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Continuity and Differentiability

Short Answer Type

501. $Prove space that space the space derivative space of space sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses comma space straight x greater than 0 space straight w. straight r. straight t. space square root of 1 minus straight x squared end root space is space equal space to space the space derivative space straight l subscript straight n space left parenthesis straight x squared right parenthesis space with space respect space to space straight x comma space left square bracket straight l subscript straight n left parenthesis straight x squared right parenthesis equals space log to the power of straight e. left parenthesis straight a squared right parenthesis right square bracket.$

Let space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses comma space straight u equals square root of 1 minus straight x squared end root Put space straight x equals cos space straight theta therefore space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 cos squared space straight theta minus 1 end fraction close parentheses equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator cos space 2 straight theta end fraction close parentheses equals sec to the power of negative 1 end exponent open parentheses sec space 2 straight theta close parentheses equals 2 straight theta therefore space straight y equals 2 space cos to the power of negative 1 end exponent straight x space space rightwards double arrow space dy over dx equals negative fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction Also space straight u equals square root of 1 minus straight x squared end root therefore space space space space space du over dx equals fraction numerator 1 over denominator 2 square root of 1 minus straight x squared end root end fraction cross times left parenthesis negative 2 straight x right parenthesis equals negative fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction Now space dy over dx equals fraction numerator begin display style dy over dx end style over denominator begin display style du over dx end style end fraction equals negative fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction cross times fraction numerator square root of 1 minus straight x squared end root over denominator negative straight x end fraction space space space space space space space space space space space space space space space space space equals 2 over straight x equals 2 straight d over dx left parenthesis log subscript straight e space straight x right parenthesis equals straight d over dx left parenthesis 2 space log subscript straight e space straight x right parenthesis equals straight d over dx left parenthesis log subscript straight e space straight x squared right parenthesis equals straight d over dx left square bracket straight l subscript straight n left parenthesis straight x squared right parenthesis right square bracket

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502.

Differentiate space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 1 plus square root of 1 minus straight x squared end root end fraction close parentheses space straight w. straight r. straight t. space sin open parentheses 2 space cot to the power of negative 1 end exponent space square root of fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction end root close parentheses

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503.

If space straight y equals tan to the power of negative 1 end exponent fraction numerator 4 straight x over denominator 1 plus 5 straight x squared end fraction plus tan to the power of negative 1 end exponent fraction numerator 2 plus 3 straight x over denominator 3 minus 2 straight x end fraction comma space prove space that space dy over dx equals fraction numerator 5 over denominator 1 plus 25 straight x squared end fraction.

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504.

Given space that space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 2 over denominator 1 minus 2 straight x end fraction close parentheses comma space minus 1 less than straight x less than 1 half. space After space using space the space property space of space inverse space trigonometric space function comma space show space that space dy over dx equals 0

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505.

If space straight theta equals cos to the power of negative 1 end exponent open parentheses straight r over straight k close parentheses minus fraction numerator square root of straight k squared minus straight r squared end root over denominator straight r end fraction comma space find space dθ over dr

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506. Differentiate the following w.r.t.x:

straight x space to the power of sin to the power of negative 1 end exponent straight x end exponent

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507.

Differentiate space the space following space straight w. straight r. straight t. straight x colon space left parenthesis sin space straight x right parenthesis to the power of cos to the power of negative 1 end exponent straight x end exponent

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508. Differentiate the following w.r.t.x:

straight x to the power of cos to the power of negative 1 end exponent straight x end exponent

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509.

If space straight y equals tan to the power of negative 1 end exponent straight x over straight y comma space then space evaluate space dy over dx.

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510.

If space left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis plus straight y to the power of cot space straight x end exponent equals 1 comma space find space dy over dx.

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