Question

Let $N=2^{101}\times \frac{{\int }_0^1{x}^{50}(1-x{)}^{50}dx}{{\int }_0^1{x}^{50}(1-x^{102}{)}^{50}dx}$ then the number of ways in which N can be resolved into two factors which are relatively prime numbers is

Answer 1

Let $I={\int }_0^1{x}^{50}(1-x^{102}{)}^{50}dx$
$x^{51}=t\Rightarrow 51x^{50}dx=dt$
$I={\int }_0^1(1-t^2{)}^{50}.\frac{dt}{51}$
$I=\frac{1}{51}{\int }_0^1(1-t{)}^{50}(1+t{)}^{50}dt$
$I=\frac{1}{51}{\int }_0^1{t}^{50}(2-t{)}^{50}dt$ $[\therefore {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right]$
$t=2Z$
$I=\frac{2}{51}{\int }_0^{\frac{1}{2}}{2}^{50}{2}^{50}{z}^{50}(1-z{)}^{50}.dz$
$I=\frac{2^{101}}{51}{\int }_0^{\frac{1}{2}}{2}^{50}(1-z{)}^{50}.dz$
So $N=\frac{2^{101}\times 2{\int }_0^{\frac{1}{2}}{x}^{50}(1-x{)}^{50}dx}{\left(\frac{2^{101}}{51}\right){\int }_0^{\frac{1}{2}}{2}^{50}(1-z{)}^{50}dz}\left(\begin{array}{c}{\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)\\ iff\left(x\right)=f(2a-x)\end{array}\right)$
N =102
So $N=2\times 3\times 17$
So N can be resolved into two factors which are relatively prime is $\frac{2\times 2\times 2}{2}=4$