Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1.

Prove that
tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x end root minus square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root plus square root of 1 minus straight x end root end fraction close square brackets space equals space straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight x comma space space fraction numerator negative 1 over denominator square root of 2 end fraction less or equal than straight x less or equal than 1


We need to prove that
tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x end root minus square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root plus square root of 1 minus straight x end root end fraction close square brackets equals space straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight X minus fraction numerator 1 over denominator square root of 2 end fraction less or equal than straight x less or equal than 1
Consider x = cos2t;

straight L. straight H. straight S. space equals space tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus cos 2 straight t end root minus square root of 1 minus cos 2 straight t end root over denominator square root of 1 plus cos 2 straight t end root plus square root of 1 minus cos 2 straight t end root end fraction close square brackets
equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 2 cost minus space square root of 2 sint over denominator square root of 2 cost plus square root of 2 sint end fraction close square brackets
equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 minus tant over denominator 1 plus tant end fraction close square brackets
equals tan to the power of negative 1 end exponent open square brackets fraction numerator tan begin display style straight pi over 4 end style minus tant over denominator 1 plus tan begin display style straight pi over 4 end style cross times tant end fraction close square brackets
equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 4 minus straight t close parentheses close square brackets
equals straight pi over 4 minus straight t
equals straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight x
equals straight R. straight H. straight S.

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2.

Find the value(s) of x for which y = open square brackets straight x left parenthesis straight x minus 2 right parenthesis close square brackets squared is an increasing function.


Given function is
straight f left parenthesis straight x right parenthesis space equals space open square brackets straight x left parenthesis straight x minus 2 right parenthesis close square brackets squared
rightwards double arrow space straight f apostrophe left parenthesis straight x right parenthesis space equals space straight x squared cross times space 2 left parenthesis straight x minus 2 right parenthesis plus left parenthesis straight x minus 2 right parenthesis squared cross times space 2 straight x
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals space 2 straight x left parenthesis straight x minus 2 right parenthesis space open square brackets straight x plus left parenthesis straight x minus 2 right parenthesis close square brackets
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals space 2 straight x left parenthesis straight x minus 2 right parenthesis left square bracket 2 straight x minus 2 right square bracket
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals space 2 straight x left parenthesis straight x minus 2 right parenthesis thin space open square brackets 2 left parenthesis straight x minus 1 right parenthesis close square brackets
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals 4 straight x left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 2 right parenthesis

Since space straight f apostrophe left parenthesis straight x right parenthesis space is space an space increasing space function comma space straight f apostrophe left parenthesis straight x right parenthesis greater than 0
rightwards double arrow space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 straight x left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 2 right parenthesis greater than 0
rightwards double arrow straight x left parenthesis straight x minus 1 right parenthesis space left parenthesis straight x minus 2 right parenthesis thin space greater than 0
rightwards double arrow 0 less than straight x less than 1 space or space straight x greater than 2
rightwards double arrow straight x space element of space left parenthesis 0 comma space 1 right parenthesis space union space left parenthesis 2 comma space infinity right parenthesis

 
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3.

If open vertical bar table row cell 3 straight x end cell cell space 7 end cell row cell negative 2 end cell cell space 4 end cell end table close vertical bar space equals open vertical bar table row 8 cell space 7 end cell row 6 cell space 4 end cell end table close vertical bar comma find the value of x.


Given that open vertical bar table row cell 3 straight x end cell cell space space 7 end cell row cell negative 2 end cell cell space 4 end cell end table close vertical bar space equals space open vertical bar table row 8 cell space space space 7 end cell row 6 cell space space space 4 end cell end table close vertical bar
We need to find the value of x

open vertical bar table row cell 3 straight x end cell cell space space 7 end cell row cell negative 2 end cell cell space 4 end cell end table close vertical bar space equals space open vertical bar table row 8 cell space space space 7 end cell row 6 cell space space space 4 end cell end table close vertical bar
rightwards double arrow space space space 12 straight x minus left parenthesis negative 14 right parenthesis space equals space 32 minus 42
rightwards double arrow space space 12 straight x plus 14 space equals negative 10
rightwards double arrow space space 12 straight x space equals space minus 10 minus 14
rightwards double arrow space 12 straight x space equals space minus 24
rightwards double arrow space space straight x space equals space minus 2

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4.

