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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.
| 34001. |
Is it necessary to solve LHS firstly? |
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Answer» No No it is not nessesiry it depends on eq. No |
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| 34002. |
Ncert solution for class 10th exercise 3.6 |
| Answer» The answer of 3.6 is available in study ranker | |
| 34003. |
f(x)=x2+px+45=144 find p.sum of squras of solution |
| Answer» If the squared difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, then ,we have to find the value of p.Let {tex}\\alpha{/tex}\xa0and {tex}\\beta{/tex}\xa0be the zeroes of the given quadratic polynomial.{tex}\\therefore{/tex}\xa0{tex}\\alpha{/tex} + {tex}\\beta{/tex} = - p and {tex}\\alpha\\beta{/tex}= 45 ...(i)Given, ({tex}\\alpha{/tex} - {tex}\\beta{/tex})2 = 144or, ({tex}\\alpha{/tex} + {tex}\\beta{/tex})2- 4{tex}\\alpha{/tex}{tex}\\beta{/tex}\xa0= 144or, (-p)2\xa0- 4 {tex}\\times{/tex}\xa045 = 144 [Using (i)]p2\xa0- 180 = 144p2 = 144 + 180 = 324{tex}\\therefore{/tex}\xa0p = ± {tex}\\sqrt{324}{/tex}= ± 18Hence,\xa0the value of p is ± 18. | |
| 34004. |
A rational number between√2 and √3 is |
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Answer» They both are irrational So the answer is not exist There r not determinable no. It can be 1.5 There are infinitely many numbers. |
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| 34005. |
If y =f(x) is represent on graph, find f(x) |
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| 34006. |
Represent √3on the number line |
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| 34007. |
18-8= |
| Answer» 10 | |
| 34008. |
If sec v-tan v=√2 tan v then prove that ..sec v + tan v=√2sec v |
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| 34009. |
If the zeros of the polynomial f(x)=x*x*x-12x*x+39x+a are (4+√17)(4-√17) |
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| 34010. |
bx+ay=2abax-by=a×a-b×b |
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| 34011. |
For ch 3 |
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| 34012. |
How to solve linear eqatuon in two variable |
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Answer» Study chapter 3 in book With the help of formulas |
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| 34013. |
Describe the meeting between The Frog and The Nightingale |
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| 34014. |
Find lcm of 2.5, 0.5, 0.175 |
| Answer» LCM of rational number ={tex}\\frac{{{\\text{LCM of numerators}}}}{{{\\text{HCF of denominators}}}}{/tex}Numbers are\xa0{tex}\\frac { 25 } { 10 } , \\frac { 5 } { 10 } , \\frac { 175 } { 1000 }{/tex}Now,\xa025 = 5{tex}\\times{/tex}5; 5 = 5{tex}\\times{/tex}1; 175 = 5{tex}\\times{/tex}5{tex}\\times{/tex}7LCM of (25, 5, 175) = 5{tex}\\times{/tex}5{tex}\\times{/tex}7 = 175Also,\xa010 = 2{tex}\\times{/tex}5; 1000 = 2{tex}\\times{/tex}2{tex}\\times{/tex}2{tex}\\times{/tex}5{tex}\\times{/tex}5{tex}\\times{/tex}5HCF of (10,10,1000) = 10LCM of (2.5, 0.5, 0.175) ={tex}\\frac { 175 } { 10 } = 17.5{/tex} | |
| 34015. |
The product of a non zero rational and an irrational no is |
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Answer» Irrational Irrational Rational |
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| 34016. |
Express sin theta in terms of sec theta |
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| 34017. |
Given that HCF (306,657 )=9, find LCM of (306,657) |
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Answer» 234u9 22338 |
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| 34018. |
Show that 9n cannot end with zero for any natural number n |
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Answer» If 9n ends with 0 then it must have 5 as a factor.But 9n has factor of (3.3)n =(3n.3n) which shows that 3 is the only factor of 9n.We know the fundamental theorem of arithematic ie. (5m.2n) and 9n does not apply on 9n.So, 9n can never end with zero. It can be written in the form of 3n^2. So to end with 0 it should have 5 as a prime factor but clearly it does not have any so it cannot end with 0 9n= 3^2n the prime factorisation does not contain 10 and it is unique so 9n deos not end with zero |
