Subject

Mathematics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

21.

A line passing through the point A with position vector straight a with rightwards arrow on top space equals space 4 straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top is parallel to the vector straight b with rightwards arrow on top space equals space 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top. Find the length of the perpendicular drawn on this line from a point p with vector stack straight r subscript 1 with rightwards arrow on top space equals straight i with hat on top plus 2 straight j with hat on top plus 3 straight k with hat on top.

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22. If space straight y space equals space straight e to the power of ax. cosbx comma space then space prove space that
space space space fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight a dy over dx plus left parenthesis straight a squared plus straight b squared right parenthesis straight y space equals space 0
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23. If space straight x to the power of straight x plus straight x to the power of straight y plus straight y to the power of straight x space equals space straight a to the power of straight b comma space space then space find space dy over dx.
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24.

If straight x equals straight a space sin space 2 straight t left parenthesis 1 plus cos space 2 straight t right parenthesis space and space straight y space equals straight b space cos space 2 straight t left parenthesis 1 minus cos space 2 straight t right parenthesis comma space then space find dy over dx space at space straight t space equals straight pi over 4.

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25.

Evaluate: integral fraction numerator left parenthesis straight x plus 3 right parenthesis straight e to the power of straight x over denominator left parenthesis straight x plus 5 right parenthesis cubed end fraction dx

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26.

Three schools X, Y, and Z organized a fete (mela) for collecting funds for flood victims in which they sold hand-held fans, mats and toys made from recycled material, the sale price of each being Rs. 25, Rs. 100 andRs. 50 respectively. The following table shows the number of articles of each type sold:
School/Article                     School X                   School Y                School Z
Hand-held fans                         30                         40                         35
      Mats                                  12                         15                         20
       Toys                                 70                         55                         75
Using matrices, find the funds collected by each school by selling the above articles and the total funds collected. Also write any one value generated by the above situation. 

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 Multiple Choice QuestionsLong Answer Type

27.

Sketch the region bounded by the curves straight y equals square root of 5 minus straight x squared end root space and space straight y space equals open vertical bar straight x minus 1 close vertical bar and find its area using integration. 

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28.

Find the particulars solution of the differential equation straight x squared dy space equals space left parenthesis 2 xy plus straight y squared right parenthesis dx comma space given that y =1 when x = 1.

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29.

Find the particulars solution of the differential equation left parenthesis 1 plus straight x squared right parenthesis dy over dx equals open parentheses straight e to the power of straight m space tan to the power of negative 1 straight x end exponent end exponent minus straight y close parentheses comma space given space that space straight y space equals space 1 space when space straight x space equals space 0.


left parenthesis 1 plus straight x squared right parenthesis dy over dx space equals space straight e to the power of straight m space tan to the power of negative 1 straight x end exponent end exponent minus straight y
rightwards double arrow dy over dx equals fraction numerator straight e to the power of mtan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction minus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction
rightwards double arrow dy over dx plus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction equals fraction numerator straight e to the power of mtan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction
straight P space equals fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction comma space straight Q equals fraction numerator straight e to the power of mtan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction
straight I. straight F. space equals space straight e to the power of integral Pdx end exponent
space space space space space equals straight e to the power of integral fraction numerator dx over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction end exponent
space space space space space space equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent

Thus the solution is
ye to the power of integral Pdx end exponent space equals integral Qe to the power of integral Pdx end exponent dx
rightwards double arrow space space straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space space equals integral fraction numerator straight e to the power of straight m space tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction straight e to the power of tan to the power of negative 1 end exponent straight x end exponent. dx
rightwards double arrow space straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space integral fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx.... left parenthesis straight i right parenthesis
integral fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space.... left parenthesis ii right parenthesis
Let space left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x space equals straight z
fraction numerator left parenthesis straight m plus 1 right parenthesis over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dz
fraction numerator dx over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dz
fraction numerator dx over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction equals space fraction numerator dz over denominator left parenthesis straight m plus 1 right parenthesis end fraction
Substituting in (ii),
  fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis end fraction integral straight e to the power of straight z dz
equals fraction numerator straight e to the power of straight z over denominator left parenthesis straight m plus 1 right parenthesis end fraction
equals fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction

Substituting in (i),

rightwards double arrow straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus straight C space.... left parenthesis iii right parenthesis
Putting space straight y space equals space 1 space and space straight x space equals space 1 comma space in space the space above space equation comma
space space rightwards double arrow space space straight y space cross times space straight e to the power of tan to the power of negative 1 end exponent 1 end exponent space equals fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent 1 end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus straight C
space rightwards double arrow 1 cross times straight e to the power of straight pi over 4 end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis end exponent begin display style straight pi over 4 end style over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus straight C
therefore straight C space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis begin display style straight pi over 4 end style end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction minus straight e to the power of straight pi over 4 end exponent
Particular solution of the D.E. is straight y cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction plus fraction numerator straight e to the power of left parenthesis straight m plus 1 right parenthesis begin display style straight pi over 4 end style end exponent over denominator left parenthesis straight m plus 1 right parenthesis end fraction minus straight e to the power of straight pi over 4 end exponent

 
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30.

Show that lines:
straight r with rightwards arrow on top space equals space straight i with hat on top space plus straight j with hat on top space plus straight k with hat on top space plus space straight lambda open parentheses straight i with hat on top minus straight j with hat on top space plus space straight k with hat on top close parentheses
straight r with rightwards arrow on top space equals space 4 straight j with hat on top space plus space 2 straight k with hat on top space plus space straight mu open parentheses 2 straight j with hat on top minus straight j with hat on top plus 3 straight k with hat on top close parentheses space are space coplanar.
Also, find the equation of the plane containing these lines. 

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