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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.
| 10501. |
Yaha Tak ki apna gender change kr diya.....boys girls k naam see id bana the hai... |
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Answer» Kabse bol rha Mera answer do meine churaya hai kya tere answer ko To Tera kya question hai bolna Nai see nai gaali sikho vo bhi free mein Bahut see idhat fake I\'d bani hui hai......aur galiyon ka to bhandar hai Mere qstn ka ans kro ? It\'s true Dimaag phir gaya hai kya ?virus faila rahi ho yaha Type krte time to galia tapkti hai Oh |
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| 10502. |
roman rumenal |
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Answer» Nhi banni m hu swaggy Harshit हो tum |
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| 10503. |
SinA/cosA+cosecA=2+sinA/cotA-cosecA |
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Answer» u have not sent answers Slve lhs and rhs separately |
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| 10504. |
1M |
| Answer» 100cm | |
| 10505. |
Find the sum of first five multiples of 2 |
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Answer» 30 Iska ans humne de diya to aage poochogi ki 2+2 kitne hote hain |
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| 10506. |
In the AP 2,x,26 find the value of x |
| Answer» x = 14 | |
| 10507. |
Find the positive root of under root 3x square +6=9 |
| Answer» Ans is 5 baise me mathematics ke question ke ans nahi deta kunki inki koi seema nahi but tumhare question ans ish liye diya ki agar poori baat padi to kabhi bhi koi bhi koi ish type ka ker sakte ho sabse pehle hum under root hatane ke liye squaring kerte hainjisse underroot hat jaye phir solve khud se karsakte hain ok thank ku jagah mein jo likha hai usme kuch likh dena tumhari marzi | |
| 10508. |
In your app the sample paper is for board? |
| Answer» | |
| 10509. |
Give example of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm |
| Answer» \tdeg p(x) = deg q(x)\t{tex}p ( x ) = 2 x ^ { 2 } - 2 x + 14{/tex}\tg(x) = 2\tq(x) = x2 - x + 7\tr(x) = 0\tClearly, p(x) = q(x) {tex} \\times {/tex}\xa0g(x) + r(x)\tdeg q(x) = deg r(x)\t{tex}p ( x ) = x ^ { 3 } + x ^ { 2 } + x + 1{/tex}\t{tex}g ( x ) = x ^ { 2 } - 1{/tex}\tq(x) = x + 1\tr(x) = 2x + 2\tClearly,\xa0{tex}p ( x ) = q ( x ) \\times g ( x ) + r ( x ){/tex}\tdeg r(x) = 0\t{tex}p ( x ) = x ^ { 3 } + 2 x ^ { 2 } - x - 2{/tex}\t{tex}g ( x ) = x ^ { 2 } - 1{/tex}\tq(x) = x + 2\tr(x) = 4\tClearly,\xa0{tex}p ( x ) = q ( x ) \\times g ( x ) + r ( x ){/tex} | |
| 10510. |
The sum of three consecutive terms of an Ap is 18 and their product is 192. Find the number |
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| 10511. |
3-2 is |
| Answer» 1 | |
| 10512. |
If sina + sin^2a + sin^3a = 1 then prove that cos^6a + 4cos^4a + 8cos^2a = 4 |
| Answer» We have,sin{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0+ sin3{tex}\\theta{/tex}\xa0= 1{tex}\\Rightarrow{/tex}\xa0sin{tex}\\theta{/tex}\xa0+ sin3{tex}\\theta{/tex}\xa0= 1 - sin2{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0sin{tex}\\theta{/tex}\xa0(1 + sin2{tex}\\theta{/tex}) = cos2{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0sin2{tex}\\theta{/tex}\xa0(1 + sin2{tex}\\theta{/tex})2 = cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - cos2{tex}\\theta{/tex}\xa0){\xa01 + (1 - cos2{tex}\\theta{/tex}\xa0) }2\xa0= cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - cos2{tex}\\theta{/tex}) (2 - cos2{tex}\\theta{/tex})2 = cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - cos2{tex}\\theta{/tex}) (4 - 4 cos2{tex}\\theta{/tex}\xa0+ cos4{tex}\\theta{/tex}) = cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa04 - 4 cos2{tex}\\theta{/tex}\xa0+ cos4{tex}\\theta{/tex}\xa0- 4 cos2{tex}\\theta{/tex}\xa0+ 4 cos4{tex}\\theta{/tex}\xa0- cos6{tex}\\theta{/tex}\xa0= cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0-cos6{tex}\\theta{/tex}\xa0+ 4 cos4{tex}\\theta{/tex}\xa0- 8cos2{tex}\\theta{/tex}\xa0+ 4 = 0\xa0{tex}\\Rightarrow{/tex}\xa0cos6{tex}\\theta{/tex}\xa0- 4cos4{tex}\\theta{/tex}\xa0+ 8cos2{tex}\\theta{/tex}\xa0= 4 | |
| 10513. |
Exercise 6.5 question number 15 |
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| 10514. |
if sintheta + costheta +sin2theta + cos2 theta=1 then prove that tantheta +cottheta +1- sec2theta=1 |
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| 10515. |
Why is 2 a prime number as well as even number |
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Answer» 2 Is a prime no. because it is divisible by 1 and itself and 2 is also an even no. A prime number is such that it is divisible by only itself and one. So Two is a prime no. because it is divisible by only two and one. A no. which is divisible by two is also an even no. Two is the only no. that is both even and prime. |
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| 10516. |
Prove that √3+√5 is an irrational number |
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Answer» Let the given no. be rational. So there exist co prime integers a and b such that. √3+√5=a/b S.B.S ( √3+√5)^2=a/bSolving this we get the answer Yes Let us assume that √2+√3 is a rational no.√3+√5=p/qP/q-√3=√5Squaring both sides(P/q-√3)2=(√5)2P2/q2-2√3p/q+3=5P2/q2-2√3p/q+3-5P2/q2-2=2√3p/qP2-2q2/q22p/q√3P2-2q2/2pq=√3.This contradicts the fact that √ 3 is a irrational no.So,aur assume was incorrect.Here ,√2+√3is irrational no. |
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| 10517. |
Is 0 an even number? |
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Answer» Yes 0 is an even number . |
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| 10518. |
Sin + cos |
| Answer» Sin²A + cos²A=1 | |
| 10519. |
5x-6y+8=07x+6y-9=0 |
| Answer» X=1/12 andy=..........??????????? | |
| 10520. |
3x +2y = 52x _ 3y= 7 |
| Answer» On a graph paper, draw a horizontal line X\'OX and a vertical line YOY\' as the x-axis and the y-axis respectively.Graph of 3x + 2y = 4{tex}3x + 2y = 4{/tex}{tex}\\Rightarrow{/tex}\xa0y ={tex}\\frac { 4 - 3 x } { 2 }{/tex}\xa0...... (i)Thus, we have the following table for\xa0{tex}3x + 2y = 4{/tex}\tx0{tex}2{/tex}{tex}-2{/tex}y2{tex}-1{/tex}{tex}5{/tex}\tNow, plot the points\xa0{tex} A (0, 2), B(2, -1) \\ and \\ C(-2, 5){/tex} on the graph paper.Join AB and BC to get the graph line ABC.Extend the graph line BC on both sidesThus, the line BC is the graph of the equation\xa0{tex}3x + 2y = 4.{/tex}Graph of 2x - 3y = 7{tex}2x - 3y = 7{/tex}{tex}\\Rightarrow y = \\frac { 2 x - 7 } { 3 }{/tex}\xa0.......(ii)\tx2-15y-1-31\tNow, plot the points\xa0{tex}P(-1,-3) \\ and \\ Q (5, 1){/tex} on the same graph paper.Join PB\xa0and QB\xa0to obtain the line PQ.Thus, the line PQ\xa0is the graph of\xa0{tex}2x - 3y = 7.{/tex}The two graph lines intersect at point\xa0{tex}B(2,-1).{/tex}{tex}\\therefore{/tex}\xa0{tex}x = 2, y = -1{/tex} is the solution of the given system of equations. | |
| 10521. |
SecA+tanA=p find the value of cosec A |
