Subject

Mathematics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsLong Answer Type

31.

Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).

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32.

A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of 80 on each piece of type A and 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

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33.

There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of The three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

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34.

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X, and hence find the mean of the distribution.

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35.

Evaluate:
integral fraction numerator 1 over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction dx


We need to evaluate integral fraction numerator dx over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction
Let space straight I space equals integral fraction numerator dx over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction
Multiply the numerator and the denominator by sec to the power of 4 straight x, we have

straight I space equals space integral fraction numerator sec to the power of 4 xdx over denominator tan to the power of 4 straight x plus tan squared straight x plus 1 end fraction
straight I space equals space integral fraction numerator sec squared straight x cross times sec squared xdx over denominator tan to the power of 4 straight x plus tan squared straight x plus 1 end fraction
Now space substitute space straight t space equals space tanx semicolon space dt space equals space sec squared xdx
Therefore comma space
straight I space equals space integral fraction numerator left parenthesis 1 plus straight t squared right parenthesis dt over denominator straight t to the power of 4 plus straight t squared plus 1 end fraction
straight I space equals space integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t squared plus begin display style 1 over straight t squared end style plus 1 close parentheses end fraction
straight I space equals integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t squared plus begin display style 1 over straight t squared end style minus 2 plus 2 plus 1 close parentheses end fraction
straight I space equals space integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t minus begin display style 1 over straight t end style close parentheses squared plus 3 end fraction
 Substitute space straight z equals straight t minus 1 over straight t semicolon space space dz space equals space open parentheses 1 plus 1 over straight t squared close parentheses dt
straight I space equals space integral fraction numerator dz over denominator straight z squared plus 3 end fraction
straight I space equals space integral fraction numerator dz over denominator straight z squared plus open parentheses square root of 3 close parentheses squared end fraction
straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight z over denominator square root of 3 end fraction close parentheses plus straight c
straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight t minus begin display style 1 over straight t end style over denominator square root of 3 end fraction close parentheses plus straight c

straight I equals space fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tanx minus begin display style 1 over tanx end style over denominator square root of 3 end fraction close parentheses plus straight c
straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tanx minus cotx over denominator square root of 3 end fraction close parentheses plus straight c

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