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CBSE Class 12 Mathematics Solved Question Paper 2010

Long Answer Type

11.

Find all points of discontinuity of f, where f is defined as following:

$f (x) = \{\begin{cases} \begin{matrix} |x| + 3, x \leq - 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$

12.

$Find \frac{dy}{dx}, if y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}}$

13.

Prove the following:

$\tan^{- 1} \sqrt{x} = \frac{1}{2} \cos^{- 1} (\frac{1 - x}{1 + x}), x \in (0, 1)$

14.
Prove the following:
$\cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65})$

Let a be in I quadrant such that

$\cos^{- 1} (\frac{12}{13}) = a So \cos a = \frac{12}{13} \Rightarrow \sin a = \sqrt{1 - {(\frac{12}{13})}^{2}} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{169 - 144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} And \tan a = \frac{5}{12} So, a = \tan^{- 1} (\frac{5}{12}) . . . . . . . . . (i) Again b \in I quadrant such that \sin^{- 1} (\frac{3}{5}) = b So, \sin b = \frac{3}{5}$

$\Rightarrow \cos b = \sqrt{1 - {(\frac{3}{5})}^{2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} And \tan b = \frac{3}{4} so, b = \tan^{- 1} (\frac{3}{4}) . . . . . . . . . . . . . . . (ii) Now, let \sin^{- 1} (\frac{56}{65}) = c where c is in I quadrant So, \sin c = \frac{56}{65}$

$\Rightarrow \cos c = \sqrt{1 - {(\frac{56}{65})}^{2}} = \sqrt{1 - \frac{3136}{4225}} = \sqrt{\frac{4225 - 3136}{4225}} = \sqrt{\frac{1089}{4225}} = \frac{33}{65} And, \tan c = \frac{56}{33} So, c = \tan^{- 1} (\frac{56}{33}) \Rightarrow \sin^{- 1} (\frac{56}{65}) = \tan^{- 1} (\frac{56}{33}) . . . . . . . . . . . (iii) Now, we need to prove \cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65})$

consider a + b

$= \cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{5}{12}) + \tan^{- 1} (\frac{3}{4}) [\begin{matrix} \cos^{- 1} (\frac{12}{13}) = \tan^{- 1} (\frac{5}{12}) and \\ \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{3}{4}) \end{matrix}] = \tan^{- 1} [\frac{\frac{5}{12} + \frac{3}{4}}{1 - (\frac{5}{12} x \frac{3}{4})}] [using, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] = \tan^{- 1} (\frac{20 + 36}{48 - 15}) = \tan^{- 1} (\frac{56}{33}) = c = \sin^{- 1} (\frac{56}{65}) [using, eq, (iii)]$

Hence proved.

15.

Let $*$ be a binary operation on Q defined by $a * b = \frac{3 ab}{5}$
Show that $*$ is commutative as well as associative. Also find its identity element, if it exists.

16.

Using elementary row operations, find the inverse of the following matrix:

$[\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$

17.

Find the equations of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

18.

Using properties of determinants show the following:

$|\begin{matrix} {(b + c)}^{2} & ab & ca \\ ab & {(a + c)}^{2} & bc \\ ac & bc & {(a + b)}^{2} \end{matrix}| = 2 abc (a + b + c)^{3}$

19.

Find the values of x for which f(x) = [x(x - 2)]2 is an increasing function. Also, find the points on the curve where the tangent is parallel to x-axis.

20.

Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base.