JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 9)
If $$I = \sum\nolimits_{k = 1}^{98} {\int_k^{k + 1} {{{k + 1} \over {x(x + 1)}}} dx} $$, then
$$I > {\log _e}99$$
$$I < {\log _e}99$$
$$I < {{49} \over {50}}$$
$$I > {{49} \over {50}}$$
Explanation
$$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $$
Clearly, $$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x{{(x + 1)}^2}}}dx} } $$
$$ \Rightarrow I > \sum\limits_{k = 1}^{98} {(k + 1)\int_k^{k + 1} {{1 \over {{{(x + 1)}^2}}}dx} } $$
$$ \Rightarrow I > \sum\limits_{k = 1}^{98} {( - (k + 1))\left[ {{1 \over {k + 2}} - {1 \over {k + 1}}} \right]} $$
$$ \Rightarrow I > \sum\limits_{k = 1}^{98} {{1 \over {k + 2}}} $$
$$ \Rightarrow I > {1 \over 3} + ... + {1 \over {100}} > {{98} \over {100}}$$
$$ \Rightarrow I > {{49} \over {50}}$$
Also, $$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $$
$$ = \sum\limits_{k = 1}^{98} {[{{\log }_e}(k + 1) - {{\log }_e}k]} $$
$$I < {\log _e}99$$
Clearly, $$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x{{(x + 1)}^2}}}dx} } $$
$$ \Rightarrow I > \sum\limits_{k = 1}^{98} {(k + 1)\int_k^{k + 1} {{1 \over {{{(x + 1)}^2}}}dx} } $$
$$ \Rightarrow I > \sum\limits_{k = 1}^{98} {( - (k + 1))\left[ {{1 \over {k + 2}} - {1 \over {k + 1}}} \right]} $$
$$ \Rightarrow I > \sum\limits_{k = 1}^{98} {{1 \over {k + 2}}} $$
$$ \Rightarrow I > {1 \over 3} + ... + {1 \over {100}} > {{98} \over {100}}$$
$$ \Rightarrow I > {{49} \over {50}}$$
Also, $$I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} } $$
$$ = \sum\limits_{k = 1}^{98} {[{{\log }_e}(k + 1) - {{\log }_e}k]} $$
$$I < {\log _e}99$$
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