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Question

Ify=sec-1((2x)(1+x2))+sin-1[x-1][x+1],thendydx=


A

1

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B

x+1x-1

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C

does not exist

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D

none of these

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Solution

The correct option is C

does not exist


Explanation for the correct option:

Step : Find the value of dydx

Given that y=sec-1((2x)(1+x2))+sin-1[x-1][x+1]

Differentiation by substitution

sincey=sec-1((2x)(1+x2))+sin-1[x-1][x+1]substitutex=tanθthen(2x)(1+x2)=2tanθ(1+tan2θ)=sin2θSince-1sinθ1-1(2x)(1+x2)1Butsec-1((2x)(1+x2)isdefinedonlyatx=-1and1Sosolutiondoesnotexist.

Hence, option (C) is correct.


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