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| 34019. |
What are irrational numbers exactly? |
| Answer» Numbers which have decimal expansion as non terminating and non repeating are called as irrational no or numbers which are not rational no they are irrational | |
| 34020. |
If two positive integers a and b are written as a=xy2 |
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| 34021. |
Is1 is a prime number |
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Answer» No, because every no. has the factor of 1 No because it not have more than one factor It do not have more than 1 factor. So it is neithre a composite number nor a prime number. No its not a prime no. It is nor prime nor a composite number No |
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| 34022. |
1/0 |
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Answer» Not defined Since denominator is 0 then the no. is imaginary. So it do not have any real value. In other word we can say that it is not defined. The answer is"not define" |
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| 34023. |
Tan13°tan21°tan30°tan69°tan77° |
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Answer» One upon root 3 Root 3 Itti jaldi aap thita wale chapter padhne lage Tan30 answer Tan(90-77)tan21tan30 tan(90-21)tan77Cot77tan21tan30cot21tan771/tan77 tan21 tan30 1/tan21 tan77 |
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| 34024. |
Factorised root 3 x square _2 root two x _2 root 33 = 0 |
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| 34025. |
Show that any positive integeris in the form 8q plus 3 or 8q plus 5where q is some integer |
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| 34026. |
Q.If alpha beta gamma are zeroes of 6x³+3x²+5x+1 then find the value of 1/alpha+1/beta+1/gamma. |
| Answer» {tex}\\alpha , \\beta \\text { and } \\gamma{/tex} are zeroes of the polynomial 6x3 + 3x2 - 5x + 1in the given polynomial, 6x3 + 3x2 - 5x + 1a=6, b=3, c=-5, d=1Sum of the roots =\xa0{tex}- \\frac {b}{a}{/tex}{tex}\\alpha + \\beta + \\gamma = - \\frac { 3 } { 6 }{/tex}{tex}\\alpha + \\beta + \\gamma = - \\frac { 1 } { 2 }{/tex}sum of the Product of the roots =\xa0{tex}\\frac {c}{a}{/tex}{tex}\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = - \\frac { 5 } { 6 }{/tex}Product of the roots =\xa0{tex}- \\frac{d}{a}{/tex}\xa0{tex}\\alpha \\beta \\gamma = - \\frac { 1 } { 6 }{/tex}{tex}\\therefore \\quad \\frac { 1 } { \\alpha } + \\frac { 1 } { \\beta } + \\frac { 1 } { \\gamma } = \\frac { \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha } { \\alpha \\beta \\gamma }{/tex}{tex}= \\frac { - 5 / 6 } { - 1 / 6 } = \\frac { - 5 } { 6 } \\times \\frac { 6 } { - 1 }{/tex}Hence,\xa0{tex}\\alpha ^ { - 1 } + \\beta ^ { - 1 } + \\gamma ^ { -1 } = 5{/tex} | |
| 34027. |
Prove that the sum of three angle of a triangle is 180° |
| Answer» Please i need answer | |
| 34028. |
Y=x2-1 find the zeros by graphical method |
| Answer» Y=x2-1find the zeros by graphical method | |
| 34029. |
Polynomial 2 A question no. 3 |
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| 34030. |
How to find lcm and hcf |
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Answer» You can find hcf and lcm by prime factorisation. You can find lcm and hcf like this :Ex 28 and 35So , 28:7×4 35:7×5 In above you can see that 7 is common than its hcf.Now,.L.c.m in above 7 repeat in both so take 7 common 7×4×5=140 Try it after practice you will understand it.? |
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| 34031. |
Show that √2 is irrational? |
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Answer» Let √2 is irrational number ,means √2 is a rational number .Let √2= p by q √2^2= p^2 by q^22 = p^2 by q^22q^2 = p^2 -------(1)q^2=p^2 by 2~2 divides p^2~2 divides p also. Let p by 2 =r ,for some integer r. p=2r ----------(2)On substituting eq.(2) in eq.(1). 2q^2=2^2.r^2q^2 = 4r^2q^2=2r^2q^2 by 2= r^2~2 divides q^2.~2 divides q also Thus, is a common factor of p and q but this contradict that p and q are coprime so our assumption is wrong .Hence ,√2 is an irrational number Let route to be a rational number so it can be written in p by Q form where p and q are coprime numbers means it has only one factor that is 1 and where is not equal to zero |
