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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. |
Consider acircle with centre at origin and radius 6 units. Let PA be a tangent to the cirdefrom P(100) Find the length ofthe tangent. Find the coordinates of A |
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| 2. |
1 . \int x e ^ { x } d x |
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| 3. |
\int x ^ { 2 } e ^ { x } d x |
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| 4. |
4. In a train, 15people travel in first class and 35% travel in second class and theremaining travel in the A.C. class. Calculate the percentage of A.C. class travellers.5, A boy eats 25% of the cake and gives away 35% of it to his friends. What percent ofthe cake is still left with him ?6. What is the percentage of vowels in the English alphabet ?7, (i) 6 % of what number is 375 ?4(ii) 0.2% of a number is 5, Find the number.(iii) 30 is 16% of a number. Find the number. |
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| 5. |
cOS XEvaluate J 1 + sin²x |
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| 6. |
\operatorname cos ^ 2 x %2B \operatorname cos ^ 2 ( x %2B \frac \pi 3 ) %2B \operatorname cos ^ 2 ( x - \frac \pi 3 ) = \frac 3 2 |
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Answer» Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(x - π/3) = {1 + cos(2x - 2π/3)}/2 ii) Hence, left side of the given one is: = {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2 = (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)] = (3/2) + (1/2)[cos(2x) + 2cos(2x)*cos(2π/3)][Since cos(A+B) + cos(A-B) = 2cosA*cosB] = (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2] = 3/2 = Right side HENCE PROVED |
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| 7. |
dxEvaluate J 1 + sin x |
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Answer» tanx - secx +c is the anti derivative of the given question |
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| 8. |
e distance between the following pairs of points31.(4, 1Im) (-(iii) (a, b),( |
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| 9. |
\operatorname { cos } 2 x \operatorname { cos } \frac { x } { 2 } - \operatorname { cos } 3 x \operatorname { cos } \frac { 9 x } { 2 } = \operatorname { sin } 5 x \operatorname { sin } |
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| 10. |
\int x ( 1 + x ) ( 1 - x ) d x |
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| 11. |
\int x \sin x d x |
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Answer» Let u = x, which implies du/dx = 1 and let dv/dx = sin(x). Integrating this to get v gives v =–cos(x). So our integral is now of the form required for integration by parts. ∫ x sin(x)dx = ∫ u(dv/dx) dx = uv –∫ v(du/dx)dx =–xcos(x)– ∫–cos(x)*1dx =–xcos(x)– ∫–cos(x) dx =–xcos(x)+ ∫cos(x)dx The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x). Finally, as with all integration without limits, there must be a constant added, which I'll callc. So the final answer is ∫ x sin(x)dx =–xcos(x)+ sin(x) + c |
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| 12. |
-4 511 ŕĽŕ¤ż A2 tan~ 0 |
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| 13. |
\int x \sqrt { 1 + x - x ^ { 2 } } d x |
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| 14. |
\int x e ^ { x } \operatorname { cos } x d x |
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| 15. |
Meena went to a bank to withdraw 2000. She asked the cashier to give her750 and 100 notes only. Meena got 25 notes in all. Find how many notes of750 and 100 she received. |
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Answer» meena got 10 note of 50 and 15 note of 100 |
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| 16. |
\int x \operatorname { log } x d x = ? |
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| 17. |
19. Write the 2's complement of 1001010 |
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Answer» At first flip the Digits to get 1's compliment as 0110101 now add 1 to it.. to get 2's compliment as.... 110110........ (1& 1 , will add up to 0 and 1 will be carried ahead) the 2's complement of 1001010 is 110110 . |
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| 18. |
At what rate per annum will Rs1650 amount to Rs2046 in two years ? |
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| 19. |
de and check your answer.2781 = 35(b) 49277 = 5117335 = 122(d) 64895 = 247the least number that should be subtracted from |
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Answer» anwer will be 333. (a) 79.45(b) 96.43(c) 60.12(d) 262.247 |