If straight R equals open curly brackets open parentheses straight x comma space straight y close parentheses colon straight x plus 2 straight y space equals space 8 close curly brackets is a relation on N, write the range of R.


The set of natural numbers, N, = {1, 2, 3, 4, 5, 6.....}
The relation is given as
straight R equals open curly brackets open parentheses straight x comma space straight y close parentheses colon straight x plus 2 straight y equals 8 close curly brackets
Thus comma space straight R space equals open curly brackets left parenthesis 6 comma space 1 right parenthesis comma space left parenthesis 4 comma space 2 right parenthesis comma space left parenthesis 2 comma space 3 right parenthesis close curly brackets
Domain equals space open curly brackets 6 comma space 4 comma space 2 close curly brackets
Range space equals open curly brackets 1 comma space 2 comma space 3 close curly brackets.

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5.

Prove that
If tan to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction close parentheses equals straight pi over 4 comma find the value of x.


 Given space that space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction close parentheses equals straight pi over 4
We space need space to space find space the space value space of space straight x.
tan to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction close parentheses equals straight pi over 4
rightwards double arrow space space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction plus fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction end style over denominator 1 minus open parentheses begin display style fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction end style close parentheses open parentheses begin display style fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction end style close parentheses end fraction close parentheses space equals space tan straight pi over 4
rightwards double arrow space fraction numerator begin display style fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction end style plus begin display style fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction end style over denominator 1 minus open parentheses begin display style fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction end style close parentheses open parentheses begin display style fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction end style close parentheses end fraction equals tan straight pi over 4
rightwards double arrow space space fraction numerator left parenthesis straight x minus 2 right parenthesis left parenthesis straight x plus 4 right parenthesis plus left parenthesis straight x plus 2 right parenthesis left parenthesis straight x minus 4 right parenthesis over denominator left parenthesis straight x minus 4 right parenthesis left parenthesis straight x plus 4 right parenthesis minus left parenthesis straight x minus 2 right parenthesis left parenthesis straight x plus 2 right parenthesis end fraction equals 1
rightwards double arrow space fraction numerator left parenthesis straight x squared plus 2 straight x minus 8 right parenthesis plus left parenthesis straight x squared minus 2 straight x minus 8 right parenthesis over denominator left parenthesis straight x squared minus 16 right parenthesis minus left parenthesis straight x squared minus 4 right parenthesis end fraction equals 1
rightwards double arrow fraction numerator 2 straight x squared minus 16 over denominator negative 12 end fraction equals 1
rightwards double arrow 2 straight x squared minus 16 equals negative 12
rightwards double arrow 2 straight x squared space equals 4
rightwards double arrow straight x squared space equals 2
rightwards double arrow space straight x equals plus-or-minus square root of 2

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6.

If A is a square matrix such that straight A squared equals straight A comma then write the value of 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed comma where I is an identity matrix.


Given that straight A squared equals straight A
We need to find the value of 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed comma Where I is the identity matrix.
Thus, 

7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed space equals space 7 straight A minus open parentheses straight I cubed plus 3 straight I squared straight A plus 3 IA squared plus straight A cubed close parentheses
space space rightwards double arrow space 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed space equals space 7 straight A minus open parentheses straight I cubed plus 3 straight A plus 3 straight A squared plus straight A squared cross times straight A close parentheses space space open square brackets straight I cubed space equals space straight I comma space straight I squared straight A equals straight A comma space IA squared equals straight A squared close square brackets
space space rightwards double arrow space 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed equals 7 straight A minus open parentheses straight I plus 3 straight A plus 3 straight A plus straight A close parentheses space space space open square brackets because straight A squared equals straight A close square brackets
space space rightwards double arrow 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed equals 7 straight A minus straight I minus 3 straight A minus 3 straight A minus straight A
space space space rightwards double arrow 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed equals 7 straight A minus straight I minus 7 straight A
space space space space rightwards double arrow 7 straight A minus left parenthesis straight I plus straight A right parenthesis cubed equals negative straight I

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7.

