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Continuity and Differentiability

Short Answer Type

571.

If space straight y equals left parenthesis cos to the power of negative 1 end exponent space straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis space straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0.

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572.

If space space space straight y equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x 2 right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0. space

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573.

If space space straight y equals open square brackets log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses close square brackets squared comma space prove space that space left parenthesis 1 space plus space straight x squared right parenthesis straight y subscript 2 space plus space straight x space straight y subscript 1 space equals space 2.

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574. $If space space straight y equals sin left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 plus straight m squared space straight y equals 0.$

space straight y equals sin left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis therefore space straight y 1 equals cos left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis. fraction numerator straight m over denominator square root of 1 minus straight x squared end root end fraction space space rightwards double arrow space square root of 1 minus straight x squared end root. straight y 1 equals straight m space cos left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis rightwards double arrow space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals straight m squared space cos squared left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis space space rightwards double arrow space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals straight m squared left square bracket 1 minus sin squared left parenthesis straight m space sin to the power of negative 1 end exponent straight x right parenthesis right square bracket therefore space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals straight m squared left parenthesis 1 minus straight y squared right parenthesis Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma left parenthesis 1 space straight x squared right parenthesis. space 2 space straight y subscript 1 space straight y subscript 2 space plus straight y subscript 1 squared left parenthesis negative 2 space straight x right parenthesis equals straight m squared left parenthesis negative 2 space straight y space straight y subscript 1 right parenthesis Dividing space both space sides space by space 2 space straight y subscript 1 comma space get comma left parenthesis 1 minus straight x squared right parenthesis space straight y subscript 2 minus straight x space straight y subscript 1 equals negative straight m squared straight y space or space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 space minus straight x space straight y subscript 1 plus straight m squared space straight y equals 0

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575. If y = e^{m sin₁ x} , prove that (1 - x²) y₂ - x y₁ = m² y.

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576. If y = e^{a cos_{-1 x}} , show that

left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0.

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577.

If space space straight f left parenthesis straight x right parenthesis equals vertical line straight x vertical line cubed comma space show space that space straight f " left parenthesis straight x right parenthesis space exists space for space all space real space straight x space and space find space it.

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578. If x = f(t), y = g (t) possess second order derivative for all t in (a, b) and f"(t) is inevitable then prove that

fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator straight f apostrophe left parenthesis straight t right parenthesis space straight g apostrophe apostrophe left parenthesis straight t right parenthesis minus straight g apostrophe left parenthesis straight t right parenthesis space straight f apostrophe apostrophe left parenthesis straight t right parenthesis over denominator left parenthesis straight f apostrophe left parenthesis straight t right parenthesis right parenthesis squared end fraction

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Long Answer Type

579.

If space left parenthesis straight x minus straight a right parenthesis squared space plus left parenthesis straight y minus straight b right parenthesis squared space equals straight c squared space comma space prove space that space space fraction numerator open square brackets 1 plus open parentheses begin display style dy over dx end style close parentheses squared close square brackets to the power of begin display style 3 over 2 end style end exponent over denominator begin display style fraction numerator straight d squared straight y over denominator dx squared end fraction end style end fraction space is space straight a space constant space independent space of space straight a space and space straight b.

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Short Answer Type

580.

Verify space Rolle apostrophe straight s space theorem space for space the space function straight f left parenthesis straight x right parenthesis equals fraction numerator 8 straight x squared over denominator 3 end fraction minus 2 straight x comma space straight x element of open square brackets 0 comma 3 over 4 close square brackets

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