2024 AMC 12A Problems/Problem 8

Problem

Let $x$ be a real number with $\sin x \neq -1$ . What is the sum of the maximum and minimum possible values of $\frac{(\sin x + \cos x + 1)^{2}}{\sin x + 1}?$

$\textbf{(A)}~2 \qquad \textbf{(B)}~3 \qquad \textbf{(C)}~4 \qquad \textbf{(D)}~6 \qquad \textbf{(E)}~8$

Solution

The expression in question is equal to $\frac{\sin^{2}x+\cos^{2}x+2\sin x\cos x+2\cos x+2\sin x+1}{\sin x+1}=\frac{2\sin x\cos x+2\sin x+2\cos x+2}{\sin x+1}=\frac{2(\sin x+1)(\cos x+1)}{\sin x+1}.$ Assuming $\sin x\neq -1$ , it is possible to cancel out $\sin x+1$ leaving $2\cos x+2$ . The minimum and maximum values of $\cos x$ are $-1$ and $1$ , respectively, giving $2\cdot(-1)+2+2\cdot 1+2=\boxed{\textbf{(C)}~4}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 8

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