Find the equations of the tangent and normal to the curve straight x squared over straight a squared minus straight y squared over straight b squared equals 1 space at space the space point space open parentheses square root of 2 straight a comma space straight b close parentheses.


Let straight x squared over straight a squared minus straight y squared over straight b squared equals 1 be the equation of the curve.
Rewriting the above equation as,

straight y squared over straight b squared equals straight x squared over straight a squared minus 1
space rightwards double arrow space space space straight y squared space equals space straight b squared over straight a squared straight x squared minus straight b squared
Differentiating the above function with respect to x, we get,
2 straight y dy over dx equals straight b squared over straight a squared 2 straight x
rightwards double arrow space space dy over dx equals straight b squared over straight a squared straight x over straight y
rightwards double arrow space space open square brackets dy over dx close square brackets subscript left parenthesis square root of 2 straight a comma space straight b right parenthesis end subscript space equals space straight b squared over straight a squared fraction numerator square root of 2 straight a over denominator straight b end fraction equals fraction numerator square root of 2 straight b over denominator straight a end fraction

Slope of the tangent is straight m equals fraction numerator square root of 2 straight b over denominator straight a end fraction
Equation of the tangent is
left parenthesis straight y minus straight y subscript 1 right parenthesis space equals space straight m left parenthesis straight x minus straight x subscript 1 right parenthesis
rightwards double arrow space space space left parenthesis straight y minus straight b right parenthesis equals fraction numerator square root of 2 straight b over denominator straight a end fraction open parentheses straight x minus square root of 2 straight a close parentheses
rightwards double arrow space space straight a left parenthesis straight y minus straight b right parenthesis equals square root of 2 straight b left parenthesis straight x minus square root of 2 straight a right parenthesis
rightwards double arrow space square root of 2 bx minus ay plus ab minus 2 ab equals 0
rightwards double arrow square root of 2 bx minus ay minus ab space equals space 0
Slope of the normal is negative fraction numerator 1 over denominator begin display style fraction numerator square root of 2 straight b over denominator straight a end fraction end style end fraction
Equation of the normal is
open parentheses straight y minus straight y subscript 1 close parentheses space equals space straight m left parenthesis straight x minus straight x subscript 1 right parenthesis
rightwards double arrow space space left parenthesis straight y minus straight b right parenthesis space equals space fraction numerator negative straight a over denominator square root of 2 straight b end fraction left parenthesis straight x minus square root of 2 straight a right parenthesis
rightwards double arrow square root of 2 straight b left parenthesis straight y minus straight b right parenthesis equals negative straight a left parenthesis straight x minus square root of 2 straight a right parenthesis
rightwards double arrow ax plus square root of 2 by minus square root of 2 straight b squared plus square root of 2 straight a squared equals 0
rightwards double arrow ax plus square root of 2 by plus square root of 2 left parenthesis straight a squared minus straight b squared right parenthesis space equals space 0


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8. If space 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 straight pi over 4 comma space xy less than 1 comma space then space write space the space value space of space straight x plus straight y plus xy.

Given 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 straight pi over 4 space and space xy less than 1.
We need to find the value of x+y+xy.

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 straight pi over 4

rightwards double arrow space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses space equals space straight pi over 4 space open square brackets because space xy less than 1 close square brackets
rightwards double arrow space tan open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses close square brackets space equals space tan open parentheses straight pi over 4 close parentheses
rightwards double arrow space space fraction numerator straight x plus straight y over denominator 1 minus xy end fraction equals 1
rightwards double arrow straight x plus straight y space equals space 1 minus xy
rightwards double arrow space straight x plus straight y plus xy equals space space 1

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9.

If the function f: R rightwards arrow R  be given by straight f left parenthesis straight x right parenthesis space equals space straight x squared plus 2 space space and space straight g space colon thin space straight R rightwards arrow space straight R be given by straight g left parenthesis straight x right parenthesis space equals fraction numerator straight x over denominator straight x minus 1 end fraction comma space straight x not equal to 1 comma find fog and gof and hence find fog (2) and gof ( −3).


