to the fact that √5 is an irrational no.Thus,7√5+3√5 is irrational number.Hence ,proved |
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| 6927. |
15/x-y+22/x+y=5,40/x-y+55/x+y=13 solve |
| Answer» A=1/11 b=1/5 | |
| 6928. |
x2 gj |
| Answer» | |
| 6929. |
today math ka exam hua |
| Answer» | |
| 6930. |
x+x |
| Answer» 2x | |
| 6931. |
40+30 |
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Answer» Of cource 70 70 |
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| 6932. |
(a+b) |
| Answer» | |
| 6933. |
tan A=cosec (90-A) |
| Answer» | |
| 6934. |
24 upon 5 divided by 11 upon 5 minus 1 upon 2 into 5.4 minus 1 upon 4 minus |
| Answer» | |
| 6935. |
24.5 / 11 upon 5 minus 1 upon 2 into 5 upon 4 minus 1.4 minus |
| Answer» | |
| 6936. |
24/5 |
| Answer» Kun khud ko dokha de rahe ho | |
| 6937. |
Please solve(secA-tanA)*2 (1+sinA)=(1-sinA)*2 means square |
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Answer» Answer agar ham bata denge to doshra phucho ge is liye khud se karo sab se phele sec ko cos me bad lo tan ko sin upon cos me is 5arike se solve karke dekho tum samjo ki mene bataya hai nahi Kudse solve karo guchhe ki akhri chabi tala khol deti hai har mat mano you can do it AT NINE o\'clock on the morning of September 11th 2001, President George Bush sat in an elementary school in Sarasota, Florida, listening to seven-year-olds read stories about goats. “Night fell on a different world,” he said of that day. And on a different America.At first, America and the world seemed to changetogether. “We are all New Yorkers now,” ran an e-mail from Berlin that day, mirroring John F. Kennedy\'s declaration 40 years earlier, “Ich bin ein Berliner”, and predicting Le Monde\'s headline the next day, “Nous sommes tous Américains”. And America, for its part, seemed to become more like other countries. Al-Qaeda\'s strikes, the first on the country\'s mainland by a foreign enemy, stripped away something unique: its aura of invulnerability, its sense of itself as a place apart, “the city on a hill”.Two days after the event, President George Bush senior predicted that, like Pearl Harbour, “so, too, should this most recent surprise attack erase the concept in some quarters that America can somehow go it alone.” Francis Fukuyama, a professor at Baltimore\'s Johns Hopkins University, suggested that “America may become a more ordinary country in the sense of having concrete interests and real vulnerabilities, rather than thinking itself unilaterally able to define the nature of the world it lives in.”Both men were thinking about foreign policy. But global terrorism changed America at home as well. Because it made national security more important, it enhanced the role of the |
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| 6938. |
Cos 2A=2 and cos S is true when what is the value of A. |
| Answer» | |
| 6939. |
find the all of the zeros |
| Answer» Given polynomial is x4 + x3 - 34x2 - 4x + 120Since, the two zeroes of the polynomial given is 2 and -2So, factors are (x + 2)(x - 2) = x2 + 2x - 2x - 4 = x2 - 4dividend = divisor × quotient + remainderdividend= x4 + x3 - 34x2 - 4x + 120divisor = x2 - 4quotient = x2 + x - 30remainder = 0So, x4 + x3 - 34x2 - 4x + 120 = (x2 - 4)(x2 + x - 30)= (x - 2)(x + 2)(x2 + 6x - 5x - 30)= (x - 2)(x + 2)(x + 6)(x - 5)Therefore, the zeroes of the polynomial = x = 2, -2, -6, 5 | |
| 6940. |
Sir passing marks 80 me 35 hai ki100 me se |
| Answer» bhai passing 33 hai | |
| 6941. |
Prove that quardratic equation |
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| 6942. |
tanα/(secα+1)+cotα/(cosecα+1) |
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| 6943. |
How many terms of the A.P. 3,5,7... Must be taken so that the sum is 120? |
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Answer» a=3d=2n=?sn=n/2(2a+(n-1)d)=120n/2(6+2n - 2)=1203n+n2-n=120n2+2n-120=0n2+12n-10n-120=0n(n+12) - 10(n+12)=0n = 10n= -12 not possiblecheck ur answer s10 = 10/2(2a+18) = 5(6+18) = 5x24=120 a=3,d=2,n=?,An=120So if we use the formulaAn=a+(n-1)d120=3+(n-1)2120=3+2n-2120=1+2n120÷2=n+160=n+160-1=nn=59 |
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| 6944. |
if tab A = n tan B and sinA =m sin B prove that cos ^ 2 A = m^2-1 /n^2-1 |
| Answer» ISKA ANSWER S.M. AYEGA | |
| 6945. |
Is there any new sample papers for maths if so then send me please |
| Answer» You can check sample paper here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 6946. |
Prove that (SinA+CosecA)2+(cosA+SecA)2=7+tan2A+Cot2A.Here , 2 is square |
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| 6947. |
What is an ap |
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Answer» Arithmetic progression is AP Ap is arithmatic progression |
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| 6948. |
What is the symbol of carbon |
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Answer» Symbol is C C |
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| 6949. |
If a,b,c are in a AP then relation between them is |
| Answer» 2b=a+c | |
| 6950. |
Solve for x (√6x+7)=2x-7 |
| Answer» | |