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CBSE Class 12 Mathematics Solved Question Paper 2010

Long Answer Type

21.

On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

22.

Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, -1, 2) and parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7}$

23.

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2 \vec{a} + \vec{b}) and (\vec{a} - 3 \vec{b})$ respectively, externally in the ratio 1:2. Also, show that P is the midpoint of the line segment R.

24.

Evaluate: $\int_{0}^{π} \frac{x}{1 + sinx} dx$

25.

Evaluate: $\int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) dx$

26.

Evaluate: $\int \frac{1 - x^{2}}{x (1 - 2 x)} dx$

27.
Find the particular solution of the differential equation satisfying the given conditions: x² dy + (xy + y² )dx = 0; y = 1 when x = 1.

$x^{2} dy + (xy + y^{2}) dx = 0 x^{2} dy = - (xy + y^{2}) dx \Rightarrow \frac{dy}{dx} = - \frac{- (xy + y^{2})}{x^{2}} . . . . . . . . . . (i)$

This is a homogeneous differential equation.

Such type of equations can be reduced to variable separable form by the substitution y = vx.

Differentiating w.r.t.x we get,

$\frac{d}{dx} (y) = \frac{d}{dx} (vx) \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} substituting the value of y and \frac{dy}{dx} in equation (i), we get : v + x \frac{dv}{dx} = \frac{- [x \times vx + {(vx)}^{2}]}{x^{2}} \Rightarrow x \frac{dv}{dx} = - v^{2} - 2 v = - v (v + 2) \Rightarrow \frac{dv}{v (v + 2)} = - \frac{dx}{x} \Rightarrow \frac{1}{2} [\frac{1}{v} - \frac{1}{v + 2}] dv = - \frac{dx}{x}$

Integrating both sides, we get:

$\frac{1}{2} [\log v - \log (v + 2)] = - \log x + \log C \Rightarrow \frac{1}{2} \log (\frac{v}{v + 2}) = \log \frac{C}{x} \Rightarrow \frac{v}{v + 2} = {(\frac{C}{x})}^{2} Substituting v = \frac{y}{x}$

$\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x} + 2} = {(\frac{C}{x})}^{2} \Rightarrow \frac{y}{y + 2 x} = \frac{C^{2}}{x^{2}} \Rightarrow \frac{x^{2} y}{y + 2 x} = D . . . . . . . . . . . . (ii)$

Now, it is given that y = 1 at x = 1.

$\Rightarrow \frac{1}{1 + 2} = D \Rightarrow D = \frac{1}{3} Substituting D = \frac{1}{3} in equation (ii), we get \frac{x^{2} y}{y + 2 x} = \frac{1}{3} \Rightarrow y + 2 x = 3 x^{2} y So, the required solution is y + 2 x = 3 x^{2} y$