NIMCET 2007 — Mathematics PYQ

To solve this problem, we will find the values of key algebraic building blocks like $xy$ and $x^2+y^2$ using the given equations, and then combine them to find $x^5+y^5$.

Step 1: Find the value of $xy$

We know the identity for the sum of cubes:

$$x^3+y^3=(x+y{)}^3-3xy(x+y)$$

Substitute the given values $x+y=1$ and $x^3+y^3=4$ into this formula:

$$4=(1{)}^3-3xy(1)$$

$$4=1-3xy$$

Rearranging the terms to solve for $3xy$:

$$3xy=1-4$$

$$3xy=-3$$

$$xy=-1$$

Step 2: Find the value of $x^2+y^2$

Using the standard squaring identity:

$$x^2+y^2=(x+y{)}^2-2xy$$

Substitute $x+y=1$ and $xy=-1$:

$$x^2+y^2=(1{)}^2-2(-1)$$

$$x^2+y^2=1+2=3$$

Step 3: Create an expression for $x^5+y^5$

Let's look at what happens when we multiply $(x^2+y^2)$ and $(x^3+y^3)$:

$$(x^2+y^2)(x^3+y^3)=x^5+x^2{y}^3+y^2{x}^3+y^5$$

Rearranging the terms:

$$(x^2+y^2)(x^3+y^3)=x^5+y^5+x^2{y}^2(x+y)$$

Now, isolate $x^5+y^5$:

$$x^5+y^5=(x^2+y^2)(x^3+y^3)-(xy{)}^2(x+y)$$

Step 4: Substitute the values to find the final answer

Substitute the values we have calculated so far ($x^2+y^2=3$, $x^3+y^3=4$, $xy=-1$, and $x+y=1$):

$$x^5+y^5=(3)(4)-(-1{)}^2(1)$$

$$x^5+y^5=12-(1)(1)$$

$$x^5+y^5=12-1$$

$$x^5+y^5=11$$

Correct Answer

The correct option is (a) 11.