Need Help(IIT Problem)

Q - If f(x)=1xlnt1+tdtf(x) = \displaystyle\int_1^x \dfrac{\ln t}{1 + t} \text{dt} where x>0x > 0 ,
Then the value of x satisfying the equation f(x)+f(1x)=2 f(x) + f\left(\dfrac{1}{x}\right) = 2 is

I attacked the question in this way,
f(x)+f(1x)=2 f(x) + f\left(\dfrac{1}{x}\right) = 2
Differentiating both sides with respect to x.
f(x)+f(1x)1x2=0...1 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)=lnx1+xf(1x)=ln(1x)1+1x=xlnx1+x 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(1x)f(x) \ and \ f'\left(\dfrac{1}{x}\right) in equation 1 ,

lnx1+x=xlnx1+x \dfrac{\ln x}{1+x} = \dfrac{x \ln x}{1 + x}
x=1\boxed{x = 1}
x = 1 is the wrong answer.Can you please help me out where i am wrong?

Note by Akhil Bansal
4 years ago

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1 vote

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

The answers on this post highlights the fallacy of differentiating an equation which is not an identity.

Prasun Biswas - 4 years ago

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Thanks both of you.

Akhil Bansal - 4 years ago

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

Then g(x)=f(x)+f(1x)1x2=ln(x)1+x+ln(x)x(1+x)=xlnx+ln(x)x(1+x)=ln(x)x.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)u = \ln(x) to find that g(x)=12(ln(x))2+C.g(x) = \dfrac{1}{2}(\ln(x))^{2} + C.

Next, since g(1)=0g(1) = 0 we see that C=0,C = 0, and so g(x)=12(ln(x))2.g(x) = \dfrac{1}{2}(\ln(x))^{2}.

Finally, for this to equal 22 we need to have (ln(x))2=4ln(x)=±2,(\ln(x))^{2} = 4 \Longrightarrow \ln(x) = \pm 2,

and so either x=e2x = e^{2} or x=1e2.x = \dfrac{1}{e^{2}}.

Brian Charlesworth - 4 years ago

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I really liked your solution,great observation that g(1)=0g(1) = 0 .Thanks!

Akhil Bansal - 4 years ago

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I am getting x=e2x=e^2

Tanishq Varshney - 4 years ago

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Correct answer is e2e^2 and e2 e^{-2}

Akhil Bansal - 4 years ago

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akhil i am connected brillant now

Kunal Garg - 3 years, 12 months ago

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Good,sabse pehle mje apna phone number de..

Akhil Bansal - 3 years, 12 months ago

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are you subscribed to brilliant

Kunal Garg - 3 years, 12 months ago

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@Kunal Garg No

Akhil Bansal - 3 years, 12 months ago

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