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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

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

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

$π$

$We have, \int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx = k \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) dx Let I = \int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx . . . (i) = \int_{0}^{π} (π - x) f (\sin^{2} (π - x) + \sec^{2} (π - x)) dx = \int_{0}^{π} (π - x) f (\sin^{2} (x) + \sec^{2} (x)) . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = π \int_{0}^{π} f (\sin^{2} (x) + \sec^{2} (x)) \Rightarrow 2 I = 2 π \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) \Rightarrow I = π \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x))$

On comparing with given integral, we get $k = π$ .

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