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Question

Let f be a function defined implicitly by the equation 1ef(x)1+ef(x)=x and g be the inverse of f. If g′′(ln3)g(ln3)=pq, where p and q are relatively prime, then the value of (p+q) is 


Solution

Given, 1ef(x)1+ef(x)=x
f and g are inverse of each other. 
So, g(f(x))=x
1ef(x)1+ef(x)=g(f(x))
g(x)=1ex1+ex

g(x)=2ex(1+ex)2=2ex(1+ex)2
g′′(x)=2[ex(1+ex)2+ex(2)(1+ex)3ex]
         =2[ex(1+ex)22(1+ex)3(ex)2]

Now, g(ln3)=38
and g′′(ln3)=316
g′′(ln3)g(ln3)=316+38=916
Hence, p+q=9+16=25
 

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