Match the terms given in Column I with the compound given in Column II.

Answer

4

Explanation

a. Let `f(x)=ax^(2)+bx ( :' f(0)=0)`
Given `int_(0)^(1)f(x)dx=1`
`=2a+3b=6`
`=(a,b)=(0,2)` and `(3,0)`
b. `f(x)=sin(x^(2))+cos(x^(2))`
`=sqrt(2)cos(x^(2)-(pi)/4)`
For maximum value `x^(2)-(pi)/4=2n pi, n epsilonZ`
`impliesx^(2)=2n pi+(pi)/4,n epsilon Z`
`implies x=+-sqrt(x/4)+-sqrt((9pi)/4)` as `x epsilon[-sqrt(13),sqrt(13)]`
c. `I=int_(-2)^(2)(3x^(2))/((1+e^(x)))dx`...............1
`=int_(-2)^(2)(3(-x)^(2))/(1+e^(-x))dx`
`:. I=int_(-2)^(2)(e^(x)(3x)^(2))/(e^(x)+1)dx`............2
Addding 1 and 2
`impliesI+I=int_(-2)^(2)(3x^(2))/((1+e^(x)))dx+int_(-2)^(2)(e^(x)(3x^(2)))/(e^(x)+1)dx`
`=int_(-2)^(2)3x^(2)dx=2int_(0)^(2)3x^(2)dx=16`
`implies I=8`
d. We have `I=(int_(1//2)^(1//2)cos2x.log((1+x)/(1-x))dx)/(int_(0)^(1//2)cos2x.log((1+x)/(1-x))dx)`
Let `f(x)=cos2xI((1+x)/(1-x))`
`:.f(-x)=cos(-2x)In((1-x)/(1+x))`
`=-cos(2x)In((1+x)/(1-x))=-f(x)`
Thus, `f(x)` is an odd function.
`impliesI=0`