Medium MCQ +4 / -1 PYQ · JEE Mains 2021

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 ${\log _e}$2 Correct answer
B e
C ${\log _e}\left( {{e \over 2}} \right)$
D e2 = 1

Solution

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)) $= {\log _e}[(1 + a)(1 + b)]$ $= {\log _e}2$

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Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range

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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 :

Solution

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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 :