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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.
| 37451. |
X square + 4x+k=0 |
| Answer» Which chapter it is in? | |
| 37452. |
In chapter 9.. how do we know that where we will have to use sin theta,cos theta and tan theta....?? |
| Answer» In book | |
| 37453. |
Compute the indefinite integral of gamma of x. |
| Answer» | |
| 37454. |
Use eculid alogrithim to find hcf 4052 and 12576 |
| Answer» Step 1: Since 12576 > 4052, apply the division lemma to 12576 and 4052, to get12576 = 4052 × 3 + 420Step 2: Since the remainder 420 ≠ 0, apply the division lemma to 4052 and 420, to\xa0get4052 = 420 × 9 + 272Step 3: Consider the new divisor 420 and the new remainder 272, and apply the\xa0division lemma to get420 = 272 × 1 + 148Consider the new divisor 272 and the new remainder 148, and apply the division\xa0lemma to get272 = 148 × 1 + 124Consider the new divisor 148 and the new remainder 124, and apply the division\xa0lemma to get148 = 124 × 1 + 24Consider the new divisor 124 and the new remainder 24, and apply the division\xa0lemma to get124 = 24 × 5 + 4Consider the new divisor 24 and the new remainder 4, and apply the division\xa0lemma to get24 = 4 × 6 + 0Hence, the HCF of 12576 and 4052 is 4.\xa0 | |
| 37455. |
X4-5x+6 |
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| 37456. |
Substitution method3x-y=39x-3y=9 |
| Answer» 3x - y = 39x - 3y = 9The given pair of linear equations is\xa03x - y = 3..............(1)9x - 3y = 9.............(2)From equation(1),\xa0y = 3x - 3...................(3)9x - 3(3x - 3) = 9{tex}\\Rightarrow{/tex}\xa09x - 9x + 9 = 9{tex}\\Rightarrow{/tex}\xa09 = 9which is true. Therefore, equation (1) and (2) have infinitely many solutions. | |
| 37457. |
Prove that 3 by2root5 is irrational number |
| Answer» Let 3/2root5 be rational no. Multiply 3/2root5 by 2/3(product of rational os always rational )therefore root 5 is rational this conteadicts the fact root 5 is irrational. So, our assumption is wrong 3/2root 5 is irrational no. | |
| 37458. |
Find the greatest number that will divide 43,91,183 so as to leave the same remainder in each case |
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Answer» We have here 43,91,183So the difference r183-91=92183-43=14091-43=48NowHcf(48,140,92)As48=2×2×2×2×392=2×2×23140=2×2×5×7Hcf = 2×2×=4And 4 is the required number How to solve it 183 |
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| 37459. |
2-84 |
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Answer» Well i guess your question is incomplete coz its not s class 10 th question at all -82 Sure u r in class 10?? -82 |
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| 37460. |
2+77588 |
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Answer» R u sure u r in 10th class because this nost the ques. Of 10th class 77590 77590 |
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| 37461. |
Find quadratic polynomial whose one zero is -5 and product of zeroes is 0 |
| Answer» One zero {tex}= -5{/tex}product of zeroes {tex}= 0{/tex}{tex}\\therefore{/tex}\xa0Other zero = {tex}\\frac { 0 } { - 5 }{/tex}{tex}= 0{/tex}Sum of zeroes {tex}= -5 + 0 = -5{/tex}Polynomial {tex}p(x) = x^2\xa0- (S)x + P{/tex}= {tex}x^2\xa0+ 5x{/tex} | |
| 37462. |
In a quadratic equation 9x^2+ax-2=0, find the value of alpha for which x =1/3 |
