Let a = λ − 2024 and b = sqrt(4 − (λ − 2024)²). Find every real λ for which both a and b are real numbers.

Step-by-step solution

1. 1

   Notice that only one of the two expressions restricts anything. $a=\lambda-2024$ is a polynomial in $\lambda$, so it is real for every real $\lambda$ — it imposes no condition at all. All the information lives in $b$, because a square root of a negative number is not real.

2. 2

   Write down the radicand condition. For $b$ to be real,

   $4-(\lambda-2024)^2\ge 0\quad\Longleftrightarrow\quad (\lambda-2024)^2\le 4$

3. 3

   Turn the squared inequality into an absolute value. For any real $t$, $t^2\le 4$ is equivalent to $|t|\le 2$, i.e. $-2\le t\le 2$. With $t=\lambda-2024$:

   $-2\le \lambda-2024\le 2$

   Adding $2024$ throughout (which never flips an inequality) gives

   $2022\le \lambda\le 2026$

4. 4

   Check the endpoints and one interior point. At $\lambda=2022$: $4-(-2)^2=0$, so $b=0$ — real, and the endpoint is included. Same at $\lambda=2026$. At $\lambda=2027$: $4-3^2=-5<0$, so $b$ would be imaginary and the value is correctly excluded. At the centre $\lambda=2024$: $b=\sqrt{4}=2$, the largest $b$ can be.

5. 5

   Read off the geometry behind the answer. Adding the two definitions,

   $a^2+b^2=(\lambda-2024)^2+\left(4-(\lambda-2024)^2\right)=4$

   so $(a,b)$ always lies on the circle of radius $2$ centred at the origin — and since a principal square root is non-negative, on its upper half. That is why $a=\lambda-2024$ can only range over $[-2,2]$: it is the horizontal coordinate of a point on a radius-$2$ circle. The answer is the interval

   $\lambda\in[2022,\,2026]$