InterviewSolution
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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.
| 36501. |
If cosecA+cotA=mAnd cosecA-cotA=nThen prove that mn=1. |
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Answer» mn=(cosecA+cotA)(cosecA-cotA)=>cosec²A-cot²A=>1 Someone give me answer |
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| 36502. |
xsin³A+ycos³A=sinA.cosA and xsinA=ycosAProve that x²+y²=1 |
| Answer» We have,xsin3{tex}\\theta{/tex}\xa0+ ycos3{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(xsin{tex}\\theta{/tex}) sin2{tex}\\theta{/tex}\xa0+ (ycos{tex}\\theta{/tex}) cos2{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0(sin2{tex}\\theta{/tex}) + (x sin{tex}\\theta{/tex}) cos2{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa0x sin{tex}\\theta{/tex}\xa0= y cos{tex}\\theta{/tex}]{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0(sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) = sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x = cos{tex}\\theta{/tex}Now, xsin {tex}\\theta{/tex}\xa0= ycos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0= y cos{tex}\\theta{/tex}\xa0[{tex}\\because{/tex} x = cos{tex}\\theta{/tex}]{tex}\\Rightarrow{/tex}\xa0y = sin{tex}\\theta{/tex}Hence, x2 + y2 = cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0= 1 | |
| 36503. |
Prove that (tanA+2)(2tanA+1)=5tanA+2sec^2A |
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| 36504. |
a^3-b^3 = |
| Answer» (a-b)(a²+ab+b²) | |
| 36505. |
5(25)+5-6=6 |
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| 36506. |
Is it compulsory that pouch should be transparent |
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Answer» Yes , it is written in admit card..If you don\'t have then buy it Yes it is wrritten on admit card Yes it is compulsory plus transparent bottle also No |
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| 36507. |
Is clipboard and pouch are allowed in board? |
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Answer» Ya both are allowed but they both should be transparent Re nawab mera naam bhi Kabir hai..... Kabir Baisla (Bansal) Yes but both should be transparent. No |
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| 36508. |
if sin thrita=a/b,then find cos thrita |
| Answer» Cos tita is b-a/b | |
| 36509. |
If sec thita -tan thita =xShow that sec thita +tan thita =1/x |
| Answer» Multiply (sec theta - tan theta) with (sec theta + tan theta)/(sec theta + tan theta). This will give you x=reciprocal of(sec theta + tan theta). Thus 1/x= sec theta + tan theta | |
| 36510. |
If the difference between two numbers is 26, and one number is three times the other. Find them |
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Answer» 13 and 39 13 and 39 13 and 39..... |
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| 36511. |
How many terms of the AP: 9,17,25,.... must be taken to give a sum of 636 |
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Answer» First term of A.P. = 9 , common difference = 8if sum of n terms 636, Sn = (n/2) [ 2×9 + (n-1)8 ] = 636Sn = n [ 10+ 8n] = 12728n2 +10n -1272 = 04n2 +5n -636 = 0By factorising LHS,(4n+53)(n-12) = 0hence n = 12 12 12 12 terms should be taken 12 |
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| 36512. |
Prove that sin/cos=tan |
| Answer» Sin=opp/hypCos=adj/hypSin/cos= (opp/hyp)/adj/hyp=Opp/adj=tanHence proved. | |
| 36513. |
How to find k term |
| Answer» By hands | |
| 36514. |
Which term of the AP 3,15,27,39.....will be 132 more than its 54th term? |
| Answer» a = 3 and d = 15 - 3 = 12Let the required term be nth term54th term = a +(n-1)d\xa0= 3 + 53x12 = 3 + 636 = 639therefore 132 + 639 = 771 will be the nth term.771 = 3 + (n-1)12768/12 +1 = ntherefore n = 64 +1 = 65therefore the 65th term will be 132 more than the 54th term.\xa0 | |
| 36515. |
https://drive.google.com/file/d/1utbNM3mqBwQkJVuHTFcF4a9At5jP4o6V/view?usp=drivesdk |
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| 36516. |
2x+6xy-14=0 |
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Answer» #jogita bss itna sa hi nikalna ha? 2x + 6xy - 14 = 02 ( x + 3xy - 7) = 0x + 3xy - 7 = 0x ( 1+ 3y) = 7 |
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| 36517. |
What is the equation for a cubic polynomial. ???? |
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Answer» K[x³-(a+b+c)x²+(ab+bc+ac)x-(abc)] I mean to find a polynomia ax^3+bx^2+cx+d=0 |
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| 36518. |