| Answer» Given,{tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also, we know that,\xa0{tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1 [{tex}\\because a^2-b^2=(a+b)(a-b){/tex}]{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex})p = 1 [using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(i)+(ii), we get,{tex}sec\\theta + tan\\theta+ sec\\theta - tan\\theta = p+ \\frac{1}{p}{/tex}{tex}\\Rightarrow 2sec\\theta = \\frac{p^2+1}{p}{/tex}{tex}\\Rightarrow sec\\theta = \\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow \\frac{1}{cos\\theta} =\\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow cos\\theta =\\frac{2p}{p^2+1}{/tex}------(iii)Now, we know that,{tex}sin\\theta = \\sqrt( 1- cos^2\\theta) {/tex}put the value of\xa0{tex}cos\\theta{/tex}\xa0from eq. (iii), we get,{tex}sin\\theta = \\sqrt(1-(\\frac{2p}{p^2+1})^2){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(1-\\frac{4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2+1)^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1+2p^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1-2p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2-1)^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\frac{p^2-1}{p^2+1}{/tex}{tex}cosec\\theta = \\frac{p^2+1}{p^2-1} [\\because cosec\\theta =\\frac{1}{sin\\theta}]{/tex}hence, {tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex} | |
| 10522. |
SecA+tanA=p |
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| 10523. |
State and prove bpt |
| Answer» | |
| 10524. |
(a+b)% |
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| 10525. |
if 9th term of AP is zero ,prove that it\'s 29th term is double of its 19th term. |
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Answer» a9=a+8d=0 ............. ia29=a+28d..................iia19=a+18d..................iiifrom i and iia= - 8da29= - 8d + 28d a29 = 20d ..... ivfrom i and iiia19 = - 8d + 18d = 10d ... vthus a29 is double of a19 (20d is double of 10d) ...... proved a9=0a + 8d = 0a= -8d ............... (I)Now, a29 = a + 28d = -8d + 28d = 20dAgain, a19= a + 18 d = -8d + 18d =10dHence, 29th term is double of its 19th term. |
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| 10526. |
√6√6√6...find the value of this no. |
| Answer» 3 | |
| 10527. |
Alpha²/beta²+beta²/alpha²=p4/q4- 4q²/q +2 |
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| 10528. |
Sin ² 25°+cos²25° |
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Answer» 1 =why bcoz sin²25 +cos²25=1 1 1 |
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| 10529. |
Kya class10 ka paper asan ayega |
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Answer» No if you work hard ,and think it is easy you will succeed ,or if not Yes , After the commanding on NCERT No , After the commanding on NCERT I saw the sample paper according to it the difficulty level will be moderate. |
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| 10530. |
If cosec theta =15÷7, and alpha+theta=900, find the value of sec and also of sec alpha . |
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| 10531. |
2x+5x |
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Answer» x(2+5) = 7x 7x 7x |
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| 10532. |
In right angle triad ABC right angled atB if tanA=1 verify that 2sinA cosA=1 |
| Answer» In {tex} \\triangle A B C{/tex},{tex}\\tan A = 1{/tex}{tex}\\Rightarrow \\quad \\frac { B C } { A C } = 1{/tex}{tex}\\Rightarrow {/tex}\xa0BC = x and\xa0AC = xUsing Pythagoras theorem,\xa0{tex}\\Rightarrow A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 2 } x{/tex}{tex}\\therefore \\quad \\sin A = \\frac { B C } { A B } = \\frac { x } { \\sqrt { 2 } x } = \\frac { 1 } { \\sqrt { 2 } } \\text { and } \\cos A = \\frac { A C } { \\sqrt { 2 } x } = \\frac { x } { \\sqrt { 2 } x } = \\frac { 1 } { \\sqrt { 2 } } {/tex}2 sin A cos A\xa0{tex}= 2 \\times \\frac { 1 } { \\sqrt { 2 } } \\times \\frac { 1 } { \\sqrt { 2 } } = 1{/tex} | |
