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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.
| 7001. |
A² +b² = 144. X² +PX +45=0, find the value of p |
| 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. | |
| 7002. |
Simplify (CosecA-SinA) |
| Answer» Sahil bro it\'s very simple convert cosecA to 1/sinA and by taking LCM. Solve its answer will be 1-sin2A/sinA =cos2A/sinA | |
| 7003. |
What is bond formation |
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| 7004. |
What is the pattern of examof sa1 |
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Answer» Search on google don\'t worry about that |
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| 7005. |
Prove√3is a irrational number |
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| 7006. |
3y+583x=56583y_3x=575 |
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| 7007. |
A right triangle hypotenuse 25cm and perimeter 60cm what is the area of triangle |
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| 7008. |
If pq is value 250 and p/q =10 then find p and q |
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Answer» q=5 ,p=50 P/q = 10 => p=10q.Now pq= 10q *q =>10q2 =250q2 = 25 => q =5.q=5 so p = 50 ? |
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| 7009. |
What are consecutive positive integers |
| Answer» 1,2,3,4,5,6,7,8,................... | |
| 7010. |
Solve a/x-b + b/x-a =2 |
| Answer» a(1)-b(x)/x+b(1)-9(x)/x=2A-bx/x+b-9x/x=2A-bx+b-9x/x=2A-bx+b-9x=2(x)A-bx+b-9x=2xA-bx+b-9x-2xA-bx+b-11x...dunno furthr...if uh knw...reply back...thnku.? | |
| 7011. |
State and prove Pythagoras theorem |
| Answer» Hypotenuse square is equal to perpendicular square + base square | |
| 7012. |
Find the maximum value of a sin£ + b cos¢ |
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| 7013. |
A fraction bears the same ratio to 1/27 as 3/7 does to 5/9 . The fraction is : |
| Answer» Answer is 1÷35 | |
| 7014. |
In triangle ABC , prove that area of triangle ABC= (1÷2) ×a×c×sinB |
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| 7015. |
If coseca a =2 then find all others trigonometry ratios |
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| 7016. |
Anil |
| Answer» Anil is something hiding and he is only a person | |
| 7017. |
What Is the value of si |
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| 7018. |
If roots of quadratic equation 2xsquare-kx+k=0 are equal then find K. |
| Answer» Ans is 8 | |
| 7019. |
(1+cota_cosseca) (1+tana+seca) = 2 |
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| 7020. |
Sin2A=2A is true when A |
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| 7021. |
Important question of trignometry |
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| 7022. |
If sin theta +cos theta =under root2 cos (90°-theta) . determine cot theta. |
| Answer» We have{tex}\\sin \\theta + \\cos \\theta = \\sqrt 2 \\cos (90^\\circ - \\theta ){/tex}{tex} \\Rightarrow \\sin \\theta + \\cos \\theta = \\sqrt 2 \\sin \\theta {/tex}\xa0{tex}\\left[ {\\because \\cos (90^\\circ - \\theta ) = \\sin \\theta } \\right]{/tex}{tex} \\Rightarrow \\cos \\theta = \\sqrt 2 \\sin \\theta - \\sin \\theta {/tex}{tex} \\Rightarrow \\cos \\theta = \\sin \\theta \\left( {\\sqrt 2 - 1} \\right){/tex}{tex} \\Rightarrow \\frac{{\\cos \\theta }}{{\\sin \\theta }} = \\sqrt 2 - 1{/tex}{tex} \\Rightarrow \\cot \\theta = \\sqrt 2 - 1{/tex}\xa0{tex}\\left[ {\\because \\frac{{\\cos \\theta }}{{\\sin \\theta }} = \\cot \\theta } \\right]{/tex} | |
| 7023. |
Are you all appearing Talentex? |
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| 7024. |
