Do you like symmetry?

Calculus Level 4

Let a real valued function $f$ satisfies $f(x)+2f(1-x)=x^{2}+2$ for all real $x$ , then the graph of $f(x)$ is symmetrical about the line $x=a$ . What is the value of $a$ ?

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1 solution

Sandeep Bhardwaj
Sep 26, 2014

The given functional equation is

$f(x)+2f(1-x)=x^{2}+2. \qquad \qquad (1)$

Replacing $x$ with $1-x$ in the above equation we get

$f(1-x)+2f(x)=(1-x)^{2}+2. \qquad \qquad (2)$

Now on multiplying $(2)$ by 2 , we have

$2f(1-x)+4f(x)=2(1-x)^{2}+4 . \qquad \qquad (3)$

Subtracting $(1)$ from $(3)$ , we get

$3f(x)=2(1-x)^{2} - x^{2} + 2.$

$\Rightarrow f(x)=\frac{x^{2}-4x+4}{3}$

$\Rightarrow f(x)=\frac{(x-2)^{2}}{3}$

So clearly $f(x)$ is symmetrical about the line $\boxed{x=2}$ .

Hello no !!!! Did everything same just went on wrong in calculations :(

Pawan pal - 5 years, 3 months ago

It's all okay to make mistakes. But don't forget to learn from them. All the best!

Sandeep Bhardwaj - 5 years, 3 months ago

Replacing $\large x$ with $\large -x$ , we get the equation $\large f(-x)+2f(1+x)=x^{2}+2$ .

Now since $\large x^2 + 2 = f(x)+2f(1-x)$ , we get the equation $\large f(-x)+2f(1+x)= f(x)+2f(1-x)$ .

This suggests the function is symmetric about x=0.

Where am I going wrong?

Okay, so I tried to prove my point by putting x=2, & x =-2 into the equation given. Clearly, I'm wrong.

I noticed in the reports that Akhil's solution was incorrect due to a similar approach in which he compared the functions on the LHL & the RHL to arrive at his answer.

I wanted to ask why the approach of comparing functions fails here ?

Pulkit Gupta - 5 years, 3 months ago

You proved that the function $F(x)=f(x)+2f(1-x)$ is symmetric around $0$ , you can't deduce from that that $f(x)$ is symmetric around $0$ .

Miloje Đukanović - 5 years ago

Do you like symmetry?

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