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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.
| 36301. |
Sec=17/8 verify that (3-4sin²)/(4cos²-3)=(3-tan²)/(1-3tan²) |
| Answer» Given\xa0{tex}\\sec A = \\frac{{17}}{8} = \\frac{{AC}}{{AB}}{/tex}Let AC = 17Kand, AB = 8KIn\xa0{tex}\\Delta ABC{/tex}, by Pythagoras theoremBC2 + AB2 = AC2BC2 + (8K)2 = (17K)2BC2 + 64K2 = 289K2BC2 = 289K2 - 64K2BC2 = 225K2{tex}BC = \\sqrt {225{K^2}} = 15K{/tex}{tex}\\therefore \\sin A = \\frac{{BC}}{{AC}} = \\frac{{15K}}{{17K}} = \\frac{{15}}{{17}}{/tex}{tex}\\cos A = \\frac{{AB}}{{AC}} = \\frac{{8K}}{{17K}} = \\frac{8}{{17}}{/tex}{tex}\\tan A = \\frac{{BC}}{{AB}} = \\frac{{15K}}{{8K}} = \\frac{{15}}{8}{/tex}LHS{tex} = \\frac{{3 - 4{{\\sin }^2}A}}{{4{{\\cos }^2}A - 3}}{/tex}{tex} = \\frac{{3 - 4 \\times {{\\left( {\\frac{{15}}{{17}}} \\right)}^2}}}{{4{{\\left( {\\frac{8}{{17}}} \\right)}^2} - 3}}{/tex}{tex} = \\frac{{3 - \\frac{{900}}{{289}}}}{{\\frac{{256}}{{289}} - 3}}{/tex}{tex} = \\frac{{\\frac{{867 - 900}}{{289}}}}{{\\frac{{256 - 867}}{{289}}}}{/tex}{tex} = \\frac{{\\frac{{ - 33}}{{289}}}}{{\\frac{{ - 611}}{{289}}}}{/tex}{tex} = \\frac{{ - 33}}{{289}} \\times \\frac{{289}}{{ - 611}}{/tex}{tex} = \\frac{{33}}{{611}}{/tex}RHS{tex} = \\frac{{3 - {{\\tan }^2}A}}{{1 - 3{{\\tan }^2}A}}{/tex}{tex} = \\frac{{3 - {{\\left( {\\frac{{15}}{8}} \\right)}^2}}}{{1 - 3 \\times {{\\left( {\\frac{{15}}{8}} \\right)}^2}}}{/tex}{tex} = \\frac{{3 - \\frac{{225}}{{64}}}}{{1 - \\frac{{675}}{{64}}}}{/tex}{tex} = \\frac{{\\frac{{192 - 225}}{{64}}}}{{\\frac{{64 - 675}}{{64}}}}{/tex}{tex} = \\frac{{\\frac{{ - 33}}{{64}}}}{{\\frac{{ - 611}}{{64}}}}{/tex}{tex}= \\frac{{ - 33}}{{64}} \\times \\frac{{64}}{{ - 611}}{/tex}{tex} = \\frac{{33}}{{611}}{/tex}Hence verified | |
| 36302. |
What\'s the best method to solve quadratic equation using completeing square method |
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Answer» Splitting middle term Formula method Division method is is best than multiplication method in completing square method Algorithm ? By applying the algorithm |
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| 36303. |
An ap consists of 50 terms of which 3rd term is 12 and the last term is 106 find the 29th term |
| Answer» a3 = 12 expand ita + 2d = 12n = 50a50 = a + 49d = 106Eliminate the equation u will get the value of d as 2 and a = 8Now, a29 = a + 28d = 8 + 28× 2 =8+56 = 64 | |
| 36304. |
Find the largest number which divides 320and457 leaving remainder 5 and 7 respectively |
| Answer» The given numbers are 320 and 457Now as 5 and 7 are remainders on division of 320 and 457 by said numberOn subtracting the reminders 5 and 7 from 320 and 457 respectively we get:320 - 5 = 315,457 - 7 = 450The prime factorization:of 315 and 405 are{tex}315 = 3 \\times 3 \\times 5 \\times 7{/tex}{tex}= 3 ^ { 2 } \\times 5 \\times 7{/tex}{tex}450 = 2 \\times 3 \\times 3 \\times 5 \\times 5{/tex}{tex}= 2 \\times 3 ^ { 2 } \\times 5 ^ { 2 }{/tex}{tex}\\therefore{/tex}\xa0H.C.F. of 315 and 450 =\xa0{tex}3 ^ { 2 } \\times 5 = 9 \\times 5 = 45{/tex}Hence the said\xa0number =45 | |
| 36305. |
How to find the missing term in a ap |
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| 36306. |
Cos cube theta + 3cos square sin squar +asin cube +3cose square sin theta |
