Question

Let $a_1, a_2, a_3, a_4$ and b be real numbers such that $b + \sum_{k=1}^{4} a_k = 8, b^2 + \sum_{k=1}^{4} a_k^2 = 16$ then b can be-

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A), (B)

2 and 3

Answer 2

Let the given equations be:

1. $b + \sum_{k=1}^{4} a_k = 8$
2. $b^2 + \sum_{k=1}^{4} a_k^2 = 16$

Let $S_1 = \sum_{k=1}^{4} a_k$ and $S_2 = \sum_{k=1}^{4} a_k^2$. From (1), $S_1 = 8 - b$. From (2), $S_2 = 16 - b^2$.

For any real numbers $a_1, a_2, a_3, a_4$, the Cauchy-Schwarz inequality states that $(\sum_{k=1}^{n} x_k)^2 \le n \sum_{k=1}^{n} x_k^2$. In this case, $n=4$, so $S_1^2 \le 4 S_2$.

Substitute the expressions for $S_1$ and $S_2$: $(8 - b)^2 \le 4(16 - b^2)$ $64 - 16b + b^2 \le 64 - 4b^2$ $b^2 - 16b + 64 \le 64 - 4b^2$ $5b^2 - 16b \le 0$ $b(5b - 16) \le 0$

This inequality holds when $b$ is between the roots of $b(5b-16)=0$, which are $b=0$ and $b=\frac{16}{5}$. So, $0 \le b \le \frac{16}{5}$. Since $\frac{16}{5} = 3.2$, the possible values of $b$ are in the interval $[0, 3.2]$.