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| 20. |
5 years. Find the sum and the rate per cent per ali24. Divide & 3000 into two parts such that the simple interest on the first part for 4 years at oper annum is equal to the simple interest on the second part for 2 years at 9% per annum1, that if one nari be lent at 9% per annum and the other a |
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| 21. |
\frac { \operatorname { cos } 4 x + \operatorname { cos } 3 x + \operatorname { cos } 2 x } { \operatorname { sin } 4 x + \operatorname { sin } 3 x + \operatorname { sin } 2 x } = \operatorname { cot } 3 x |
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Answer» We have to prove (Cos4x+Cos3x+Cos2x)/(Sin4x+Sin3x+Sin2x) =cot3x. So, Using the formula for the first and third terms of numerator : (CosA+CosB)=Cos((A+B)/2)*Cos((A-B)/2). =Cos((4x+2x)/2)*Cos((4–2)/2) =Cos3x*Cosx Using the formula for the first and third terms of denominator : (SinA+SinB)=Sin((A+B)/2)*Cos((A-B)/2). =Sin((4+2)/2)*Cos((4–2)/2) =Sin3x*Cosx So, Left hand side of given expression becomes =(Cos3x + Cos3x*Cosx)/(Sin3x + Sin3x*Cosx) =Cos3x(1+Cosx)/Sin3x(1+Cosx) =cot3x =Right hand side of the given expression. |
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| 22. |
Meena went to a bank to withdraw 2000. She asked the cashier to give her50 and? 100 notes only. Meena got 25 notes in all. Find how many notes of50 and 100 she received |
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| 23. |
find k for which quadratic polynomial 2x²-kx+k=0 have equal roots. |
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Answer» 2x^2 -kx +k =0 . equal roots condition B^2 -4AC=0 From equation A= 2 ;B= K ; C= k k^2 - 4 (2)(k)=0 k^2 -8k=0 K^2=8k k=8. or k(k-8)=0 k=0 thanks |
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| 24. |
Question 2:Farm the pair of linear equations in the following problems, and find their solutions (if they() If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces( Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as(i) The sum of the digits of a two-digit number is 9. Also, nine times this number is(iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50exist) by the e im nation method:to 1. It becomes % if we only add 1 to the denominator. What is the fraction?old as Sanu. Haw old are Nuri and Sonu?twice the number abtained by reversing the order of the digts. Find the number.and Rs 100 notes anly. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs100 she received.A lending library has a fxed charge for the first three days and an additional chargefor each day therea fter. Saritha paid Rs 27 for a book kept for seven deys, while Susy |
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Answer» there are two metal bars of lengthsc120cm and125cm. Short bar of equal length are cut from both metal bars. find the largest possible length of each short bar? no percentage lesson 9B 12 I'll write a two-digit number whose sum is 14 and subtract 29 from the number. The numerator will be equal. Let's create a symmetry and solve what will be the two-digit number. |
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| 25. |
9cosx - sin xdxEvaluate J 5 cos +4x +4sin x |
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| 26. |
If one of the zeroes of the quadratic polynomial (-1) 2 + kx + 1 is-3, then the value of kis. |
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Answer» Given p(x) = (k -1)x² + kx + 1 As (-3) is zero then p(-3) = 0 So, p(-3) = (k -1) (-3)² + k × (-3) + 1 = 0 9( k -1) - 3k + 1 = 0 9k - 9 - 3k + 1 = 0 6k = 8 k = 4/3 |
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| 27. |
2.If the sum of the zeroes of the quadratic polynomial 3x-kx + 6 is 3 the value of k is: |
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| 28. |
| (1 - cos 2x) dx |
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| 29. |
cos(3x +5)dx |
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| 30. |
\int x ( \sqrt { x } + a ) d x |
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| 31. |
\int \frac { 1 - \operatorname { cos } 2 x } { 1 + \operatorname { cos } 2 x } d x |
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| 32. |
An open metallic bucket is in the shape of a frustum of a cone. If thediameters of the two circular ends of the bucket are 45 cm and 25 cm andthe vertical height of the bucket is 24 cm, find the area of the metallicsheet used to make the bucket. Also find the volume of the water it canhold. (Use π -) |
