JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 5)
If 0 < a, b < 1, and tan$$-$$1a + tan$$-$$1b = $${\pi \over 4}$$, then the value of
$$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$$ is :
$$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$$ is :
$${\log _e}$$2
e
$${\log _e}\left( {{e \over 2}} \right)$$
e2 = 1
Explanation
tan$$-$$1a + tan$$-$$1b = $${\pi \over 4}$$ 0 < a, b < 1
$$ \Rightarrow {{a + b} \over {1 - ab}} = 1$$
a + b = 1 $$-$$ ab
(a + 1)(b + 1) = 2
Now $$\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3} + ....} \right]$$
$$ = {\log _e}(1 + a) + {\log _e}(1 + b)$$
($$ \because $$ expansion of loge(1 + x))
$$ = {\log _e}[(1 + a)(1 + b)]$$
$$ = {\log _e}2$$
$$ \Rightarrow {{a + b} \over {1 - ab}} = 1$$
a + b = 1 $$-$$ ab
(a + 1)(b + 1) = 2
Now $$\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3} + ....} \right]$$
$$ = {\log _e}(1 + a) + {\log _e}(1 + b)$$
($$ \because $$ expansion of loge(1 + x))
$$ = {\log _e}[(1 + a)(1 + b)]$$
$$ = {\log _e}2$$
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