IIT JEE Advanced 2006 Paper 1

(A) We have x + y = |a| and ax – y = 1. Therefore, $\frac{x}{\frac{1}{a}} - \frac{y}{1}$ = 1 (a > 0) Now, x + y = a and $\frac{x}{\frac{1}{a}}$ - y = 1 For intersection, at the first quadrant, we have a ∊ $(m_{AB} , m_{AC})$ $m_{AB}$ = $\frac{0+1}{a-0}$ = $\frac{1}{a}$ And a ∊ (1 , ∫) $\frac{1}{a}$ = a ⇒ $a^2$ = 1 ⇒ a = 1 , - 1 Therefore, a = 1. $m_{AB}$ = 1 and $m_{AC}$ = ∞. Hence, the value of $\left(\frac{2ac}{3}\right)$ = $\frac{2}{3}$ Therefore, (A)→(Q). (B) If the point (α, β, γ) lies on the plane x + y + z = 2, then α + β + γ = 2. Now, $\hat{a}$ = $\frac{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}}{\sqrt{\alpha^2+\beta^2\gamma^2}}$ $\hat{k} \times (\hat{k} \times \hat{a})$ = $(\hat{k} \cdot \hat{a})\hat{k}$ - $(\hat{k} \cdot \hat{k})\hat{a}$ = $\frac{\gamma\hat{k}}{\sqrt{\alpha^2+\beta^2+\gamma^2}}$ - $\frac{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}}{\sqrt{\alpha^2+\beta^2+\gamma^2}}$ = - $\frac{\alpha\hat{i}+\beta\hat{j}}{\sqrt{\alpha^2+\beta^++\gamma^2}}$ = 0 Therefore, αi + β j = 0 ⇒ α = 0 = β ⇒ γ = 2 Therefore, (B)→(P). (C) We have I = $\left|\int\limits_{0}^{1}(1-y^2)dy\right|$ + $\left|\int\limits_{1}^{0}(y^2-1)dy\right|$ = $\left|\int\limits_{0}^{1}(1-y^2)dy\right|$ + $\left|-\int\limits_{0}^{1}(y^2-1)dy\right|$ $\left|\int\limits_{0}^{1}(1-y^2)dy\right|$ + $\left|\int\limits_{0}^{1}(1-y^2)dy\right|$ = $2\left|\int\limits_{0}^{1}(1-y^2)dy\right|$ = $2\int\limits_{0}^{1}(1-y^2)dy$ = 2 $\left(y-\frac{y^3}{3}\right)\Big|_{0}^{1}$ = $2\left(1-\frac{1}{3}\right)$ = $\frac{4}{3}$ $I_2$ = $\left|\int\limits_{0}^{1}\sqrt{1-x}dx\right| + \left|\int\limits_{-1}^{0}\sqrt{1+x}dx\right|$ = = $2\left|\int\limits_{0}^{1}\sqrt{1-x}dx\right|$ $2\int\limits_{0}^{1}\sqrt{1-x}dx$ = - 4 $\int\limits_{1}^{0}t^2$ dt = 4 $\int\limits_{0}^{1} t^2$ dt = $\frac{4}{3}$ ⇒ $I_1$ = $I_2$ Therefore, 1 – x = $t^2$ dx = −2t dt Therefore, (C)→(R). (D) We have sinAsinBsinC + cosAcosB = 1 Therefore, sin C = $\frac{1-\cos A\cos B}{\sin B\sin A}$ 0 < sin C ≤ 1 0 < $\frac{1-\cos A\cos B}{\sin B\sin A}$ ≤ 1 1 – cosAcosB ≤ sinAsinB 1 ≤ cosAcosB + sinAsinB cos(A – B) ≥ 1 Therefore, it is possible only if cos(A – B) = 1 A – B = 0 A = B Therefore, sin C = $\frac{1-\cos^2 A}{\sin^2 a}$ = 1 Therefore, (D)→(S).