Q80 / 120 · JEE Advanced 2006 · Mathematics
Match the following.
- A(i) (a), (ii) (b), (iii) (c)
- Correct answerB(i) (b), (ii) (a), (iii) (c)
- C(i) (a), (ii) (c), (iii) (b)
- D(i) (b), (ii) (c), (iii) (a)
Solution
$$
\begin{aligned}
& \text { (i) } \sum_{i=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 i^2}\right)=t \Rightarrow \sum_{i=1}^{\infty} \tan ^{-1}\left(\frac{2}{4 i^2-1+1}\right) \\
& \Rightarrow \sum_{i=1}^{\infty} \tan ^{-1}\left\{\frac{(2 i+1)-(2 i-1)}{1+(2 i-1)(2 i+1)}\right\} \\
& \Rightarrow\left(\tan ^{-1} 3-\tan ^{-1} 1\right)+\left(\tan ^{-1} 5-\tan ^{-1} 3\right)+\ldots+\left\{\tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)\right\} \\
& \therefore \quad \quad t=\lim _{n \rightarrow \infty}\left(\tan ^{-1}(2 n+1)-\tan ^{-1} 1\right)=\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{2 n}{1+2 n+1}\right)=\frac{\pi}{4} \\
& \therefore \quad \tan t=1 .
\end{aligned}
$$
(ii) We have, $\cos \theta_1=\frac{1-\tan ^2 \frac{\theta_1}{2}}{1+\tan ^2 \frac{\theta_2}{2}}=\frac{a}{b+c}$
$$
\Rightarrow \quad \tan ^2 \frac{\theta}{2}=\frac{b+c-a}{b+c+a}
$$
Also, $\quad \cos \theta_3=\frac{1-\tan ^2 \frac{\theta_3}{2}}{1+\tan ^2 \frac{\theta_3}{2}}=\frac{c}{a+b}$
$$
\Rightarrow \quad \tan ^2 \frac{\theta_3}{2}=\frac{a+b-c}{a+b+c}
$$
$\Rightarrow \tan ^2 \frac{\theta_1}{2}+\tan ^2 \frac{\theta_3}{2}=\frac{2 b}{a+b+c}=\frac{2 b}{3 b}=\frac{2}{3}, \quad$ [as, $a, b, c$ are in AP $\Rightarrow 2 b=a+c$ ]
(iii) Line through $(0,1,0)$ and perpendicular to plane $x+2 y+2 z=0$ is given by $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-0}{2}=r$
$\therefore P(r, 2 r+1,2 r)$ be the foot of perpendicular on the straight line, then
$$
\begin{array}{ll}
& r \cdot 1+(2 r+1) \cdot 2+(2 r) \cdot 2=0 \\
\Rightarrow \quad r= & \quad r\left(-\frac{2}{9}, \frac{5}{9},-\frac{4}{9}\right) \\
\therefore & \quad P(r) \\
\therefore & \text { Required perpendicular distance }=\sqrt{\frac{4+25+16}{81}}=\frac{\sqrt{5}}{3} \text { unit. }
\end{array}
$$
Hence, the correct answer is option: 'B'.