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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. |
33*sqrt((-tan(x) %2B 1)/(tan(x) %2B 1)) |
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| 2. |
Find a if the 17th and 18th terms of the expansion (2 + a)^50 are equal |
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| 3. |
Find a if the 17th and 18th terms of the following expansion are equal: (2 + a)^50 |
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| 4. |
A die is dhrown 6 times If gening an odd mumber is a succes what isprobability of5 successesGi) at least 5 soccesses?(i) at most 5 successes? |
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| 5. |
The mumber of sides of two regular polygons are as 5 4 and the dieretheir angles is 9. Find the number of sides in the polygonsb |
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Answer» Let the no. of sides of the 1st polygon be 5n and that of the 2nd one be 4n. Then the sum of internal angles of the 1st polygon is (5n-2)x180 degrees and that of the 2nd is (4n-2)x180. Since they're regular polygons all internal angles are same. So internal angles of the 1st polygon are ((5n-2)x180)/5n degrees and that of the 2nd is ((4n-2)x180)/4n degrees. Then from question we can say : (((5n-2)x180)/5n) - (((4n-2)x180)/4n)=9 180 {((5n-2)/5n)-((4n-2) /4n) }=9 180 { 4(5n-2) - 5(4n-2) / 20n } = 9 solving we get n=2. So no of sides of the 1st polygons = 5n = 5 * 2 = 10 no of sides of the 1st polygons = 4n = 4 * 2 = 8 |
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| 6. |
21. (a) If tan θ . v2-1, prove that sin θ cos θ:212(b) If ptan θ-q, firnd the value of p sin θ + qcos θpcos θ + qsin θ(c) If5cos A Ξ 1 2sin A, prove that tan 2 A-sin' A-sin' A sech |
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| 7. |
r. Find a if the 17thand 18th terms of the following expansion are equal : ( 2a + a) ^5n |
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Answer» 50C17*2^33*a^16=50C18*2^32*a^179,847,379,391,150*2=18,053,528,883,775*aso a=1 |
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| 8. |
an-1=420then evaluate na tha |
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Answer» please like my answer plzzzzzzzzz🙏🙏🙏 |
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| 9. |
bread this 16 cmbogad na tenth 20 mIS 16 of POP IOof the board is 2000, WhoisVolume ? |
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Answer» volume=length×breath×height = 20×16×.1 = 32.0 = 32 cm^3 32 cc is the correct answer to this question . volume = length × breath × height = 20 ×16 × 0.1 = 32.0 = 32 cmis the correct answer |
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| 10. |
1 %2B tan(theta)^2/(sec(theta) %2B 1)=sec(theta) |
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| 11. |
(-tan(theta) %2B sec(theta) %2B 1)/(tan(theta) %2B sec(theta) %2B 1)=(-sin(theta) %2B 1)/cos(theta) |
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Answer» Let theta = x LHS:secx - tanx + 1/secx + tanx + 1 = secx - tanx + (sec^2x - tan^2x)/ (secx + tanx + 1) = (secx - tanx) + (secx - tanx)(secx + tanx)/(secx + tanx + 1) = (secx - tanx)(secx + tanx + 1)/ (secx + tanx + 1) = secx - tanx = 1/cosx - sinx/cosx = (1 - sinx) /cosx = RHS Hence proved |
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| 12. |
(sec(A) %2B 1)/tan(A) %2B tan(A)/(sec(A) %2B 1) |
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Answer» 2sinA tanA/(1+secA) + (1+secA)/tanA = tan^2 A + (1+secA)^2 / tanA. (1 + secA) = (tan^2 A + 1 + sec^2 A + 2secA)/tanA. (1 + secA) [using 1 + tan^2 A = sec^2 A] = (2sec^2 A + 2secA)/ tanA. (1 + secA) = 2secA(1 + secA) / tanA. (1 + secA) = 2secA/tanA = 2/cosA * cosA/sinA = 2/sin A = 2 cosec A |
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| 13. |
हि S— fl\ ./N,\ 5o3 H |
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Answer» 45 45 is the right answer of the following |
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| 14. |
(v) Find the product of roots of the equation (lo9 x)2 -2log-5o |
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| 15. |
के... जवान ७. 30+ ०0९९6 गिरPIRDALURD 5o a3y *7 |
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Answer» Let √ 5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime] √ 5=p/q=> root 5 * q = p squaring on both sides=> 5*q*q = p*p ------> 1p*p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p*p = 25c*c --------- > 2 sub p*p in 15*q*q = 25*c*cq*q = 5*c*c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso√5 is an irrational |
