Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1.

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. 


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
rightwards double arrow space space space space straight A space equals space 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 minus space open square brackets table row 1 cell space space 2 end cell cell space minus 1 end cell row 0 cell space space 4 end cell cell space 9 end cell end table close square brackets
rightwards double arrow space space space space straight A space equals space open square brackets table row cell 9 minus 1 end cell cell space space minus 1 minus 2 end cell cell space space space 4 plus 1 end cell row cell negative 2 minus 0 end cell cell space space space space space 1 minus 4 space end cell cell space space 3 minus 9 end cell end table close square brackets
rightwards double arrow space space space straight A space equals space open square brackets table row 8 cell space minus 3 end cell cell space space space space 5 end cell row cell negative 2 end cell cell space space space 3 end cell cell space minus 6 end cell end table close square brackets
207 Views

2.

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


179 Views

 Multiple Choice QuestionsLong Answer Type

3.

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.


Let the award money given for honesty, regularity and hard work be Rs. x, Rs. y and Rs. z respectively.
Since total cash award is Rs. 6,000.
therefore space straight x plus straight y plus straight z space equals space 6 comma 000 space... left parenthesis 1 right parenthesis
Three times the award money for hard work and honesty amounts to Rs.11,000.
therefore space space straight x plus 3 straight z space equals space 11 comma 000
rightwards double arrow space space straight x plus 0 cross times straight y plus 3 straight z space equals space 11 comma 000 space... left parenthesis 2 right parenthesis
Award money for honesty and hard work is double that given for regularity. 
therefore space straight x plus straight z space equals space 2 straight y
rightwards double arrow space space straight x minus 2 straight y plus straight z space equals space 0 space space space... left parenthesis 3 right parenthesis
The above system of equations can be written in matrix form AX = B as:

open square brackets table row 1 cell space space space 1 end cell cell space space space 1 end cell row 1 cell space space space 0 end cell cell space space space 3 end cell row 1 cell space minus 2 end cell cell space space space 1 end cell end table close square brackets space open square brackets table row x row y row z end table close square brackets space equals space open square brackets table row 6000 row 11000 row 0 end table close square brackets
Here, 

straight A space equals space open square brackets table row 1 cell space space space 1 end cell cell space space 1 end cell row 1 cell space space space 0 end cell cell space space 3 end cell row 1 cell space minus 2 end cell cell space space 1 end cell end table close square brackets comma space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets space space and space straight B space equals space open square brackets table row 6000 row 11000 row 0 end table close square brackets
open vertical bar straight A close vertical bar space equals space 1 left parenthesis 0 plus 6 right parenthesis minus 1 left parenthesis 1 minus 3 right parenthesis plus 1 left parenthesis negative 2 minus 0 right parenthesis space equals space 6 not equal to 0
Thus, A is non-singular. Hence, it is invertible.

Adj space straight A space equals space open square brackets table row 6 cell space minus 3 end cell cell space space space space space 3 end cell row 2 cell space space space 0 end cell cell space space minus 2 end cell row cell negative 2 end cell cell space space 3 end cell cell space minus 1 end cell end table close square brackets
therefore space space straight A to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator open vertical bar straight A close vertical bar end fraction left parenthesis adj space straight A right parenthesis space equals space 1 over 6 open square brackets table row 6 cell space minus 3 end cell cell space space 3 end cell row 2 cell space space 0 end cell cell negative 2 end cell row cell negative 2 end cell cell space 3 end cell cell negative 1 end cell end table close square brackets
straight X space equals space straight A to the power of negative 1 end exponent straight B space equals space 1 over 6 open square brackets table row 6 cell space space minus 3 end cell cell space space space 3 space end cell row 2 cell space space space 0 end cell cell negative 2 end cell row cell negative 2 end cell cell space space 3 end cell cell negative 1 end cell end table close square brackets open square brackets table row 6000 row 11000 row 0 end table close square brackets equals 1 over 6 open square brackets table row cell 36000 minus 33000 plus 0 end cell row cell 12000 plus 0 minus 0 end cell row cell negative 12000 plus 33000 minus 0 end cell end table close square brackets equals 1 over 6 open square brackets table row 3000 row 12000 row 21000 end table close square brackets
rightwards double arrow open square brackets table row straight x row straight y row straight z end table close square brackets space equals open square brackets table row 500 row 2000 row 35000 end table close square brackets


Hence, x = 500,  y = 2000,  and z = 3500.
Thus, award money given for honesty, regularity and hardwork is Rs. 500, Rs.2000 and Rs. 3500 respectively. 
The school can include awards for obedience. 

