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# Need Help(IIT Problem)

Q - If $$f(x) = \displaystyle\int_1^x \dfrac{\ln t}{1 + t} \text{dt}$$ where $$x > 0$$ ,
Then the value of x satisfying the equation $$f(x) + f\left(\dfrac{1}{x}\right) = 2$$ is

I attacked the question in this way,
$f(x) + f\left(\dfrac{1}{x}\right) = 2$
Differentiating both sides with respect to x.
$f'(x) + f'\left(\dfrac{1}{x}\right)\cdot \dfrac{-1}{x^2} = 0 \quad \quad ...1$

Applying leibniz theorem in the given integral,
$f'(x) = \dfrac{\ln x}{1+x} \quad \quad f'\left(\dfrac{1}{x}\right) = \cfrac{\ln\left(\frac{1}{x}\right)}{1 + \frac{1}{x}} = \dfrac{-x \ln x}{1 + x}$
Substituting values of $$f(x) \ and \ f'\left(\dfrac{1}{x}\right)$$ in equation 1 ,

$\dfrac{\ln x}{1+x} = \dfrac{x \ln x}{1 + x}$
$\boxed{x = 1}$

Note by Akhil Bansal
1 year, 3 months ago

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The error in your method is that you differentiated both sides of a non-identical equation. Only functions (implicit or explicit) and identities can be differentiated (this is one of the basic rules of differentiation).

The statement $$f(x)+f(1/x)=2$$ is an equation in terms of $$x$$ which holds true for a finite set of values of $$x$$, i.e., the equality here is non-identical. As such, you can't differentiate it w.r.t $$x$$.

The answers on this post highlights the fallacy of differentiating an equation which is not an identity. · 1 year, 3 months ago

Thanks both of you. · 1 year, 3 months ago

I think that you first need to look at the function $$g(x) = f(x) + f\left(\dfrac{1}{x}\right)$$ in general.

Then $$g'(x) = f'(x) + f'\left(\dfrac{1}{x}\right)*\dfrac{-1}{x^{2}} = \dfrac{\ln(x)}{1 + x} + \dfrac{\ln(x)}{x(1 + x)} = \dfrac{x\ln{x} + \ln(x)}{x(1 + x)} = \dfrac{\ln(x)}{x}.$$

Integrate using the substitution $$u = \ln(x)$$ to find that $$g(x) = \dfrac{1}{2}(\ln(x))^{2} + C.$$

Next, since $$g(1) = 0$$ we see that $$C = 0,$$ and so $$g(x) = \dfrac{1}{2}(\ln(x))^{2}.$$

Finally, for this to equal $$2$$ we need to have $$(\ln(x))^{2} = 4 \Longrightarrow \ln(x) = \pm 2,$$

and so either $$x = e^{2}$$ or $$x = \dfrac{1}{e^{2}}.$$ · 1 year, 3 months ago

I really liked your solution,great observation that $$g(1) = 0$$ .Thanks! · 1 year, 3 months ago

akhil i am connected brillant now · 1 year, 3 months ago

Good,sabse pehle mje apna phone number de.. · 1 year, 3 months ago

are you subscribed to brilliant · 1 year, 3 months ago

No · 1 year, 3 months ago

I am getting $$x=e^2$$ · 1 year, 3 months ago

Correct answer is $$e^2$$ and $$e^{-2}$$ · 1 year, 3 months ago