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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. |
Springs of spring constant K,3K,9K,27K,….243K are connected in series. A mass M is attached to the last spring and the system is allowed to make oscillation. The time period will be |
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Answer» `2pisqrt((M)/(K))` `(1)/(K_(0))=(1)/(K)+(1)/(3K)+(1)/(9K)+.......+(1)/(243K)` `:. K_(0)=(243K)/(364)` `T=2pisqrt((m364)/(243K))=(4pi)/(9)sqrt((91M)/(3K))` |
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| 2. |
If a line with y - intercept 2 is perpendicular to the lines 3x - 2y=6, then its x - intercept is |
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Answer» 1 |
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| 3. |
Evaluate the following integrals (x) int_(pi/6)^(pi/3)cot^(2) x dx |
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| 4. |
What is the greatest possible perimeter of a right angled triangle with integer side lengths if one of the A sides has length 12? |
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| 5. |
State the converse, inverse and contrapositive of if triangle ABC is equilateral, then its three angles are congruent propositions. Stating it as a conditional, wherever necessary. |
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Answer» SOLUTION :If `Delta` ABC is equilateral , then its three ANGLES are CONGRUENT. CONVERSE :If the three angles of the `Delta` ABC are congruent, then it is equilateral. If `Delta`ABC is not equilateral, then its three angles are not congruent. Contrapositive : If the three angles of the Delta ABC are not congruent, then it is not equilateral. |
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| 6. |
int(dx)/(tanx+cotx)=.... |
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Answer» `(COS2X)/(4)+C` |
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| 7. |
If sum_(i=1)^(n) x_(i)^(2) =300 and sum_(i=1)^(n) x_(i) =50 thenpossible value of n is : |
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Answer» 4 `=(300)/n -((50)/(n))^(2)=100/(n^2)(3n-25) gt C` `implies n gt 25/3`. |
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| 8. |
A(2, 1, 2), B(1, 0, 0) and C(1 + sqrt(3), sqrt(3), - sqrt(6)) are the vertices of a triangle. If the length of the median drawn to the side BC is equal to lambda sqrt(9 - 2) sqrt(3 + 2) sqrt(6). then lambda equal to |
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Answer» 4 |
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| 9. |
The sum of the series sum_(n =8)^(17)1/((n + 2)(n + 3)) is equal to |
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Answer» `1/17` |
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| 10. |
int(dx)/(sin^(5)xcos^(5)x) is equal to |
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Answer» `(1)/(4)(TAN^(4)x-cot^(4)x)+2(tan^(2)x-cot^(2)x)+C` |
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| 11. |
Evaluate the following integrals int(x^2)/(sqrt(1-x^(2)))e^(m sin^(-1)x)dx |
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| 12. |
Out of the following which statement is false ? |
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Answer» All the DIAGONAL ELEMENTS of a symmetric matrix are zero. |
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| 13. |
Definite integration as the limit of a sum : lim_(ntooo)[(1+(1)/(n^(2)))(1+(2^(2))/(n^(2)))(1+(3^(2))/(n^(2)))......(1+(n^(2))/(n^(2)))]^(1/n)=....... |
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Answer» `2E^((pi-4)/(2))` |
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| 14. |
The value(s) of x for which f(x)=(e^(sinx))/(4-sqrt(sqrt(x^(2)-9)) is continuous, is (are): |
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Answer» 3 |
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| 15. |
Mr. X is selected for interview for 3 posts. For the first post there are 5 candidates, for the second there are 4 and for the third there are 6. If the selection of each candidate is equally, likely, find the chance that Mr. X will be selected for atlest one post |
| Answer» Answer :D | |
| 16. |
int(10x^(9)+10^(x)log_(e)10dx)/(x^(10)+10^(x)) equals |
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Answer» `10^(X)-x^(10)+C` |
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| 17. |
If f : R rarr Rbe given by f(x) = (3-x^(3))^(1/3) then fof (x) is |
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Answer» `X^(1/3)` |
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| 18. |
If function f(x) is continuous at x= 0, f(x)= {((sin (4x))/(9x)",",x ne 0),(k^(2),x= 0):} then k= …….. |
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Answer» `-(3)/(2)` |
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| 19. |
Let T_(n) denotes the n ^(th) term of a G.P. with common ratio 2 and (log _(2) ( log _(512) T _(100))))=1. If three sides of a triangle ABC are the values of (T _(1) + T _(2)), T_(2) and T_(3) then area of the triangle is (sqrt 2160)/(N), where N is a positive integer. Find the remainder when N is divided by 2 ^(10). |
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| 20. |
int_(-pi//2)^(pi//2) sin^2 x log ((2- sin x)/(2 + sin x)) dx is : |
