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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. |
$ \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right) $(a) $ \frac{\pi}{3} $(c) $ \frac{\pi}{6} $(b) $ \frac{\pi}{4} $(d) $ \frac{\pi}{2} $ |
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| 2. |
. For the equations (a)JE-2 + 4z + 3 (b) 3x2 +3z +2-0 (c)Sy2-Gy-9-x : 0complete the square and reduce to one of the standard forms y-b A(z -a) or -a A(y - b). |
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| 3. |
\frac{5}{x-1}+\frac{1}{y-2}=2 \quad \text { tai } \frac{6}{x-1}-\frac{3}{y-2}=1 |
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Answer» 5/x-1+1/y-2=2 eq 16/x-1-3/y-2=1 eq 2let 1/x-1=u, 1/y-2=vmultiply eq 1 by 3, we get15u=3v=6 eq 3add eq 2 and 321u=7u=1/3x-1=3x=4b=1/3y-2=3y=5 |
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| 4. |
8.Two dice are throm simultaneously. If Xdenotes the number of sixes.Find the mean texpectation) of X.PART D |
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| 5. |
y-2Tai=3 |
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Answer» (y - 2)/7 - (y - 3)/3 = 3 Take LCM of 7 and 3 = 21 (y - 2)*3 - (y - 3)*7 = 21*3 3y - 6 - 7y + 21 = 63 -4y = 63 - 15 y = - 48/4 = - 12 Therefore,Value of y = - 12 |
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| 6. |
tai angleA (B |
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| 7. |
At matt tuo tai means |
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Answer» At most implies maximum. At most two tails imply maximum two tails.That is, the number of tails can be 0, 1, 2, but not more than 2. |
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| 8. |
(7) Let all the students of a class is an Universal set. Let set A be the students whosecure 50% or more marks in Maths. Then write the complement of setA. |
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Answer» Set of students who scored less than 50% marks = U-A=100-50=50% |
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| 9. |
2. Multiply and reduce2/3 and 2*2/3 |
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Answer» Mixed fraction into improper fraction 2 2/3 --------> 8/3. so, 2/3 × 8/3 =16/9. |
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| 10. |
MATT2.Multiply and reduce |
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Answer» 2/3×8/316/91.77 |
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| 11. |
Express 12 year 3rth in seconds |
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Answer» 386316000 is the correct answer. |
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| 12. |
5. A company earned 8,39,54,825 in the year 2010. Next year its earning increased by13,58,475. What was its earning in the year 2011? .The difference between two numbers je 170 507 Th11a0 A |
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Answer» In year 2010 company earned Rs 83954825 In year 2011 earning increased by 1358475 Then, its earning in the year 2011 = 83954825 + 1358475= Rs 85313300 |
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| 13. |
Find the number of radians between the hands of a clock at half past the34 बजे किसी घटी के दोनों सड़यों के बीच के कोण का मान रेडियन में निक।hand in einer |
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Answer» 75° is the right answer 75° is the right answer 60 is the write answer 75° is the correct answer of the following question. 75° is the right answer 75° is the correct answer of the given question |
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| 14. |
The angles of a triangle are in A.P and thegreatest angle is 84°. Find all the three angles in radians. |
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| 15. |
How to find gof |
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Answer» we can find gof by(gof)(x)=g(f(x)) gof(x)=g(f(x)) is the right answer |
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| 16. |
\left. \begin{array} { l } { \text { If } f ( x ) = \operatorname { log } _ { e } ( \frac { 1 + x } { 1 - x } ) , g ( x ) = \frac { 3 x + x ^ { 3 } } { 1 + 3 x ^ { 2 } } \text { and } } \\ { \text { gof } ( t ) = g ( f ( t ) ) , \text { then what is gof } ( \frac { e - 1 } { e + 1 } ) } \\ { ( a ) 2 } & { ( b ) 1 } & { ( c ) 0 } & { ( d ) } \end{array} \right. |
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| 17. |
