`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

`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)`

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)`

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...

Correct Answer - `I_(1)=piI_(2)` `I_(1)=int_(0)^(1)(pi-x)f(sin^(3)x+cos^(2)x)dx` `:. 2I=pi int_(0)^(pi)f(sin^(3)x+cos^(2)x) dx` `=2pi int_(0)^(pi//2) f(sin^(3)x+cos^(2)x)dx` or `I_(1)=pi int_(0)^(pi//2) f(sin^(3)x+cos^(2)x)dx=piI_(2)`

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)`.

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`

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`

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`

Correct Answer - A `I_(3)=int_(0)^(pi)e^(x)(sinx)^(3)dx` `=e^(x)(sinx)^(3)|_(0)^(pi)-3int_(0)^(pi)(sinx)^(2)cosx e^(x)dx` `=0-3(sinx)^(2)cosx e^(x)|_(0)^(pi)+3int_(0)^(pi)(2sinx cos x cosx)` `-sin x sin^(2)x)e^(x)dx` `=0+6int_(0)^(pi)sin x cos^(2) xe^(2) dx-3 int_(0)^(pi) sin^(3) xe^(x) dx` `=6int_(0)^(pi) sinx(1-sin^(2)x)e^(x)dx-3int_(0)^(pi)sin^(3)xe^(x)dx` `=6int_(0)^(pi) sinxe^(x)dx-9int_(0)^(pi) sin^(3)x e^(x)dx=6I_(1)-9I_(3)` or...

Correct Answer - C `AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]` `=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi...