JEE Novice - (20)

Algebra Level 3

f ( x + y ) = f ( x ) + f ( y ) + 2 x y \Large f(x+y) = f(x) + f(y)+2xy

The function f ( x ) f(x) satisfies the above property ∀ x , y ∈ N \forall x,y \in \mathbb{N} . Given that f ( 1 ) = 4 f(1)=4 , find the value of f ( 8 ) f(8) .

This question is a part of JEE Novices .


The answer is 88.

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Nihar Mahajan by Nihar Mahajan , 20, India

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4 solutions

Chew-Seong Cheong
May 27, 2015

f ( x + y ) = f ( x ) + f ( y ) + 2 x y f ( 1 ) = 4 ⇒ f ( 2 ) = f ( 1 + 1 ) = f ( 1 ) + f ( 1 ) + 2 ( 1 ) ( 1 ) = 4 + 4 + 2 = 10 f ( 4 ) = f ( 2 + 2 ) = f ( 2 ) + f ( 2 ) + 2 ( 2 ) ( 2 ) = 10 + 10 + 8 = 28 f ( 8 ) = f ( 4 + 4 ) = f ( 4 ) + f ( 4 ) + 2 ( 4 ) ( 4 ) = 28 + 28 + 32 = 88 \begin{aligned} f(x+y) & = f(x)+f(y)+2xy \\ f(1) & = 4 \\ \Rightarrow f(2) & = f(1+1) = f(1)+f(1)+2(1)(1) = 4+4+2 = 10 \\ f(4) & = f(2+2) = f(2)+f(2)+2(2)(2) = 10+10+8 = 28 \\ f(8) & = f(4+4) = f(4)+f(4)+2(4)(4) = 28+28+32 = \boxed{88} \end{aligned}

For general function, let y = 1 y = 1 :

f ( x + 1 ) = f ( x ) + f ( 1 ) + 2 x = f ( x ) + 2 x + 4 ⇒ f ( x + 1 ) − f ( x ) = 2 x + 4 Δ x = 2 x + 4 ⇒ f ( x + 1 ) = f ( 1 ) + ∑ k = 1 x Δ x = 4 + ∑ k = 1 x ( 2 x + 4 ) = 4 + x ( x + 1 ) + 4 x = x 2 + 5 x + 4 = ( x + 1 ) 2 + 3 ( x + 1 ) ⇒ f ( x ) = x 2 + 3 x \begin{aligned} f(x+1) & = f(x) + f(1) + 2x = f(x) + 2x + 4 \\ \Rightarrow f(x+1) - f(x) & = 2x + 4 \\ \Delta_x & = 2x + 4 \\ \Rightarrow f(x+1) & = f(1) + \sum_{k=1}^x {\Delta_x} \\ & = 4 + \sum_{k=1}^x {(2x+4)} \\ & = 4 + x(x+1) + 4x \\ & = x^2 + 5x + 4 \\ & = (x+1)^2 + 3(x+1) \\ \Rightarrow f(x) & = x^2+3x \end{aligned}

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Moderator note:

What is the general solution to this functional equation?

Nice solution. :)

Nihar Mahajan - 6 years ago

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I found two or three of the JEE Novice solutions could be wrong. Please check again.

Chew-Seong Cheong - 6 years ago

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Oh my bad! Can you give me links please?

Nihar Mahajan - 6 years ago

The way you used for finding the general solution was pretty impressive...

Vighnesh Raut - 6 years ago

Brilliant method to find the function.

Mehul Arora - 5 years, 4 months ago

In response to Challenge Master

f ( x + y ) = f ( x ) + f ( y ) + 2 x y f(x+y) = f(x) + f(y)+2xy Differentiating treating y y as a constant. f ′ ( x + y ) = f ′ ( x ) + 2 y f^{ ' }\left( x+y \right) =f^{ ' }\left( x \right) +2y Putting x = 0 x=0 and y = x y=x , f ′ ( x ) = f ′ ( 0 ) + 2 x f^{ ' }\left( x \right) =f^{ ' }\left( 0 \right) +2x Integrating with respect to x x , f ( x ) = f ′ ( 0 ) x + x 2 + k f\left( x \right) =f^{ ' }\left( 0 \right) x+{ x }^{ 2 }+k Putting x = 0 x=0 f ( 0 ) = f ′ ( 0 ) × 0 + 0 + k f\left( 0 \right) =f^{ ' }\left( 0 \right) \times 0+0+k f ( 0 ) = k f\left( 0 \right) =k Putting x = 0 x=0 in the original equation, f ( 0 ) = f ( 0 ) + f ( 0 ) f(0)=f(0)+f(0) f ( 0 ) = 0 f(0)=0 Using this result in the previous statement, k = 0 k=0 Therefore our function is f ( x ) = x f ′ ( 0 ) + x 2 f(x)=xf^{ ' }\left( 0 \right) +{ x }^{ 2 } Now putting x = 1 x=1 , f ( 1 ) = f ′ ( 0 ) + 1 f(1)=f^{ ' }\left( 0 \right) +1 Using the the value given in the question we get, f ′ ( 0 ) = 3 f^{ ' }\left( 0 \right) =3

Therefore our function is f ( x ) = x 2 + 3 x f\left( x \right) ={ x }^{ 2 }+3x

Abhishek Sharma - 6 years ago

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Good solution!

Chew-Seong Cheong - 6 years ago

You are given that the function is from the integers to the integers, which you cannot differentiate. The issues with your solution are:
1) You are assuming that any such function from the integers can be extended to the reals.
2) You are assuming that any such function on the reals is differentiable.

Well, you can "integer differentiate" such a function, but that doesn't follow the exact same rules that you are used to. For that, see Adarsh's solution below.