| Answer» -5/3 | |
| 37463. |
Terminating or not terminating |
| Answer» | |
| 37464. |
A boat takes 12 h to go downstream and 24h to go upstream. Find time taken in still water. |
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Answer» It is incorrect 12h+24h=36h?? |
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| 37465. |
Find k if the following equations have two equal roots 1. 2x×x +kx-3 =0 |
| Answer» 2x2 + kx - 3 = 0Comparing equation with ax2 + bx + c = 0, we geta = 2, b = k and c = -3Discriminant = b2 - 4ac= (k)2 - 4(2) (-3)= k2 + 24For equal roots,Discriminant = 0k2 + 24 = 0k2 = - 24k = -√24 = -2√6\xa0 | |
| 37466. |
Verification of completing squere method |
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| 37467. |
Show that n2-1 is divisible by 8 if nisan odd positive integers |
| 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. | |
| 37468. |
Divide 4xcube + 3xsquare + 5x-6 by 2xsquare +3x+1 |
| Answer» | |
| 37469. |
Find zeroes of polynomial 2xcube + 3xsquare - 5x -6 |
| Answer» 2xcube+2xsquare+xsquare+x-6x-62x^2(x+1)+x(x+1)-6(x+1)2x^2+4x-3x-6(x+1)2x(x+2)-3(x+6)(x+1)(2x-3)(x+6)(x+1)Therefore the zeros are- (3/2),(-6),(-1). | |
| 37470. |
When two quadratic equations have equal discriminants and have a common root 1the the other roots |
| Answer» Than the other root also 1 | |
| 37471. |
a+b a-b= |
| Answer» | |
| 37472. |
Find the zeroes of the polynomial 7y2-11/3y-2/3 |
| Answer» 7y2\xa0-\xa0{tex}\\frac { 11 } { 3 } y - \\frac { 2 } { 3 }{/tex}=\xa0{tex}\\frac 13{/tex}(21y2\xa0- 11y - 2)=\xa0{tex}\\frac 13{/tex}(21y2\xa0- 14y + 3y - 2)=\xa0{tex}\\frac 13{/tex}[7y(3y - 2) + 1(3y - 2)]=\xa0{tex}\\frac 13{/tex}(3y - 2)(7y + 1){tex}\\Rightarrow y = \\frac { 2 } { 3 } , \\frac { - 1 } { 7 }{/tex}\xa0are zeroes of the polynomial. | |
| 37473. |
What is odd and even number |
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Answer» The odd odd no. is written in the form of 2m+1 and even no. is 2m Odd number are 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 5153 55-57 5961 6365 67 69etcAnd even number are 2 4 6 810 2012 The number which is not divisible by 2 is odd number n number which is divisible by 2 is even number |
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| 37474. |
Find the roots of the quadratic equation by the method of completing the square 4x²+3x+5=0 |
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Answer» No dear route exist because of minus sign under root No real root exist because of minus sign under root |
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| 37475. |
Please ask me quistion |
| Answer» ?., | |
| 37476. |
Find the coordinates of circumcentre |
| Answer» \xa0Let A(3, 0), B(-1, -6) and C(4, -1) be the given points.Let O(x, y) be the circumcentre of the triangle.OA = OB = OCOA2 = OB2(x - 3)2 + (y - 0)2 = (x + 1)2 + (y + 6 )2{tex}\\Rightarrow{/tex}\xa0x2 + 9 - 6x + y2 = x2 + 1 +2x + y2 + 36 + 12y{tex}\\Rightarrow{/tex}\xa0x2 - 6x + y2 - x2 - 2x - y2 - 12y = 1 + 36 - 9{tex}\\Rightarrow{/tex}\xa0-8x - 12y = 28{tex}\\Rightarrow{/tex}\xa0-2x - 3y = 7{tex}\\Rightarrow{/tex}\xa02x + 3y = -7 ........(i)Again,OB2 = OC2(x + 1)2 + (y + 6)2 = (x - 4)2 + (y + 1)2{tex}\\Rightarrow{/tex}\xa0x2 + 1 + 2x + y2 + 36 +12y = x2 +16 - 8x + y2 + 1 + 2y{tex}\\Rightarrow{/tex}\xa0x2 + 2x + y2 + 12y - x2 + 8x + y2 - 2y = 16 + 1 - 1 - 3610x + 10y = -20x + y = -2 ....... (ii)Solving (i) and (ii), we getx = 1, y = -3Hence circumcentre of the triangle is (1, -3)Circumradius\xa0{tex}= \\sqrt { ( 1 + 1 ) ^ { 2 } + ( - 3 + 6 ) ^ { 2 } }{/tex}{tex}= \\sqrt { ( 2 ) ^ { 2 } + ( 3 ) ^ { 2 } }{/tex}{tex}= \\sqrt { 4 + 9 }{/tex}{tex}= \\sqrt { 13 }{/tex}\xa0units. | |