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| 36307. |
Find all the zeros of the polynomial 2x^4-9x^3+5x^2+3x-1, if two of its zeros are 2+√3, 2-√3. |
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| 36308. |
The distance of the (4,-7 ) from y- axis |
| Answer» 7 units | |
| 36309. |
If one zero of polynomial (a²+9)x²+13x+6a is reciprocal of the other,find the value of a ? |
| Answer» Let {tex} \\alpha{/tex}\xa0and\xa0{tex} \\frac { 1 } { \\alpha }{/tex} be the zeros of\xa0(a2\xa0+ 9)x2\xa0+ 13x\xa0+ 6a.Then, we have{tex} \\alpha \\times \\frac { 1 } { \\alpha } = \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa01 =\xa0{tex} \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa0a2\xa0+ 9 = 6a⇒ a2 - 6a + 9 = 0⇒\xa0a2\xa0- 3a - 3a + 9 = 0⇒\xa0a(a - 3) - 3(a - 3) = 0⇒\xa0(a - 3) (a - 3) = 0⇒\xa0(a - 3)2\xa0= 0⇒\xa0a - 3 = 0⇒ a = 3So, the value of a in given polynomial is 3. | |
| 36310. |
What is Euclid \'s lemma |
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Answer» a>b a=b×q+r 0= If a and b are two numbers and a is greater then b |
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| 36311. |
If we say x3 is a cubic polynomial then what we say x4 |
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Answer» -2 ,-2 Quadetic polynomial |
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| 36312. |
-2×-2 |
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Answer» 4 4 Its 4 4 4 |
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| 36313. |
What is 0^0=? |
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| 36314. |
2+4+8+&5= |
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Answer» 19 & 19+& 19 19& 19 |
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| 36315. |
Diagonal ofsquare |
| Answer» Diagonal =\xa0{tex}\\sqrt 2 \\times{/tex}side | |
| 36316. |
Types of fractions |
| Answer» Mixed, like, unlike, proper fraction. | |
| 36317. |
Prove APOLLIONUS Theorem |
| Answer» b^2+c^2=2(m^2+d^2) | |
| 36318. |
√1-sinA/√1+sinA=cosA/1+sinA |
| Answer» 1-sinA. 1-sinA/1+sinA. 1-sinA = RESIPROCAL1-SIN2A/1-SIN2A=REMOVE SQURE ROOT 1+SINA/1-SINA=1+sina/ cosa | |
| 36319. |
Polynomals 2,3 question no 5 |
| Answer» 2.3 exercise or what | |
| 36320. |
Find k if (k, 2k),(3kk, 3k), (3,1) are collinear |
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Answer» sry sry k = - 3/4 k = -5/4 |
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| 36321. |
If a no. when divided by 143 leaves remainder 31 what wil be the remainder if same is divided by 13 |
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| 36322. |
In a right angle triangle ABC,AD is perpendicular to BC,prove that AD^2=3BD^2 |
| Answer» According to the question,\xa0{tex}\\triangle ABC {/tex}\xa0is an equilateral triangle.In {tex}\\triangle{/tex}ABD, using Pythagoras theorem,{tex}\\Rightarrow{/tex} AB2 = AD2 + BD2{tex}\\Rightarrow{/tex} BC2 = AD2 + BD2, (as AB = BC = CA){tex}\\Rightarrow{/tex} (2 BD)2 = AD2 + BD2, (perpendicular is the median in an equilateral triangle){tex}\\Rightarrow{/tex} 4BD2 - BD2 = AD2{tex}\\therefore{/tex}\xa03BD2 = AD2 | |
| 36323. |
Is optional exercise of class 10th is must for point of view of bord or not? |