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| 34032. |
3x-5y-19=0-7x+3y+1=0 |
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| 34033. |
My maths is very weak |
| Answer» Do more practice | |
| 34034. |
If cot theta =3\\4, show that[sectheta -cosec theta \\SEC Theta +cosec theta]in root=1\\root7 |
| Answer» Let us draw a triangle ABC such that,\xa0{tex}\\angle{/tex}B =\xa090°.Let\xa0{tex}\\angle{/tex}A =\xa0{tex}\\theta{/tex}°.We have,\xa0{tex}\\cot \\theta = \\frac { 3 } { 4 }{/tex}Then,\xa0{tex}\\cot \\theta = \\frac { \\text { Base } } { \\text { Perpendiaular } } = \\frac { A B } { B C } = \\frac { 3 } { 4 }{/tex}Let AB\xa0= 3 and BC = 4,By Pythagoras\' theorem, we know thatAC2\xa0= AB2 + BC2= 32\xa0+ 42\xa0= 9\xa0+ 16\xa0= 25{tex}\\Rightarrow \\quad AC = 5{/tex}Now,{tex}\\sec \\theta = \\frac { \\text { Hypotenuse } } { \\text { Base } } = \\frac { A C } { A B } = \\frac { 5 } { 3 }{/tex}{tex}\\text{cosec} \\theta = \\frac { \\text { Hypotenuse } } { \\text { Perpendicular } } = \\frac { A C } { B C } = \\frac { 5 } { 4 }{/tex}L.H.S =\xa0{tex}\\sqrt { \\frac { \\sec \\theta - \\text{cosec} \\theta } { \\sec \\theta + \\text{cosec} \\theta } }{/tex}{tex}= \\sqrt { \\frac { 5 / 3 - 5 / 4 } { 5 / 3 + 5 / 4 } }{/tex}{tex}= \\sqrt { \\frac { \\frac { 20 - 15 } { 12 } } { \\frac { 20 + 15 } { 12 } } }{/tex}{tex}= \\sqrt { \\frac { 5 } { 35 } }{/tex}{tex}= \\sqrt { \\frac { 1 } { 7 } }{/tex}{tex}= \\frac { 1 } { \\sqrt { 7 } }{/tex}= R.H.Stherefore,\xa0{tex}\\sqrt { \\frac { \\sec \\theta - \\text{cosec} \\theta } { \\sec \\theta + \\text{cosec} \\theta } }{/tex}{tex}= \\frac { 1 } { \\sqrt { 7 } }{/tex}Hence proved. | |
| 34035. |
What is the value of the root 3 |
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Answer» 1.732 is the value of root3 1.732 |
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| 34036. |
Obtain all other zeroes of the polynomial x⁴-17x²-36x-20 if to of its zeroes are 5 and -2 |
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| 34037. |
-2b×-2b |
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Answer» Sk harsh -2b×-2b=+4bsquare -2b×-2b =2b square |
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| 34038. |
Prove that root2+1/root2 is irrational number |
| Answer» √2+1/√2 =3/√2 =3√2/2Let us assume that √2+1/√2 is rational. Then there must exist two positive co primes a and b such that√2+1/√2=a/b=>3√2/2=a/b2a/3b=√2 which is a contradiction.Hence, √2+1/√2 is irrational. | |
| 34039. |
In chapter triangles exercise 6.2 2nd main 2nd question how to solve |
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| 34040. |
2÷35×643 |
| Answer» 36.74286 | |
| 34041. |
657×2 |
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Answer» 1314 1314 1314 1314 |
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| 34042. |
Find the number nearest to 110000 call greater than the 100000 exactly divisible by 8,15 and 21 |
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| 34043. |
Find the HCF and LCM of 10224 and 1608 using prime factorization method |
| Answer» answer? | |
| 34044. |
A number when divided by 52 gives 32 as quotient and 20 as remainder .find the number? |
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| 34045. |
HCF and LCM of two numbers are 9 and 90 respectively. if one number is 18 .find the number? |
| Answer» 45 | |
| 34046. |
How to write the answer of question to get full marks in each question |
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| 34047. |
What is value of cosec 30. |
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Answer» Cosec 30 \' = 22=2 Cosec 30°=2 2 |
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| 34048. |
Factorise x Square+7x+10 |
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Answer» (x+2)(x+5) (x+2)(x+5) |
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| 34049. |
Hcf of 25 and 25500 and represent in linear combination |
| Answer» 5 | |
| 34050. |
If (x+a)is a factor of polynomials x² -px +q & x² +mx +n . Then prove that a=n-p/m-p |
| Answer» Sorry prove that a=n-q/m-p | |