Subject

Mathematics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

21.

The scalar product of the vector straight a with rightwards arrow on top equals space straight i with hat on top plus straight j with hat on top plus straight k with hat on top with a unit vector along the sum of vectors straight b with rightwards arrow on top equals 2 straight i with hat on top plus 4 straight j with hat on top minus 5 straight k with hat on top space space and space straight c with rightwards arrow on top space equals space straight lambda straight i with hat on top plus 2 straight j with hat on top plus 3 straight k with hat on top is equal to one. Find the value of straight lambda and hence find the unit vector along straight b with rightwards arrow on top plus straight c with rightwards arrow on top.

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22. Evaluate colon
integral subscript 0 superscript straight pi fraction numerator 4 straight x space sinx over denominator 1 plus cos squared straight x end fraction dx
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23.

Evaluate:
integral fraction numerator straight x plus 2 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx


We need to evaluate the integral
integral fraction numerator straight x plus 2 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
Let space straight I space equals space integral fraction numerator straight x plus 2 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
Consider space the space integral space as space follows colon
fraction numerator straight x plus 2 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction space equals fraction numerator straight A begin display style straight d over dx end style left parenthesis straight x squared plus 5 straight x plus 6 right parenthesis plus straight B over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction
rightwards double arrow straight x plus 2 space equals space straight A left parenthesis 2 straight x plus 5 right parenthesis plus straight B
rightwards double arrow straight x plus 2 equals left parenthesis 2 straight A right parenthesis straight x plus 5 straight A plus straight B
Comparing the coefficients, we have
2A = 1;  5A+B = 2
Solving the above equations, we have

straight A equals 1 half space and space straight B equals space minus 1 half
Thus comma space
straight I space equals space integral fraction numerator straight x plus 2 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
space space equals space integral fraction numerator begin display style fraction numerator 2 straight x plus 5 over denominator 2 end fraction end style minus begin display style 1 half end style over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
space space equals 1 half integral fraction numerator 2 straight x plus 5 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx minus 1 half integral fraction numerator 1 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
straight I space equals 1 half straight I subscript 1 space minus space 1 half straight I subscript 2 comma
where space straight I subscript 1 space equals integral fraction numerator 2 straight x plus 5 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx

and space straight I subscript 2 equals space integral fraction numerator 1 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
Now space consider space straight I subscript 1 colon
straight I subscript 1 space equals space integral fraction numerator 2 straight x plus 5 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
Susbstitute
straight x squared plus 5 straight x plus 6 space equals straight t semicolon space left parenthesis 2 straight x plus 5 right parenthesis dx space equals space dt
straight I subscript 1 space equals integral fraction numerator dt over denominator square root of straight t end fraction space equals 2 square root of straight t
equals 2 square root of straight x squared plus 5 straight x plus 6 end root
Now consider I2:
straight I subscript 2 space equals space integral fraction numerator 1 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx
equals space integral fraction numerator 1 over denominator square root of straight x squared plus 5 straight x plus open parentheses begin display style 5 over 2 end style close parentheses squared plus 6 minus open parentheses begin display style 5 over 2 end style close parentheses squared end root end fraction dx
equals integral fraction numerator 1 over denominator square root of open parentheses straight x plus begin display style 5 over 2 end style close parentheses squared plus 6 minus begin display style 25 over 4 end style end root end fraction dx
equals space integral fraction numerator 1 over denominator square root of open parentheses straight x plus begin display style 5 over 2 end style close parentheses squared minus begin display style 1 fourth end style end root end fraction dx
straight I subscript 2 space equals log open vertical bar straight x plus 5 over 2 minus square root of straight x squared plus 5 straight x plus 6 end root close vertical bar plus straight C
Thus comma space straight I equals 1 half straight I subscript 1 minus 1 half straight I subscript 2
straight I space equals space square root of straight x squared plus 5 straight x plus 6 end root minus 1 half log open vertical bar straight x plus 5 over 2 minus square root of straight x squared plus 5 straight x plus 6 end root close vertical bar plus straight C


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24.

An experiment succeeds thrice as often as it fails. Find the probability that in the next five trials, there will be at least 3 successes.

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25. If space straight y space equals space Pe to the power of ax plus Qe to the power of bx show that
straight d to the power of 2 straight y end exponent over dx squared minus left parenthesis straight a plus straight b right parenthesis dy over dx plus aby space equals 0
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26. If space straight x equals cost left parenthesis 3 minus 2 cos squared straight t right parenthesis space and space straight y equals sint left parenthesis 3 minus 2 sin squared straight t right parenthesis comma space find space the space value space of space dy over dx at space straight t equals straight pi over 4.
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27.

Find the particular solution of the differential equation log open parentheses dy over dx close parentheses equals 3 straight x plus 4 straight y comma space given space that space straight y space equals space 0 comma space when space straight x equals space 0.

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28.

Find the value of p, so that the lines:
are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, -4) and parallel to line l1.

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 Multiple Choice QuestionsLong Answer Type

29.

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x- y + z = 0. Also find the distance of the plane obtained above, from the origin.

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30.

Find the distance of the point (2, 12, 5) from the point of intersection of the line 
straight r with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses space and space the space plane space straight r with rightwards arrow on top. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0.

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