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Answer» Area of metallic sheet used = CSA of Frustum + CSA of Cylinder + CSA of Base CSA of Frustum - Diameter of the bigger circular end = 45 cm Radius = 45/2 = 22.5 cm Diameter of the smaller circular end = 25 cm Radius = 25/2 = 12.5 cm Height of the frustum = Total height of the bucket - Height of the circular base ⇒40 - 6 = 34 cm Slant Height = l√h² + (r1² - r2²)² ⇒√34² + (22.5 - 12.5)² ⇒ √1156 + (10)² ⇒√1156 + 100 ⇒√1256 ⇒ Slant Height = 35.44 cm CSA of Frustum =π(r1 + r2)l ⇒ 22/7*(22.5 + 12.5)*35.44 ⇒ 22/7*35*35.44 = 3898.4 cm² Area of Circular Base - Base is a circular part with radius 25/2 = 12.5 cm Area of circular base =πr² ⇒ 22/7*12.5*12.5 491.07 cm² CSA of Cylinder = 2πrh ⇒ 2*22/7*12.5*6 ⇒ 471.428 Area of metallic sheet used = 3898.4 cm² + 491.07 cm² + 471.428 = 4860.898 cm² |
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| 33. |
100÷10×123+3333 |
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Answer» the correct answer is4563 this correct answer is :4563 the answer of the question is 4563 |
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| 34. |
ple 19 Write(3 a+4 b+5 c)^{2} in expanded form. |
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| 35. |
久c,Note BookkPage No.Date:2. |
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Answer» (14/25)×(35/21)×(17/49)=(14×35×17)/(25×21×49)=(2×5×17)/(25×3×7)=170/525=34/105 |
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| 36. |
ple 23:At what rate per annum willR1378 in 2 years?6360 yield an interest of |
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Answer» 1378 = 6360*5*r/2 r = 2756/31800 = 0.086% |
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| 37. |
C-9178 +511DATE: |
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Answer» 1a) 73,95,831 Seventy three lakhs ninety five thousand eight hundred and thirty one b) 8,70,17,002 Eight crores seventy lakhs seventeen thousand and two please answer the second question |
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| 38. |
uvite.fo.1C 6.14dec i m a오tlapao.ㅡㅡㅡㅡDate: I |
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Answer» 6.1A=6.1*10^-10=0.000061*10^-5 |
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| 39. |
2000. She asked the cashier to give the cashand F100 notes only. Rubina got 25 notes in all. Can you tell how many notes each of 50ple-8Rubina went to a bank to withdraw00 she received? |
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| 40. |
find the cube root of 729 |
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Answer» the cube root of 729 is 9 9 is the right answer 9 is the correct answer 9 is the cube root of 729 |
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| 41. |
The force applied on a body of mass 6 kg is directly proportional to the accelerationproduced in the body. Write an equation to express this observation and draw the graphof the equation. |
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| 42. |
sin(sin(IIC76. The value of LtX+0X21)02) II_3) II24 ) of theseమిగతా ప్రశ్నలు రేపటి పేజీల్లో |
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Answer» 3 no is the correct answer Differentiate it you will get correct answer |
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| 43. |
iIc le.ctors.. It one zero of the quadratic polynomial kx +7x +k is 2. Find thevalue of k. |
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| 44. |
\begin{array}{l}{\text { Evaluate the following integral }} \\ {\text { i) } \int \cos 2 x \cos x d x}\end{array} |
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| 45. |
\begin{array}{c}{\text { (or integral) of the following fu }} \\ {\text { 2. } \cos 3 x \text { . }} \\ {\text { 5. } \sin 2 x-4 e^{3 x}} \\ {\text { egrals in Exercises } 6 \text { to } 20 \text { : }} \\ {\text { 7. } \int x^{2}\left(1-\frac{1}{x^{2}}\right) d x} \\ {\text { 10. } \int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x}\end{array} |
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| 46. |
\int x \cos 2 x d x |
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| 47. |
Identify which of the following pairs of angles are complementary and which arsupplementary(i) 65°, 115,63°, 270Gii) 112, 680(ii) |
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| 48. |
42 \int \int \frac \operatorname cos 2 x ( \operatorname sin x %2B \operatorname cos x ) ^ 2 d x |
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| 49. |
\int x \sin x \cos x d x |
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| 50. |
\int \cos x \cos 2 x \cos 3 x |
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