Prove thatIf  find the value of x. from Class 12 CBSE Previou

Subject

Mathematics

Class

CBSE Class 12

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Sample Papers

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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).

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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.


 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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