Match Column I with Column II: Column I Column II (A) ^{{t=1}{∞}} ^{-1}≤ft(1/2i²) = t , then tan t is (P) 2/3 (B) Sides a, b, c of a ΔABC are in A.P. and cos θ₁ = a/b+c , θ₂ = b/a+c , θ₃ = c/a+b , then ²≤ft(θ₁/2)+²≤ft(θ₃/2) is (Q) 1 (C) A line is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance of this line from the origin is (R) {{5}}{3} (D) If sinAsinBsinC + cosAcosB = 1, then the value of sinC is (S) 1 [JEE Adv 2006 P1]

Correct Answer is : (A)→(Q); (B)→(P); (C)→(R)

(A)→(R); (B)→(P); (C)→(Q)

(A)→(Q); (B)→(P); (C)→(R)

(A)→(P); (B)→(Q); (C)→(R)

(A)→(Q); (B)→(P); (C)→(P)

Correct Answer is : (A)→(Q); (B)→(P); (C)→(R)

Test Index

IIT JEE Advanced 2006 Paper 1

Section: Mathematics

Question : 120 of 120

Marks: +1, -0

Column I	Column II
(A) $\sum^{\underset{t=1}{\infty}} \tan^{-1}\left(\frac{1}{2i^2}\right)$ = t , then tan t is	(P) $\frac{2}{3}$
(B) Sides a, b, c of a ΔABC are in A.P. and cos $\theta_1$ = $\frac{a}{b+c}$ , $\cos\theta_2$ = $\frac{b}{a+c}$ , $\cos\theta_3$ = $\frac{c}{a+b}$ , then $\tan^2\left(\frac{\theta_1}{2}\right)+\tan^2\left(\frac{\theta_3}{2}\right)$ is	(Q) 1
(C) A line is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance of this line from the origin is	(R) $\frac{\sqrt{5}}{3}$
(D) If sinAsinBsinC + cosAcosB = 1, then the value of sinC is	(S) 1

Solution:

(A) We have

\sum^{\underset{t=1}{\infty}} \tan^{-1}\left(\frac{2}{4i^2}\right)

\sum^{\underset{t=1}{\infty}} \tan^{-1}\left(\frac{2}{1+4i^2-1}\right)

\sum^{\underset{t=1}{\infty}} \tan^{-1}\left(\frac{2}{1+(2i-1)(2i+1)}\right)

\sum^{\underset{t=1}{\infty}} \tan^{-1}\left(\frac{(2i+1)-(2i-1)}{1+(2i-1)(2i+1)}\right)

\sum^{\underset{t=1}{\infty}} \left(\tan^{-1}(2i+1)-\tan^{-1}(2i+1)\right)

\lim\limits_{n\rightarrow\infty} \tan^{-3} 3 \tan^{-1} 1 + \tan^{-1} 5 - \tan^{-1} 3

+ L +

\tan^{-1}(2n+1)-\tan^{-1}(2n-1)

\lim\limits_{n\rightarrow\infty}\left(\tan^{-1}(2n+1)-\tan^{-1}\right)

\frac{\pi}{2}-\frac{\pi}{4}

\frac{\pi}{4}

Therefore, (A)→(Q).

(B) We have

cos

\theta_1

\frac{a}{b+c}

\frac{1-\tan^2\left(\frac{\theta_1}{2}\right)}{1+\tan^2\left(\frac{\theta_1}{2}\right)}

\frac{1+\tan^2\left(\frac{\theta_1}{2}\right)}{1-\tan^2\left(\frac{\theta_1}{2}\right)}

\frac{b+c}{a}

On applying componendo and dividendo, we get

\frac{2\tan^2\left(\frac{\theta_1}{2}\right)}{2}

\frac{b+c-a}{b+c+a}

\tan^2\left(\frac{\theta_1}{2}\right)

\frac{b+c-a}{b+c+a}

Similarly,

\tan^2\frac{\theta_3}{2}

\frac{a+b-c}{a+b+c}

\tan^2\frac{\theta_1}{2} + \tan^2\frac{\theta_3}{2}

\frac{b+c+a}{a+b+c}

\frac{a+b-c}{a+b+c}

Now, we see that a, b, c are in A.P. Therefore,

2b = a + c

\tan^2\frac{\theta_1}{2}+\tan^2\frac{\theta_2}{2}

\frac{2b}{a+b+c}

\frac{2b}{3b}

\frac{2}{3}

Therefore, (B)→(P).

\frac{x-0}{1}

\frac{y-1}{0}

\frac{z-0}{2}

and the line through the origin is

\frac{x}{a}

\frac{y}{b}

\frac{z}{c}

= λ

which is perpendicular to given line a + 2b + 2c = 0. Now, P(aλ, bλ, cλ) lies on given line

aλ =

\frac{b\lambda-1}{2}

\frac{c\lambda}{2}

⇒ a =

\frac{c}{2}

⇒ c = 2a

Therefore,

bλ - 1 = cλ ⇒ λ =

\left(\frac{1}{b-c}\right)

That is,

a + 2b + 4a = 0 ⇒ b =

-\frac{5a}{2}

2aλ = bλ - 1

λ =

\left(\frac{1}{b-2a}\right)

λ =

\frac{1}{b-c}

\frac{1}{-\frac{5a}{2}-2a}

-\frac{2}{9a}

Therefore,

P (aλ , bλ , cλ) =

P\left[\left(-\frac{2}{9a}\right),-\frac{5a}{2}\left(-\frac{2}{9a}\right),2a\left(-\frac{2}{9a}\right)\right]

P\left(\frac{2}{9},-\frac{5}{9},-\frac{4}{9}\right)

Therefore, the perpendicular distance is

\sqrt{\frac{4}{81}+\frac{25}{81}+\frac{16}{81}}

\sqrt{\frac{45}{81}}

\sqrt{\frac{5}{9}}

\frac{\sqrt{5}}{3}

Therefore, (C)→(R).

Go to Question:

Prev Question