932 Views

 Multiple Choice QuestionsShort Answer Type

4.

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.
281 Views

5.

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


We know that:
sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction equals 2 tan to the power of negative 1 end exponent straight x space for space open vertical bar straight x close vertical bar less or equal than 1 space... left parenthesis 1 right parenthesis
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 equals 2 tan to the power of negative 1 end exponent straight y space for space straight y greater than 0 space... left parenthesis 2 right parenthesis
therefore space space 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 equals 2 tan to the power of negative 1 end exponent straight x plus 2 tan to the power of negative 1 end exponent straight y.
rightwards double arrow space 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
equals tan 1 half left parenthesis 2 tan to the power of negative 1 end exponent straight x plus 2 tan to the power of negative 1 end exponent straight y right parenthesis
equals tan left parenthesis tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y right parenthesis
equals tan open parentheses tan to the power of negative 1 end exponent fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses space space open square brackets because tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y equals 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 for space xy less than 1 close square brackets
equals fraction numerator straight x plus straight y over denominator 1 minus xy end fraction

218 Views

6.

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


open square brackets table row cell straight a minus straight b end cell cell 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 space 13 end cell end table close square brackets
Equating the corresponding elements, we get, 
rightwards double arrow straight a minus straight b space equals space minus 1 comma space space 2 straight a plus straight c space equals space 5 comma space space space 2 straight a minus straight b space equals space 0 comma space space space 3 straight c plus straight d space equals space 13
Now consider the equations, 
a - b = -1   and 2a - b = 0
Subtracting first equation from second, we get: a = 1
171 Views

7.

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.


Let space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis space equals space straight y
space space rightwards double arrow space tan space straight y space equals space 1 space equals space tan open parentheses straight pi over 4 close parentheses space rightwards double arrow space space space straight y space equals space straight pi over 4
space rightwards double arrow space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis space equals space straight pi over 4
space cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses space equals space straight z
rightwards double arrow space cos space straight z space equals space minus 1 half equals negative cos straight pi over 3 equals cos open parentheses straight pi minus straight pi over 3 close parentheses equals cos open parentheses fraction numerator 2 straight pi over denominator 3 end fraction close parentheses
rightwards double arrow space straight z space equals space fraction numerator 2 straight pi over denominator 3 end fraction space rightwards double arrow space cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses equals fraction numerator 2 straight pi over denominator 3 end fraction
therefore space 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 equals straight pi over 4 plus fraction numerator 2 straight pi over denominator 3 end fraction equals fraction numerator 11 straight pi over denominator 12 end fraction
1188 Views

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 R1, 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.
247 Views

9.

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


We know: 
2 tan to the power of negative 1 end exponent straight x space equals space tan to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction
rightwards double arrow space 2 tan to the power of negative 1 end exponent 1 fifth space equals space tan to the power of negative 1 end exponent fraction numerator 2 open parentheses begin display style 1 fifth end style close parentheses over denominator 1 minus open parentheses begin display style 1 fifth end style close parentheses squared end fraction space equals space tan to the power of negative 1 end exponent fraction numerator begin display style 2 over 5 end style over denominator begin display style 24 over 25 end style end fraction equals tan to the power of negative 1 end exponent 5 over 12
therefore space space tan open parentheses 2 tan to the power of negative 1 end exponent 1 fifth close parentheses space equals space tan open parentheses tan to the power of negative 1 end exponent 5 over 12 close parentheses equals 5 over 12

264 Views

10.

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. 


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 plus 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
rightwards double arrow space space left parenthesis x plus 1 right parenthesis thin space left parenthesis straight x plus 2 right parenthesis space minus space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis space equals space 12 plus 1
rightwards double arrow space straight x squared plus 3 straight x plus 2 minus straight x squared plus 4 straight x minus 3 space equals space 13
rightwards double arrow space space 7 straight x minus 1 space equals space 13
rightwards double arrow 7 straight x space equals space 14
rightwards double arrow space straight x space equals space 2
193 Views