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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.
| 40201. |
Sin theta= Cos (theta-6) |
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Answer» Given: sintheta = cos(theta-6)Now sintheta = cos(90- theta)So, cos(90- theta) = cos(theta-6)So, 90 - theta = theta-6 90 + 6 = theta + theta96 = 2theta96/2 = theta48 = theta. Sin theta=sin 6Theta=6 |
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| 40202. |
x² - a² > 0 Then x < -a or x > a Explain why? |
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Answer» Then how x < -a ? Since x^2 - a^2 > 0So, x^2 > a^2 So, this implies x > a. |
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| 40203. |
What is the CSA of cylinder |
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Answer» 2πrh 2 pie r.h |
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| 40204. |
Cos(15)= |
| Answer» U can find it by using this formula: cos(A−B)=cos(A)cos(B)+sin(A)sin(B)[Divide cos15 as cos[45-30]] | |
| 40205. |
Find HCF of 144 &90 |
| Answer» 90 = 2 x\xa03 x\xa03 x\xa05 = 2 x\xa032\xa0x\xa05144 = 2 x\xa02 x\xa02 x\xa02 x\xa03 x\xa03 = 24\xa0x\xa032HCF (90, 144) = 2 x\xa032 = 18 | |
| 40206. |
12.2 question 2 |
| Answer» Ncert book | |
| 40207. |
Find the value of m so that the quadratic equation mx(x-7) + 49=0 has two equal roots. |
| Answer» M=4 | |
| 40208. |
If nth term of an ap is (2n+1), what is the sum of its first three terms ? |
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Answer» The sum is 15 3,5,7 |
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| 40209. |
Find value of cot square theta minus one upon tan square theta |
| Answer» Answer is 0 | |
| 40210. |
The tangent at any point of a circle is perpendicular to the radius the point of contact |
| Answer» Let APB be the tangent and take O as centre of the circle.Let us suppose that MP{tex}\\bot{/tex}AB does not pass through the centre.Then,{tex}\\angle OPA = 90^\\circ{/tex} [{tex}\\because{/tex} Tangent is perpendicular to the radius of circle]But {tex}\\angle MPA = 90^\\circ{/tex} [Given]{tex}\\therefore \\angle OPA = \\angle MPA{/tex}This is only possible when point O and point M coincide with each other.Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle. | |
| 40211. |
If2sinA=1then find the value of A |
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Answer» Given that 2sin A=1 sinA =1/2 sin A=sin30 A=30 2sinA =1 sinA =1/2 sinA = sin30 A = 30 |
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| 40212. |
If (x+1) is a factor of 2xcube+axsquare +1 then find the value of a& b given that 2a-3b=4 |
| Answer» Since {tex}(x + 1){/tex} is a factor of {tex}2x^3 + ax^2 + 2bx + 1{/tex}{tex}\\Rightarrow{/tex}{tex}x = -1{/tex} is a zero of {tex}2x^3 + ax^2 + 2bx + 1{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 {/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a - 2b - 1 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0a - 2b = 1 ...(i)Given that {tex}2a - 3b = 4{/tex} ...(ii)Multiplying equation (i) by 2, we get{tex}2a - 4b = 2{/tex} ...(iii)Subtracting equation (iii) from (ii), we getb = 2Substituting b = 2 in equation (i), we havea - 2(2) = 1{tex}\\Rightarrow{/tex}\xa0a - 4 = 1{tex}\\Rightarrow{/tex}\xa0a = 5Hence, a = 5 and b = 2. | |
| 40213. |
What is sin2thita |
| Answer» Its 2sinA means 2XsinA | |
| 40214. |
Solve the equation for x (X+3/X-2) |
| Answer» | |
| 40215. |
Find hcf of 52and 117and express it in the form 52x+117y |
| Answer» | |
| 40216. |
If sinB =12/13,then find cot B |
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Answer» hiaja 5/13 |
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| 40217. |