If the L.C.M (12, 28) 84, find the H.CF(02,28) |
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Answer» We know LCM × HCF = Product of two numbers 84 × HCF = 12 × 28 HCF = (12×28)/84 = 28/7 = 4 HCF = 4 Hit like and share this app |
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| 18. |
Find gof if f(x) = 8x3 and g(x)x 3 . |
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Answer» GIVEN,f(x) = 8x^3 and g(x) = x^1/3 Then,gof = g(f(x))= (8x^3)^1/3= (2x)^3*1/3= 2x |
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| 19. |
1. In an equilateral triangle ABC, D is a point on side BC, such that BDBC. Prove that[CBSE-13-MAVDELI[CBSE-12-MA2-049] |
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Answer» ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC . Now, ∆ABE and ∆AEC ∠AEB = ∠ACE = 90° AE is common side of both triangles , AB = AC [ all sides of equilateral triangle are equal ]From R - H - S congruence rule , ∆ABE ≡ ∆ACE ∴ BE = EC = BC/2 Now, from Pythagoras theorem , ∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2) From equation (1) and (2) AB² - AD² = BE² - DE² = (BC/2)² - (BE - BD)² = BC²/4 - {(BC/2) - (BC/3)}² = BC²/4 - (BC/6)² = BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9 ∵AB = BC = CA So, AB² = AD² + 2AB²/9 9AB² - 2AB² = 9AD²Hence, 9AD² = 7AB² |
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| 20. |
12500 students appeared in an exam. 50% of theboys and 70% of the girls cleared theexamination. If the total percent of studentsqualifying is 60%, how many girls appeared in |
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Answer» 6250 the correct answer of the given question |
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| 21. |
in examination the ratio of passes to faliures is 4:1. had 30 less appeared and 20 less passed, the ratio would be 5:1 . how many students appeared for the examination?? |
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| 22. |
During the day the temperature in a town was 40°C. At night the temperature came downby 10°F. What was the night-time temperature in Fahrenheit scale? |
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Answer» 40 degree -10 degree = 30 degree F aso answer is 30 degree F was the temperature at night |
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| 23. |
Yangchen sleeps at half past 9 at night and gets up at quarter to 5 delily i tbemoming But one night,she was disturbed between 1am to 2.23 a.m How mutime does she sleep on that particular night? |
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| 24. |
S5. HOTS IFS, denotes the sum of first nterms ofcommonan AP then show that thedifference d of the AP is given by |
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| 25. |
1. (0) If f (x) log (I +x) and g(x)e, then value of (gof) (x) is |
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Answer» f(x) = log(1+x) g(x) = e^x (gof)(x) = g(f(x)) = g(log(1+x)) = e^(log(1+x)) = 1+x If you find this answer helpful then like it. |
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| 26. |
84 Find the number of integers between 1 and200, each of which has 10 divisors. |
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Answer» If there is an even number of divisors, then the product of an even number of negatives is positive. The product of an odd number of negatives is negative. The only numbers that have an odd number of divisors are perfect squares, so it suffices to find all the negative perfect squaresThe largest one is then(14)^2=196, which has factors1,2,4,7,14,28,49,98,196Therefore, there are1414integers. unable to understand |
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| 27. |
\begin{array} { l } { \text { Ex. 16. If } f ( x ) = x ^ { 2 } - 1 \text { and } g ( x ) = \sqrt { x } , \text { find } } \\ { \text { (a) } ( \text { fog } ) ( x ) } \\ {\ text { (gof)} { x } }end{array} |
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Answer» a) f(g) = f(√x) = (√x)² - 1 = x - 1 b)g(f) = g(x²-1) = √(x²-1) |
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| 28. |
1. If/(x) = log (1+x)and g(x)-er, then value of(HSB 2018gof (x) is(A) el tx(C) log x1 + x(D) none of these. |
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| 29. |
ŕ¤ŕ¤ž e Sz gt ool oy !9. Prove the following identity(tan 6 +2)(2 tan 0 + )=51tan 6 +2 sec? 0 |
