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JEE Mathematics Solved Question Paper 2015

Multiple Choice Questions

21.
The solution of the differential equation $\frac{dy}{dx} + \sin (\frac{y + x}{2}) + \sin (\frac{y - x}{2}) = 0$ is

$\log (\tan (\frac{y}{2})) = C - 2 \sin (x)$

$\log (\tan (\frac{y}{4})) = C - 2 \sin (\frac{x}{2})$

$\log (\tan (\frac{y}{2} + \frac{π}{4})) = C - 2 \sin (x)$

$\log (\tan (\frac{y}{2} + \frac{π}{4})) = C - 2 \sin (\frac{x}{2})$

$\log (\tan (\frac{y}{4})) = C - 2 \sin (\frac{x}{2})$

$Given, \frac{dy}{dx} + \sin (\frac{y + x}{2}) + \sin (\frac{y - x}{2}) = 0 This differential equation can be rewritten as \frac{dy}{dx} = \sin (\frac{x - y}{2}) + \sin (\frac{x + y}{2}) = 2 \cos (\frac{x}{2}) \sin (\frac{- y}{2}) = - 2 \cos (\frac{x}{2}) \sin (\frac{y}{2})$

$\Rightarrow \int \frac{dy}{2 \sin (\frac{y}{2})} = - \int \cos (\frac{x}{2}) dx \Rightarrow \frac{1}{2} \int \csc (\frac{y}{2}) dy = - \frac{\sin (\frac{x}{2})}{(\frac{1}{2})} + C \Rightarrow \frac{1}{2} \frac{\log (\csc (\frac{y}{2}) - \cot (\frac{y}{2}))}{(\frac{1}{2})} = - \frac{\sin (\frac{x}{2})}{\frac{1}{2}} + C \Rightarrow \log (\tan (\frac{y}{4})) = C - 2 \sin (\frac{x}{2})$

22.

The part of straight line y = x between x = 1 and x = 2 is revolved about X-axis, then the curved surface area of the solid thus generated is

$2 \sqrt{2} π$
$3 \sqrt{2}$
$3 \sqrt{2} π$
None of these

23.

If $\int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx = k \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) dx$ , then the value of k is

$\frac{π}{2}$
$π$
$- \frac{π}{2}$
None of these

24.

A body of 6 kg rests in limiting equilibrium on an inclined plane whose slope is 30°. If the plane is raised to slope of 60°, then force (in kg-wt) along the plane required to support it is