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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.

207 Views

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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8.
Using properties of determinants prove the following:
$open vertical bar table row 1 cell space space straight x end cell cell space space straight x squared end cell row cell straight x squared end cell cell space 1 end cell straight x row straight x cell space straight x end cell 1 end table close vertical bar space equals space left parenthesis 1 minus straight x cubed right parenthesis squared$

increment space equals space open vertical bar table row 1 cell space space straight x end cell cell 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 squared end cell 1 end table close vertical bar

Applying

straight R subscript 1 rightwards arrow straight R subscript 1 plus straight R subscript 2 plus straight R subscript 3 comma space

we have

increment space equals space open vertical bar table row cell 1 plus straight x plus straight x squared end cell cell 1 plus straight x plus straight x squared end cell cell 1 plus straight x plus straight x squared end cell row cell straight x squared end cell 1 straight x row straight x cell straight x to the power of 21 end cell blank end table close vertical bar equals space left parenthesis 1 plus straight x plus straight x squared right parenthesis space open vertical bar table row 1 cell space space 1 end cell cell space space space 1 end cell row cell straight x squared end cell cell space space 1 end cell cell space space straight x end cell row straight x cell space space straight x squared end cell cell space space space 1 end cell end table close vertical bar

Applying space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 space and space straight C subscript 3 minus straight C subscript 1 comma space we space have colon increment space equals space left parenthesis 1 plus straight x plus straight x squared right parenthesis space open vertical bar table row 1 cell space space space 0 end cell cell space space 0 end cell row cell straight x squared end cell cell space 1 minus straight x squared end cell cell space space straight x minus straight x squared end cell row straight x cell space 1 plus straight x squared end cell cell space 1 minus straight x end cell end table close vertical bar equals left parenthesis 1 plus straight x plus straight x squared right parenthesis thin space left parenthesis 1 minus straight x right parenthesis left parenthesis 1 minus straight x right parenthesis space open vertical bar table row 1 cell space 0 end cell cell space 0 end cell row cell straight x squared end cell cell 1 plus straight x end cell cell space straight x end cell row straight x cell negative straight x end cell cell space 1 end cell end table close vertical bar equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x right parenthesis open vertical bar table row 1 cell space 0 end cell cell space 0 end cell row cell straight x squared space end cell cell 1 plus straight x end cell cell space straight x end cell row straight x cell negative straight x end cell cell space 1 end cell end table close vertical bar

Expanding along R_1,we have:_{$increment equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x right parenthesis thin space left parenthesis 1 right parenthesis space open vertical bar table row cell 1 plus straight x end cell cell space space space straight x end cell row cell negative straight x end cell cell space space 1 end cell end table close vertical bar space space space equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x right parenthesis left parenthesis 1 plus straight x plus straight x squared right parenthesis space space space equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x cubed right parenthesis space space space equals left parenthesis 1 minus straight x cubed right parenthesis squared Hence space proved.$}

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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.

281 Views

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.

932 Views