IIT JEE Advanced 2006 Paper 1

(A) We have,$y=(\sin x)^{\cos x}$ $\ln y = \cos x \times \ln (\sin x)$ Now, $y' = y(\cos x \cot x - \sin x \ln \sin x)$ $y' = (\sin x)^{\cos x} (\cos x \cot x - \sin x \ln \sin x)$ Therefore, Therefore, $(A) \rightarrow (P)$ (B) For the curves, we have $y^2 = \frac{-x}{4}$ $y^2 = -\frac{1}{5}(x-1)$ That is, $-\frac{x}{4} = -\frac{1}{5}(x-1)$ $5x = 4x - 4$ $x = -4$ $y^2 = 1 \Rightarrow y = -1, 1$ Therefore, the area bounded is given by, $\int\limits_{-1}^{1} (1 - 5y^2 + 4y^2) \, dy$ $= 2 \int\limits_{0}^{1} (1 - y^2) \, dy$ $= 2 \left( y - \frac{y^3}{3} \right) \Big|_{0}^{1}$ $= 2 \left(1 - \frac{1}{3}\right) = \frac{4}{3}$ Therefore (B) → (S) (C) The point of intersection of the curve is $y^3 = 3^{x-1} (\ln x)$ and $y = x^{x} - 1$ is $(1,0)$. Now, for the first curve, we have $y' = 3^{x-1} \left( \frac{1}{x} \right) + (3^{x-1} \ln x \ln 3)$ $y'_{(1,0)} = 1 + 0 = 1$ and for the second curve, we have $y' = (y+1)(1 + \ln x)$ $y'_{(1,0)} = 1$ Therefore, the angle between the curves is 0 . Therefore, $(C) \rightarrow (Q)$. (D) We have $\frac{dy}{dx} = \frac{2}{x+y} \Rightarrow x+y = t$ $1 + \frac{dy}{dx} = \frac{dt}{dx}$ That is, $\frac{dt}{dx} - 1 = \frac{2}{t} \Rightarrow \frac{dt}{dx} = \frac{2}{t} + 1 = \frac{t+2}{t} \Rightarrow \frac{t}{t+2} \, dt = dx$ $\Rightarrow \int \left( \frac{t+2-2}{t+2} \right) \, dt = \int dx + c$ $\Rightarrow \int \left( 1 - \frac{2}{t+2} \right) \, dt = \int dx + c$ $\Rightarrow t - 2 \ln (t+2) = x + c$ $\Rightarrow (x+y) - 2 \ln (x+y+2) = x + c$ $\Rightarrow 1 - 2 \ln 3 = 1 + c \left[ \frac{dy}{dx} = \frac{2}{x+y} \text{ passes through } (1,0) \right]$ $\Rightarrow c = -2 \ln 3$ $\Rightarrow y - 2 \ln (x+y+2) = -2 \ln 3$ $\Rightarrow y = 2 \ln \left( \frac{x+y+2}{3} \right)$ Therefore, $x+y+z = 3 e^{y/2}$ Therefore, $(D) \rightarrow (R)$.