`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 decreasing order in which values `I_1,I_2,I_3` lie.
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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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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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`I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx`, `I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1
Correct Answer - C `I_(1)=int_(0)^(pi//2)(sinx_cosx)/(1+sinx cos) dx` `=int_(0)^(pi//2) ("sin"((pi)/2-x)-"cos"((pi)/2-x))/(1+"sin"((pi)/2-x)"cos"((pi)/2-x))` `=int_(0)^(pi//2)(cosx-sinx)/(1+sinx cosx)dx=-I_(1)` or `I_(1)=0` `I_(3)=0` as `sin^(3)x` is odd `I_(4)=int_(0)^(1)In((1-x)/x)dx` `=int_(0)^(1)In((1-(1-x))/(1-x))dx` `=int_(0)^(1)In x/(1-x)dx=-I_(4)` or `I_(4)=0` `I_(2)=int_(0)^(2pi)cos^(6)x dx=2int_(0)^(pi)cos^(6)x dx!=0`
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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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`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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`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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`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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Evaluate : (i) `int_(-1)^(1)e^(x)dx` , (ii) `int_(-pi//4)^(pi//4)|sinx|dx` , (iii) `int_(-pi//4)^(pi//4)(x+pi//4)/(2-cos2x)dx` (iv) `int_(-1)^(1)sin^(
Correct Answer - (i) `2e-2` , (ii) `2-sqrt(2)`, (iii) `(pi^(2))/(6sqrt(3))` , (iv) `0` , ( v) `0`
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Evaluate : (i) `int_(0)^(2pi) {sin(sinx)+sin(cosx)}dx`, (ii) `int_(0)^(pi) (dx)/(5+4cos2x)` (iii) `int_(0)^(pi//2) (2lnsinx-ln sin2x)dx` , (iv) `int_(
Correct Answer - (i) `0` , (ii) `pi/3` , (iii) `- (pi)/(2) ln 2` , (iv) `piln 2`
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