Given that straight f left parenthesis straight x right parenthesis space equals space straight x squared plus 2 space and space straight g left parenthesis straight x right parenthesis space equals space fraction numerator straight x over denominator straight x minus 1 end fraction
Let us find fog:

space fog space equals space straight f open parentheses straight g left parenthesis straight x right parenthesis close parentheses
rightwards double arrow space space fog space equals space open parentheses straight g left parenthesis straight x right parenthesis close parentheses squared plus 2
rightwards double arrow space fog space equals open parentheses fraction numerator straight x over denominator straight x minus 1 end fraction close parentheses squared plus 2
rightwards double arrow fog space equals fraction numerator straight x squared plus 2 left parenthesis straight x minus 1 right parenthesis squared over denominator left parenthesis straight x minus 1 right parenthesis squared end fraction
rightwards double arrow fog space equals fraction numerator straight x squared plus 2 left parenthesis straight x squared minus 2 straight x plus 1 right parenthesis over denominator straight x squared minus 2 straight x plus 1 end fraction
rightwards double arrow fog space equals fraction numerator 3 straight x squared minus 4 straight x plus 2 over denominator straight x squared minus 2 straight x plus 1 end fraction
Therefore, left parenthesis fog right parenthesis space left parenthesis 2 right parenthesis space equals space fraction numerator 3 cross times 2 squared minus 4 cross times 2 plus 2 over denominator 2 squared minus 2 cross times 2 plus 1 end fraction

rightwards double arrow left parenthesis fog right parenthesis thin space left parenthesis 2 right parenthesis space equals space fraction numerator 12 minus 8 plus 2 over denominator 4 minus 4 plus 1 end fraction space equals 6
Now space let space us space find space gof colon
gof space equals space straight g open parentheses straight f left parenthesis straight x right parenthesis close parentheses
rightwards double arrow space space gof space equals space fraction numerator straight f left parenthesis straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis minus 1 end fraction
rightwards double arrow gof space equals space fraction numerator straight x squared plus 2 over denominator straight x squared plus 2 minus 1 end fraction
rightwards double arrow space gof space equals fraction numerator straight x squared plus 2 over denominator straight x squared plus 1 end fraction

 Therefore comma space left parenthesis gof right parenthesis thin space left parenthesis negative 3 right parenthesis space equals space fraction numerator left parenthesis negative 3 right parenthesis squared plus 2 over denominator left parenthesis negative 3 right parenthesis squared plus 1 end fraction equals space fraction numerator 9 plus 2 over denominator 9 plus 1 end fraction equals 11 over 10

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10. If space open square brackets table row cell straight x minus straight y end cell cell space straight z end cell row cell 2 straight x minus straight y end cell cell space straight w end cell end table close square brackets space equals space open square brackets table row cell negative 1 end cell cell space 4 end cell row 0 cell space 5 end cell end table close square brackets comma space find space the space value space of space straight x plus straight y.

Given space that space open square brackets table row cell straight x minus straight y end cell straight z row cell 2 straight x minus straight y end cell straight w end table close square brackets space equals space open square brackets table row cell negative 1 end cell cell space 4 end cell row cell space space 0 end cell cell space 5 end cell end table close square brackets
We need to find the value of x+y.

open square brackets table row cell straight x minus straight y end cell straight z row cell 2 straight x minus straight y end cell straight w end table close square brackets space equals space open square brackets table row cell negative 1 end cell cell space 4 end cell row cell space space 0 end cell 5 end table close square brackets

Two matrices A and B are equal to each other, if they have the same dimensions and the same elements

 straight a subscript ij space equals space straight b subscript ij comma space space space for space straight i space equals space 1 comma 2 comma.... comma straight n
and space straight j space equals space 1 comma space 2.... comma space straight m.
rightwards double arrow space space
space straight x minus straight y space equals negative 1 space.... left parenthesis 1 right parenthesis
2 straight x minus straight y equals 0..... left parenthesis 2 right parenthesis
Equation space left parenthesis 2 right parenthesis minus left parenthesis 1 right parenthesis space is space straight x space equals space 1
Substituting space the space value space of space straight x space equals space 1 space in space equation space left parenthesis 1 right parenthesis comma space we space have
1 minus straight y space equals space minus 1
rightwards double arrow straight y space equals space 2
Therefore comma space straight x plus straight y space equals space 1 plus 2 space equals space 3
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