If the function f: R  R  be given by  be given by  find 

Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1.

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.

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2. 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.
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3. 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.
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4.

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.

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5.

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.

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6.

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.

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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.

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8.

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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9.

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

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10.

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.

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