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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.
| 10551. |
Show that for odd positive integer to be a perfect square it should be in the form of 8a+1 |
| Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q. | |
| 10552. |
formula |
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| 10553. |
If the difference between the roots of the equation x^2 + px + 8 = 0 is 2, then p = ? |
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| 10554. |
if angle of elevation of sun is 60 degree ..find height of tower which casts a shadow of 30 meter |
| Answer» let AB = tower htAC = shadow = 30 mLC = 60tan c = AB / ACtan\xa060 = AB / 30root 3 = AB / 30AB = tower height = 30 root 3 = 30{tex} \\sqrt{3}{/tex} m | |
| 10555. |
2π5 |
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| 10556. |
Find the values of k for quadratic equation 2x2 – x + k = 0, so that they have two equal roots. |
| Answer» a=2, b=-1. C= k b2 -4ac=0. (since the real roots are equal ) (-1)^2-4(2)(K) =01-8k=0-8k=-1K=-1/-8K=1/8 | |
| 10557. |
x+y=14; x-y=4 |
| Answer» x is equal to 9 y is equal to 5 | |
| 10558. |
The larger of two supplementary angles exceeds the smaller by 18degree ,find them |
| Answer» let one angle=xother angle=180-x (as they are supplementary, their sum = 180)as per question (bigger angle - smaller angle = 18)180-x-x=182x=162x=81other angle=180-81=99Hence supplementary angles are 81 and 99 | |
| 10559. |
(cosecA-SinA)(SecA-cos |
| Answer» 1/sinA-sinA×1/cosA-cosA=1-sin^A/sinA×1-cos^A/cosA=cos^A/sinA×sin^A/cosA=cosAsinA | |
| 10560. |
How many real route can a quadratic polynomial can have |
| Answer» 2 | |
| 10561. |
cosA÷1+ sinA + 1+sinA÷cosA=2secA |
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| 10562. |
How to learn trigonometric ratios |
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Answer» Think creatively learn them by writing Just do practice as much as possible. Avoid tricks. |
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| 10563. |
If cos 0° =sin0°÷tan0°Then. 0÷0 should be 1.But 0÷0 is an indeterminate term? Why |
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Answer» Cos 0°= 1=sin0°÷tan0°=0/undefine Here,In place of undefine. 1,2,3,4..... till undefined can also come Becauz 0 can\'t be denominator |
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| 10564. |
Find the value of k for which system of eq. X-2y=3 and3x+ky=1 has a unique solution |
| Answer» The given system of equations isx - 2y - 3 = 0 .................... (i)3x + ky - 1 = 0 ................ (ii)These equations are of the forma1x + b1y+ c1\xa0= 0 ........ (iii)a2x + b2y +c2= 0 ........ (iv)Compare (i) and (iii) and (ii) and (iv), we geta1 = 1, b1= -2, c1= -3\xa0and a2\xa0= 3, b2\xa0= k, c2= -1For a unique solution, we must have\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}{tex}\\therefore \\quad \\frac { 1 } { 3 } \\neq \\frac { - 2 } { k } \\Rightarrow k \\neq - 6{/tex}Hence, the given system of equations will have a unique solution for all real values of k, other than - 6. | |
| 10565. |
When we can say that two points are collinear |
| Answer» When they fall in same line | |
| 10566. |
How many 4 digit no divisible by 10 and natural number |
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| 10567. |
If 2k+1,6,3k+1 are in A.P. then find the value of kk. |
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Answer» Find the value of common difference between6-(2k+1)=3k+1-66-1-2k=-5+3k5-2k=-5+3k-2k-3k=-5-5-5k=-25k=-25/-5k=5Hence value of k is 5. 5/2=K |
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| 10568. |
Tan+sin/tan-sin =Sec+1/sec-1 prove |
| Answer» {tex}\\text { L.H.S. } = \\frac { \\tan \\theta + \\sin \\theta } { \\tan \\theta - \\sin \\theta } = \\frac { \\frac { \\sin \\theta } { \\cos \\theta } + \\sin \\theta } { \\frac { \\sin \\theta } { \\cos \\theta } - \\sin \\theta }{/tex}{tex}= \\frac { \\sin \\theta \\left( \\frac { 1 } { \\cos \\theta } + 1 \\right) } { \\sin \\theta \\left( \\frac { 1 } { \\cos \\theta } - 1 \\right) } = \\frac { \\sec \\theta + 1 } { \\sec \\theta - 1 } = R . H S{/tex} | |
| 10569. |
Ex5.1 |
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| 10570. |
If i study this app i will pass in class 10 woth good number |
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Answer» Of cause you will get good marks Yess |
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| 10571. |
The smallest composite number is? |
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Answer» What a stupid question Right answer will 4 2 |
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| 10572. |
2x/x-3+1/2x+3+3x+9(x-3)2x+3 |
| Answer» Given, {tex}\\frac{2x}{x-3}+\\frac{1}{2x+3}+\\frac{3x+9}{(x-3)(2x+3)}=0{/tex}{tex}\\Rightarrow\\frac{{2x(2x + 3) + x - 3 + 3x + 9}}{{(x - 3)(2x + 3)}} = 0{/tex}{tex} \\Rightarrow \\frac{{4{x^2} + 6x + x - 3 + 3x + 9}}{{2{x^2} + 3x - 6x - 9}} = 0{/tex}{tex} \\Rightarrow \\frac{4{x^2} + 10x +6 }{{2{x^2} - 3x - 9}} = 0{/tex}Cross multiplying equation , we get ,{tex} \\Rightarrow 4{x^2} + 10x +6 = 0{/tex}{tex} \\Rightarrow 4{x^2} + 6x+4x +6 = 0{/tex}{tex} \\Rightarrow x(4{x} + 6)+1(4x +6) = 0{/tex}{tex}\\Rightarrow(x+1)=0{/tex} or {tex} (4x+6)=0{/tex}{tex}\\Rightarrow x=-1{/tex}{tex}\\Rightarrow x=-\\frac{3}{2}{/tex}Hence , the roots of the given quadratic equation are -1 and {tex}-\\frac{3}{2}{/tex} | |
| 10573. |
How to find the completing sqare |
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| 10574. |
Find 10th and 18th term of AP 1,4,7,10,.......... |
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Answer» a=1d=3a10=a+9d=1+27=28a18=a+17d=1+51=52 10th term is 28 and 18th term is 54 |
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| 10575. |
3+4 |
| Answer» 7 | |
| 10576. |
Use euclid division lemma to show that the cube of any positive integer is of the form 9m, 9m+1,9m+8 |
| Answer» Let a = 3q + r\xa0{tex}: 0 \\leq r < 3{/tex}\xa0{tex}\\therefore \\quad a = 3 q ; \\text { then } a ^ { 3 } = 27 q ^ { 3 } = 9 m ; \\text { where } m = 3 q ^ { 3 }{/tex}{tex}\\text { when } a = 3 q + 1 ; \\text { then } a = 27 q ^ { 2 } + 27 q ^ { 2 } + 9 q + 1{/tex}{tex}= 9 \\left( 3 q ^ { 3 } + 3 q ^ { 2 } + q \\right) + 1{/tex}{tex}= 9 m + 8 \\quad \\left( \\text { where } m = 3 q ^ { 3 } + 3 q ^ { 2 } + q \\right){/tex}{tex}\\text { when } a = 3 q + 2 ; \\text { then } a ^ { 3 } = ( 3 q + 2 ) ^ { 2 }{/tex}{tex}= 27 q ^ { 3 } + 54 q ^ { 2 } + 36 q + 8{/tex}{tex}= 9 m + 8 \\quad \\left( \\text { where } m = 3 q ^ { 3 } + 6 q ^ { 2 } + 4 q \\right){/tex}{tex}= 9 m + 8 \\quad \\left( \\text { where } m = 3 q ^ { 3 } + 6 q ^ { 2 } + 4 q \\right){/tex}Hence, cubes of any positive integer is\xa0either of the from 9m, (9m + 1) or (9m + 8). | |
| 10577. |
Kx square + root 2 x - 4=0X=✓2 |
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| 10578. |
(a+b+c)3 |
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Answer» a3+b3+c3+3a2b+3a2c+3b2c+3b2a+3c2a+3c2a+6abc jab yee bhi nahi malum to nothing you understand |
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| 10579. |
a formula of cube |
| Answer» (a+b)3=a3+b3+3a2b+3ab2 | |
| 10580. |
2x+4y=5 |
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| 10581. |
Exwmple 2 in circles |
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| 10582. |
Find area of the triangle whose vertices are (2,4),(_1,0)(2,_4) |