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Answer» 1 |
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| 21. |
Tower of Hanoi, a famous Beauxbatons puzzle which consists of three towers (pegs) and more than one rings, is as depicted − These rings are of different sizes and stacked upon in an ascending order, i.e. the smaller one sits over the larger one. There are other variations of the puzzle where the number of disks increase, but the tower count remains the same. Rules: The mission is to move all the disks to another tower without violating the sequence of arrangement. A few rules to be followed for Tower of Hanoi are − 1.Only one disk can be moved among the towers at any given time.2.Only the "top" disk can be removed. 3.A larger disc cannot be placed on a smaller disc. For instance if we take 3 discs in 1st peg saying it as peg A it will take 7 steps to take it to 2nd peg. The following variation of the famous TOWER OF HANOI puzzle was offered to Fleur Delacour by Madame Maxime in order to get the location of the next puzzle by Madame Maxime which was locked inside a box.The rules of the puzzle are essentially the same: disks are transferred between pegs one at a time. At no time may a bigger disk be placed on top of a smaller one. The difference is that now for every size there are two disks: one green and one red. Also, there are now two towers of disks of alternating colors. The goal of the puzzle is to arrange all red discs on one tower, and all green discs on another. The biggest discs at the bottom of the towers are assumed to swap positions. For instance: Now Fleur is given 6 discs of each colour( n = 6)arranged in the same way as the initial configuration. What are least number of moves she will need in order to get the location of the next puzzle ? |
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Answer» Solution :`N =6` No of moves is 45 Let the DISKS be numbered 1,…,`N_(1)`,…,N from smallest to LARGEST. WITHOUT loss of generality (WLOG), suppose they are all on peg A at the start. Disks N and `N_(−1)` are different colors, so at a minimum you MUST move disk `N_(−1)` off of disk N and onto another peg. WLOG, suppose you decide to move disk `N_(−1)` to peg B. In order to do that, you must first move disks `1,…,N_(−21),…,N_(−2 )` to peg C. So now you have disk N on peg A, disk `N_(−1)` on peg B, and all the other disks on peg C. You have to get disk `N_(−2)` onto peg A. But in order to do this you have to move all the smaller disks from peg C to peg B . Do all of this, so now you have disks N and `N_(−2)` on peg A and all the other disks on peg B. Fortunately, you do not have to move disk `N_(−3)` again, because it is already exactly where you want it (on top of disk `N_(−1)`). In fact, you now have the same problem you started out with, but with N−2 disks on peg B instead of N disks on peg A, and you need to move disk `N_(−3)` to peg A we can MAKE a recursive algorithm out of this, with the recursion occurring each time the number of disks to move is reduced by 2. Likewise, we can make a recursive formula to compute how many moves the algorithm will require. |
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| 22. |
Let A_(1), A_(2)………A_(26) be a regular polygon with 26 sides inscribed in a circle of radius R. Now A_(1)^('), A_(7)^('), A_(g)^('), be the projections of the orthocentre H ot /_\A_(1)A_(7)A_(9) onto sides A_(7)A_(9), A_(1)A_(9), A_(1)A_(7) respectively. Then |
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Answer» `HA_(1)^(')-HA_(7)^(')+HA_(9)^(')=(2R)/3` `HA_(7)^(')=-2RcosA_(1)cosA_(7)` `HA_(9)^(')=-2RcosA_(1)cosA_(7)` `/_A_(1)=(pi)/13, /_A_(7)=(9pi)/13, /_A_(9)=(3pi)/13` so USING, we get `HA_(1)^(')-HA_(7)^(')+HA_(9)^(')=R/2` |
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| 23. |
You are given the following data: Correlation coefficient between x and y=0.66. Find the equations of the lines of regression. |
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Answer» Regression equation of x or y : x = 0.91 y - 41. 35 |
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| 24. |
If the sum of two roots of the equation x^(3)+ax^(2)+bx +c=0 is zero, then the value of ab equals |
| Answer» ANSWER :a | |
| 25. |
Let I_(m","n)= int sin^(n)x cos^(m)x dx. Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) "" (ii) I_(n+2","m) (iii) I_(n","m-2) "" (iv) I_(n","m+2) (v) I_(n-2","m+2)"" I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2), then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos xin m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1. Now, choose the exponent of sin xfor m - 1of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x. Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation I_(4","2) and I_(4","4) is |
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Answer» `I_(4","2)=1/3 (sin^(5)X cos^(3)x+8I_(4","4))` |
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| 26. |