Calvin Lin Staff - 6 years ago
Jesse Nieminen
May 27, 2015

if y = 1 then f(x + 1) = 4 + f(x) + 2x

if x = y = 1 then f(2) = 4 + 4 + 2 = 8

if y = 2 then f(x + 2) = 8 + f(x) + 4x

f(x + 2) - f(x + 1) = 2x + 4

f(x+2) = f(x+1) + 2x + 4

f(m) = f(m-1) + 2(m-1) + 4

f(m-1) = f(m-2) + 2(m-2) + 4

f(m) = f(m-2) + 2(m-2) + 4 + 2(m-1) + 4

f(m) = f(m-n) + 4n + 2mn - n(n+1)

f(m)=4 + 4(m-1) + 2(m(m-1)) - m(m-1)

f(m) = 4 + 4m - 4 + m^2 - m = m^2+3m

f(x) = x^2 + 3x

f(8) = 64 + 24 = 88

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Good job with identifying the function :)

Calvin Lin Staff - 6 years ago
Adarsh Kumar
May 28, 2015

Just put x = 1 x=1 and the equation given changes to f ( y + 1 ) = f ( 1 ) + f ( y ) + 2 y ⟹ f ( y + 1 ) − f ( y ) = 2 y + 4 f(y+1)=f(1)+f(y)+2y\\ \Longrightarrow f(y+1)-f(y)=2y+4 .Now,start by putting y = 7 y=7 so as to get f ( 8 ) f(8) ,and keep on putting values of y y till 1 1 .Now,it becomes, f ( 8 ) − f ( 7 ) = 4 + 2 × 7 f ( 7 ) − f ( 6 ) = 4 + 2 × 6 . . . . f ( 2 ) − f ( 1 ) = 4 + 2 × 1 f(8)-f(7)=4+2\times 7\\ f(7)-f(6)=4+2\times 6\\ .\\.\\.\\. f(2)-f(1)=4+2\times 1 .Adding the above equations we are left with f ( 8 ) − f ( 1 ) = 4 × 7 + 2 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) ⟹ f ( 8 ) = 4 + 28 + 56 = 88 f(8)-f(1)=4\times 7+2(1+2+3+4+5+6+7)\\ \Longrightarrow f(8)=4+28+56\\ =88 .And we are done! Response to the Challenge Master's Note: From above we have, f ( y + 1 ) − f ( y ) = 4 + 2 y f ( y ) − f ( y − 1 ) = 4 + 2 ( y − 1 ) . . . f ( 2 ) − f ( 1 ) = 4 + 2 × 1 f(y+1)-f(y)=4+2y\\ f(y)-f(y-1)=4+2(y-1)\\ .\\.\\.\\ f(2)-f(1)=4+2\times 1 .Now,adding the above equations,we get f ( y + 1 ) − f ( 1 ) = 4 × y + 2 × 1 2 × y × ( y + 1 ) ⟹ f ( y + 1 ) = 4 + 4 y + y ( y + 1 ) f(y+1)-f(1)=4\times y + 2 \times \dfrac{1}{2} \times y\times(y+1)\\ \Longrightarrow f(y+1)=4+4y+y(y+1) .Now,substitute y + 1 = x y+1=x and the expression would become, f ( x ) = 4 + 4 x − 4 + x 2 − x = x 2 + 3 x f(x)=4+4x-4+x^2-x\\ =x^2+3x . And we are done! @Calvin Lin

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Moderator note:

Can you find the general solution to the functional equation? You're on the right step already!

@Calvin Lin please see the modification.

Adarsh Kumar - 6 years ago

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You should take it a step further to define it in terms of f ( x ) f(x) . Namely do the substitution x = y + 1 x = y + 1 to conclude that f ( x ) = x 2 + 3 x f(x) = x^2 + 3x .

Calvin Lin Staff - 6 years ago

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Ok sir! Thanx!

Adarsh Kumar - 6 years ago
Roman Frago
Aug 7, 2015

We can say that f ( x ) f(x) is quadratic, so let y = f ( x ) = ( x + a ) 2 + b y=f(x)=(x+a)^2+b

f ( x + y ) = ( x + y + a ) 2 + b f(x+y)=(x+y+a)^2+b

f ( x + y ) = x 2 + y 2 + a 2 + 2 x y + 2 a x + 2 a y + b f(x+y)=x^2+y^2+a^2+2xy+2ax+2ay+b

f ( x + y ) = x 2 + 2 a x + a 2 + y 2 + 2 a y + a 2 + 2 x y − a 2 + b f(x+y)=x^2+2ax+a^2+y^2+2ay+a^2+2xy-a^2+b

f ( x + y ) = ( x + a ) 2 + b + ( y + a ) 2 + b + 2 x y − a 2 − b f(x+y)=(x+a)^2+b+(y+a)^2+b+2xy-a^2-b

f ( x + y ) = f ( x ) + f ( y ) + 2 x y − ( a 2 + b ) f(x+y)=f(x)+f(y)+2xy-(a^2+b) ; thus, b = − a 2 b=-a^2

4 = ( 1 + a ) 2 − a 2 4=(1+a)^2-a^2 ; a = 3 2 a=\frac{3}{2}

f ( x ) = ( x + 3 2 ) 2 − 9 4 f(x)=(x+\frac{3}{2})^2-\frac{9}{4}

f ( 8 ) = ( 8 + 3 2 ) 2 − 9 4 = 88 f(8)=(8+\frac{3}{2})^2-\frac{9}{4}=88

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Adarsh Kumar is right, f ( x ) = x 2 + 3 x f(x)=x^2+3x .

Roman Frago - 5 years, 10 months ago

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