If 0<a,b<1, and tan^(-1)a+tan^(-1)b=(π)/4, then the value of a+b-(a^2+b^2)/2+(a^3+b^3)/3-(a^4+b^4)/4+… is :

Correct answer: (D) log_e2. tan^(-1)a+tan^(-1) b=(π)/4 0<a,b<1 ⇒(a+b)/(1-ab)=1 a+b=1-ab a+1b+1=2 Now a-a^22+a^33+…+b-b^22+b^33+… =log_e1+a+log_e1+b (∵ expansion of log_e1+x) =log_e1+a1+b =log_e2

Q60 / 90 · JEE Main 2021 (26 Feb Shift 2) · Mathematics

If $0 < a, b < 1$ , and $\tan^{- 1} a + \tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - (\frac{a^{2} + b^{2}}{2}) + (\frac{a^{3} + b^{3}}{3}) - (\frac{a^{4} + b^{4}}{4}) + \dots$ is :

A
$\log_{e} (\frac{e}{2})$
B
$e$
C
$e^{2} - 1$
Correct answer
D
$\log_{e} 2$

Solution

$\tan^{- 1} a + \tan^{- 1} b = \frac{π}{4} 0 < a, b < 1$

$\Rightarrow \frac{a + b}{1 - a b} = 1$

$a + b = 1 - a b$

$(a + 1) (b + 1) = 2$

Now $[a - \frac{a^{2}}{2} + \frac{a^{3}}{3} + \dots] + [b - \frac{b^{2}}{2} + \frac{b^{3}}{3} + \dots]$

$= \log_{e} (1 + a) + \log_{e} (1 + b)$

( $∵$ expansion of $\log_{e} (1 + x)$ )

$= \log_{e} [(1 + a) (1 + b)]$

$= \log_{e} 2$

Hence, the correct answer is option: 'D'.

If 0<a,b<1, and tan-1a+tan-1b=π4, then the value of a+b-a2+b22+a3+b33-a4+b44+… is :

Solution

Relevant Questions

If $0 < a, b < 1$ , and $\tan^{- 1} a + \tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - (\frac{a^{2} + b^{2}}{2}) + (\frac{a^{3} + b^{3}}{3}) - (\frac{a^{4} + b^{4}}{4}) + \dots$ is :