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CBSE Class 12 Mathematics Solved Question Paper 2015

Short Answer Type

11.

Find the angle between the vector $straight i with hat on top minus straight j with hat on top space and space straight j with hat on top space minus space straight k with hat on top.$

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12.

Find the distance of a point (2, 5, −3) from the plane $straight r with rightwards arrow on top. open parentheses 6 straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top close parentheses space equals 4.$

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13.

Find the differential equation of the family of lines passing through the origin.

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14.

Find the integrating factor for the following differential equation: $straight x space log space straight x dy over dx space plus space straight y space equals space 2 logx space$

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15.

Evaluate: $integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator sin squared straight x over denominator sinx plus cosx end fraction dx.$

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16.

Evaluate $integral subscript negative 1 end subscript superscript 2 left parenthesis straight e to the power of 3 straight x end exponent plus 7 straight x minus 5 right parenthesis dx$

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17.

Evaluate:
$integral fraction numerator straight x squared over denominator straight x to the power of 4 plus straight x squared minus 2 end fraction dx$

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18.
In a set of 10 coins, 2 coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides

Let E₁ E₁ and A be the events defined as follows:
E₁ = Selecting a coin having head on both the sides
E₁= Selecting a coin not having head on both the sides
A = Getting all heads when a coin is tossed five times
We have to find P(E1/A).
There are 2 coins having heads on both the sides.
$straight P left parenthesis straight E subscript 1 right parenthesis space equals space fraction numerator straight C presuperscript 2 subscript 1 over denominator straight C presuperscript 10 subscript 1 end fraction equals space 2 over 10$
There are 8 coins not having heads on both the sides.

$straight P left parenthesis straight E subscript 2 right parenthesis equals space fraction numerator straight C presuperscript 8 subscript 1 over denominator straight C presuperscript 10 subscript 1 end fraction space equals 8 over 10 straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis space equals space left parenthesis 1 right parenthesis to the power of 5 space equals space 1 straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis space equals space open parentheses 1 half close parentheses to the power of 5$
By Baye's Theorem, we have

$straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis space equals space fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis end fraction space space space space space space equals space fraction numerator open parentheses begin display style 2 over 10 end style close parentheses left parenthesis 1 right parenthesis over denominator open parentheses begin display style 2 over 10 end style close parentheses left parenthesis 1 right parenthesis plus open parentheses begin display style 8 over 10 end style close parentheses open parentheses begin display style 1 half end style close parentheses to the power of 5 end fraction space space space space space space space equals space fraction numerator 2 over denominator 2 plus open parentheses begin display style 8 over 32 end style close parentheses end fraction equals space 8 over 9$