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CBSE Class 12 Mathematics Solved Question Paper 2012

Long Answer Type

21.
Using elementary operations, find the inverse of the following matrix:
$(\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix})$

Consider the given matrix.

$Let A = [\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}]$

We know that, A = I_n A

Perform sequence of elementary row operations on A on the left hand side and the term I_n on the right hand side till we obtain the results,

I_n = BA

Thus, B = A^{- 1}

$Here, I_{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] Thus, we have, [\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{1} \leftrightarrow R_{2} [\begin{matrix} 1 & 2 & 3 \\ - 1 & 1 & 2 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{2} \to R_{2} + R_{1} R_{3} \to R_{3} - 3 R_{1} [\begin{matrix} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 3 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{1} \to R_{1} + R_{2} [\begin{matrix} 1 & 5 & 8 \\ 0 & 3 & 5 \\ 0 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 1 & 2 & 0 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{1} \to R_{1} + R_{3}$

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 3 & 5 \\ 0 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{2} \to \frac{R_{2}}{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{3} \to R_{3} + 5 R_{2} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3} \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & \frac{- 4}{3} & 1 \end{matrix}] A [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & - 4 & 3 \end{matrix}] A$

$R_{2} \to R_{2} - \frac{5}{3} R_{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{matrix}] A$

Thus, the inverse of the matrix A is given by

$[\begin{matrix} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{matrix}]$

22.

Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

23.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}| = 5, |\vec{b}| = 12$ and $|\vec{c}| = 13 and \vec{a} + \vec{b} + \vec{c} = 0$ find the value of $\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} .$

24.

Solve the following differential equation:

$2 x^{2} \frac{dy}{dx} - 2 x y + y^{2} = 0$

25.

How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?

26.

Evaluate: $\int$ sin x sin 2x sin 3x dx

27.

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

28.

Find the equation of the line passing through the point (-1,3,-2) and

perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3} and \frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5} .$

29.

Find the particular solution of the following differential equation:

$x + 1 \frac{dy}{dx} = 2 e^{- y} - 1; y = 0 when x = 0 .$

30.

A manufacturer produces nuts and bolts. It takes 1 hours of work on machine A and 3 hours on machine B to product a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of `17.50 per package on nuts and `7 per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operates his machines for at the most 12 hours a day? From the above as a linear programming problem and solve it graphically.