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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.
| 2351. |
What is TRIGONOMRTRY |
| Answer» Trigonometry (from Greek trigōnon, triangle and metron, measure) is a branch of mathematics that studies relationships involving lengths and angles of triangles. | |
| 2352. |
ABC is an isosceles right angled triangle AB =BC=1unit. Find the radius of semicircle BD |
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Answer» Bc is √2 1/√2 |
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| 2353. |
x2+6x_(a2+2a_8)=0 |
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| 2354. |
Show that reciprocal of any irrational no.is irrational |
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| 2355. |
Ax+bx= |
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| 2356. |
If cos theta + sin theta = root 2 cos theta then show that |
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| 2357. |
Find the value of p, for which one Root of the equation pxsquare- 14x+18=0is 6 times the other |
| Answer» Let\xa0{tex}\\alpha{/tex}\xa0and\xa0{tex}6\\alpha{/tex}\xa0be the roots of equation.We have, {tex}px^2-14x+8=0{/tex} where a= p, b = -14, c = 8Sum of zeroes{tex} = -\\frac ba = -\\frac{-14}{p}{/tex}{tex}\\alpha +6\\alpha=\\frac{14}{p}{/tex}{tex}7\\alpha = \\frac{14}{p}{/tex}{tex}\\alpha = \\frac2p{/tex}............(i)Also, Product of the zeroes\xa0{tex} = \\frac 8p=\\frac ca{/tex}{tex}\\alpha \\times 6\\alpha = \\frac 8p{/tex}{tex}6\\alpha^2=\\frac 8p{/tex}From (i){tex}6(\\frac{2}{p})^2=\\frac 8p{/tex}{tex}6\\times \\frac {4}{p^2}=\\frac 8p{/tex}{tex}\\frac{6}{p^2}=\\frac2p{/tex}{tex}\\frac 62=\\frac{p^2}{p}{/tex}{tex}Hence, \\ p=3{/tex} | |
| 2358. |
Find the hcf of 81 and 237 and express it as linear combination of 81 and 237 |
| Answer» Since, 237 > 81On applying Euclid\'s division algorithm, we get237 = 81 {tex}\\times{/tex}\xa02 + 75 ..........(i)81 = 75 {tex}\\times{/tex}\xa01 + 6 ..........(ii)75 = 6 {tex}\\times{/tex}\xa012 + 3 .........(iii)6 = 3 {tex}\\times{/tex}\xa02 + 0 .............(iv)Hence, HCF (81,237) = 3.In order to write 3 in the form of 81x + 237y,Now,{tex}\\style{font-family:Arial}{\\begin{array}{l}3=75-6\\times12(\\;from(iii))\\\\=75-(81-75\\times1)\\times12\\\\=75-(81\\times12-75\\times12)\\\\=75-81\\times12+75\\times12\\\\=75(1+12)-81\\times12\\\\=75\\times13-81\\times12\\\\=13\\times(237-81\\times2)-81\\times12\\;(\\;Substuting\\;75\\;from(i))\\\\=13\\times237-13\\times81\\times2-81\\times12\\\\=237\\times13-81\\times(26+12)\\\\=81(-38)+237\\times13\\\\=81x+237y\\end{array}}{/tex}Hence, x = -38 and y = 13 | |
| 2359. |
Formula of perimeter |
| Answer» Sum of all sides in fig | |
| 2360. |
find the discriminent of the quadratic equation 4upon3x square_2x + 3 upon 4 =0 |
| Answer» 0 | |
| 2361. |
1/cot^2A +1/1+tan^2A = 1/1-sin^2A -1/cosec^2A |
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| 2362. |
4 + (-1) +2+.....+x =437 |
| Answer» (-4) + (-1) + 2 + 5 + ---- + x = 437.Now,-1 - (-4) = -1 + 4 = 32 - (-1) = 2 +\xa01 = 35 - 2 = 3Thus, this forms an A.P. with a = -4, d = 3,l = xLet their be n terms in this A.P.Then,Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] {/tex}{tex}\\Rightarrow 437 = \\frac { n } { 2 } [ 2 \\times ( - 4 ) + ( n - 1 ) \\times 3 ]{/tex}{tex}\\Rightarrow{/tex}\xa0874 = n[-8 + 3n - 3]{tex}\\Rightarrow{/tex}874 = n[3n - 11]{tex}\\Rightarrow{/tex}874 = 3n2\xa0- 11n{tex}\\Rightarrow{/tex}3n2\xa0- 11n - 874 = 0{tex}\\Rightarrow{/tex}3n2\xa0- 57n + 46n - 874 = 0{tex}\\Rightarrow{/tex}3n(n - 19) + 46(n - 19) = 0{tex}\\Rightarrow{/tex}3n + 46 = 0 or n = 19{tex}\\Rightarrow n = - \\frac { 46 } { 3 }{/tex}\xa0or n\xa0= 19Numbers of terms cannot be negative or fraction.{tex}\\Rightarrow{/tex}\xa0n = 19Now, Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ a + l ]{/tex}{tex}\\Rightarrow 437 = \\frac { 19 } { 2 } [ - 4 + x ]{/tex}{tex}\\Rightarrow - 4 + x = \\frac { 437 \\times 2 } { 19 }{/tex}{tex}\\Rightarrow - 4 + x = 46{/tex}{tex}\\Rightarrow x = 50{/tex} | |
