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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.
| 21601. |
If secA + tanA = p, then find the value of cosecA |
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| 21602. |
What is relation between mean mode median |
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Answer» 3median =2mean + mode There is an empirical relationship between the three measures of central tendency: 3Median = Mode + 2Mean. |
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| 21603. |
What are the tips for practicing maths to get more than 90% marks? |
| Answer» Just practice more and more...Best of luck..???? | |
| 21604. |
1/x+1/a+1/b |
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| 21605. |
एक बेलन एक संकु औऱ एक अर्धगोला तीनो के आधार औऱ ऊंचाई समान है इनके आयतनों का अनुपात क्या होगा? |
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Answer» A cylinder is equal to the base and height of a cone and a hemisphere three, what is the ratio of their volume Plz write in English and proper language... |
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| 21606. |
Prove that equation x²(a²+b²) + 2x(ac+bd)+(c²+d²)=0 has no real roots, if ad≠bc |
| Answer» According to question{tex}{/tex}, the given equation isx{tex}^2{/tex}(a2 + b2)+ 2x (ac + bd) +(c2 + d2)= 0 .\xa0{tex}{/tex}Let D be the discriminant of this equation.Therefore,D = 4(ac + bd)2 - 4(a2 + b2)(c2 + d2){tex} \\Rightarrow{/tex}\xa0D = 4 [(ac+ bd)2 - (a2 + b2) (c2 + d2)]{tex} \\Rightarrow{/tex}\xa0D = 4 [a2c2 + b2d2 + 2ac.bd - a2c2 - a2d2 - b2c2 - b2d2]{tex} \\Rightarrow{/tex}D = 4 [2ac.bd - a2d2 - b2c2] = - 4[a2d2 + b2c2 - 2ad.bc] = -4(ad - bc){tex}^2{/tex}It is given that ad {tex} \\neq{/tex}\xa0bc.{tex} \\Rightarrow{/tex} ad - bc\xa0{tex} \\neq{/tex}\xa00{tex} \\Rightarrow{/tex} (ad - bc){tex}^2{/tex} > 0{tex} \\Rightarrow{/tex} - 4 (ad - bc){tex}^2{/tex} < 0{tex} \\Rightarrow{/tex} D < 0.Therefore, given equation has no real root. | |
| 21607. |
m/nx² +n/m =1-2x. Solve for x. Please solve fully |
| Answer» According to the question,{tex}\\frac{m}{n}{x^2} + \\frac{n}{m} = 1 - 2x{/tex}{tex}\\Rightarrow \\frac{m}{n}{x^2} + 2x + \\frac{n}{m} - 1 = 0{/tex}{tex} \\Rightarrow {x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2}}}{{{m^2}}} - \\frac{n}{m} = 0{/tex} [multiplying both sides by \'n\' and dividing both sides by \'m\']{tex}\\Rightarrow {x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}To factorize {tex}{x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2} - mn}}{{{m^2}}}{/tex}, we have to find two numbers {tex}\'a\'\\ and\\ \'b\'{/tex} such that.{tex}a + b = \\frac{{2n}}{m}{/tex} and {tex}a b = \\frac { n ^ { 2 } - m n } { m ^ { 2 } }{/tex}Clearly, {tex}\\frac{{n + \\sqrt {mn} }}{m} + \\frac{{n - \\sqrt {mn} }}{m} = \\frac{{2n}}{m}{/tex} and {tex}\\frac{{\\left( {n + \\sqrt {mn} } \\right)}}{m}\\times\\frac{{\\left( {n - \\sqrt {mn} } \\right)}}{m} = \\frac{{{n^2} - mn}}{{{m^2}}}{/tex} ({tex}\\therefore a = \\frac{{n + \\sqrt {mn} }}{m}{/tex} and {tex}b = \\frac{{n - \\sqrt {mn} }}{m}{/tex}){tex}\\Rightarrow {x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}{tex} \\Rightarrow {x^2} + \\frac{{\\left( {n + \\sqrt {mn} } \\right)}}{m}x + \\frac{{\\left( {n - \\sqrt {mn} } \\right)}}{m}x{/tex}{tex} + \\frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}{tex} \\Rightarrow x\\left[ {x + \\frac{{n + \\sqrt {mn} }}{m}} \\right] + \\frac{{n - \\sqrt {mn} }}{m}{/tex}{tex}\\left[ {x + \\frac{{n + \\sqrt {mn} }}{m}} \\right] = 0{/tex}{tex} \\Rightarrow \\left( {x + \\frac{{n - \\sqrt {mn} }}{m}} \\right)\\left( {x + \\frac{{n + \\sqrt {mn} }}{m}} \\right) = 0{/tex}{tex} \\Rightarrow x + \\frac{{n - \\sqrt {mn} }}{m} = 0{/tex} or {tex}x + \\frac{{n + \\sqrt {mn} }}{m} = 0{/tex}{tex} \\Rightarrow x = \\frac{{- n - \\sqrt {mn} }}{m}{/tex} or {tex}x = \\frac{{ - n + \\sqrt {mn} }}{m}{/tex} | |
