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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.

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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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space space space space space Let space straight y equals 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 therefore space log space straight y equals log 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 equals cos to the power of negative 1 end exponent straight x. space log space sin space straight x therefore space 1 over straight y dy over dx equals left parenthesis cos to the power of negative 1 end exponent straight x right parenthesis. fraction numerator cos space straight x over denominator sin space straight x end fraction minus fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction. log space sin space straight x therefore space dy over dx equals open parentheses sin space straight x close parentheses to the power of cos to the power of negative 1 end exponent straight x end exponent open square brackets cot space straight x. space cos to the power of negative 1 end exponent straight x minus fraction numerator log space sin space straight x over denominator square root of 1 minus straight x squared end root end fraction close square brackets

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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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