| 2363. |
sin2A=2sinA |
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Answer» A= 0° it is false statement Never No they both ate not equal |
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| 2364. |
CSA of sphere |
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Answer» Gou 4/3πr2 4×22/7×r2 CSA AND TSA ARE SAME IN SPHERE 4×22÷7×r×r |
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| 2365. |
Prove √5 irrational |
| Answer» Let √5=a/b. (Where a nd b are co prime integers)Squaring both sides5=a^2/b^2a^2=5b^2. ------(1)5 is a factor of a^2So 5 is afactor of a alsoLet a=5cSquaring both sidesa^2=25c^25b^2=25c^2. -------(from1)b^2=5c^25 is a factor of b^2So 5 is a factor of b also since 5 is a factor of both a and b This is contradiction Our supposition is wrong √5 is irrational | |
| 2366. |
All formula of maths 10th class ncert |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 2367. |
Find 4 numbers in GP |
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| 2368. |
How do attend probablity sums |
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| 2369. |
Find the eleventh term from the last term of the AP:27,23,19.................-65. |
| Answer» -32 | |
| 2370. |
Is board examination is very hard What can I study RS agg. Or NCERTExamples |
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| 2371. |
is preboard exam marks jion in 10th |
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| 2372. |
Sinthith+costhitha |
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| 2373. |
What is parallelogram |
| Answer» In parallelogram the opposites sides and angles are equal But his diagonal are not equal | |
| 2374. |
2+2+1 |
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Answer» 5 2+1+2 stupid |
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| 2375. |
0/0=?write the numerical value. |
| Answer» undefined. | |
| 2376. |
Model test paper 14 u like |
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| 2377. |
A die is thrown once what is the probability of not getting a prime number |
| Answer» There r 3 prime nos - 2, 3, 5 so nos that r not prime r 1, 4, 6 so P is 3/6 = 1/2 | |
| 2378. |
What is circumference of circle |
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Answer» 2πr 2πr |
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| 2379. |
(a+b) (a+b) (x) (x) - 4abx-(a-b)(a-b) |
| Answer» We have,{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}In order to factorize {tex}(a + b)^2x^2 - 4abx - (a - b)^2{/tex}, we have to find two numbers \'l\' and \'m\' such that.l + m = -4ab and lm = -(a + b)2(a - b)2Clearly, (a - b)2 + [-(a + b)2] = -4ab and\xa0lm = -(a + b)2(a - b)2 l = (a - b)2 and m = -(a + b)2Now,{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}(a + b)^2x^2 - (a + b)^2x + (a - b)^2x - (a - b)^2 = 0{/tex}{tex}\\Rightarrow{/tex} (a + b)2x [x - 1] + (a - b)2[x - 1] = 0{tex}\\Rightarrow{/tex} (x - 1)[(a + b)2x + (a - b)2] = 0{tex}\\Rightarrow{/tex} x - 1 = 0 or (a + b)2x + (a - b)2 = 0{tex}\\Rightarrow{/tex} x = 1 or {tex}x = - \\frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}}{/tex} | |
| 2380. |
Prove that n3-n is divisble by6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) | |
| 2381. |
16/X-1=15/X+1 |
| Answer» X = -31 | |
| 2382. |
How to use Complete square method |
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| 2383. |
If cos a + sin a=root 2 cos a, prove that cos a - sin a = root 2 sin a |
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| 2384. |
How to find that the given sides of an triangle is possible or not |