| 37477. |
If one root of the quadratic equation 2x^+2x+k=0is -1÷3 then find k |
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| 37478. |
Trignomatery |
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| 37479. |
Is 3×7×51+21 are a composite number |
| Answer» 3×7×51+21=21×51+21=(51+1)21 | |
| 37480. |
7x2+19x-6=0 find the quadratic equaction root of fsctorisation |
| Answer» 7x2 +21x-2x-6=07x(x+3)-2(x+3)=0(X+3) (7x-2)=0X=-3. X=2/7 | |
| 37481. |
I not understand ch-8 trigonomatry of 10th |
| Answer» Do more practice | |
| 37482. |
If a and b are two positive integer such that a=14b , then find the HCF (a,b) |
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Answer» 41ab 41ab |
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| 37483. |
check whether 3n can end with digit 0 for any natural number n. |
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Answer» Only when r=0- No |
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| 37484. |
Show that 2+3root5 is an irrational number. Given that root5 is an irrational number. |
| Answer» 986676 17 | |
| 37485. |
If sn=11n-4nsquare find a and 9.10 or 9.101 and an |
| Answer» | |
| 37486. |
If the sin Alfa and cos Alfa are root of equation a^2+bx+c then prove that a^2+2ac=b^2 |
| Answer» The given equation is {tex}ax^2 + bx + c = 0{/tex}sin\xa0{tex}\\alpha{/tex}\xa0and cos\xa0{tex}\\alpha{/tex}\xa0are roots of the given equation.{tex}\\therefore{/tex}\xa0sin\xa0{tex}\\alpha{/tex}\xa0+ cos\xa0{tex}\\alpha{/tex}\xa0=\xa0-{tex}\\frac{b}{a}{/tex}\xa0...(i)and {tex}sin{/tex}\xa0{tex}\\alpha{/tex}{tex}.cos{/tex}\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{c}{a}{/tex}...(ii)Squaring both sides of equation (i),\xa0{tex}\\Rightarrow {/tex}(sin\xa0{tex}\\alpha{/tex}\xa0+ cos\xa0{tex}\\alpha{/tex})2 =\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}sin^2{/tex}\xa0{tex}\\alpha{/tex}\xa0{tex}+ cos^2{/tex}\xa0{tex}\\alpha{/tex}\xa0+\xa02 sin\xa0{tex}\\alpha{/tex}cos\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}1 + 2{tex}\\frac{c}{a}{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{a + 2c}{a}{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}a^2 + 2ac = b^2{/tex} | |
| 37487. |
Find k for which( p-q)*2 + 2(p2-q2)x +k =0 |
| Answer» Yde37887 | |
| 37488. |
India lost the match |
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Answer» Some people are also saying that dhoni is going to retire after world Cup 2019 And what about kohli\'s performance...????? Yeah?? YesNew Zealand take advantage of rainfall Yesss. Coz of dhoni he underperformed |
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| 37489. |
(cos0°+sin45°+sin30°)(sin90°-cos45°+cos60°). Plz give the answer.....plz |
