Subject

Mathematics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

21.

If straight x equals straight a space sint space and space straight y space equals space straight a open parentheses cost plus log space tan straight t over 2 close parentheses space find space fraction numerator straight d squared straight y over denominator dx squared end fraction

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22.

Evaluate: integral fraction numerator sin left parenthesis straight x minus straight a right parenthesis over denominator sin left parenthesis straight x plus straight a right parenthesis end fraction dx

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23.

Evaluate:
integral fraction numerator 5 straight x minus 2 over denominator left parenthesis 1 plus 2 straight x plus 3 straight x squared end fraction dx

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24.

Evaluate:
integral fraction numerator straight x squared over denominator left parenthesis straight x squared plus 4 right parenthesis left parenthesis straight x squared plus 9 right parenthesis end fraction dx


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25.

Evaluate: integral subscript 0 superscript 4 open parentheses open vertical bar straight x close vertical bar plus open vertical bar straight x minus 2 close vertical bar plus open vertical bar straight x minus 4 close vertical bar close parentheses dx

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26.

If straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are two vectors such that open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar equals open vertical bar straight a with rightwards arrow on top close vertical bar comma then prove that vector 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top is perpendicular to vector straight b with rightwards arrow on top.

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27.

Find the coordinates of the point, where the line fraction numerator straight x minus 2 over denominator 3 end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction intersects the plane straight x minus straight y plus straight z minus 5 space equals space 0. Also find the angle between the line and the plane. 

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28.

Find the vector equation of the plane which contains the line of intersection of the planes. straight r with rightwards arrow on top. open parentheses straight i with hat on top plus 2 straight j with hat on top plus 3 straight k with hat on top close parentheses minus 4 space equals 0 and straight r with rightwards arrow on top. open parentheses 2 straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses plus 5 space equals 0 and which is perpendicular to the plane straight r with rightwards arrow on top. space open parentheses 5 straight i with hat on top plus 3 straight j with hat on top minus 6 straight k with hat on top close parentheses plus 8 space equals space 0

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29.

A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A?

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 Multiple Choice QuestionsLong Answer Type

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30.

Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.


The shaded area OBAO represents the area bounded by the curve x2 = 4y and the line x = 4y – 2.


Let A and B be the points of intersection of the line and parabola.
Co-ordinates of point A are open parentheses negative 1 comma space 1 fourth close parentheses. space Co-ordinates of point B are (2, 1).
Area OBAO = Area OBCO + Area OACO   ...(1)
Area OBCO = 
equals integral subscript 0 superscript 2 fraction numerator straight x plus 2 over denominator 4 end fraction dx minus integral subscript 0 superscript 2 straight x squared over 4 dx
equals 1 fourth open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript 0 superscript 2 minus 1 fourth open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 2
equals 1 fourth open square brackets 2 plus 4 close square brackets minus 1 fourth open square brackets 8 over 3 close square brackets
equals 3 over 2 minus 2 over 3 equals 5 over 6
Area OACO = 
  equals integral subscript negative 1 end subscript superscript 0 fraction numerator straight x plus 2 over denominator 4 end fraction dx minus integral subscript negative 1 end subscript superscript 0 straight x squared over 4 dx
equals 1 fourth open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript negative 1 end subscript superscript 0 space minus space 1 fourth open square brackets straight x cubed over 3 close square brackets subscript negative 1 end subscript superscript 0
equals 1 fourth open square brackets negative fraction numerator left parenthesis negative 1 right parenthesis squared over denominator 2 end fraction minus 2 left parenthesis negative 1 right parenthesis close square brackets minus 1 fourth open square brackets negative open parentheses negative 1 close parentheses cubed over 3 close square brackets
equals 1 fourth open square brackets negative 1 half plus 2 close square brackets minus 1 fourth open square brackets 1 third close square brackets
equals 3 over 8 minus 1 over 12 equals 7 over 24
Therefore, required area = open parentheses 5 over 6 plus 7 over 24 close parentheses space equals 9 over 8 sq. units

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