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 ,

\[ \dfrac{\ln x}{1+x} = \dfrac{x \ln x}{1 + x} \]

\[\boxed{x = 1}\]

x = 1 is the wrong answer.Can you please help me out where i am wrong?

## Comments

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TopNewestThe 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. – Prasun Biswas · 1 year, 9 months ago

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– Akhil Bansal · 1 year, 9 months ago

Thanks both of you.Log in to reply

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 \(x = \dfrac{1}{e^{2}}.\) – Brian Charlesworth · 1 year, 9 months ago

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– Akhil Bansal · 1 year, 9 months ago

I really liked your solution,great observation that \(g(1) = 0\) .Thanks!Log in to reply

akhil i am connected brillant now – Kunal Garg · 1 year, 9 months ago

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– Akhil Bansal · 1 year, 9 months ago

Good,sabse pehle mje apna phone number de..Log in to reply

– Kunal Garg · 1 year, 9 months ago

are you subscribed to brilliantLog in to reply

– Akhil Bansal · 1 year, 9 months ago

NoLog in to reply

I am getting \(x=e^2\) – Tanishq Varshney · 1 year, 10 months ago

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– Akhil Bansal · 1 year, 9 months ago

Correct answer is \(e^2\) and \( e^{-2}\)Log in to reply

@Brian Charlesworth , @Sandeep Bhardwaj , @Prasun Biswas , @Calvin Lin , @Pi Han Goh , @Tanishq Varshney , @Otto Bretscher , @Jon Haussmann – Akhil Bansal · 1 year, 10 months ago

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