JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 2)
Let f be a real valued function, defined on R $$-$$ {$$-$$1, 1} and given by
f(x) = 3 loge $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$.
Then in which of the following intervals, function f(x) is increasing?
f(x) = 3 loge $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$.
Then in which of the following intervals, function f(x) is increasing?
($$-$$$$\infty $$, $$-$$1) $$\cup$$ $$\left( {[{1 \over 2},\infty ) - \{ 1\} } \right)$$
($$-$$$$\infty $$, $$\infty $$) $$-$$ {$$-$$1, 1)
($$-$$$$\infty $$, $${{1 \over 2}}$$] $$-$$ {$$-$$1}
($$-$$1, $${{1 \over 2}}$$]
Explanation
f(x) = 3 loge $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$
$$f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0$$
$$ = {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0$$
$$ = {{2(3(x - 1) + (x + 1))} \over {{{(x - 1)}^2}(x + 1)}} = {{4(2x - 1)} \over {{{(x - 1)}^2}(x + 1)}} > 0$$
_16th_March_Evening_Shift_en_2_1.png)
$$ \Rightarrow x \in ( - \infty , - 1) \cup \left[ {{1 \over 2},\infty ) - \{ 1\} } \right.$$
$$f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0$$
$$ = {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0$$
$$ = {{2(3(x - 1) + (x + 1))} \over {{{(x - 1)}^2}(x + 1)}} = {{4(2x - 1)} \over {{{(x - 1)}^2}(x + 1)}} > 0$$
$$ \Rightarrow x \in ( - \infty , - 1) \cup \left[ {{1 \over 2},\infty ) - \{ 1\} } \right.$$
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