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Continuity and Differentiability

Short Answer Type

421.

Differentiate space straight w. straight r. straight t. straight x colon space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 1 plus 2 sin space straight x over denominator 2 plus cos space straight x end fraction close parentheses

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422.

Differentiate space sin to the power of negative 1 end exponent open parentheses fraction numerator straight a plus straight b space cos space straight x over denominator straight b plus straight a space cos space straight x end fraction close parentheses space straight w. straight r. straight t. straight x comma space using space chain space rule

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423.

Prove space that space straight d over straight d open square brackets square root of 1 minus straight x squared end root sin to the power of negative 1 end exponent straight x minus straight x close square brackets equals negative fraction numerator straight x space sin to the power of negative 1 end exponent straight x over denominator square root of 1 minus straight x squared end root end fraction

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424.

Prove space that space straight d over dx left square bracket 2 straight x space tan to the power of negative 1 end exponent straight x minus log left parenthesis 1 plus straight x squared right parenthesis right square bracket equals 2 space tan to the power of negative 1 end exponent straight x.

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425.

Prove space that space straight d over dx open square brackets straight x over 2 square root of straight a squared minus straight x squared end root plus straight a squared over 2 sin to the power of negative 1 end exponent straight x over straight a close square brackets equals square root of straight a squared minus straight x squared end root

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426.

Differerntiate space cos to the power of negative 1 end exponent open parentheses fraction numerator 3 cos space straight x minus 4 sin space straight x over denominator 5 end fraction close parentheses space straight w. straight r. straight t. straight x.

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427.

Find space dy over dx space if space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 5 straight x plus 12 square root of 1 minus straight x squared end root over denominator 13 end fraction close parentheses

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428. $If space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses comma space find space dy over dx.$

straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses space space equals cos to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space open square brackets because space sec to the power of negative 1 end exponent straight x equals cos to the power of negative 1 end exponent open parentheses 1 over straight x close parentheses close square brackets space space equals straight pi over 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space sin to the power of negative 1 end exponent straight theta plus cos to the power of negative 1 end exponent straight theta equals straight pi over 2 close square brackets therefore space dy over dx equals 0

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

tan to the power of negative 1 end exponent open square brackets open parentheses fraction numerator 1 minus cos space straight x over denominator 1 plus cos space straight x end fraction close parentheses to the power of 1 half end exponent close square brackets

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

tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 plus sin space straight x over denominator 1 minus sin space straight x end fraction end root close parentheses

79 Views