| 21608. |
-2x^2 - x +1=0 ....solve the quadratic equation by completion of square method |
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Answer» Thnx..... but i want the answer with compeleting square method 2x^2 - 2x+x-12x(x-1)+1(x-1)(2x+1)(x-1)X=-1/2,x=1 |
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| 21609. |
Prove that √3 + √5 is an irrational number |
| Answer» What do i mean by modi ke relative | |
| 21610. |
√5 irrational |
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Answer» Sol- let, √5 be rational no. Then, it is in the form of p/q where q not equals to zero and P and q are co-prime. √5=p/q On squaring both side-5=p²/q²5q²=p² ____eqation (1)Since, 5 divides p²Therefore, 5 divides pAgain, Let, p=5cOn squaring both side, P²=25c²5q² =25c² from equation 1q² =5c²Since, 5 divides q²Therefore, 5 divides q Since, 5 is the common factor .Therefore, this contradiction Therefore, √5 is an irrational no. Yes |
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| 21611. |
In board which book l should study with |
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Answer» Ncert(example question also) Ncert, previous year questions, all theorems,. Sample paper NCERT |
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| 21612. |
Solve for x :1/a+b+c = 1/a+1/b+1/x |
| Answer» ON solving the above equation we have a quadratic equation (a+b)x^2+(a+b)^2x+ab(a+b)on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)we got -a and -b as roots. | |
| 21613. |
What is theorem? |
| Answer» A, proved statement.......... | |
| 21614. |
1+1=12 |
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Answer» Yes,1+1=12 but only because of typing mistake ?????????? 1+1+10=12 You want to add 10 in this equation ?????? |
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| 21615. |
If cos =4/5 ,find all other trigonometric ratio of angle |
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Answer» sin θ = 3/5tanθ = 3/4cosec θ = 1/sinθ = 5/3Sec θ = 1/cos = 5/4Cot θ = 1/tan = 4/3 Here p=3 ,b=4 and h=5 now use trigo ratios sin=p/h ,tan=p/b and so on....... |
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| 21616. |
4sin square 45 degree +tan square 60 degree +cosec square 30 degree |
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Answer» Please reply sis Yes I done it As sin 45 degree is 1/√2 so sin square 45degree is (1/√2)^2= 1/2 and its 4times become 2.......... Eq(1)Similarly tan sqare 60 become 3 ..... Eq(2)And cosec square 30 become 2.... Eq (3)\xa0On adding all these equations as required in question\xa0We get2+3+2 = 7 |
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| 21617. |
What is the formula of finding the diagonal of a cuboid and of cube...plz help |
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Answer» Cuboid => √(l^2 + b^2 + h^2) The main diagonal of any cube can be found my multiplying the length of one side by the square root of 3.Diagonal of cube = √ 3 × side |
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| 21618. |
In triangle PQR right angled at Q , PQ =3 cm and PR =6cm. Determine angle QPR and angle PRQ |
| Answer» Given PQ=3cm and PR=6cm.So, PQ/PR =sinRSin R=3/6=1/2And sin at 30° is 1/2 so angle PRQ = 30 ° therefore angle QPR = 60 °[Bcs sin P = QR/PR | |
| 21619. |
What is the segment of a circle |
| Answer» The region of a circle cut by the chord | |
| 21620. |
Aap sabhi ka board ka centre kha pda hai kuch pta chla kya kisi ko admit card mil aa yha pr? |