X=1/x+a+b=1/x+1/a+1/b |
| Answer» Shift 1/x to lhs nd solve both sides | |
| 36519. |
Show that exactly one of the polynomials n , n + 2 or n + 4 is divisible by 3. |
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Answer» https://Brainly.in We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0 |
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| 36520. |
What is meant by frustum |
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Answer» We will take a right circular cone and remove a portion of it. There are so many ways in which we can do this. But one particular case that we are interested in is the removal of a smaller right circular cone by cutting the given cone by a plane parallel to its base. You must have observed that the glasses, used for drinking water, are of this shape. This remaining portion of cone is called Frustum of a cone. A cone which is cut through parallel line THE REMAINING OBJECT IS FRUSTUM |
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| 36521. |
Surface areas nd volumes class 10 cbse frustum |
| Answer» Volume 1/ 3 Pi H (R1 square + R2 square + R1 R2) | |
| 36522. |
Can two numbers have 18 as their hcf and 380 as their lcm? |
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Answer» Wlcm Thnx for ur support venkateshwar.. No it is not possible No, this case is not possible because HCF of the number must be a factor of LCM but here, 18 is not a factor of 380... No fhis not possible |
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| 36523. |
Find the value for k. Kx^2-14x+8=0is2Or next question is x^2+5kx+16=0are equal and real roots |
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Answer» Both questions are different ? But i think first question discriminant pe nhi krna h na For first eqation K = 49/8 and for second eqation k=8/5 First answer, the value of k = 5 second answer, k= 8/5 |
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| 36524. |
If the point p(x,y) is equidistant from the points Q(a+b,b-a) and R(a-b,a+b), then prove that bx=ay. |
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Answer» Use identity (a+b+c)^2 Break it into distance formula |
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| 36525. |
Find 20 to term of an AP is 3,8,13,.......,253 tell me how to do |
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Answer» Put A=3,and d= 5 in this formula An=A+(n-1)d , 20th term is 98 98 |
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| 36526. |
Sintheta+ costheta |
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| 36527. |
Find the distance of a point p(x.y)from the origin |
| Answer» √x^2 + y^2 | |
| 36528. |
Find the sum of all two digit odd positive numbers |
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Answer» 2475 2475 |
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| 36529. |
Given ^ ABC,~^PQR. If Ab=1then find arΔ ABC Δpqr Pq. 3 |
| Answer» Ans 1/9 by using theorem ratio of areas of similar traingles is equal to ratio of square of corressponding sides :) | |
| 36530. |
solve for x and y : px + qy = p-q and qx - py = p + q.......☺☺ |
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Answer» Multiply (i) by p and (ii) by q and add them and solve by elimination method ..... Then tha value of x= 1 and y=-1 X=1 and y=-1 x=1 nd y=-1 |
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| 36531. |
In the given figure angle 1 equal to angle 2 |
| Answer» Fig kit se | |
| 36532. |
In the figure, DE || BC, DE=3cm, BC=9cm and ar(∆ADE)=30cm square . Find ar(trap.BCED). |
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Answer» Ans=9/72 Where is the fig |
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| 36533. |
Realy i am also getting tence |
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Answer» ? Anushka is right |
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| 36534. |
1/x+3-1/x-8=11/x ;x is not equal to -3,8 |
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Answer» Is this answer right? Just take lcm amd do calculations Its very difficult to do so 2+-2√7 |
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| 36535. |
ABCDE is a polygon whose vertices are a - 10 B 400, 1407 and - 6, to find the area of polygon |
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| 36536. |
AB, BC and BD are tangents. If AB is 5cm. Find BD |
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Answer» Incomplete question Because it is a theorem that. A tangents drawn from an same external point of a circle are equal It means ans is 5 cm No... Three tangents from one point is possible? Muskan |