| Answer» We know that, cos0°=1 =sin90°, sin45°=(1/√2)= cos45° & sin30°=(1/2)=cos60°, putting these values in the given expression, we get:-{tex}\\left( {\\cos 0^\\circ +\\sin 45^\\circ +\\sin 30^\\circ }\\right)\\left( {\\sin 90^\\circ -\\cos 45^\\circ +\\cos 60^\\circ } \\right){/tex}{tex} = \\left( {1 + \\frac{1}{{\\sqrt 2 }} + \\frac{1}{2}} \\right)\\left( {1 - \\frac{1}{{\\sqrt 2 }} + \\frac{1}{2}} \\right){/tex}{tex} = \\left( {\\frac{{2\\sqrt 2 + 2 + \\sqrt 2 }}{{2\\sqrt 2 }}} \\right)\\left( {\\frac{{2\\sqrt 2 - 2 + \\sqrt 2 }}{{2\\sqrt 2 }}} \\right){/tex}{tex} = \\left( {\\frac{{3\\sqrt 2 + 2}}{{2\\sqrt 2 }}} \\right)\\left( {\\frac{{3\\sqrt 2 - 2}}{{2\\sqrt 2 }}} \\right){/tex}{tex} = \\frac{{{{\\left( {3\\sqrt 2 } \\right)}^2} - {{\\left( 2 \\right)}^2}}}{8}{/tex}\xa0[Identity (a + b)(a - b) = a2 - b2]{tex} = \\frac{{18 - 4}}{8}{/tex}{tex} = \\frac{{14}}{8} = \\frac{7}{4}{/tex} | |
| 37490. |
Please any gyes exolain me linear equation in two varibles |
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| 37491. |
Pair of linaer equation in two varibals explain |
| Answer» | |
| 37492. |
Anyone know about gowebrachnasagar.com? |
| Answer» | |
| 37493. |
What is the general form of ap |
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Answer» a+(n-1)d is the general form of AP Tn=a+(n-1)d an=a+(n-1)d General form of A.P :-an=a+(n-1)d a+(n-1)d a+(n-1)d |
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| 37494. |
Chapter8 |
| Answer» Tan48 tan23 tan42 tan67 =1 | |
| 37495. |
If sec A =15/7 & A+B=90°, find the value of cosec B |
| Answer» In your queA + B = 90A = 90 - B. ( equ. i)Sec A =15/7Sec ( 90 - B ) = 15/7Cosec B = 15/7cosec B = 15/7 | |
| 37496. |
Key answers of the class 10std board maths paper |
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Answer» Of which year ???? what are you saying |
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| 37497. |
3x^2+40x-675=0 |
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Answer» I dont know sorry you can solve this by Sridharacharya formula |
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| 37498. |
Solve for x and y using Elimination method : 6(9x+6x)=3a+3b;7(bx-ay)=3a-2a |
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| 37499. |
In an equvilateral triangle ABC ,D is a point on side BC such that 4BD=BC.Prove that 16AD^2=13BC^2 |
| Answer» In equilateral {tex}\\triangle{/tex}ABC. 4BD = BCConstruction: Draw AE\xa0{tex}\\perp{/tex} BC.\xa0{tex}\\therefore{/tex} BE = {tex}\\frac{1}{2}{/tex}BC.In right {tex}\\triangle{/tex}AED, AD2 = DE2 + AE2\xa0{tex}\\Rightarrow{/tex} AE2 = AD2 - DE2 ..(i)In right {tex}\\triangle{/tex}AEB, AB2 = AE2 + BE2{tex}\\Rightarrow{/tex}\xa0AB2 = AD2 - DE2 + BE2 [using (i)]{tex}\\Rightarrow{/tex}\xa0AB2 + DE2 - BE2 = AD2{tex}\\Rightarrow{/tex}\xa0AB2 + (BE - BD)2 - BE2 = AD2{tex}\\Rightarrow{/tex}\xa0AB2\xa0+ BE2\xa0+ BD2\xa0- 2BE.BD - BE2\xa0= AD2{tex}\\Rightarrow{/tex}\xa0AB2 + ({tex}\\frac{1}{2}{/tex}BC)2 - {tex}2 \\times \\frac{1}{2}{/tex}BC{tex}\\times \\frac{1}{4}{/tex}BC = AD2{tex}\\Rightarrow{/tex}\xa0AB2\xa0+\xa0{tex}\\frac{1}{16}{/tex}BC2\xa0-\xa0{tex}\\frac{1}{4}{/tex}BC2\xa0= AD2{tex}\\Rightarrow{/tex}\xa0BC2\xa0- {tex}\\frac{3}{16}{/tex}BC2\xa0= AD2\xa0[{tex}\\because{/tex} AB = BC]{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 13 \\mathrm { BC } ^ { 2 } } { 16 }{/tex} = AD2{tex}\\Rightarrow{/tex}\xa013BC2\xa0= 16AD2 | |
| 37500. |
1+20 |
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Answer» 201 &. 21 21 obviously ? Get admitted in nursery 21 21 |
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