Solve the following pair of linear equations by the cross multiplication method 8x +5y=9, 3x +2y=9? |
| Answer» The given pair of linear equations8x + 5y = 9 ...(1) ...(1)3x + 2y = 4 ...(2) ...(2)\tby substituting method\tFrom equation (2), 2y = 4 - 3x\t{tex}\\Rightarrow \\;y = \\frac{{4 - 3x}}{2}{/tex}\xa0...(3)\tSubstitute this value of y in equation (1), we get\t{tex}8x + 5\\left( {\\frac{{4 - 3x}}{2}} \\right) = 9{/tex}\t{tex}\\Rightarrow{/tex} 16x + 20 - 15x = 18\t{tex}\\Rightarrow{/tex} x + 20 = 18\t{tex}\\Rightarrow{/tex} x = 18 - 20\t{tex}\\Rightarrow{/tex} x = -2\tsubstituting this value of x in equation (3), we get\t{tex}y = \\frac{{4 - 3( - 2)}}{2} = \\frac{{4 + 6}}{2} = \\frac{{10}}{2} = 5{/tex}\tSo the solution of the given pair of linear equations is x = -2, y = 5.\tby cross-multiplication method\tLet us write the given pair of linear equation is\t8x + 5y - 9 = 0 ...(1)\t3x + 2y - 4 = 0 ...(2)\tTo solve the equation (1) and (2) by cross multiplication method,\twe draw the diagram below:\t\tThen,\t{tex}\\frac{x}{{(5)( - 4) - (2)( - 9)}} = \\frac{y}{{( - 9)(3) - ( - 4)(8)}}{/tex}{tex} = \\frac{1}{{(8)(2) - (3)(5)}}{/tex}\t{tex}= \\frac{x}{{ - 20 + 18}} = \\frac{y}{{ - 27 + 32}} = \\frac{1}{{16 - 15}}{/tex}\t{tex}\\Rightarrow \\frac{x}{{ - 2}} = \\frac{y}{5} = \\frac{1}{1}{/tex}\t{tex}\\Rightarrow{/tex} x = -2 and y = 5\tHence, the required solution of the given pair of linear equations is x = -2, y = 5.\tVerification : substituting x = -2, y = 5, we find that both the equation (1) and (2) are satisfied as shown below:\t8x + 5y = 8(-2) + 5(5) = -16 + 25 = 9\t3x + 2y = 3(-2) + 2(5) =- 6 + 10 = 4\tHence, the solution is correct. | |
| 7025. |
Hormones |
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| 7026. |
1=11 2=223=334=445=556=6611=? |
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Answer» A1=11D=11N=11Then we have, =>an=a+(n-1)d a11=11+(11-1)11 a11=11+110 a11=121 Its 121? 121 I think 1111 ? |
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| 7027. |
Root 3 is irrational |
| Answer» Yes | |
| 7028. |
What is x plus 2 |
| Answer» X+2 | |
| 7029. |
How we convert class intervaL into excLusive form |
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| 7030. |
Formula of mean mod and median |
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| 7031. |
What is the square of 81 |
| Answer» 6561 | |
| 7032. |
Value of cos 200\' |
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| 7033. |
when any thing is perpendicular to the side then the side is biseactor . |
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| 7034. |
How to find squreroot |
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| 7035. |
1+tanA÷1+cotA |
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| 7036. |
Sin25cos65+cos25sin65 |
| Answer» Sin25cos(90-25)+ cos25sin (90-25)Sin25sin25+ cos25 cos25Sin²25+ cos25( sin²A+ cos²A=1)=1 | |
| 7037. |
CosA_sinA |
| Answer» Sin a | |
| 7038. |
3+3-6+6/2 =? |
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Answer» 3 3 3 |
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| 7039. |
Find the area by the using 1/2 a (-5,-3) b(-4,-6) c (2,-1) d (1,2) |
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| 7040. |
How to understand triangles chapter easily. |
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Answer» Mere too demag main hi nahi aate sir dukhne laagta hai? Sabse sahi advice me rd sharma ko padhne ki nahi kehta magar usme se tumhe saari theorem ker leni chahiye totle 32 hai ko bhi question hila sakte ho |
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| 7041. |
Pythagoras property |