| Answer» 94 plus | |
| 21621. |
Given that SinA-cosA=√2 to prove, sinA+cosA=√2 |
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Answer» It is correct Check ur question |
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| 21622. |
Cos60.cos30+sin60.sin30 |
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Answer» Answer is √3/2 Answer is 2 √3/2=sin60°=cos30° |
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| 21623. |
If the 3rd term and the 9th term of an AP are 4 and -8 then which term of these AP is 0. |
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Answer» Sry the crrct ans is 5 3rd term |
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| 21624. |
Volume surface area |
| Answer» What!! | |
| 21625. |
Prove that 8.4 exercise question no 5 last question |
| Answer» Palala 1+tanA and 1+cotA ko secA and cosecA mai change karo aur yai secA/cosecA banega to secA=1/cosA and cosecA=1/sinA mai change kar dana hoga aur yai sinA/cosA=tanA hoga | |
| 21626. |
Class 10 maths sample paper of 2018-19 academic with solutions |
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| 21627. |
Eb ki baar maths ka paper ncert ma aaya ga ya ncert sa baar sa aaya ga |
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Answer» Yup u can try.....but ncert is enough Jo cbse guide ki app ma jo social science ma hindi medium ma important diya hua una yaad kar sakta jisa no. Badiya aajaya S.s ka or baatiyo ek baar Only ncert based |
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| 21628. |
If tsa of a cylinder is 462 m2and csa is equal to 1/3tsa then find the volume of cylinder |
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Answer» 539cm³ Ya its 539 cm cube ,calculation mistake 539 cm^3 549 cm cube |
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| 21629. |
a+(1\\a)=2sinx find value of a for which this questn is possible |
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Answer» L.H.S. => a+1\\a =1+1\\1 = 2R.H.S.=> 2sinxNow, 2= 2sinx=> Sinx = 1.•. X = 90° Solve krke dikhao lekin 1 Maybe |
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| 21630. |
a+(1\\a)=2sinQ find Q |
| Answer» 90° | |
| 21631. |
I have doubt in ncert chapter 10 example 4 (page no.231) |
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Answer» Maths Which subject |
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| 21632. |
Find the root of the following quadratic equation: |
| Answer» _2,5 | |
| 21633. |
Show that the number 9n cannot end with the digit 2 for any natural number |
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| 21634. |
T s A of sphere |
| Answer» 4 πr square | |
| 21635. |
What is a successor of -18 |
| Answer» _17 | |
| 21636. |
What is fullfom of (143)? |
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| 21637. |
X2+11x-60 |
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Answer» X=4,-15 x2 + 11x - 60= x2 + 15x - 4x - 60= x ( x + 15) - 4 ( x +15)= (x + 15) (x - 4) X=4 and -15 |
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| 21638. |
How to take more marks in mathematics |
| Answer» You have to learn all formulas of ncert textbook and then solve sample papers ,list 2 sample paper | |
| 21639. |
Tan÷ 1-cot + cot÷ 1-tan=1+ sec×cosec |