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| 30. |
049Ifs+ y = 7 and xy = 12, then the value of (x2 + y2) is |
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Answer» x+y=7xy=12 (x+y)²=x²+y²+2xy7²=x²+y²+2(12)49-24=x²+y²x²+y²=25 |
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| 31. |
Question 3(a)Prove the following identity\frac{\sec A}{\sec A-1}+\frac{\sec A}{\sec A+1}=2 \csc ^{2} A |
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| 32. |
दिसमीकरण .* - 8: + £ न 0 का एक मूल 3 है, तो ८ का मान क्या है?(NDA 2011 1)) -15 (0) 9 15 ( |
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| 33. |
एक उतथापक किसी खान कूपक में ७ मीटर प्रति मीटर की दर से नीचे जाता है Yeh Dil Mein Hai Tumhi Se 5 metre Upar Se Neeche Jana Shuru karta hai to 205 metre Neeche Jaane Mein Kitna Samay lagega |
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Answer» bhai question me galti hai |
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| 34. |
If'a, b, c are non-coplanar vectors, then prove that the fourpoints -a+4b-3c, 3a + 2bse,-3a + 8b-se,-3a+2b+e are coplanar |
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| 35. |
20. Prove the following identity, where angle A is an acute angle:(sin A + cosec A)? + (cos A + sec A) 7 + tan?A + cot?A |
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Answer» (sinA+cosecA)²+(cosA+secA)² =sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA =sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA =1+cosec²A+sec²A+2+2 =5+(1+cot²A)+(1+tan²A) =7+tan²A+cot²A Like my answer if you find it useful! |
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| 36. |
पक (2 116. हक(5 a5a, 18 A0SEuetias Slivision dlgesiyसनक कक हक "ह 1686, 4% |
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| 37. |
a(se |
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Answer» true.Every rhombus is a kite, and any quadrilateral that is both akite and parallelogram is arhombus. Arhombusis a tangential quadrilateral. That is, it has an inscribed circle that is tangent to all four sides. |
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| 38. |
in math L Jcunc12 1 |
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| 39. |
116. The number of even divisors of 100will be2(2) 6(3) 5(4) 8 |
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| 40. |
If f:R ---->R and g: R---->R are definedby f(x)=2x^2+3 and g(x)= 3x - 2, then findi) (fogMx)ii ) (gof)(x)iii) (fof) (0)iv) (go(fof)(3) |
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| 41. |
2x +313) Ifs()3x-23x-2prove that fof is identity function. |
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Answer» Given, f(x) = 2x-3 / 3x-2 y =2x-3/ 3x-2 To find out f(y), substitute the value of y in f(x).f(y) = [2 (2x-3 / 3x-2) -3] / [3 (2x-3/3x-2) -2] = [2(2x-3) -3(3x-2)] / [3(2x-3) -2(3x-2)] = [4x - 6 - 9x + 6] / [6x -9 - 6x + 4] = [-5x] / [-5] = xhence fof is X an identity function. |
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| 42. |
iff:A+R:f(x)=(3-2P,show that (fof)(x)=2. |
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Answer» f(x)=(3-x^3)^1/3fof(x)=(3-(3-x^3)^1/3 ^3)^1/3=(x^3)^(1/3)=x thanks brother bt we have to prove that it is equal to x^3 ....I am also getting x.... |
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| 43. |
f ( x ) = ( a - x ^ { n } ) ^ { \frac { 1 } { n } } \text { then } \text { fof } ( x ) |
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| 44. |
17 Fastfofcot A -cos Acot A + cosAcosecA -1cosecA +1 |
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| 45. |
1.6x2-13x+5-0 |
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Answer» 6x² - 13x + 5 = 0 6x² - 3x - 10x + 5 = 0 3x ( 2x - 1) - 5 ( 2x - 1) = 0 (3x - 5) ( 2x - 1 ) = 0 So, x = 5/3 and 1/2 |
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| 46. |
6x2-13x + 6 = 0 |
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| 47. |
if perimeter is 100 meter and width is 20 metres what willbe the length? |
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Answer» Perimeter = 2(length+breadth)100=2(20+length)50-20 = length= 30meter |
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| 48. |
x2-13x+20=0 |
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| 49. |
X+ 13x t 15 0 |
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| 50. |
3x2-13x + 12 = 0 |
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