`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`
`I_3> I_4`
(d) `I_3A. `I_(1)gtI_(2)`
B. `I_(2)gtI_(1)`
C. `I_(3)gtI_(4)`
D. `I_(3)ltI_(4)`
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 - 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)^(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)`