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Answer» They are as same as other questions in chapters so it is not important to solve them.? But it is given in book that part is not for examination point of view Yes they are likely to come in board exams |
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| 36324. |
If p(x,y) is a point equidistance from the points A(6,-1)and B(2,3)shoe thatx-y=3 |
| Answer» Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point.Since point P is equidistant from points A and B,therefore,PA = PB\xa0{tex}\\Rightarrow{/tex}\xa0(PA)2\xa0= (PB)2{tex}\\Rightarrow{/tex}(x - 6)2\xa0+ (y + 1)2\xa0= (2 - x)2\xa0+ (3 - y)2{tex}\\Rightarrow{/tex}\xa036 + x2 - 12x + y2 + 1 +2y = 4 + x2 - 4x + 9 + y2 - 6y{tex}\\Rightarrow{/tex}\xa0-12x + 4x + 2y + 6y = 4 + 9 - 1 - 36{tex}\\Rightarrow{/tex}\xa0-8x + 8y = -24-8 (x - y)=-24{tex}\\Rightarrow{/tex}\xa0x - y = 3Hence, x - y = 3 | |
| 36325. |
How to create equation from fraction situation |
| Answer» By taking LCM | |
| 36326. |
If x+1 is a factor of 2x cube+ax square+2bx+1,find aand b ,if 2a-3b=4 |
| Answer» b=2&a=5 | |
| 36327. |
How to put value in graphical method |
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| 36328. |
What is consistent $ inconsistent |
| Answer» The lines which have solution like coincident and interesting lines are called consistent lines and the lines which have no solution like parallel lines are said to be inconsistent lines | |
| 36329. |
A×b= |
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Answer» ab Ab ab ab |
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| 36330. |
The median is 20.75 find x and y if th total frequency is 100 |
| Answer» Idiot | |
| 36331. |
Sin30 value |
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Answer» Sin30°=1/2 as u can see in the table 1/2..?????? Its 1/2 1/2 1/2 |
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| 36332. |
The square root of 4th term is the first term then find 4 terms of an AP |
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| 36333. |
Wgat is the formula of nth term |
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Answer» a+(n-1)d a + (n-1)×d A+(n -1)d A+(n-1)d |
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| 36334. |
Different between terminating and non terminating numbers |
| Answer» In terminating numbers the natural numberAre present before decimal point but in non terminating number the number are not present before decimal point | |
| 36335. |
2x/a+y/b=4 x/a-y/b=4 |
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| 36336. |
If cos theta+sin theta =1prove that cos theta _ sin theta +_ 1 |
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| 36337. |
1+cotA/secA =sin^2A/1-cosA |
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| 36338. |
2x^2-7x+3=0 |
| Answer» 2x^2–x–6x+3=0x(2x–1)–3(2x–1)=0(2x–1)(x–3)=0So:2x–1=0 OR. x–3=0Nowx=1/2 or x=3 | |
| 36339. |
Prove 7√11/3 is irrational. |
| Answer» Let 7√11/3 be rational 7√11/3= r 7√11=3r√11=3r/7. .......(1)Here is r is rational 3r is rational 3r/7 is rational Therefore √11 is rational, which is a contradiction .So our supposition was wrong7√11/7 is irrational. | |
| 36340. |
In an Ap, if the common difference (d)=-4,and the seventh term is 4,then find the first term. |
| Answer» T7= 4, d=-4Or, a+6(-4)= 4 a= 4+24= 28 | |
| 36341. |
Please give me important questions |
