Given Im=∫1e (log⁡x)mdx, If ImK+Im−2L= e then values of K and L are

Im=∫1e (log⁡x)mdx=x(log⁡x)m1e−m∫1e (log⁡x)m−1dx=e−mx(log⁡x)m−11e−(m−1)∫1e (log⁡x)m−2dx=e−me+m(m−1)Im−2=(1−m)e+m(m−1)Im−2So Im1−m+mIm−2=e. Thus K=1−m and L=1m.

Given Im=∫1e (log⁡x)mdx, If ImK+Im−2L= e then values of K and L are

Im=∫1e (log⁡x)mdx=x(log⁡x)m1e−m∫1e (log⁡x)m−1dx=e−mx(log⁡x)m−11e−(m−1)∫1e (log⁡x)m−2dx=e−me+m(m−1)Im−2=(1−m)e+m(m−1)Im−2So Im1−m+mIm−2=e. Thus K=1−m and L=1m.

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

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$1 - m, \frac{1}{m}$

$\frac{1}{1 - m}, \frac{m (m - 2)}{m - 1}$

$\frac{m}{m - 1}, m - 2$

$\frac{1}{1 - m}, m$

answer is A.

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Detailed Solution

$I_{m} = \int_{1}^{e} (\log x)^{m} d x = {x (\log x)^{m}|}_{1}^{e} - m \int_{1}^{e} (\log x)^{m - 1} d x$

$\begin{array}{l} = e - m [{x (\log x)^{m - 1}|}_{1}^{e} - (m - 1) \int_{1}^{e} (\log x)^{m - 2} d x] \\ = e - m e + m (m - 1) I_{m - 2} = (1 - m) e + m (m - 1) I_{m - 2} \end{array}$