Animated Solution for Mathematics - Quadratic Equations: Let $\lambda \in \mathbb{R}$ and let the equation $E$ be $|x{|}^2-2|x|+|\lambda -3|=0$. Then the largest element in the set $S=\{x+\lambda :x$ is an integer solution of $E\}$ is

Imagine you are standing before a complex equation, one that seems to have too many moving parts. You see $|x{|}^2-2|x|+|\lambda -3|=0$.

It looks intimidating, doesn't it? But here is the secret of the JEE Advanced: most complex-looking problems are just simple structures wearing a disguise. Let's peel back the layers.

First, notice the structure. We have $|x{|}^2$ and $-2|x|$. This is the classic signature of a quadratic expression waiting to be completed.

If we treat $|x|$ as a single entity—let's call it $u$—the equation becomes $u^2-2u+|\lambda -3|=0$. Suddenly, the fog clears.

We can complete the square by adding and subtracting $1$. This transforms our equation into:

Rearranging this, we get:

This is the heart of the problem. Because the left side is a perfect square, it must be greater than or equal to zero.

Therefore, the right side, $1-|\lambda -3|$, must also be non-negative. This gives us a crucial constraint:

This simple inequality tells us that $\lambda$ is trapped in the interval $[2,4]$.

Now, we introduce the most important constraint: $x$ is an integer. If $x$ is an integer, then $|x|$ is an integer, and consequently, $(|x|-1)$ must be an integer.

Let $k=|x|-1$. Our equation becomes:

Since $k$ is an integer, $k^2$ must be a perfect square integer. But look at the right side: $1-|\lambda -3|$.

Since $|\lambda -3|\ge 0$, the maximum value of the right side is $1$. So, $k^2$ can only be $0$ or $1$.

This is the breakthrough! We have reduced an infinite set of possibilities down to just two manageable cases.

The Final Descent

In Case 1, we set $k^2=0$. This implies $(|x|-1{)}^2=0$, so $|x|=1$, which gives $x=1$ or $x=-1$.

Substituting this back, we find $1-|\lambda -3|=0$, leading to $|\lambda -3|=1$, so $\lambda =4$ or $\lambda =2$.

Calculating $x+\lambda$ for these pairs, we get values: $1+4=5$, $-1+4=3$, $1+2=3$, and $-1+2=1$.

In Case 2, we set $k^2=1$. This implies $(|x|-1{)}^2=1$, so $|x|-1=\pm 1$.

This gives $|x|=2$ or $|x|=0$. Solving for $x$, we get $x=2,-2,0$.

Substituting back, we find $1-|\lambda -3|=1$, which means $|\lambda -3|=0$, so $\lambda =3$.

Calculating $x+\lambda$ here, we get: $2+3=5$, $-2+3=1$, and $0+3=3$.

When we gather all our results, the set $S$ is simply $\{1,3,5\}$.

The largest element is clearly 5.

You see? By breaking the problem into logical, bite-sized pieces—completing the square, respecting the integer constraint, and testing the cases—we didn't just solve the problem; we mastered it. Keep this mindset, and no equation will ever be too daunting again.