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.
| 2251. |
Hello ... |
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| 2252. |
Kya kisi ko boyfriend chahiye |
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| 2253. |
पपधध |
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| 2254. |
when is the cbse 10th board exams are gonna happen?? |
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| 2255. |
Prove that (cot A +sec A) *2 -(tanB -cosec A) *2 = 2(cot.secB.cosecA |
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| 2256. |
In triangle abc,oa=3cm,ob=4cm,angle aob=90° ac=12cm and bc =13cm, prove that angle cab=90° |
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| 2257. |
If sinA+sinA=Cos squareA prove that cos power 6A-4cos powerA-8cos squareA=4 |
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| 2258. |
If sinA=√3/2 find value of 2cotsquare-1 |
| Answer» According to the question, sin A\xa0{tex}= \\frac { \\sqrt { 3 } } { 2 }{/tex}{tex}sin A= sin 60°{/tex}{tex}\\Rightarrow{/tex}\xa0A = 60°Now 2 cot2\xa0A - 1\xa0= 2cot2\xa060° - 1{tex}= 2 \\times \\left( \\frac { 1 } { \\sqrt { 3 } } \\right) ^ { 2 }{/tex}\xa0- 1{tex}= - \\frac { 1 } { 3 }{/tex} | |
| 2259. |
1upon secA+tanA-1 upon cosA=1 upon cosA-1 upon secA-tanA |
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| 2260. |
Thales therm proved by similarity rule in short from |
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| 2261. |
If one zero\'s of 4xx-9-8kx is negative of other find K |
| Answer» Let, the two zeroes of the polynomial f(x) = 4x2 - 8kx - 9 be α and -α.Comparing f(x) = 4x2\xa0- 8kx - 9 with ax2+bx+c we geta = 4; b = -8k and c = -9.Sum of the zeroes = α + (-α) ={tex}-\\frac ba=\\;-\\frac{-8k}4 {/tex}0 = 2kk = 0 | |
| 2262. |
If the quadratic equation(k+1)x-2(k-1)x+1= 0have real and equal roots then find the value of k |
| Answer» D=b2 - 4ac=0(k-1)2 - 4/k+1=0(k-1)2 = 4(k+1)k2 - 2k+1 = 4k+4k2 - 6k -3=0there appears to some error in the qn please check and resubmit , thanks | |
| 2263. |
2 4 6. 12 it is aAp |
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Answer» a=2d=2an=12an=a+(n-1)d=122+2n-2=12n=6yes it form an AP No |
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| 2264. |
Prove the distance formula. |
| Answer» (Take a triangle on graph name it as PQR)In triangle PQRPQ^2 = PR^2 + QR^2PQ^2 = (x2 - x1)^2 + ( y2 - y1 )^2PQ = √(x2 - x1)^2 + (y2 - y1)^2 | |
| 2265. |
2+3= |
| Answer» 5 | |
| 2266. |
Do all prepared for board exam |
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| 2267. |
a + bq=ra=?b=?q=?r=? |
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| 2268. |
a÷ax-1+b÷bx-1=a+b inwhich xis not equal to 1÷a,1÷b |
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| 2269. |
To Cbse Mentor Please provide us with data sheet as soon as possible |
| Answer» Check datesheet here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 2270. |
Distance between (2,3),(4,1) |
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Answer» 8 units Because Cbse uncle is not responding 0 |
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| 2271. |
From where you are coming |
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| 2272. |
Trigonometric identities come how many marks |
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Answer» Trigonometric identities are never asked in exams bt sometimes they can ask as their approvement... Application comes majorly |
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| 2273. |
Sol root 3 |
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| 2274. |
can(x-2)be the remainder on division of a polynomial p (x) by (2x+3)? justify your answer. |
| Answer» No | |
| 2275. |
16 kiska perfect square h eska 2 answer h btao kon kon se |
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Answer» 4,-4 4, -4 |
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| 2276. |
The sum of first n terms of an AP is given by Find the sixteenth termof the AP. |
