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CBSE Class 12 Mathematics Solved Question Paper 2013

Short Answer Type

Write the principal value of $tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis plus cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses.$

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Write the value of $open parentheses 2 tan to the power of negative 1 end exponent 1 fifth close parentheses$

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Find the value of a if $open square brackets table row cell straight a minus straight b end cell cell space space space 2 straight a plus straight c end cell row cell 2 straight a minus straight b end cell cell space 3 straight c plus straight d end cell end table close square brackets space equals space open square brackets table row cell negative 1 end cell cell space space space 5 end cell row 0 cell space 13 end cell end table close square brackets$

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If $open vertical bar table row cell straight x plus 1 end cell cell space space straight x minus 1 end cell row cell straight x minus 3 end cell cell space straight x plus 2 end cell end table close vertical bar space equals space open vertical bar table row 4 cell space space minus 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma$ then write the value of x.

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If $open square brackets table row cell space 9 end cell cell space minus 1 end cell cell space space space 4 end cell row cell negative 2 end cell cell space space space 1 end cell cell space space space 3 end cell end table close square brackets space equals space straight A plus open square brackets table row 1 cell space space 2 end cell cell negative 1 end cell row 0 cell space 4 end cell 9 end table close square brackets comma$ then find the matrix A.

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Find the value of the following:
$tan 1 half open square brackets sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction close square brackets comma space open vertical bar straight x close vertical bar space less than 1 comma space space straight y greater than 0 space and space xy less than 1$

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Prove that $tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses space equals straight pi over 4$

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Using properties of determinants prove the following:
$open vertical bar table row 1 cell space space straight x end cell cell space space straight x squared end cell row cell straight x squared end cell cell space 1 end cell straight x row straight x cell space straight x end cell 1 end table close vertical bar space equals space left parenthesis 1 minus straight x cubed right parenthesis squared$

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9.
Show that the function $straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 3 close vertical bar comma space straight x element of bold R bold comma$ is continuous but not differentiable at x=3.

straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 3 close vertical bar space equals space open vertical bar table row cell 3 minus straight x comma space space space straight x less than 3 end cell row cell straight x minus 3 comma space straight x greater or equal than 3 end cell end table close vertical bar

Let c be a real number.
Case I: c<3 Then f(c) = 3-c.

limit as straight x rightwards arrow straight c of straight f left parenthesis straight x right parenthesis space equals space limit as straight x rightwards arrow straight c of left parenthesis 3 minus straight x right parenthesis space equals space 3 minus straight c. Since comma space limit as straight x rightwards arrow straight c of straight f left parenthesis straight x right parenthesis space equals space straight f left parenthesis straight c right parenthesis comma space straight f space is space continous space at space all space negatives space real space numbers.

CaseII: c = 3. Then f(c) = 3 - 3 = 0

limit as straight x rightwards arrow straight c of straight f left parenthesis straight x right parenthesis space equals space limit as straight x rightwards arrow straight c of left parenthesis straight x minus 3 right parenthesis space equals space 3 minus 3 space equals space 0

Since

limit as straight x rightwards arrow straight c of straight f left parenthesis straight x right parenthesis space equals space straight f left parenthesis 3 right parenthesis comma space

f is continuous at x = 3.

Case III: C>3. Then f(c) = c - 3

limit as straight x rightwards arrow straight c of straight f left parenthesis straight x right parenthesis space equals space limit as straight x rightwards arrow straight c of left parenthesis straight x minus 3 right parenthesis space equals space straight c minus 3.

Since,

limit as straight x rightwards arrow straight c of left parenthesis straight x minus 3 right parenthesis space equals space straight c minus 3.

Therefore, f is a continuous function.
Now, we need to show that

straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 3 close vertical bar comma space straight x space element of space bold R bold space is space not space differentiable space at space straight x space equals space 3.

Consider the left hand limit of f at x = 3

limit as straight h rightwards arrow 0 to the power of minus of fraction numerator straight f left parenthesis 3 plus straight h right parenthesis minus straight f left parenthesis 3 right parenthesis over denominator straight h end fraction space equals space limit as straight h rightwards arrow 0 to the power of minus of fraction numerator open vertical bar 3 plus straight h minus 3 close vertical bar minus open vertical bar 3 minus 3 close vertical bar over denominator straight h end fraction equals limit as straight h rightwards arrow 0 to the power of minus of fraction numerator open vertical bar straight h close vertical bar minus 0 over denominator straight h end fraction equals limit as straight h rightwards arrow 0 to the power of minus of fraction numerator negative straight h over denominator straight h end fraction equals 1 left parenthesis straight h less than 0 space rightwards double arrow space open vertical bar straight h close vertical bar space equals space minus straight h right parenthesis

Consider the right hand limit of f at x = 3

limit as straight h rightwards arrow 0 to the power of plus of fraction numerator straight f left parenthesis 3 plus straight h right parenthesis minus straight f left parenthesis 3 right parenthesis over denominator straight h end fraction limit as straight h rightwards arrow 0 to the power of plus of fraction numerator open vertical bar 3 plus straight h minus 3 close vertical bar minus open vertical bar 3 minus 3 close vertical bar over denominator straight h end fraction space equals limit as straight h rightwards arrow 0 to the power of plus of fraction numerator open vertical bar straight h close vertical bar minus 0 over denominator straight h end fraction equals limit as straight h rightwards arrow 0 to the power of plus of straight h over straight h equals 1

left parenthesis straight h greater than 0 space rightwards double arrow space open vertical bar straight h close vertical bar space equals space straight h right parenthesis

Since the left and right hand limits are not equal, f is not differentiable at x = 3.

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Long Answer Type

10.

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

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