The total number of different terms in the product `(.^(101)C_(0) - .^(101)C_(1)x+.^(101)C_(2)x^(2)-"….."-.^(101)C_(101)x^(101))(1+x+x^(2)+"…."+x^(100))^(101)` is `"____"`.
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Give the atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in the number of total p- and s- elec
Correct Answer - 36 First gas which contains d-electrons is Kr. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6` Total number of d-electrons = 10 Total number of p-electrons =6+6+6=18 Total...
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Two dice are thrown together and the total score is noted. The event E, F and G are a total 4, a total of 9 or more, and a total divisible by 5, respe
Two dices are thrown together i.e., sample spaces (S) =36`rArr`n(S)=36 E=A total of 4 = {(2,2),(3,1),(1,3)} `rArr n(E )=3` F= A total of 9 or more `={(3,6),(6,3),(4,5),(4,6),(5,4),(6,4),(5,5),(5,6),(6,5),(6,6)}` `rArr n(F)=10` G...
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Suppose a population A has 100 observations `101,102,................,200` and another population B has 100 observations `151,152,........250.` If `V_
Correct Answer - A `sigma^(2)=(sum d_(i)^(2))/(n)` (Here deviations are taken from the mean.) Since both A and B have 100 consecutive integers, they have the same standard deviation and hence variance....
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Consider any set of observations `x_(1),x_(2),x_(3),..,x_(101)`. It is given that `x_(1) lt x_(2) lt x_(3) lt .. Lt x_(100) lt x_(101)`, then the mean
Correct Answer - B Mean deviation is minimum when it is considered about an item equidistant from the beginning and the end, i.e., the median. In this case, the median is...
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Prove that `int_0^(102)(x-1)(x-2)(x-100)` `x(1/((x-1)+1/((x-2))+1/((x-100))dx=101 !-100 !`
Correct Answer - NA `I=int_(0)^(102)(x-1)(x-2)……(x-100)` `xx(1/((x-1))+1/((x-2))+…+1/((x-100)))dx` `-int_(0)^(102)d/(dx)((x-1)(x-2)………(x-100))dx` `=[(x-1)(x-2)…………(x-100)]_(0)^(102)` `=101!-100!`
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`"I f"I_1=int_(-100)^(101)(dx)/((5+2x-2x^2)(1+e^(2-4x)))` `"and"I_2=int_(-100)^(101)(dx)/(5+2x-2x^2),t h e n(I_1)/(I_2)"i s"` 2 (b) `1/2` (c) 1 (d) `-
Correct Answer - B `I_(1)=int_(-100)^(101)(dx)/((52x-2x^(2))(1+e^(2-4x)))` `=int_(-100)^(101)(dx)/((5+2(1-x)-2(1-x)^(2)(1+e^(2-4(1-x))))` or `2I_(1)=int_(-100)^(101)(dx)/(5+2x-2x^(2))=I_(2)` or `(I_(1))/(I_(2))=1/2`
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Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`
To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...
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The sum of the series `(.^(101)C_(1))/(.^(101)C_(0)) + (2..^(101)C_(2))/(.^(101)C_(1)) + (3..^(101)C_(3))/(.^(101)C_(2)) + "….." + (101..^(101)C_(101)
Correct Answer - 5151 Here `T_(r) = (r^(.1001)C_(r))/(.^(100)C_(r-1))= (r.(1001-r+1))/(r) = 102 -r` `:.` Given sum `= underset(r=1)overset(101)sum(102-r)` `= 101+100+99+"...."+2+1` `=5151`
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If the sum of the first `100` terms of an `AP` is `-1` and the sum of even terms lying in first `100` terms is `1`, then which of the following is not
Correct Answer - D `(d)` `x_(1)+x_(2)+x_(3)+……+x_(100)=(100)/(2)(x_(1)+x_(100))=-1` `impliesx_(1)+x_(100)=-(1)/(50)` `x_(2)+x_(4)+…+x_(100)=(50)/(2)(x_(1)+d+x_(100))=1` `impliesx_(1)+x_(100)+d=(1)/(25)` `impliesd=(3)/(50)` ltbr `x_(1)+x_(1)+99d=(-1)/(50)` `impliesx_(1)=(-149)/(50)` `x_(100)=x_(1)+99d` `=(-149)/(50)+99xx(3)/(50)` `=(74)/(25)`
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There is certain number of rows of chairs in a room .The number of chairs in each row is thrice the total number of rows. Find the number of charis in
Correct Answer - 81,27
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