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Continuity and Differentiability

Short Answer Type

271. Differentiate the following w.r.t. x :
x^x + x^a + a^x + a^a for some fixed a > 0 and x > 0

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272.

Find space dy over dx comma space if space straight y to the power of straight x plus straight x to the power of straight y plus straight x to the power of straight x equals straight a to the power of straight b.

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273.

If space straight y equals 1 plus fraction numerator straight c subscript 1 over denominator straight x minus straight c subscript 1 end fraction plus fraction numerator straight c subscript 2 straight x over denominator left parenthesis straight x minus straight c subscript 1 right parenthesis left parenthesis straight x minus straight c subscript 2 right parenthesis end fraction comma space prove space that space dy over dx equals straight y over straight x open parentheses fraction numerator straight c subscript 1 over denominator straight c subscript 1 minus straight x end fraction plus fraction numerator straight c subscript 2 over denominator straight c subscript 2 minus straight x end fraction close parentheses

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274. $If space straight y equals fraction numerator ax over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator bx over denominator left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator straight c over denominator straight x minus straight c end fraction plus 1. prove space that space fraction numerator straight y apostrophe over denominator straight y end fraction equals 1 over straight x open parentheses fraction numerator straight a over denominator straight a minus straight x end fraction plus fraction numerator straight b over denominator straight b minus straight x end fraction plus fraction numerator straight c over denominator straight c minus straight x end fraction close parentheses$

space straight y equals fraction numerator ax squared over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator bx over denominator left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator straight c over denominator straight x minus straight c end fraction plus 1 space space space equals fraction numerator ax squared over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator bx over denominator left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus open parentheses fraction numerator straight c over denominator straight x minus straight c end fraction plus 1 close parentheses space space space equals fraction numerator ax squared over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator bx over denominator left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator straight x over denominator straight x minus straight c end fraction space space space equals fraction numerator ax squared over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator bx plus straight x left parenthesis straight x minus straight b right parenthesis over denominator left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction space space space equals fraction numerator ax squared over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction plus fraction numerator straight x squared over denominator left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction equals fraction numerator ax squared plus straight x squared left parenthesis straight x minus straight a right parenthesis over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction therefore space straight y equals fraction numerator ax squared over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction therefore space log space straight y equals log open square brackets fraction numerator straight x cubed over denominator left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis end fraction close square brackets rightwards double arrow space log space straight y equals log space straight x cubed minus log left square bracket left parenthesis straight x minus straight a right parenthesis left parenthesis straight x minus straight b right parenthesis left parenthesis straight x minus straight c right parenthesis right square bracket therefore space log space straight y equals 3 log space straight x minus log left parenthesis straight x minus straight a right parenthesis minus log left parenthesis straight x minus straight b right parenthesis minus log left parenthesis straight x minus straight c right parenthesis Differentiating space straight w. straight r. straight t. straight x comma space we space get comma

space space space space space 1 over straight y dy over dx equals 3 over straight x minus fraction numerator 1 over denominator straight x minus straight a end fraction minus fraction numerator 1 over denominator straight x minus straight b end fraction minus fraction numerator 1 over denominator straight x minus straight c end fraction therefore space 1 over straight y dy over dx equals open parentheses 1 over straight x minus fraction numerator 1 over denominator straight x minus straight a end fraction close parentheses plus open parentheses 1 over straight x minus fraction numerator 1 over denominator straight x minus straight b end fraction close parentheses plus open parentheses 1 over straight x minus fraction numerator 1 over denominator straight x minus straight c end fraction close parentheses therefore space 1 over straight y dy over dx equals fraction numerator straight x minus straight a minus straight x over denominator straight x left parenthesis straight x minus straight a right parenthesis end fraction plus fraction numerator straight x minus straight b minus straight x over denominator straight x left parenthesis straight x minus straight b right parenthesis end fraction plus fraction numerator straight x minus straight c minus straight x over denominator straight x left parenthesis straight x minus straight c right parenthesis end fraction therefore space 1 over straight y dy over dx equals 1 over straight x open square brackets fraction numerator straight a over denominator straight a minus straight x end fraction plus fraction numerator straight b over denominator straight b minus straight x end fraction plus fraction numerator straight c over denominator straight c minus straight x end fraction close square brackets therefore space space space space space space space space fraction numerator straight y apostrophe over denominator straight y end fraction equals 1 over straight x open square brackets fraction numerator straight a over denominator straight a minus straight x end fraction plus fraction numerator straight b over denominator straight b minus straight x end fraction plus fraction numerator straight c over denominator straight c minus straight x end fraction close square brackets

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275. If u, v, w are differentiable function ofx, then show that

straight d over dx left parenthesis straight u. straight v. straight w right parenthesis equals du over dx. vw plus straight u. dv over dx. straight w plus straight u. straight v dw over dx

in two ways - first by repeated application of product rule, second by logarithmic differentiation.

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276.

Differentiate (x²-5x+8)(x³+7x+ 9) in three ways mentioned below :
(i) by using product rule.
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Also verify that three answers so obtained are the same.

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