Prove that `int_0^(102)(x-1)(x-2)(x-100)` `x(1/((x-1)+1/((x-2))+1/((x-100))dx=101 !-100 !`
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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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Find the mistake of the following evaluation of the integral `I=int_0^pi(dx)/(1+2sin^2x)` `I=int_0^pi(dx)/(cos^2x+3sin^2x)` `=int_0^pi(sec^2x dx)/(1+3
Here the anti derivative `1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]=F(x)` is discontinuous at `x=pi//2` in the interval `[0,pi]`. Since `F((pi^(+))/2)=lim_(hto0)F((pi)/2+h)` `=lim_(hto0)(1/(sqrt(3)))tan^(-1){sqrt(3)"tan"(1/2pi+h)}` `=lim_(hto0)(1//sqrt(3))"tan"^(-1){-sqrt(3)coth}` `=(1/(sqrt(3))) tan^(-1)(-oo)=-pi//(2sqrt(3))` and `F(1/2pi-0)=pi//(2sqrt(3))!=F(1/2pi+0)` the second fundamental theorem of integral calculus is not...
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Prove that `int_0^1tan^(-1)(1/(1-x+x^2))dx=2int_0^1tan^(-1)x dxdot` Hence or otherwise, evaluate the integral `int_0^1tan^(-1)(1-x+x^2)dx`
`int_(0)^(1)"tan"^(-1)1/(1-x+x^(2))dx=int_(0)^(1)"tan"^(-1)(x+(1-x))/(1-x(1-x))dx` `=int_(0)^(1)[tan^(-1)x+tan^(-1)(1-x)]dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)(1-x)dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)tan^(-1)[1-(1-x)]dx` `=2int_(0)^(1)tan^(-1)x dx`................1 Now, `I=int_(0)^(1)tan^(-1)(1-x+x^(2))dx` `=int_(0)^(1)cot^(-1)(1/(1-x+x^(2)))dx` `=int_(0)^(1)[(pi)/2-tan^(-1)(1/(1-x+x^(2)))]dx` `=(pi)/2-2int_(0)^(1)x dx`[from equation 1] `=(pi)/2-2{x tan^(-1) x-1/2 log(1+x^(2))}_(0)^(1)` `=log_(e)2`
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`IfI_I=int_0^(pi//2)cos(sinx)dx ,I_2=int_0^(pi/2)sin(cosx)d ,a n dI_3=int_0^(pi/2)cosx dx ,` then find the order in which the values `I_1,I_2,I_3,` ex
Correct Answer - `I_(1)gtI_(3)gtI_(2)` We know that `cosx` is decreasing function in `(0,pi//2)`. Also `xgtsinx` for `xepsilon(0,pi//2)` Thus, `cosx lt cos (sinx)` Further, `xgtsinx` and `cosxepsilon(0,1)` for `xepsilon(0,pi//2)` `:.cosxgtsin(cosx)` Thus, `sin...
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The numbers of possible continuous `f(x)` defined in `[0,1]` for which `I_1=int_0^1f(x)dx=1,I_2=int_0^1xf(x)dx-a ,I_3=int_0^1x^2f(x)dx=a^2i s//a r e`
Correct Answer - D Since `a^(2)I_(1)-2aI_(2)+I_(3)=0` `int_(0)^(1)(a-x)^(2)f(x)dx=0` Hence, there is no such positive function `f(x)`.
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`IfI_1=int_0^(pi/2)(cos^2x)/(1+cos^2x)dx ,I_2=int_0^(pi/2)(sin^2x)/(1+sin^2x)dx` `I_3=int_0^(pi/2)(1+2cos^2xsin^2x)/(4+2cos^2xsin^2x)dx ,t h e n` `I_1
Correct Answer - C `I_(1)int_(0)^(pi//2)(cos^(2)x)/(1+cos^(2)x)dx` `=int_(0)^(pi//2)(cos^(2)(pi//2-x))/(1+cos^(2)(pi//2-x))dx` `=int_(0)^(pi//2) (sin^(2)x)/(1+sin^(2)x)dx=I_(2)` Also `I_(1)+I_(2)=int_(0)^(pi//2)((sin^(2)x)/(1+sin^(2)x)+(cos^(2)x)/(1+cos^(2)x))dx` `=int_(0)^(pi//2)(sin^(2)x+sin^(2)xcos^(2)x+cos^(2)x+sin^(2)xcos^(2)x)/(1+sin^(2)x+cos^(2)x+sin^(2)xcos^(2)x)` `=int(1+2sin^(2)xcos^(2)x)/(2+sin^(2)x cos^(2)x)dx=2I_(3)` `:. 2I_(1)=2I_(3)` or `I_(1)=I_(3)` or `I_(1)=I_(2)=I_(3)`
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`int_0^x{int_0^uf(t)dx}d ui se q u a lto` `int_0^x(x-u)f(u)d u` `int_0^x uf(x-u)d u` `xint_0^xf(u)d u` (d) `xint_0^x uf(u-x)d u`
Correct Answer - A::B L.H.S `=int_(0)^(x) {int_(0)^(u)f(t)dt}du` Integrating by parts choose 1 as the second function. Then, L.H.S`={uint_(0)^(u)f(t)dt}_(0)^(x)-int_(0)^(x)f(u)u du` `=x int_(0)^(x)f(t)dt-int_(0)^(x)f(u)u du` `=x int_(0)^(x)f(u)du-int_(0)^(x)f(u) udu-int_(0)^(x)f(u)(x-u)du` `=R.H.S`.
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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 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
Correct Answer - 102 `(.^(101)C_(0).^(101)C_(1)x+.^(101)C_(2)x^(2)-"...."-.^(101)C_(101)x^(101))(1+x+x^(2)+"...." +x^(100))^(101)` `= (1-x)^(101).((1-x^(101))/(1-x))^(101)` `= (1-x^(101))^(101)` `:.` Number of different terms `= 102`
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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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