`"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) `-1/2`
A. `2`
B. `1/2`
C. `1`
D. `-1/2`
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`L e tI_1=int_(pi/6)^(pi/3)(sinx)/x dx ,I_2=int_(pi/6)^(pi/3)("sin"(sinx)/(sinx)dx ,I_3=int_(pi/6)^(pi/3)(sin(tanx)/(tanx)dx` Then arrange in the decr
`f(x)=(sinx)/x` is a decreasing function and `(sinx)/xgt0` for all `x` in `(0,pi)`. Since `sin lt xlt tanx`, `(sin(sinx))/(sinx)gt(sinx)/xgt(sin(tanx))/(tanx)` for `(pi)/6lt xlt (pi)/3` `:.I_(2)gtI_(1)gtI_(3)`
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`IfI_1=int_0^1 2x ,I_2=int_0^1 2^x^3dx ,I_3=int_1^22^x^2dx ,I_4=int_1^2 2^x^3dx ,` then which of the following is/are ture? `I_1> I_2` (b) `I_2> I_2`
Correct Answer - A::D For `0lt xlt 1, x^(3)gtx^(3)` or `2^(x^(2))gt2^(x^(3))` or `int_(0)^(1)2^(x^(2))dxgt int_(0)^(1)2^(x^(3))dx` Hence `I_(1)gtI_(2)` Also for `1ltxlt2, x^(2)ltx^(3)` or `2^(x^(2))lt2^(x^(3))` or `int_(1)^(2)2^(x^(2))dxlt int_(1)^(2)2^(x^(3))dx` or `I_(3)ltI_(4)`
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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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`Iff(x)=(e^x)/(1+e^x),I_1=int_(f(-a))^(f(a))xg(x(1-x)dx ,a n d` `I_2=int_(f(-a))^(f(a))g(x(1-x))dx ,t h e nt h ev a l u eof(I_2)/(I_1)i s` `-1` (b) `-
Correct Answer - C `f(x)=(e^(x))/(1+e^(x))` `:.f(a)=(e^(a))/(1+e^(a))` and `f(-a)=(e^(-a))/(1+e^(-a))=(e^(-a))/(1+1/(e^(a)))=1/(1+e^(a))` `:. f(a)+f(-a)=(e^(a)+1)/(1+e^(a))=1` Let `f(-a)=alpha` or `f(a)=1-alpha` Now `I_(1)=int_(alpha)^(1-alpha)xg(x(1-x))dx` `=int_(alpha)^(1-alpha)(1-x)g((1-x)(1-(1-x))dx` `=int_(alpha)^(1-alpha) (1-x)g(x(1-x))dx` `:. 2I_(1)=int_(alpha)^(1-alpha) g(x(1-x))dx=I_(2)` or `(I_(2))/(I_(1))=2`
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If `I_1=int_0^1(e^x)/(1+x) dx`aand `I_2=int_0^1 x^2/(e^(x^3)(2-x^3)) dx` then `I_1/I_2` is
Correct Answer - C `I_(1)=int_(0)^(1)(e^(x)dx)/(1+x), I_(2)=int_(0)^(1)(x^(2)dx)/(e^(x^(2))(2-x^(3)))` In `I_(2)` put `1-x^(3)=t` `:. I_(2)=1/3 int_(1)^(0) (-dt)/(e^(1-t)(1+t))=1/(3e)int_(0)^(1)(e^(t)dt)/(1+t)=1/(3e)I_(1)` `:. (I_(1))/(I_(2))=3e`
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`L e tI_1=int_(-2)^2(x^6+3x^5+7x^4)/(x^4+2)dxa n d` `I_2=int_(-3)^1(2(x+1)^2+11(x+1)+14)/((x+1)^4+2)dtdot` Then the value of `I_1+I_2i s` 8 (b) `(200)
Correct Answer - C In `I_(2)` put `x+1=t`. Then `I_(2)=int_(-2)^(2)(2t^(2)+11t+14)/(t^(4)+2)dt=int_(-2)^(2)(2x^(2)+11x+14)/(x^(4)+2)dx` `:. I_(1)+I_(2)=int_(-1)^(2)(x^(6)+3x^(5)+7x^(4)+2x^(2)+11x+14)/(x^(4)+2) dx` `int_(-2)^(2)((x^(2)+3x+7)(x^(4)+2)+5x)/(x^(4)+2)dx` `=int_(-2)^(2)(x^(2)+3x+7)dx+5int_(-2)^(2)x/(x^(4)+2)dx` `=2 int_(0)^(2)(x^(2)+7)dx=100/3`
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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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If in triangle `A B C ,sumsinA/2=6/5a n dsumI I_1=9` (where `I_1,I_2a n dI_3` are excenters and `I` is incenter, then circumradius `R` is equal to `(1
Correct Answer - A We know that `II_(1) = 4R sin A//2` Now, `sum II_(1) = 9` `rArr 4R sum sin.(A)/(2) = 9` or `4R ((6)/(5)) = 9 " or "...
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