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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the derívative of tne following functions fromfirst principles:sin +1(it) cos x |
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Answer» it's wrong |
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| 2. |
tne valeAn almirah is sold for T5,225 after ailowingdiscount of 5% hnd ilM |
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| 3. |
25. For what value for x, the points A (1, 2) B(-4, 7) and C (x, -1) are colline26. Find the value Ofy such thot po Op where P, Q and R are the noints |
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Answer» For collinear, area of triangle=0(1/2)[1(7+1)+(-4)(-1-2)+x(2-7)]=0(1/2)[8+12-5x]=020=5xx=4 |
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| 4. |
3 Find the area of a trapezium whose parallel sides are of length 16 dm and 22dm sndwhose height is 12 dm, |
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| 5. |
3. Find the sum of the lengths of the bases of a trapezium whose altitude is 11 cm and whose arcais 0.55 m2me. |
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Answer» Area of trapezium =1/2 (sum of parallel sides) × (perpendicular distance between them) Given, Area = .55 m^2, altitude = 11 cm or .11 mLet sum of the length of bases = n Then we get, .55 = n*.11n = .55/.11 = 5 Therefore, sum of the length of bases = 5 m or 500 cm |
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| 6. |
and 2 SeaiesemaindleахCuamg |
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| 7. |
4.L,XY - 5A AMT - AAHE. In A AMT, AM = 6.3 cm. ZTAM = 50°, AT = 5.6 cm.AM 7АН ;Construct A AHE.Construction of a tangent to a circle at a point on the circle1) Using the centre of the circle. |
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| 8. |
14. An aeroplane flying horizontal 1 km above the ground is observed at an elevatio60°. After 10 seconds its elevation is observed to be 30°. Find the speed ofaeroplane. |
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| 9. |
EXERCISE 1.2âExpress each number as a product of its prime factors:) 140(ii) 156(ii) 3825(iv) 5005 |
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Answer» 140 = 2×2×5×7 156 = 2×2×3×13 3825 = 3×3×5×5×17 5005 = 5×7×11×13 |
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| 10. |
An aeroplane flying horizontally 1 kmabove the ground is observed at an elevationof 60 and after 10s the elevation is observed to be 30° Then find uniform speedof the aeroplane(in km/h) |
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| 11. |
आकृति 6.170 और (i) में.DE ABC है। में EC और (1) में AD ज्ञात कीजिए1.5cm/icm7.2cm11.8cmEM3.cm5.4cm |
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| 12. |
Cos[ (Sin^-1x)]^2 |
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| 13. |
10. |1x + 2|dx is equal to(2) 28 (6) 30 (c) 29(2) 27 |
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| 14. |
1. The relation between HCF and ICM of 12 and 20 will be |
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Answer» Relation: LCM ×HCF= Product of two numbers LCM of 12 and 20: 12 = 2*2*3 20 = 2*2*5 LCM = 2*2*3*5 = 60 HCF of 12 and 20 2*2 = 4 LCM × HCF = Product of two given numbers 60× 4 = 12× 20 ⇒240 = 240 |
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| 15. |
2, the HF of beas-2And the value o da:and ICM of 661 |
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Answer» Since, LCM×HCF=product of two numbers 60×2=6×a 120=6a 6a=120 a=120/6=20 |
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| 16. |
2) ॥2 1८ 1५०” ‘9 पा 9T &‘Qug-g)+v=Erque-qz+v =हे... हि Y |
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| 17. |
D26, (V&) (16)v2 |
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| 18. |
Find the equation of the line joining A (3, 1) and B (9, 3) using determinants? |
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| 19. |
al4. Using properties of determinants, prowthatx +2y x9y2(x+y).x + 2Or |