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| 16. |
可兩 Flo :(a) 25%(b) 20%(c) 30%(d) 35%5oqマ |
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| 17. |
(C3, Value of 5o 25 is equal to-12. A |
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Answer» 11/50+7/25(11+14)/5025/501/2 |
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| 18. |
5 frde esst mumber anich wnen divided by 25.40 amd 5Oafter being increased byisay |
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Answer» 607 lcm of 25,40,60 is 600add remainder=7ans 600+7=607 |
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| 19. |
Prove the iollowing sin 2 0$0.10 3 212cos () + cos¢sino +sinShow that:= cot (Findt for the following A.P.s - 4,9,14.1Evaluate (i)7 (ii)10:Show that log 3603 iog 27 + 2Sectifou: question of (3) mfulowing in the forn |
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Answer» i) 7! = 7×6×5×4×3×2×1 = 5040 ii) 10!/5! = (10×9×8×7×6) = 30240 |
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| 20. |
nd a, if the 17th and 18th terms of the expansion (2 + a)5o are equalin |
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| 21. |
(tan(A) %2B sec(A) - 1)/(tan(A) - sec(A) %2B 1)=cos(A)/(-sin(A) %2B 1) |
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Answer» sec A-tanA=1/cosA + sinA/cos A=1+Sin A/cosA=Cos A /1-SinA |
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| 22. |
(tan(A) %2B sec(A) - 1)/(tan(A) - sec(A) %2B 1)=(sin(A) %2B 1)/cos(A) |
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Answer» To prove: (tanA+secA-1)/(tanA-secA+1) =(1+sinA)/cosA LHS:(tanA+secA-1)/(tanA-secA+1)= (tanA+secA-(sec²A-tan²A ) ) /(tanA-secA+1) ={ tanA+secA- [(secA + tanA)(secA-tanA) ] } / (tanA-secA+1) = { (tanA+secA) (1-sec A+ tan A) } /(/(tanA-secA+1) =(1/cosA) + (sinA/cosA) =(1+ sinA ) / cosA =RHS |
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| 23. |
(cos(theta) %2B 1)/(-cos(theta) %2B 1) |
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| 24. |
// 3 + 212V2 + 12-18. Show that= |
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| 25. |
3-21+57Yyshow that a = b and ba= a.18) 1fx+iy=(a+ib)', show that a = 4(a19) If a = 1-i, show that (5a-7b) = 0.2+ib2+31)(2-31)Nalatib20) If xtiya Nc+ida+b?prove that (x+y)2 = c+d2mber1+i+i")21) If (a+ib) = 1 , then prove that (a+b)(v tisk T-id3that T-in Titis is te212)Show that |
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Answer» It is right answer this question this are easy questions |
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| 26. |
005 225? + थञ 165" =1 3)a0 |
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Answer» cos255= cos (270-15) = -sin 15Sin 165=sin (180-15)=sin15So,cos255+sin 165=-sin15+sin15=0 genius |
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| 27. |
Find the number of sides of a regular polygon, if each of its interior165°3.angle |
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| 28. |
13) If the cost price of a gold chain was722,000 and the percentage profit was896, find its selling price. |
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Answer» Selling price = (100+8)/100 * 22000= 108/100 * 22000= Rs. 23760 |
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| 29. |
The slant height of a square pyramid is 25 cm andits surface area is 896 cm2. What is its volume? |
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Answer» 1 thanks a lot |
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| 30. |
if y -6 x2 then the value of fox+ )oy will be |
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Answer» wrong answer |
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| 31. |
7 (5८८2 =13 /KZ\/» Cetoa4 ‘]‘ |
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| 32. |
Factorise: a4 - 8a2b2 + 16b4 -256. |
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Answer» a*a*a*a-2*4a*a*b*b+4b*b*4*b*b-16*16 =(a*a-4b*b)(a*a-4b*b)-16*16 =(a*a-16-4b*b)(a*a+16-4b*b) thanks |
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| 33. |
A4 =| 5O X\ o20 A30% |
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Answer» 9x9=81150x150=2250030x30=900 |
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| 34. |
tan(theta)/(-cot(theta) %2B 1) %2B cot(theta)/(-tan(theta) %2B 1)=sec(theta*(cosec*theta)) %2B 1 |
