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Differential Calculus

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4 Apr 2009 14:47:21 IST

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simple........(continuity)

None

if f(x)= 1/{(x-1)(x-2)} and g(x)= 1/x², then the point of discontinuity of f(g(x)) are :

a) {-1,0,1,k}

b) {-k , -1,0,1,k}

c){0,1}

d) {0,1,k}

where ......

kindly give solution...

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Comments (3)

A$PIRing HiGh - dEv aDitYa

Blazing goIITian

Joined: 13 Mar 2008

Posts: 540

4 Apr 2009 15:06:42 IST

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ans will be b)

A$PIRing HiGh - dEv aDitYa

Blazing goIITian

Joined: 13 Mar 2008

Posts: 540

4 Apr 2009 15:12:19 IST

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g(x) will be discontinuous at x=0. f(x) is discontinuous at X=1 and X=2. So if X=1 x=+1 and -1 and if X=2 ,x=+k and -k. So, the composite fn is discontinuous at 5 values of x namely {-k,-1,0,1,k}. Cheeres!!! Plz rate if understood!!!!

sneha

Blazing goIITian

Joined: 17 Mar 2008

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4 Apr 2009 15:13:44 IST

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to find f(g(x)), replace x by 1/x² in f(x)

hence we get

f(g(x)) =

now clearly, this function is discontinuous if denominator = 0

hence points of discontinuity are x =1,-1,k,-k where k is

also, we have to take care that g(x) is defined. hence x cannot be 0

so, the answer is {1,-1,0,k,-k}

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