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| 36342. |
90 = v/540*360 |
| Answer» 17496000 | |
| 36343. |
How to solve problems in a short period of time in exams? |
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Answer» By understanding the question properly and the answering it so that after completing the paper we dont have to recheck itt By hard working |
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| 36344. |
Show that square of any positive integer cannot be of the from 5q plus2 or5q plus3 |
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| 36345. |
(cos0°+sin45°+sin30°)(sin90°-cos45°+cos60°) |
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Answer» What will be the answer Substitute the value and you will get your answer |
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| 36346. |
√3/√2(2+2√3) |
| Answer» Substitute the value and you will get your answer | |
| 36347. |
9x^2-3(a^2+b^2)x+a^2b^2=0 Using quadratic formula solve for x. |
| Answer» D = b2 - 4ac{tex}= ( - 9 ( a + b ) ) ^ { 2 } - 4 \\times 9 \\times \\left( 2 a ^ { 2 } + 5 a b + 2 a b ^ { 2 } \\right){/tex}= 81(a + b)2 - 36(2a2 + 5ab + 2b2)= 9[9(a2 + b2 + 2ab - 8a2 - 20ab - 8b2)]= 9[a2 + b2 - 2ab]= 9(a - b)2{tex}x = \\frac { - b \\pm \\sqrt { D } } { 2 a } = \\frac { 9 ( a + b ) \\pm \\sqrt { 9 ( a - b ) ^ { 2 } } } { 2 \\times 9 }{/tex}{tex}{ \\Rightarrow x = 3 \\frac { [ 3 ( a + b ) \\pm ( a - b ) ] } { 2 \\times 9 } }{/tex}{tex}{ \\Rightarrow x = \\frac { ( 3 a + 3 b ) \\pm ( a - b ) } { 6 } }{/tex}{tex}\\Rightarrow x = \\frac { 3 a + 3 b + a - b } { 6 } \\text { or } x = \\frac { 3 a + 3 b - a + b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 4 a + 2 b } { 6 } \\text { or } x = \\frac { 2 a + 4 b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 2 a + b } { 3 } \\text { or } x = \\frac { a + 2 b } { 3 }{/tex} | |
| 36348. |
By elimination method - x+1/2 + y-1/3=8 and x-1/3 +y+1/2=9 |
| Answer» The given equations may be written as{tex} \\frac { x + 1 } { 2 } + \\frac { y - 1 } { 3 } = 8 {/tex}⇒ 3(x + 1) + 2 (y - 1) = 48⇒ 3x + 3 + 2y - 2 = 48⇒ 3x + 2y + 1= 48⇒ 3x+2y = 47 ... (i){tex}\\frac { x - 1 } { 3 } + \\frac { y + 1 } { 2 } = 9{/tex}⇒ 2(x -1 ) + 3(y +1) = 54⇒ 2x - 2 + 3y + 3 = 54⇒ 2x + 3y + 1 = 54⇒ 2x + 3y = 53. ... (ii)Multiplying (i) by 2 and (ii) by 3 and subtracting, we get(4\xa0- 9)y = 94-159{tex} \\Rightarrow - 5 y = - 65 {/tex}{tex}\\Rightarrow y = \\frac { - 65 } { - 5 } {/tex}{tex}\\Rightarrow y = 13{/tex}Putting y = 13 in (i), we get3x\xa0+ (2 {tex} \\times{/tex}\xa013) = 47{tex} \\Rightarrow{/tex} 3x + 26 = 47{tex} \\Rightarrow{/tex} 3x = (47 - 26){tex} \\Rightarrow{/tex}3x = 21{tex} \\Rightarrow \\quad x = \\frac { 21 } { 3 } = 7{/tex}So, x = 7Hence, x = 7 and y = 13 | |
| 36349. |
Solve for x - 1/x+1 +2/x+2=4/x+y,x≠-1,-2,-4? |
| Answer» {tex}\\frac { x + 1 } { x - 1 } + \\frac { x - 2 } { x + 2 } = 4 - \\frac { 2 x + 3 } { x - 2 } {/tex}{tex}\\frac { x ^ { 2 } + 3 x + 2 + x ^ { 2 } - 3 x + 2 } { x ^ { 2 } + x - 2 }{/tex}\xa0=\xa0{tex}\\frac { 4 x - 8 - 2 x - 3 } { x - 2 }{/tex}(2x2\xa0+ 4) (x - 2) = (2x -11 ) (x2\xa0+ x - 2)or, 5x2\xa0+ 19x- 30 = 0or, (5x - 6) (4x - 5) = 0or, x = {tex}\\frac{5}{4}{/tex},\xa0{tex}\\frac 65{/tex} | |
| 36350. |
Q3 check 3 |
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