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| 35. |
(A*cosec %2B cot(A) - 1)/(-A*cosec %2B cot(A) %2B 1)=(cos(A) %2B 1)/sin(A) |
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Answer» Cosec A + cot A - 1 / cot A - cosec A + 1we know that,cosec ² A - cot ² A = 1substituting this in the numeratorcosec A + cot A -(cosec ² A - cot ² A) / (cot A - cosec A + 1)x²-y²= (x+y)(x-y)cosec A + cot A - (cosec A + cot A) (cosec A - cot A) / (cot A - cosec A + 1)taking common(cosec A + cot A)(1-cosec A + cot A) / (cot A - cosec A + 1)cancelling like terms in numerator and denominatorwe are left with cosec A + cot A= 1/sin A + cos A/sin A= (1+cos A) / sin A |
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| 36. |
Find the Desvate of 212 |
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| 37. |
213+212Show that2-1 |
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Answer» thank you so much 🙂🙂 |
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| 38. |
Solve: 212-1+32x-1) |
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Answer» Thank you so much |
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| 39. |
for 165 days3. In what period willsimple interest?5,200 amounts to? 7,384 at 12% per annua |
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Answer» plz do like if you like the answer |
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| 40. |
896÷5 |
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| 41. |
(a(1 + x)n. ě322a1 +a2a3 + a4a + a23 |
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Answer» Bhai chayya aap a dkho |
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| 42. |
-y i ही.ं १ 2 NA4 मिट = 383 850333 3 > |
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Answer» If you like the solution, Please give it a 👍 |
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| 43. |
ही —~ e \C ) 2 A b Sl —_‘—*—MQ— ve o <B 2 M—MMW f = 0.yt dle(A)- o 820 (c)o.celn)d |
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Answer» P(A) + P(A') = 1 P(A') = 1 - 0.45 = 0.55 (c) is correct |
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| 44. |
प्र र्् नडीन्ड, then value ofYtX +—=is |
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Answer» x(3 - 2/x) = 3/x3x - 2 = 3/x3x - 3/x = 23(x - 1/x) = 2x - 1/x = 2/3 Squaring both sides (x - 1/x)^2 = x^2 + 1/x^2 - 2 (2/3)^2 = x^2 + 1/x^2 - 2 4/9 + 2 = x^2 + 1/x^2 Therefore,Value of x^2 + 1/x^2 = (4+18)/9= 22/9 |
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| 45. |
Find all the zeroes of the polynomiala4 + 13-34r2-4x+ 120, if two of its zeroes are2 and - 2. |
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| 46. |
R SRS ST ARy YTрео F vt реорео 2 |
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Answer» x^2 + 1/x^2= (x+ 1/x)^2 - 2= [2*(3^0.5)]^2 - 2= (12) ^ 2 - 2= 144 - 2= 142 |
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| 47. |
\frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} 0}=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} 0} |
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| 48. |
(-tan(theta)^2 %2B 1)/(cot(theta)^2 - 1)=tan(theta)^2 |
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Answer» Let theta = x LHS:1 - tan^2 x/cot^2 x - 1 = (1 - sin^2 x/cos^2 x) / ( cos^2 x/sin^2 x - 1) = (cos^2 x - sin^2 x)/cos^2 x / (cos^2 x - sin^2 x)/sin^2 x = sin^2 x/cos^2 x = tan^2 x = RHS Hence proved I don't know sorry.but plz thanks me |
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| 49. |
हा. -—xan g+t E- =MD D |
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| 50. |
\left. \begin{array} { l } { ( 1 + \operatorname { tan } \theta + \operatorname { sec } \theta ) ( 1 + \operatorname { cot } \theta - \operatorname { cosec } \theta ) } \\ { ( \operatorname { sin } \theta + \operatorname { cos } \theta ) ^ { 2 } + ( \operatorname { sin } \theta - \operatorname { cos } \theta ) ^ { 2 } } \\ { ( \operatorname { sec } ^ { 2 } \theta - 1 ) ( \operatorname { cosec } ^ { 2 } \theta - 1 ) } \end{array} \right. |
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Answer» (1+cot A -cosec A)(1+ tan A + sec A) Lhs =(1 +cos A/sin A - 1/sin A)(1 + sin A/cos A +1/cos A) =(sin A+cos A -1/sin A)(cos A +sin A+1/cos A) =(sin A+cos A-1)(sin A+cos A+1)/sin Acos A =[(sin A+cos A)²-(1)²]/sin Acos A =[sin²A+cos²A +2sin Acos A- 1]/sin Acos A =[1-1+2sin Acos A]/sin Acos A =2sin Acos A/sin Acos A =2 =Rhs thanks |
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