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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

31.

The value of $\int_{- 1}^{2} 4 x^{2} |x| d x$ is equal to

32.

The area of the region bounded by y² = 16 - x², y = 0, x = 0 in the first quadrant is (in, square units)

$8 π$
$6 π$
$2 π$
$4 π$

33.

The area bounded by the lines y - 2x = 2, y = 4 and the (Y-axis is equal to in square units)

34.

The value of $\int_{2}^{4} (x - 2) (x - 3) (x - 4) dx$ is equal to

$\frac{1}{2}$
2
3
0

35.

The general solution of the differential equation $(x + y + 3) \frac{dy}{dx} = 1$ is

x + y + 3 = Ce^y
x + y + 4 = Ce^y
x + y + 3 = Ce^-^y
x + y + 4 = Ce^-^y

36.

The differential equation representing the family of curves y² = a(ax + b), where a and b are arbitrary constants, is of

order 1, degree 1
order 1, degree 3
order 2, degree 3
order 2, degree 1

37.

The solution of the differential equation $\frac{x \frac{dy}{dx} - y}{\sqrt{x^{2} - y^{2}}} = 10 x^{2}$ is

$\sin^{- 1} (\frac{y}{x}) - 5 x^{2}$ = C
$\sin^{- 1} (\frac{y}{x}) = 10 x^{2}$ + C
$\frac{y}{x} = 5 x^{2} + C$
$\sin^{- 1} (\frac{y}{x}) = 10 x^{2} + Cx$

38.
The general solution of the differential equation xdy - ydx = y²dx is

$y = \frac{x}{C - x}$

$x = \frac{2 y}{C + x}$

$y = (C + x) (2 x)$

$y = \frac{2 x}{C + x}$

$y = \frac{x}{C - x}$

We have,

xdy - ydx = y²dx

$\Rightarrow x \frac{dy}{dx} - y = y^{2} \Rightarrow \frac{x \frac{dy}{dx} - y}{y^{2}} = 1 \Rightarrow - (\frac{y - x \frac{dy}{dx}}{y^{2}}) = 1 \Rightarrow - \frac{d}{dx} \frac{x}{y} = 1 [∵ \frac{d}{dx} (\frac{x}{y}) = \frac{y - x \frac{dy}{dx}}{y^{2}}] \Rightarrow \frac{d}{dx} \frac{x}{y} = - 1 \Rightarrow \frac{x}{y} = - x + C \Rightarrow y = \frac{x}{C - x}$