| Answer» Given ,\xa0{tex}{S_n} = 5{n^2} - 3n{/tex}Now,\xa0{tex}{a_n} = {S_n} - {S_{n - 1}}{/tex}{tex} = 5{n^2} - 3n - [5{(n - 1)^2} - 3(n - 1)]{/tex}{tex} \\Rightarrow {a_n} = 10n - 8{/tex} ............ (i)Clearly,\xa0{tex}{a_{16}} = 10 \\times 16 - 8 = 152{/tex}Now, for finding AP, put n = 1, 2, 3, 4 ... in Eq(i).\xa0So, from Eq(i), we have{tex}{a_{1}} = 2{/tex}{tex}{a_{2}} =1 2{/tex}{tex}{a_{3}} = 22{/tex}Hence, The AP is 2,12, 22....Aliter:{tex}a_1=s_1=5(1)^2-3(1)=2{/tex}{tex}a_2=s_2-s_1=[5(2)^2-3(2)]-[5(1)^2-3(1)]=12{/tex}{tex}d=a_2-a_1=12-2=10{/tex}{tex}\\therefore{/tex}\xa0AP is 2,12,22...16th term is :{tex}a_{16}=2+15\\times10=152{/tex} | |
| 2277. |
An easy formula to find the area of a segment of a circle? |
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| 2278. |
I have problems in using signs |
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| 2279. |
20 of 80 |
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| 2280. |
Prove root 2 is irritational |
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| 2281. |
18=18+2{k+8/40-k} |
| Answer» 18=18+2(8+k/40-k) 18-18=2k+16/40-k0=2k+16/40-k40-k=2k+1640-16=2k+k24=3k8=k?????? | |
| 2282. |
Hello .. Ishi |
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| 2283. |
In exercise 13.5 Q.no 4It is given to prove equivalent but in this it is proved not equvalent |
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| 2284. |
Find the centre of circle passing through (5,-8) (2,-9)(2,1) |
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| 2285. |
10.1 theoram |
| Answer» It\'s so simple . You will try it home | |
| 2286. |
three marks |
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| 2287. |
The volume of a hemisphere is 2425 1/2 cm3. Find its C.S.A |
| Answer» As, volume of hemisphere = 2425{tex}\\frac { 1 } { 2 } c m ^ { 3 }{/tex}{tex}\\Rightarrow \\frac { 2 } { 3 } \\pi r ^ { 3 } = 2425 \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow \\frac { 2 } { 3 } \\times \\frac { 22 } { 7 } r ^ { 3 } = \\frac { 4851 } { 2 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 4851 \\times 3 \\times 7 } { 2 \\times 2 \\times 22 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 441 \\times 3 \\times 7 } { 2 \\times 2 \\times 2 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 21 ^ { 3 } } { 2 ^ { 3 } }{/tex}{tex}\\Rightarrow r = \\frac { 21 } { 2 } c m{/tex}So, the curved surface area of the hemisphere ={tex}2 \\pi r ^ { 2 }{/tex}{tex}= 2 \\times \\frac { 22 } { 7 } \\times \\frac { 21 } { 2 } \\times \\frac { 21 } { 2 } = 693 \\mathrm { cm } ^ { 2 }{/tex} | |
| 2288. |
Positive integer |
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| 2289. |
Find quotient when x square +x by x |
| Answer» x+1. | |
| 2290. |
How to transver download file mobile to computer |
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| 2291. |
(X-2)(x+2)=12 find polynomial |
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Answer» X2-16= 0 Solving theequation x^2+2x-2x+4+2x-2x cancelX^2+4 is reqired polynomial |
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| 2292. |
27^999/7 |
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| 2293. |
Area of equilatral triangle |
| Answer» √3\\4 side square | |
| 2294. |
1. Sin a .cos a2.sin c .cos c |
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| 2295. |
Area of major segment |
| Answer» πr`2 theta/360 | |
| 2296. |
hpo |
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| 2297. |
If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term . Find the AP |
| Answer» AP is 7, 12, 17, 22, 27, 32....... | |
| 2298. |
Area of quadrant |
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Answer» 1/4pie r sq. 1/4×pie r sq 1/4 π r2 πr^2/4 |
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| 2299. |
What is meant by diminished |
| Answer» Diminished means very small | |
| 2300. |
What is homologus series |
| Answer» Homologous series is a family of organic compound which have same chemical properties and successive compound differ by ch2 unit. | |