Describe me triangle |
| Answer» A triangle is a polygon with three sides | |
| 40218. |
Mathematics made triangle Theorem proof Karne Ka Tarika |
| Answer» | |
| 40219. |
Area theoram State and proove |
| Answer» | |
| 40220. |
Why py value are 3.14 |
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Answer» Bcz of divide of 22/7 The ratio of circumference to diameter of a circle is approx this |
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| 40221. |
Given geomatric progressions find the common ratio r and an is _1,_3,_9,_27..... |
| Answer» 1:3 | |
| 40222. |
Cossin |
| Answer» | |
| 40223. |
solve for x and yX+6/y=6 :3x-8/y=5 |
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Answer» Full solution bhej mohit kumar X=6 and y=2 |
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| 40224. |
Plz give rd sharma solutions |
| Answer» | |
| 40225. |
Chapter no.1 ex-1 |
| Answer» | |
| 40226. |
Will there be two sets of paper in maths in 2019 board |
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Answer» Yes ,low grade and high grade Yes Yes no |
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| 40227. |
Prove that the pointsA (0,1)B (-2,3) C(6,7) and D(8,3) are the vertices of a rectangle ABCD |
| Answer» According to properties of rectangle its opposite sides are equal AB=CD and AD=BC and diagonals are equal AC=BD so here opposite sides are not equal and diagonals are not equal. | |
| 40228. |
If sin. Theta =cos theta then find the theta |
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Answer» Sin45=cos45=1by root2so theeta=45 We know,Sin(90°-θ) = cos θAnd, Sin θ = cos θ [ Given ]Then, Sin θ = sin (90°-θ)On comparing, θ =90° - θ 2θ = 90° θ = 45° |
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| 40229. |
Write the all formulae of surface area and volume |
| Answer» Check formulae in the revision notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 40230. |
What is the sum of all the natural numbers |
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Answer» Yes it\'s -1/12 Ans is infinite number _ 1/12 |
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| 40231. |
What is the distance of point p(x,y) from the origin. |
| Answer» {tex}\\sqrt{x^2 + y^2}{/tex} | |
| 40232. |
In an ap the first term is 8 n term is 33 and sum of first n term is 123 find n and d |
| Answer» Given First term (a) = 8and, nth term (an) = 33{tex}\\Rightarrow{/tex} a + (n - 1)d = 33{tex}\\Rightarrow{/tex} 8 + (n - 1)d = 33{tex}\\Rightarrow{/tex} (n - 1)d = 33 - 8{tex}\\Rightarrow{/tex} (n - 1)d = 25 .....(i)and, Sum of first n terms = 123{tex} \\Rightarrow \\frac{n}{2}\\left[ {a + {a_n}} \\right] = 123{/tex}{tex}\\Rightarrow \\frac{n}{2}\\left[ {8 + 33} \\right] = 123{/tex}{tex} \\Rightarrow \\frac{n}{2} \\times 41 = 123{/tex}{tex} \\Rightarrow n = \\frac{{123 \\times 2}}{{41}}{/tex}{tex}\\Rightarrow{/tex} n = 6Put value of n in equation (i)(6 - 1)d = 25{tex}\\Rightarrow{/tex} 5d = 25{tex}\\Rightarrow d = \\frac{{25}}{5} = 5{/tex} | |
| 40233. |
How many terms of the ap: 9,17,25... Must be taken to give a sum of 636? |
| Answer» Use the formula Sn=n/2[2a+(n-1)×d] and you will get the value of n. | |
| 40234. |
Product of HCF and LCM of two natural no. Is 19800 find the two no.s |
| Answer» | |
| 40235. |
Is in boardexamnation in math the question comes only from ncert book |
| Answer» NCERT book and its extra qestion come in board exaamination | |
| 40236. |
Let m be a natural number for how many values of m 4m+1 is a perfect square |