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| 36537. |
9ab =7ab |
| Answer» It is not possible | |
| 36538. |
sin(45+theta)-cos(45-theta) . Find its value |
| Answer» {tex}\\begin{array}{l}\\sin(45+\\mathrm\\theta)-\\cos(45-\\mathrm\\theta)\\\\=\\sin(45+\\mathrm\\theta)-\\cos\\lbrack90-(45+\\mathrm\\theta)\\rbrack\\\\=\\sin(45+\\mathrm\\theta)-\\sin(45+\\mathrm\\theta)\\\\=0\\end{array}{/tex} | |
| 36539. |
(a-b)\' |
| Answer» Complete ur que. | |
| 36540. |
Page. number 283, example 7 please tell me how to take median class?? |
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Answer» Median class is take of the class which has highest frequency in the table like in question 1 of exercise 14.3 median class is 125 to 145 as its frequency is 20 which highest amount them Median class Below 140 , 140-145 , 145-150, 150-155 , 155-160 , 160-165 Hmm |
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| 36541. |
Prove that root 2 is an irrational |
| Answer» Firstly , let assume that root 2 is rational then,Root 2=p/q , now square both sideRoot 2square=p square/q square2=p square/q square =Q square =p square (This is first equation.)Now, p is divisible by (2)Let p =2k. (Here we use any word like i use k i also use x,y,z ). (This is second equation,).From (1 )&(2)Q square=(2k) square/2Q sq=4k square ( only this square is put on the k value )Q sq /2=k sqP/q is rational The common factor of p&q should be 1 Let we proved the common factor of p&q is( 2 ) this contradction occur due to our wrong assumption so,Root 2 is irrational , thanks for ask this question | |
| 36542. |
Which term of AP 15 , 27 , 39 .....Will be 132 more than its 54th term?? |
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Answer» N =65 65th term |
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| 36543. |
if N th term of an AP is (2n +1 ), what is the sum of first three term |
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Answer» Ans. I also knows but ? Is steps of this solution 15 |
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| 36544. |
Sum of first p, q, r terms of an A.P are a, b, c. Prove that a/p (q-r) + b/q (r-p) +c/r (p-q)=0 |
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Answer» Bhai answer pucha h suggestion nhi R.d sharma uthao |
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| 36545. |
(1+cotA+tanA)(sinA-cosA)=? |
| Answer» (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA........(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} - \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} - \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} - \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} - \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A - \\cos A \\times \\cot A{/tex}.......(2)From (1) & (2),(1 + cotA + tanA) (sinA - cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } - \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA - cosA.cotA\xa0 | |
| 36546. |
Yooo i am back....mera pehla account ban ho gya to alag wala hai i am back |
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Answer» Ooooo, cool Thnx Congrats.. |
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| 36547. |
If hcf of 65 and 117 is expressible in the form 65m-117 then find the value of m |
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Answer» Ji We have 117=65×1+52 65=52×1+13And 52=13×4+0Hence, H.C.F=13Therefore, 65m-117=13 m=130/65 m=2Now 65=13×5 117=3×3×13L.C.M=13×5×3×3=585 Do not take tension |
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| 36548. |
How much questions that cbse will ask from ncert? |
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Answer» If you preprare ncert than you score 90+ How much i don\'t know but 35 to 40 marks comes from it Only 60% |
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| 36549. |
Sidharachahua formula |
| Answer» (-b+-√b^2-4ac)/2a | |
| 36550. |
Fundamental theorem of arthmetic |
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Answer» It says that every composite number cn be expressed as product of primes and this factorisation is unique. Fundamental theorem of arthmetic states that x=p/q be a rational no.such that prime factorisation of q is in the form 2n5m, where n and m are non negative integers.Then x has a decimal expansion which terminates. |
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