Question

Given I_m = $\int_{1}^{e}{(\log x)^{m}}$dx . If $\frac{I_{m}}{K}$ + $\frac{I_{m - 2}}{L}$ = e then values of K and L are –

• A. 1 – m ,$\frac{1}{m}$
• B. $\frac{1}{1 - m}$, m
• C. $\frac{1}{1 - m}$,$\frac{m(m - 2)}{1 - m}$
• D. $\frac{m}{m - 1}$, m – 2

Answer 1

Answer: (A)

1 – m ,$\frac{1}{m}$

Answer 2

I_m = $\int_{1}^{e}{(\log x)^{m}}$dx = x $\int _ { 0 } ^ { \pi } \sin x d x$ – m $\int_{1}^{e}{(\log x)^{m - 1}dx}$

= e – m $\left\lbrack \left. \ x(\log x)^{m - 1} \right|_{1}^{e} - (m - 1)\int_{1}^{e}{(\log x)^{m - 2}dx} \right\rbrack$

= e – me + m (m – 1) I_{m – 2} = (1 – m) e + m (m – 1) I_{m – 2}

So $\frac{I_{m}}{1 - m}$ + m I_{m – 2} = e. Thus K = 1 – m

and L = $\frac{1}{m}$.