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Answer» ? Square of hypotenuse = square of the other two sides of a triangle |
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| 7042. |
What is area theorem? |
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| 7043. |
sin 45.sin30+cos 45.cos30= |
| Answer» \xa0We know that Sin30°={tex}\\frac12{/tex}, Sin45°= {tex}\\frac{1}{\\sqrt2}{/tex} = Cos45°, Cos30°={tex}\\frac{\\sqrt{3}}{2}{/tex} ,putting these values in given expression, we get :-{tex}\\sin 45^\\circ \\sin 30^\\circ + \\cos 45^\\circ \\cos 30^\\circ {/tex}{tex} = \\frac{1}{{\\sqrt 2 }} \\times \\frac{1}{2} + \\frac{1}{{\\sqrt 2 }} \\times \\frac{{\\sqrt 3 }}{2}{/tex}{tex} = \\frac{1}{{2\\sqrt 2 }} + \\frac{{\\sqrt 3 }}{{2\\sqrt 2 }}{/tex}{tex} = \\frac{{1 + \\sqrt 3 }}{{2\\sqrt 2 }}{/tex} | |
| 7044. |
What are the perimeter of oval |
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| 7045. |
In a triangle ABC right angled at b ab=3cm Determine |
| Answer» Steps of construction:\tDraw a line segment BC = 4 cm.\tAt B, construct MBC = 90°.\tCut-off BA = 3 cm from BM.\tJoin AC.\tThus, right-angled {tex}\\triangle{/tex}ABC is obtained.\tBelow BC, make an acute CBX.\tAlong BX, mark off 7 points R1, R2, R3, R4, R5, R6, R7\tsuch that BR1 = R1R2 = R2R3 = R3R4 = R4R5 = R5R6 = R6R7\tJoin R5C.\tFrom R7, draw R7C1\xa0{tex}\\|{/tex} R5C, meeting BC produced at C1.\tFrom C1, draw C1A1\xa0{tex}\\|{/tex} CA, meeting BA produced at A1.\tThen, {tex}\\triangle{/tex}A1BC1 is the required triangle similar to {tex}\\triangle{/tex}A1BC1 such that each side of ABC is\xa0{tex}\\frac{7}{5}{/tex} times\tthe corresponding side of {tex}\\triangle{/tex}ABC. | |
| 7046. |
Sum of 2 no. Is 15.sum of their reciprocal is 3/10 find the no. |
| Answer» Let the numbers be x and 15 - x.According to question,{tex}\\frac { 1 } { x } + \\frac { 1 } { 15 - x } = \\frac { 3 } { 10 }{/tex}{tex}\\Rightarrow \\quad \\frac { 15 - x + x } { x ( 15 - x ) } = \\frac { 3 } { 10 }{/tex}{tex}\\Rightarrow \\quad \\frac { 15 } { x ( 15 - x ) } = \\frac { 3 } { 10 }{/tex}Cross multiply,{tex}\\Rightarrow{/tex}{tex}150=3x(15-x){/tex}{tex}\\Rightarrow{/tex}{tex}150=45x-3x^2{/tex}{tex}\\Rightarrow 150=3(15x-x^2){/tex}{tex}\\Rightarrow{/tex} x2 - 15x + 50 = 0{tex}\\Rightarrow{/tex} x2 - 10x - 5x + 50 = 0{tex}\\Rightarrow{/tex} x(x - 10) - 5(x -10) = 0{tex}\\Rightarrow{/tex} (x - 10)(x - 5) = 0{tex}\\Rightarrow{/tex} x - 10 = 0 or, x - 5 = 0 {tex}\\Rightarrow{/tex} x = 10 or, x = 5Therefore, numbers are 10 and 5 | |
| 7047. |
How can we draw a more than ogive |
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| 7048. |
Cot-A+tan A ,then what is the value of sinA |
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| 7049. |
NÇERT or R.DSharma which book should I practice |
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Answer» First complete ncert and then practise RD. It would be better to practise both RD sharma Rd Sharma ncert R.D sharma is best... If you wants good marks in exam then you should read through NCERT and if you want practice or knowledge in mathematics then you should read through R.D. SHARMA. and I also suggest you to read R.D. Sharma Rd sharma |
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| 7050. |
Tomorrow will be my exam so plzz tell me that by which book I should practice NÇERT or R.D sharma |
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Answer» Kisi se mat karo viswaas rakho apne mehnat per 6 month ki mehnat rang layegi 1din ki nahi unsolved ya solved paper dekho RD Now you prefer NCERT..... |
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