Question

Solve for x:

2sin(e^x-1+½π-x)=π^x²-1+π^1-x². Not as hideous as it looks!

(HINT: Consider the maximum and minimum values of each side of the equation.)

Answer 1

Hello Sir,

Range of  2sin(e^x-1+½π-x) is in [-2,2] since it is a sine function.  

Range of π^x²-1+π^1-x² is (0,inf] since it is sum of exponential function. 

On taking 1st derivative of RHS we get:

-2 π^(1 - x^2) xlog(π) + 2 π^(-1 + x^2) xlog(π)

So the critical numbers are:

0,1,-1

Putting the value of 0,1 and -1 in RHS we get π + 1/π, 2 and 2 respectively.

So the minimum value of RHS is 2.

Since Maximum of LHS is 2 and Minimum of RHS is 2,

Edit:

And it happens when x= -1 or x = 1

But @x =-1 LHS is less than 2,

So the correct answer is x = 1

Comments

Yes, the range of the left-hand side is -2 to 2; but the right-hand side can never be zero. Check your estimate of the range and correct it, and you'll probably arrive at the solution. You will not need calculus to find the solution.

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Sir,

Please check again.

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The answer is x=1, although your reasoning still contains errors. But it's good that you had a go at answering this question. The maximum value of the LHS equals the minimum value of the RHS--that was the important bit (the exponents on the RHS have to be minimised to zero, and this can only happen if x=±1. And on the LHS the maximum value is realised when we have sin(π/2) implying that e^x-1=x, which occurs when x=1, that is, x-1=ln(x), because 1-1=ln(1)=0.) You got there in the end! And, of course, we didn't need π on the RHS because any base to the power of zero is 1.

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Sir, though your answer was genius and way out of the box, but I still think my reasoning does not have any error, correct me if I am wrong.

My thinking was,

@RHS Minimum value is 2 and is obtained at x= -1 or 1

@LHS, Can have 2 as maximum value. But since it is a sinusoidal wave, so it is periodic and can have many maximum values,

But still LHS have only two points where it can meet RHS (ie -1 and 1)  so I checked them both and found x=1 is the answer. Please check the graph,

https://i.imgur.com/KVDlC4Z.png

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The error I was referring to is in the the range you applied to the RHS {0,inf}. The range is actually [2,∞) expressed as an interval. The sum of two exponential quantities cannot be zero. π^1-x²→0, but π^x²-1→∞, as x gets larger in magnitude. If you draw two graphs: y=LHS and y=RHS you will find they touch at one point only, i.e., when x=1, when LHS is max and RHS is min. But, hey, you arrived at the solution! Thanks for persevering.