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CBSE Class 12 Mathematics Solved Question Paper 2013

Short Answer Type

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

1188 Views

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

264 Views

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

171 Views

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.

193 Views

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

218 Views

7.
Prove that $tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses space equals straight pi over 4$

We know that:
$tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y space equals space tan to the power of negative 1 end exponent fraction numerator straight x plus straight y over denominator 1 minus xy end fraction comma space xy less than 1$
We have:
$tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses equals open square brackets tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses close square brackets equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 half end style plus begin display style 1 fifth end style over denominator 1 minus begin display style 1 half end style cross times begin display style 1 fifth end style end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses space space space open parentheses because space space 1 half cross times 1 fifth less than 1 close parentheses equals tan to the power of negative 1 end exponent open parentheses 7 over 9 close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses$
$equals tan to the power of negative 1 end exponent fraction numerator begin display style 7 over 9 end style plus begin display style 1 over 8 end style over denominator 1 minus begin display style 7 over 9 end style cross times begin display style 1 over 8 end style end fraction equals tan to the power of negative 1 end exponent fraction numerator 56 plus 9 over denominator 72 minus 7 end fraction space open parentheses because 7 over 9 cross times 1 over 8 less than 1 close parentheses equals tan to the power of negative 1 end exponent 65 over 65 equals tan to the power of negative 1 end exponent 1 equals straight pi over 4 Hence comma space tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses equals straight pi over 4$

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

247 Views

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