IIT JEE Advanced 2010 Paper 2

(A)→(T) Let the line be $\frac{x}{a}$ = $\frac{y}{b}$ = $\frac{z}{c}$ and it intersects the lines. ⇒ S.D. = 0 ⇒ a + 3b + 5c = 0 and 3a + b - 5c = 0 ⇒ a : b : c :: 5r : - 5r : 2r On solving with given lines, we get points of intersection: P = (5 , - 5 , 2) and Q = $\left(\frac{10}{3}, -\frac{10}{3}, \frac{8}{3}\right)$ ⇒ $PQ^2$ = $d^2$ = 6. (B)→(P), (R) We have $\tan^{-1}(x+3)-\tan^{-1}(x-3)$ = $\sin^{-1}\left(\frac{3}{5}\right)$ ⇒ $\tan^{-1}\frac{(x+3)-(x-3)}{1+(x^2-9)}$ = $\tan^{-1}\frac{3}{4}$ ⇒ $\frac{6}{x^2-8}$ = $\frac{3}{4}$ Hence, $x^2-8$ = 8 or x = ± 4. (C)→(Q), (S) As $\overset{\rightarrow}{a}$ = $\mu\overset{\rightarrow}{b}+\overset{\rightarrow}{4c}$ we have $\mu\left(\left|\overset{\rightarrow}{b}\right|\right)$ = $-\overset{\rightarrow}{4b}\cdot\overset{\rightarrow}{c}$ Also, $\left|\overset{\rightarrow}{b}\right|^2$ = $\overset{\rightarrow}{4a}\cdot\overset{\rightarrow}{c}$ and $\left|\overset{\rightarrow}{b}\right|+\overset{\rightarrow}{b}\cdot\overset{\rightarrow}{c}-\overset{\rightarrow}{d}\cdot\overset{\rightarrow}{c}$ = 0 Also, $2\left|\overset{\rightarrow}{b}+\overset{\rightarrow}{c}\right|$ = $\left|\overset{\rightarrow}{b}-\overset{\rightarrow}{a}\right|$. Solving and eliminating $\overset{\rightarrow}{b}\cdot\overset{\rightarrow}{c}$ and eliminating $\left|\overset{\rightarrow}{a}\right|$, we get $(2\mu^2-10\mu)\left|\overset{\rightarrow}{b}\right|^2$ = 0 Therefore, µ = 0 and 5. (D)→(R) We have I = $\frac{2}{\pi}\int\limits_{-\pi}^{\pi}\frac{\sin\left(\frac{9x}{2}\right)}{\sin(x-2)}\,dx$ = $\frac{2}{\pi}\times 2\int\limits_{0}^{\pi}\frac{\sin\left(\frac{9x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\,dx$ $\frac{x}{2}$ = θ ⇒ dx = 2dθ When x = 0 , θ = 0 When x = π , θ = π/2 Therefore, I = $\frac{8}{\pi}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin(9\theta)}{\sin\theta}\,d\theta$ = $\frac{8}{\pi}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin(9\theta)-\sin(7\theta)}{\sin\theta}\,d\theta$ + $\frac{\sin(7\theta)-\sin(5\theta)}{\sin\theta}$ + $\frac{\sin(5\theta)-\sin(3\theta)}{\sin\theta}$ + $\frac{\sin(3\theta)-\sin\theta}{\sin\theta}$ + $\frac{\sin\theta}{\sin\theta}\,d\theta$ = $\frac{16}{\pi}\int\limits_{0}^{\frac{\pi}{2}}$ (cos 8θ + cos 6θ + cos 4θ + cos 2θ + 1) dθ + $\frac{8}{\pi}\int\limits_{0}^{\frac{\pi}{2}}d\theta$ = = $0+\frac{8}{\pi}+\left[\frac{\pi}{2}-0\right]$ = 4