| 10533. |
Explain why 132333435637 is a composite number |
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Answer» Thanq Because it has more than one common factor we have no. 132333435637is divided by 13so, can be written as 13*179495049 it shows that the no. has more than two factors which satisfy the composite no. property .So it is. compo. no. |
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| 10534. |
I want to know the best site for sample papers. |
| Answer» cbseguide.com | |
| 10535. |
State and prove pythagoras theorem . |
| Answer» | |
| 10536. |
What is the formula of TSA of right circular cylinder |
| Answer» 2πr(r+h) | |
| 10537. |
Formula of TSA of hemisphere |
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Answer» TSA of hemisphere = 3πr2 TSA of hemisphere= 3πr2 |
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| 10538. |
Square of any positive integer is 4m or 4m+1 |
| Answer» let x be any possitive integer then by Euclid\'s Algorithm a = bq+r and a = x, b = 4we have x = 4q +r -------------------(1)where q is an integer >= 0 and 0 <= r <4since 0<=r<4 , r=0 ,1,2,3 ,\xa0from eqn (1) for r= 0 , x = 4q for r=1, x=4q+1 for r=2,x=4q+2therefore x is of th form 4q, 4q+1,4q+2now, x=4q=x2=16q2 ---------------------2x=4q+1=x2=16q2+8q+1------------------3x=4q+2=x2= 16q2+ 16q+4--------------4from equation 2 we have x2=16q2=4(4q2)4(4q2)=4m where m=4q2from 3rd\xa0equation, we have, x2= 4m+1 where m=4q2+ 2qfrom 4thequation , we have, x2=4m+1 where m= 4q2+4q+1hence,the square of any positive integer is either of the form 4m or 4m+1 for some integer m.\xa0\xa0 | |
| 10539. |
How i am study mathematics |
| Answer» | |
| 10540. |
KYA SAHI ME BOARD EXAM NCERT SE HI AAYEGA ?????? |
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Answer» Yes upto a great extent question will come from ncert but some question will also come outside ncert that will be of basic knowledgeIf you study the ncert properly then you will be able to answer maximum question. Yes ,all board exam in ncert book but only basic |
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| 10541. |
The sum of two no 15 if sum of their reciprocal is3/10 find the no |
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Answer» let nos are x and yx+y=15 x = 15 - y1/x + 1/y=3/101/15-y + 1/y=3/10(y+15-y)10 = 3y(15-y)150=45y-3y2y2-15y+50=0y(y-10) - 5(y-10)=0y=10 , 5x=5 , 10\xa0 Sol:Let one of the numbers be x.Other number is (15 - x)Sum of their reciprocals = 3/101/x + 1/(15 - x) = 3/10(15 ) / (15x - x2) = 3/10150 = 45x - 3x23x2 - 45x + 150 = 0x2 - 15x + 50 = 0(x - 10)(x - 5) = 0x = 10 or 5Therefore, the numbers are 10 and 5.If x = 15 |
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| 10542. |
the root of the equation x2 +x_p (p+1)=0 |
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| 10543. |
the root of the equation x2+x_p (p+1) |
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| 10544. |
Activities 1 |
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| 10545. |
Part 10 ofQ5 from imp Q of intro of trignomentry how |
| Answer» Gold | |
| 10546. |
Explain the substitution method |
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| 10547. |
2+4-0*0 |
| Answer» 6 | |
| 10548. |
If tan A, 4/5 find the value of cos A - sin A / cos A +sin A |
| Answer» 1/9 is the answer | |
| 10549. |
If Tn+1=4n+5 then find Tn. |
| Answer» 1 | |
| 10550. |
Class10 ka board ka exam as an ayega kya |
| Answer» | |