`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) `-

`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 - 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`

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 `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 - 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`

Correct Answer - 9 `f(x)=x+x int_(0)^(1)int_(0)^(1)tf(t)dt+int_(0)^(1)t^(2)f(t)dt` `:.f(x)=x(1+A)+B`, where `A=int_(0)^(1)tf(t)dt` an `B=int_(0)^(1)t^(2)f(t)dt` Now `A=int_(0)^(1)t[t(1+A)+B]dt=(t^(3))/3(1+A)|_(0)^(1)+B/2t^(2)|_(0)^(1)=(1+A)/3+B/2` or `4A-3B=2`.............1 Again `B=int_(0)^(1)t^(2)[t(1+A)+B]dt=(t^(4)(1+A))/4+(Bt^(3))/3]_(0)^(1)` `=(1+A)/4+B/3`.............2 or `8B-3A=3` Solving equations 1 and 2 we have `B-18/23=f(0)`.

Correct Answer - (i) `pi` , (ii) `pi/4` , (iii) `4 + ln 5` , (iv) `8/21`

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 "...