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| 20. |
Using properties of determinants prove the following:cl=(a-b) (b-c) (c _ a) (a + b + c) |
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| 21. |
ë |
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Answer» X+4 X=-4 |
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| 22. |
ë+4 |
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Answer» Abe galat hai |
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| 23. |
ind the equation of the line joining thepoints P(2,-3) and Q4, 1) usingdeterminants. |
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| 24. |
ëĽx2 |
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Answer» 2/4 × 6/2= (1/2) × 3 = 3/2 |
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| 25. |
sin-1 x + tan-1x =-2 |
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Answer» No answer |
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| 26. |
+2 MATHEMATICS1)Discuss the continuity of the function f given by f(x) = (x at x = 0 |
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| 27. |
52. Range of cos^-1x is |
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Answer» range of cos^(-1)x=[0,pie] |
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| 28. |
cori (2-1) =(A) 2 cos-1x (B) cos1x (C) 2cosx (D) |
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Answer» cos^-1(2x-1) Let x= cos^2A , or cosA=+/-x =cos^-1(2cos^2A-1) =cos^-1.cos2A =2A =2.cos^-1(X)^1/2 ,hence option c |
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| 29. |
1910] 18 RibB LBD4 8 '8 छह बा =08 M b BT =Y T= AT ody~2avY 2tz '€% g - S LR |
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Answer» ar(ABC)/ar(PQR) = (AB/PQ)^29/4 = (18/PQ)^23/2 = 18/PQPQ = 12cm |
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| 30. |
, (० ०िसहल + |g LY (a (8+ 9-0)!दी पल * |X |1 |
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Answer» (x^b/x^c)^[b + c - a] * (x^c/x^a)^[c + a - b] * (x^a/x^b)^[a + b - c] = x^(b-c)(b + c - a) * x^(c-a)(c + a - b) * x^(a-b)(a + b - c) = x^[b^2 + bc - ab - bc - c^2 + ac] * x^[c^2 + ac - bc - ac - a^2 + ab] * x^[a^2 + ab - ac - ab - b^2 + bc] = x^[b^2 - b^2 + c^2 - c^2 + a^2 - a^2 + ac - ac + ab - ab + bc - bc] = x^0 = 1 |
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| 31. |
EXAMPLE 11. Using the properties of determinants, prove thaty+2222 +*V*x + y= 4xyz. |
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| 32. |
13. Using properties of determinants, provethat11 1+3x1 + 3y- 9 (3xyz + xy+ yz +zx). |
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| 33. |
In the given figure, triangles ABC and DCB are right-2 angled at A and D respectively and AC DB.ed at A and D respectively and AC DIBProve that △ABC ADCB. |
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| 34. |
1/2x-1/y=-11x+1/2y=8 |
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| 35. |
SUBJECT: MATHEMATICSSECTION ACf (x-2) is a factor of x^3 +2x^2 +4a find a. |
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| 36. |
1x+2% (1+2) |
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Answer» (x+2)^2= x^2+4x+4will be the answer to this question |
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| 37. |
sin^-1x + sin^-1(1-x)=cos^-1x |
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| 38. |
Comprehension II#x is the root of the equation x2-1x-1 = 0, thenChoose the correct answer:1. The value of x51 is(2) -1(4) -i(3) i2.The value of x20 +20 may be(2) 1(4) -i(3) ix2013 13 20132013 may be equal to(2) 1(3) -2i |
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Answer» ty There was a mistake in the last part. I corrected it in this. yeah in third line how you got x=I into some value |
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| 39. |
2. Name theroseperty under multiplicsed-4X 1 = 1X-413= 1X-4 = - 4놈 |
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Answer» Communicative property is the property used. "HOPE IT HAS HELPED YOU" This is commutative property. At the end when anything multiplied by 1 it will keep same term. Commutative property is right |
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| 40. |
11. The mean of the following data is 42. Find the missing frequencies aand y if the sum of frequencies is 100.Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80Frequency 7 10x 13y 10 149[CBSE 2014 |
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| 41. |
If d is HCF of 40 and 65, find the integres x and y which satisfyd=40x+65y(do in simplest form which is understanding) |
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Answer» First we will find the hcf. Since 65>40 Therefore, 64=40×1+25 40=25×1+15 25=15×1+10 15=10×1+5 10=5×2+0 Since remainder is 0 Therefore hcf is 5 To do it in linear combination we have 5=15-(10×1) 5=15-(25-15×1) 5=40-25×1-(25-40×1-25)... And so on till we get 5=40(5)+65(-3) Now comparing 40x+65y with our ans we get that , x=5 and y=-3 |
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| 42. |
5. Find the acute angle 0, satisfying theequation sec2e+ tan2θ 3. |
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| 43. |
(vii) 13x + 1 ly = 70; 1 1x + 13y = 74 |
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| 44. |
Find the particular solution of differential equation:odytan x+ y = 2x tan x + x,x+0. given that y = 0 when= |
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| 45. |
dy-0, then |
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| 46. |
Expand using Binomial Theorem\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4} ; x \neq 0 |
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| 47. |
(ii) Squaring (I) andlexa)-(ii) Squaring (I) and[(x + a), (Ď2nusing binomial theorem, ex1, (3x + 2 |
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Answer» As per Pascal's Triangle coefficients will be1 4 6 4 1 (3x+2y)^4=1⋅(3x)^4(2y)^0+4(3x^3)(2y)+6(3y)^2(2y)^2+4(3x)(2y)^3+1(3x)^0(2y)^4 81x^4+216x^3y+216x^2y^2+96xy^3+16y^4 |
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| 48. |
Prove that if the chords of congruent circles subtend equal angles at their centers, then the chords are equal. |
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| 49. |
8. If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th termsb-ac 4arespectively, then prove that |
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Answer» (x + y)^n a = 6th term = C(n,5) x^(n−5) y^5b = 7th term = C(n,6) x^(n−6) y^6c = 8th term = C(n,7) x^(n−7) y^7d = 9th term = C(n,8) x^(n−8) y^8 =================================== b^2 − ac= [C(n,6) x^(n−6) y^6]^2 − C(n,5) x^(n−5) y^5 * C(n,7) x^(n−7) y^7= [n!/(6! (n−6)!)]^2 x^(2n−12) y^12 − n!/(5!(n−5)!) * n!/(7!(n−7)!) x^(2n−12) y^12= x^(2n−12) y^12 (n!/(5!(n−7)!))^2 [1/(36(n−6)²) − 1/(42(n−5)(n−6))]= x^(2n−12) y^12 (n!/(5!(n−7)!))^2 [(n+1)/(252(n−6)²(n−5))] c^2 − bd= [C(n,7) x^(n−7) y^7]^2 − C(n,6) x^(n−6) y^6 * C(n,8) x^(n−8) y^8= [n!/(7! (n−7)!)]^2 x^(2n−14) y^14 − n!/(6!(n−6)!) * n!/(8!(n−8)!) x^(2n−14) y^14= x^(2n−14) y^14 (n!/(6!(n−8)!))^2 [1/(49(n−7)²) − 1/(56(n−6)(n−7))]= x^(2n−14) y^14 (n!/(6!(n−8)!))^2 [(n+1)/(392(n−7)²(n−6))] ------------------------------ (b^2 − ac) / (c^2 − bd) = x^(2n−12) y^12 (n!/(5!(n−7)!))^2 [(n+1)/(252(n−6)²(n−5))]————————————————————————— =x^(2n−14) y^14 (n!/(6!(n−8)!))^2 [(n+1)/(392(n−7)²(n−6))] x^2 (6!(n−8)!)^2 (392(n−7)²(n−6))———————————————— =y^2 (5!(n−7)!)^2 (252(n−6)²(n−5)) x^2 (6)^2 (98(n−7)²)————————————— =y^2 (n−7)^2 (63(n−6)(n−5)) x^2 (36) (98)————————— =y^2 (63(n−6)(n−5)) 56 x^2———————(n−6)(n−5) y^2 =================================== 4a/(3c) = 4 C(n,5) x^(n−5) y^5—————————— =3 C(n,7) x^(n−7) y^7 4 n! / (5! (n−5)!) x^2—————————— =3 n! / (7! (n−7)!) y^2 4 (7! (n−7)!) x^2———————— =3 (5! (n−5)!) y^2 4 (7*6) x^2———————— =3 (n−5)(n−6) y^2 56 x^2——————— = (b^2 − ac) / (c^2 − bd)(n−5)(n−6) y^2 |
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| 50. |
11111(44,The value ofis/1 02(a) 331(b) -331(c) - 131(d) 311+√31)it is given that o |
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