| Answer» {tex}\\begin{array}{l}\\left(2,3\\right),\\;\\left(-1,0\\right),\\;\\left(2,-4\\right)\\\\Area\\;of\\;Triangle\\;=\\;\\frac12\\left[2\\;\\left(0+4\\right)\\;+\\left(-1\\right)\\left(\\;-4-3\\right)+2\\left(3-0\\right)\\right]\\end{array}{/tex}{tex}( 2,3 ) , ( - 1,0 ) , ( 2 , - 4 ){/tex}Area of Triangle\xa0{tex}= \\frac { 1 } { 2 } [ 2 ( 0 + 4 ) + ( - 1 ) ( - 4 - 3 ) + 2 ( 3 - 0 ) ]{/tex}{tex}= \\frac{1}{2}(8 + 7 + 6){/tex}{tex}= \\frac{{21}}{2}{/tex} square units. | |
| 10583. |
3/3*0+1+5+6/2+52-5/2-10/2-2=? |
| Answer» 3 | |
| 10584. |
Divide 57 into two parts whose product is 680. |
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Answer» x×(x-57)=680 17 and 40 |
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| 10585. |
in quadratic equations |
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| 10586. |
in quadratic chapter |
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| 10587. |
If α and βare the root of 4square -5x+1 then find out the value αsquareβ and βsquare α |
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| 10588. |
Prove that:SinA (1+ tanA) + cosA (1+ cotA ) = secA + cosecA |
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| 10589. |
When will our board exams start |
| Answer» In Feb | |
| 10590. |
3cosA=1 find cosecA =? |
| Answer» 2under root 2. | |
| 10591. |
Hcf of 129and4 |
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Answer» 1 1 |
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| 10592. |
If cosec a = 2 then find all the trigonometry ratios |
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| 10593. |
Why We use k if value of sin is given and we have to find other trigonometry ratios |
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| 10594. |
In triangle abc ac=12cm bc=6cm and ab=6root 3 find the measure of angle b |
| Answer» Triangle ABC Assuming ABC is a right angled triangleBy Pythagoras theorem we getAC*2=BC*2+AB*2 12*2=6*2+6√3*2144=36+(36)3144=36+108144=144Hence it satisfies Pythagoras theoremSo our assumption is right ∆Also angle b=90° | |
| 10595. |
For what value of p are 2p+1,13,5p-3 three consecutive terms of an ap |
| Answer» Given 2p+1, 13 , 5p-3 are in AP.To find: value of PIf a,b,c are in AP then we know that\xa02b=( a+c)Here a= 2p+1b=13c=5p-32 (13) =(2p+1+5p-3)26=7p-226+2=7p28=7pp=28/7 =4Therefore the value ofP is 4. | |
| 10596. |
What is the value cos90 |
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Answer» 0? 0 0 |
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| 10597. |
What type of book we have to learn that all questions are samexpected In board |
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| 10598. |
1+cot^2/1+cosec=1/sin |
| Answer» LHS= \u200b{tex}\\frac{1}{\\cot ^{2} \\theta}+\\frac{1}{1+\\tan ^{2} \\theta}{/tex}= tan2\xa0{tex}\\theta{/tex}\xa0+\xa0{tex}\\frac{1}{\\sec ^{2} \\theta}{/tex}= tan2\xa0{tex}\\theta{/tex}\xa0+ cos2\xa0{tex}\\theta{/tex}= (sec2{tex}\\theta{/tex}\xa0- 1) +cos2{tex}\\theta{/tex}= sec2{tex}\\theta{/tex}\xa0- ( 1 - cos2{tex}\\theta{/tex}\xa0)= sec2{tex}\\theta{/tex}\xa0- sin2{tex}\\theta{/tex}{tex}=\\frac{1}{\\cos ^{2} \\theta}{/tex}\xa0- sin2{tex}\\theta{/tex}{tex}= \\frac{1}{1-\\sin ^{2} \\theta}-\\frac{1}{\\ cosec ^{2} \\theta}{/tex}\u200b\u200b\u200b\u200b\u200b= R.H.S\xa0Hence proved. | |
| 10599. |
Vshdnyf |
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| 10600. |
3√5 is irrational |
| Answer» Let,three root five is rational number.We can find co-prime integer p and q (not equal to zero).Three root five = p/qRoot five =p/seven qSince, p and q are integer , p/seven q is rational,and root five is rational.But this contradicts the fact that root five is an irrational number.So,we conclude that seven root five is an irrational number. | |