InterviewSolution
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Match the statements/expressions given in List ( with the value given in List II. |
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Answer» 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 Addding1 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` |
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