| Answer» LHS =\xa0{tex}\\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta }{/tex}{tex}= \\frac { \\frac { \\sin \\theta } { \\cos \\theta } } { 1 - \\frac { \\cos \\theta } { \\sin \\theta } } + \\frac { \\frac { \\cos \\theta } { \\sin \\theta } } { 1 - \\frac { \\sin \\theta } { \\cos \\theta } }{/tex}\xa0{tex}\\left[ \\because \\tan \\theta = \\frac { \\sin \\theta } { \\cos \\theta } , \\cot \\theta = \\frac { \\cos \\theta } { \\sin \\theta } \\right]{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } + \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\cos \\theta - \\sin \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } - \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 3 } \\theta - \\cos ^ { 3 } \\theta } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}\xa0{tex}= \\frac { ( \\sin \\theta - \\cos \\theta ) \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta + \\sin \\theta \\cos \\theta \\right) } { ( \\sin \\theta - \\cos \\theta ) \\sin \\theta \\cos \\theta }{/tex}\xa0[ a3\xa0- b3\xa0= (a - b)(a2\xa0+ ab + b2) ]{tex}= \\frac { 1 + \\sin \\theta \\cos \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\frac { 1 } { \\sin \\theta \\cos \\theta } + 1 = 1 + \\sec \\theta cosec \\theta{/tex}\xa0= RHStherefore, RHS = LHS | |
| 21640. |
Find the value of k for which one root of the quadratic equation kx^-14x +8=0 is six times the other |
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Answer» K=3 3 K=3 |
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| 21641. |
Using Euclid\'s algorithms find the HCF of 423and 54 |
| Answer» Hcf=9 | |
| 21642. |
Hey guys do you have any link of model question paper |
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Answer» S.s , science , maths Which subject do u want?? |
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| 21643. |
All the formulas of book of all the chapters? |
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| 21644. |
How to justify the construction |
| Answer» By proving similarity between triangles through AA then showing the ratio of sides by SSS | |
| 21645. |
Exercise 13.1 1st question |
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| 21646. |
If tanA=cotA.Then find the value of secA. |
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Answer» TAN A= COT ATHEN, TANA= TAN(90-A)THEREFORE, A=90-A90=2AA=45THEREFORE SECA=SEC45=ROOT 2HOPE YOU LIKE MY APPROACH!!!!!? √2 |
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| 21647. |
What is co. Primes? |
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Answer» Co-prime number is a set of numbers or integers which have only 1 as their common factor. That is their highest common factor (HCF) will be 1. It is also known as relatively prime or mutually prime numbers.Consider a set of two numbers, if they have no positive integer that can divide both, other than 1, the pair of numbers is co-prime.Ex 1: 21 and 2221 and 22:\tThe factors of 21 are 1, 3, 7 and 21.\tThe factors of 22 are 1, 2, 11 and 22.Here 21 and 22 have only one common factor that is 1.Hence their HCF is 1 and are co-prime. No which have there common hcf . |
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| 21648. |
Show that root p + root q is irrational, where p &q are primes. |
| Answer» Rootp+rootq=a÷bSquaringP+Q+2rootPQ=a÷b whole squareRootPQ=1/2((a/b)2-P-Q)Now p and q are prime positive numbers so RootP and Root Q is irrationalAlso RootPQIrrational=rationalSo contridiction | |
| 21649. |
Find the roots of following quadratic equation - ( x+3) (x+1)=3(X- 1/3) |
| Answer» (x+3) (x-1) = x-1/3x2\xa0- x + 3x - 3 = x-1/3x2\xa0+ 2x - 3 = x-1/33x2\xa0+ 6x- 9 = x-13x2+ 5 x - 8 = 03x2+ 8x - 3x - 8 = 0x(3x + 8) - 1(3x + 8) = 0(3x + 8) (x - 1 ) = 0x = -8/3 , 1So the roots are -8/3 and 1 | |
| 21650. |
Find the value of K when the distence between the point (3,2k) and (4,1) is √10 unit . |
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Answer» Using distant formula√ |
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