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Answer» If one of the side of triangle is greater than the sum of other two sides, only then a triangle is possible If the given Side will satisfy this equation the triangle is possible By using this formula square of longer side = the sum of remaining two side |
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| 2385. |
Trignomatry |
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| 2386. |
if nth term of an AP is (2n+1) find the sum of first n terms of AP |
| Answer» Here, an = 2n + 1Put n = 1, 2. Then,a1 = 2(1) + 1 = 3{tex}\\therefore {/tex}\xa0a2 = 2(2) + 1 = 5a = 3d = a2 - a1 = 5 - 3 = 2{tex}\\therefore {/tex}\xa0Sum of first n-terms of the AP = Sn{tex} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex} = \\frac{n}{2}\\left[ {2(3) + (n - 1)2} \\right]{/tex}= n (3 + n - 1)= n (n + 2) | |
| 2387. |
Probability questions |
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| 2388. |
a÷x-a + b÷x-b + 2c÷x-c |
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| 2389. |
x^2+{(x^2)÷(1+x)^2=3Solve for x |
| Answer» 1 | |
| 2390. |
What is easy method to proving a trigonometry identity problem |
| Answer» Convert all ratios into cos and sin. | |
| 2391. |
Find the zeroes of polynomial x3-12x2+39x-28 if zeroes r in ap |
| Answer» 3 | |
| 2392. |
If p times the pth term of an A.P. is equal to q times the qth term, find (p+q)th term of the A.P. |
| Answer» 0 | |
| 2393. |
What is angle of elevation |
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Answer» The angle formed by the line of sight with the horizontal when it is above the horizontal level....i.e. the case when we raise our head to look at the object The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer\'s eye, the line of sight. |
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| 2394. |
27,23,19,...-65find.its.11.term |
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| 2395. |
Explain why 7x11x13+13 and 7x6x5x4x3x2x1+5 are composite number |
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Answer» Answer u want is in this app.....aap vahan dekh lo Bcz it has more than two factors |
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| 2396. |
o is any point inside a rectangle ABCD. Prove that OB^2+OD^2=OA^2+OC^2 |
| Answer» Let ABCD be the given rectangle and let O be a point within it. Join OA, OB, OC and OD.Trough O, draw {tex}E O F \\| A B.{/tex}\xa0Then, ABFE is a rectangle.In right triangles {tex}\\Delta{/tex}OEA and {tex}\\Delta{/tex}OFC, we have\xa0OA2 = OE2+ AE2 and OC2 = OF2\xa0+ CF2{tex}\\Rightarrow{/tex}\xa0OA2 + OC2 = (OE2 + AE2) +(OF2+ CF2){tex}\\Rightarrow{/tex}\xa0OA2 + OC2\xa0= OE2\xa0+ OF2\xa0+ AE2 + CF2\xa0...... (1)Now, in right triangles {tex}\\Delta{/tex}\xa0OFB and {tex}\\Delta{/tex}\xa0ODE, we haveOB2= OF2+FB2 and OD2= OE2+DE2{tex}\\Rightarrow{/tex}\xa0OB2 + OD2 = (OF2 + FB2) + (OE2+ DE2){tex}\\Rightarrow{/tex}\xa0OB2+ OD2= OE2+ OF2+DE2+BF2{tex}\\Rightarrow{/tex}\xa0OB2+ OD2\xa0= OE2+ OF2+ CF2+AE2 [{tex}\\because{/tex}\xa0DE= CF and AE = B F] ......(2)From (i) and (ii), we getOA2 + OC2 = OB2 + OD2 | |
| 2397. |
(a+b) cube |
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| 2398. |
prove : 1+ CosA + SinA/1+CosA - SinA=1+SinA/CosA |
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| 2399. |
Prove that cos²A+sin²A=1 |
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Answer» Cos A=B/H sinA=P/HCos²A+sin²AB²/H² + P²/+H² B²+P²/H² H²/H² =1 In triangle ABC,right Angled at B we have,AB^2+BC^2=AC^2Dividing each term of (1),we getAB^2/AC^2 + BC^2/AC^2 = AC^2/AC^2That is -(cosA)^2+ sinA |
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| 2400. |
Find mean first 12 composite numbers |
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