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Q. Let exexex+ex=ln1+x1x , then find x .

NTA AbhyasNTA Abhyas 2022

Solution:

Let f(x)=exexex+ex
y=exexex+ex
y=e2x1e2x+1
e2xy+y=e2x1
e2x(y1)=1y
e2x=1+y1y
ex=1+y1y
x=ln1+y1y
f1(y)=ln1+y1y
f1(x)=ln1+x1x
Now, given equation is
exexex+ex=ln1+x1x
f(x)=f1(x)
The solution of f(x)=f1(x) is always obtained on y=x line
ln1+x1x=x
e2x=1+x1x
Solution
Only one solution x=0 .