If 2f(x) + f(x^2 - 1) = 1, for all real numbers x, then what is the

Question

If  2f(x) + f(x^2 - 1) = 1 , for all x, then what is the value of  f(-\sqrt{2})  ?

A. -1/3
B. 0
C. 1/3
D. 2/3
E. 1

M40-147

Bunuel · Accepted Answer

We are asked to find  f(-\sqrt{2}) .

Substitute  x = -\sqrt{2}  in the given equation  2f(x) + f(x^2 - 1) = 1 :
 2f(-\sqrt{2}) + f(1) = 1 
So, to get  f(-\sqrt{2}) , we now need  f(1) .

Since the equation holds for all  x , it must also hold for  x = 1 . Substitute  x = 1 :
 2f(1) + f(0) = 1

We now need  f(0) . Substitute  x = 0 :
 2f(0) + f(-1) = 1

We now need  f(-1) . Substitute  x = -1 :
 2f(-1) + f(0) = 1

From the last two equations:
 2f(0) + f(-1) = 1 
 2f(-1) + f(0) = 1 
Subtracting gives  f(0) = f(-1) .

Substitute this into either equation:  3f(0) = 1 , so  f(0) = f(-1) = \frac{1}{3} .

Substitute  f(0) = \frac{1}{3}  into  2f(1) + f(0) = 1  to get  f(1) = \frac{1}{3} .

Finally, from the first equation:
 2f(-\sqrt{2}) + \frac{1}{3} = 1 , so  f(-\sqrt{2}) = \frac{1}{3} .

Answer: C.

gmatophobia · Answer

We can substitute the value of x = - \sqrt{2}  in the equation. Once we have obtained the function whose value is not known, that variable can be substituted in the base equation to check if we have sufficient information to solve the question.

Ex: 2f(x) + f(x^2 - 1) = 1

at x = - \sqrt{2}

2f(-\sqrt{2}) + f((-\sqrt{2})^2 - 1) = 1

f(-\sqrt{2})  =  \frac{1-f(1)}{2}

Now, we can substitute x = 1 in the base equation and solve further.

x = 1

2f(1) + f((1)^2 - 1) = 1

f(1) = \frac{1-f(0)}{2}

This process needs to continue, till we have all known values (of course, we will soon see a pattern!)

A detailed solution is attached.

IMO C

chetan2u · Answer

Such questions require you to substitute values of x in successive steps till you find that you can get a value of some function. 
 2f(x) + f(x^2 - 1) = 1

(1)  x=(-\sqrt{2}) 
 2f(-\sqrt{2}) + f((-\sqrt{2)}^2 - 1) = 1 
 2f(-\sqrt{2}) + f(1) = 1

(2)  x=(1) 
 2f(1) + f(1^2 - 1) = 1 
 2f(1) + f(0) = 1

(3)  x=(0) 
 2f(0) + f(0^2 - 1) = 1 
 2f(0) + f(-1) = 1 ………(3)

(4)  x=(-1) 
 2f(-1) + f((-1)^2 - 1) = 1 
 2f(-1) + f(0) = 1 ……..(4)

Multiply (3) with 2 and subtract 4 from it. 
 3f(0)=1……f(0)=\frac{1}{3}

Substitute this in (2)
 2f(1)+\frac{1}{3}=1 
 f(1)=\frac{1}{3}

Substitute this in (1)
 2f(-\sqrt{2})+\frac{1}{3}=1 
 f(-\sqrt{2})=\frac{1}{3}

C

Regor60 · Answer

1. 2f(-√2) + f(1) = 1

2. 2f(1) + f(0) = 1

3. 2f(0) + f(-1) = 1

4. 2f(-1) + f(0) = 1

Solve for f(-1) using 3 and 4:

f(0) = f(-1)

Substitute above into either 3 or 4:

3f(0) = 1 and f(0) = 1/3

Substitute into 2:

2f(1) + 1/3 = 1 so

f(1) = 1/3

Substitute into 1:

2f(-√2) + 1/3 = 1 so

2f(-√2) = 2/3 so

f(-√2) = 1/3

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shubhim20 · Answer

please elaborate the process of substituting in 3 & 4

Regor60 · Answer

This is just solving the two equations in two unknowns.

The first step is to develop an equation that includes the f(-√2).

The next objective is to minimize the number of new equations developed that link the second part of equation 1 and that can be solved using substitutions

If 2f(x) + f(x^2 - 1) = 1, for all real numbers x, then what is the

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