Let I_(m","n)= int sin^(n)x cos^(m)x dx. Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) "" (ii) I_(n+2","m) (iii) I_(n","m-2) "" (iv) I_(n","m+2) (v) I_(n-2","m+2)"" I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2), then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos xin m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1. Now, choose the exponent of sin xfor m - 1of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x. Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(6","2) is |
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Answer» `I_(4","2)=1/5 (sin^(3)XCOS^(3)X+8 I_(6","2))` |
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| 27. |
Let I_(m","n)= int sin^(n)x cos^(m)x dx. Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) "" (ii) I_(n+2","m) (iii) I_(n","m-2) "" (iv) I_(n","m+2) (v) I_(n-2","m+2)"" I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2), then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos xin m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1. Now, choose the exponent of sin xfor m - 1of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x. Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(2","2) is |
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Answer» `I_(4","2)=(1)/(6)(-sin^(3)x cos^(3)x+3I_(2","2))` |
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| 28. |
Ifalpha, beta , gammaare rootsof theequationx^3+ ax^2 + bx +c=0thenalpha^(-1)+ beta^(-1) + gamma^(-1) = |
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Answer» `a/c` |
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| 30. |
Let the position vectors of two points A and B be a+b+c and a-2b+3c, respectively. If the points P and Q divide AB in the ratio 1 : 3 internally and externally respectively, then 3|AB|= |
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Answer» `4 |PQ|` |
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| 31. |
Let P( n)be thestatement representthe sumof threesuccessivenaturalnumbersAA n inN ,thenthesmallestvalueofnforwhichP(n)is divisibleby 9, is |
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Answer» 1 |
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| 32. |
Let f : R^(+)toR satisfies the functional equation f(xy)=e^(xy^(-x^(-y))(e^(y)))f(x)+e^(x)f(y))AA x,yin R^(+). If f'(1)=e,determine f(x). |
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| 33. |
Write the value of int(d(x^2+1))/(x^2+5). |
| Answer» SOLUTION :`INT(d(x^2+1))/(x^2+5)int(d(x^2+1))/((x^2+1)+4)=In(x^2+1+4)+C=In(x^2+5)+c` | |
| 34. |
What is the inverse of :[[1,0,0],[0,1,0],[0,0,1]] |
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Answer» SOLUTION :`I=[[1,0,0],[0,1,0],[0,0,1]]` is a `3xx3` UNIT MATRIX `:.I^2=I` `impliesI^-I=I` |
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| 35. |
((1+cos""(pi)/(8)-isin""(pi)/(8))/(1+cos""(pi)/(8)+isin""(pi)/(8)))^(8) is equal to |
| Answer» ANSWER :B | |
| 36. |
The vertices of a triangle are at the vertices of an octagon. The number of such triangles are |
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Answer» 40 |
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| 38. |
int_1^3 e^x(x+1)dx |
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Answer» SOLUTION :`int_1^3 e^X(x+1)dx` [Integrating by parts CHOOSING (x+1) as first and `e^x`as second FUNCTION] `[[(x+1)e^x]-inte^xdx]_1^3` `=[x*e^x+e^x-e^x]_1^3` `=[x*e^x]_1^3` `=3e^3-e` |
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| 39. |
If A.M and G.M between two positive numbers a and b are 10 and 8 respectively, find the numbers. |
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Answer» |
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| 41. |
Find thevalueofK sothat thefunctionf(x) = {{:(kx+1,"if"x le 5),(3x-5,"if"x ge5):}at x=5is acontinuousfunction. |
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| 42. |
Find the volume of the parallelopiped whose sides are given by the vectors. hati+hatj+hatk, hatk, 3hati-hatj+2hatk. |
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Answer» SOLUTION :Let `VECA = hati+hatj+HATK, vecb = hatk, vecc = 3hati-hatj+2hatk` VOLUME of the PARALLELOPIPED = `(vecaxxvecb).vecc` 
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| 44. |
Let A be the least number such that 10A is a perfect square and 35 A is perfect cube. Then the number of positive divisors of A is : |
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| 45. |
Integrate the following functions with respect to x. (a tan x - b cot x)^(2) |
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| 46. |
Prove that int_(0)^(1) ("arc cosx")^(n) dx = n ((pi)/(2))^(n-1) - n (n - 1) int_(0)^(1) ("arc cos x")^(n-1) dx (n gt 1) |
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| 47. |
Evaluate the following integrals int[(1)/(logx)-(1)/((logx)^(2))]dx |
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| 48. |
int_(0)^(pi//2)(cos x)/(1+sin x + cos x)dx= |
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Answer» `(pi)/(4)+(1)/(2).LOG 2` |
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| 49. |
If cos ^(-1)\ (x)/(a) + cos^(-1)\ (y)/(b) = theta, then (x^(2))/(a^(2)) +(y^(2))/(b^(2))=? |
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Answer» `(XY)/(ab) cos THETA + cos^(2) theta` |
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| 50. |
How many different 5-digited numbers can be made, the sum of whose digits is even. |
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