| Answer» Let a be any positive integer.Applying Euclid’s division lemma with divisor = 2, we get{tex}\\begin{array}{l}a=2q+r\\;\\;\\;\\;\\;\\;\\;\\;\\;0\\leq r<2\\\\So\\;r=0,1\\\\\\end{array}{/tex}When r = 0,a = 2qSo a2= (2q)2 = 4q2 = 4m--------(1) ( where\xa0m = q2, which is an integer)When r = 1Then a= 2q+1a2= (2q + 1) 2 = 4q2 + 4q + 1 = 4(q2 + q ) +1 = 4m+1 --------(2) (where m = q2 + q, which is an integer)From (1) and (2) We Can conclude that\xa0The square of any positive integer is of the form 4m or 4m +1 for some integer m. | |
| 40237. |
Maths board exam of class 10 |
| Answer» Check Question Papers from here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 40238. |
In triangle PQR ST perallel QR PS/SQ=3/5 and PR=28cm.find PT |
| Answer» given PS/SQ=3/5 SO PQ=3+5=8 ST parallel QR PT= in triangle PQR (by bpt corollary PQ/PS=PR/PT 8/3=28/PT PT=28×3÷8 PT=21/2 | |
| 40239. |
Prove that√3 is irrrational number |
| Answer» Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. | |
| 40240. |
How to make the table of tignomantry ratio |
| Answer» √0/4,√1/4,√2/4,√3/4,√4/4√0/4=0 sin0°=0√1/4=sin30°=1/2√2/4=sin45°=1/√2√3/4=sin60°=√3/2√4/4=sin90°=1CosCos0°=1Cos30°=√3/2Cos45°=1/√2Cos60°=1/2Cos90°=0Just u have to recprocial the sin value.TanTan=sin/cosTan0°=0/1=0Tan30°=1/2/√3/2=1/√3Tan45°=1/√2/1/√2=1Tan60°=√3Tan90°=not definedCosecCosec=1/sinCosec0°=not difinedCosec30°=3Cosec45°=√2Cosec60°=2/√3Cosec90°=0SecSec1/cosSec0°=not difinedSec30°=2/√3Sec45°=√2Sec60°=2Sec90°=0CotCot=1/tanCot0°=not difinedCot30°=√3Cot45°=0Cot60°=1/√3Cot90°=1I hope this would help u... | |
| 40241. |
(x+1)² =2(x-3) |
| Answer» (x + 1) 2 = 2(x -3)x2 + 2x + 1 = 2x - 6x2 + 1 + 6 = 0x2 + 7 = 0x2 = - 7x = - √ | |
| 40242. |
If a=2and b=3 then a×b =ab then 2×3=23......is it right....simple logic |
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Answer» No, this logic is wronga×b=ab2×3=6 2*3=6 No |
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| 40243. |
If a×b=ab,then 2×3=6.consider a=2 and b=6. |
| Answer» Very simple a=2 and b=6,a×b=ab,2×6=12 | |
| 40244. |
Kis kisko Maths se dar lgta h ya jabardasti pdhna pdta h |
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Answer» Maths koi bhooth to h nhi agr ap usko ache se smajenge to darne ki jarurat nhi h. maths ko agar dhiyan se padhe to easy h yar Math mat pado hishab mai galti kar diya karna 10 ka 1 Rs. Lagana Are you a foolish |
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| 40245. |
(Sectheta +tantheta)2=cosectheta+1/cosectheta-1 |
| Answer» Cbse science | |
| 40246. |
Sec²theta + cosec²theta = sec²theta cosec ²theta |
| Answer» We have,LHS = cosec2{tex}\\theta{/tex}\xa0+ sec2{tex}\\theta{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { 1 } { \\sin ^ { 2 } \\theta } + \\frac { 1 } { \\cos ^ { 2 } \\theta }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\cos ^ { 2 } \\theta + \\sin ^ { 2 } \\theta } { \\cos ^ { 2 } \\theta \\sin ^ { 2 } \\theta } = \\frac { 1 } { \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta } = \\frac { 1 } { \\sin ^ { 2 } \\theta } \\times \\frac { 1 } { \\cos ^ { 2 } \\theta } = \\operatorname { cosec } ^ { 2 } \\theta \\sec ^ { 2 } \\theta{/tex}\xa0= RHS | |
| 40247. |
Tan²theta *cos²theta = 1- cos²theta |
| Answer» tan^2theta = sin^2theta/cos^2thetaThis implies tan^2theta × cose^2theta = sin^2theta& 1 - cos^2 theta also = sin^2thetaLHS = RHS , Hence Proved | |
| 40248. |
Solve for x , 2(2x+3÷x-3)-25(x-3÷2x-3)=5 |
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| 40249. |
If Sec theta +tan theta = p, find cosec theta |
| 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} | |
| 40250. |
Volume of frustum of cone = |
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Answer» 1/3pie×h(r1^2+r2^2+r1×r2) The Formula v=πh/3(R²+Rr+r²) |
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