Questão 10 (Matemática)

**Evaluating x in quadrants** I'm considering how to handle x, specifically how it relates to the first and third quadrants. I need to check that there are no hidden exclusions. I think using mod 2π for handling periodic issues is the way to go. So, for the first quadrant, I can take x in the range of (0, π/2) and for the third quadrant, it should be (π, 3π/2). Lastly, I'll finalize with k=3 while ensuring everything functions correctly.

**Evaluating integer k** I'm considering whether I can determine integer k as 3. I want to check the definitions in the first and third quadrants and whether x is real and periodic. It seems like I might simplify f in the third quadrant to (1+√2)(-sin-cos). To establish that 3 can be achieved, I think about the range. Specifically, I’m looking into whether |sin| + |cos| can equal a certain value. It looks feasible, so I need to explore continuity and the function range further.**Defining square root identity and quadrants** I want to explain the square root identity, which involves the property that takes a number and shows its relationship to its square. Next, I should define the quadrants, which relate to the coordinate plane and how angles or points are positioned within it. It’s interesting to think about how these concepts connect in mathematics and how I can compose a clear response to better explain them to the user. Let's explore this together!

I simplify the square roots using the double angle identities: 1+cos2x becomes √2|cos x| and 1−cos2x becomes √2|sin x|, so f(x) reduces to √2(|cos x|+|sin x|) − sin x − cos x. Now I'm examining the first quadrant where both sine and cosine are positive. In that region f(x) simplifies to (√2−1)(sin x+cos x), which ranges over (√2−1, 2−√2] — roughly (0.414, 0.586] — containing no integers. In the third quadrant where both are negative, f(x) becomes (√2+1)(−(sin x+cos x)), ranging over (√2+1, 2+√2] ≈ (2.414, 3.414], which does include the integer 3. I check the boundary cases where x sits on the axes and confirm they still don't produce integer values. Verifying k=3: solving −(√2+1)(sin x+cos x)=3 gives sin x+cos x = −3(√2−1) ≈ −1.243, which indeed falls within the valid range [−√2, −1), so the solution holds.

I'm working through the first quadrant case where sine and cosine are both nonnegative, simplifying f to (√2−1)(sin x + cos x), and figuring out the range of sin x + cos x depending on whether the interval is open or closed at the axes. Now checking the third quadrant, where both sine and cosine are negative: f simplifies to (√2+1)(|sin x|+|cos x|), giving a range of roughly 2.414 to 3.414. The only integer in range is 3, and I verify it's actually attainable since |sin x|+|cos x| = 3(√2−1) ≈ 1.243 falls within the valid range (1, √2]. So the answer is k = 3.

I'm breaking f(x) into cases by quadrant, since |cos x| and |sin x| change sign: in the first quadrant both are positive, giving f = (√2−1)(sin x + cos x), while in the third quadrant both are negative, giving f = (√2+1)|sin x + cos x|. I'm continuing to work through the remaining quadrants similarly. Checking the first quadrant range, sin x + cos x ranges over (1, √2], so f lands in (√2−1, 2−√2], roughly 0.414 to 0.586 — no integers there. For the third quadrant, |sin x + cos x| gives the same range (1,√2], so f lands in (√2+1, 2+√2], roughly 2.414 to 3.414, which contains the integer 3. So k = 3.

The problem is from ITA 2027, second phase, Mathematics. I'm simplifying the function using the identities $1 + \cos(2x) = 2\cos^2(x)$ and $1 - \cos(2x) = 2\sin^2(x)$. This gives me $\sqrt{1 + \cos(2x)} = \sqrt{2}|\cos(x)|$ and $\sqrt{1 - \cos(2x)} = \sqrt{2}|\sin(x)|$. I factored the expression into $f(x) = \sqrt{2}(|\sin(x)| + |\cos(x)|) - (\sin(x) + \cos(x))$. Now I'm identifying the first and third quadrants on the trigonometric circle. I'm now setting up the first and third quadrant cases. In Q1 both sine and cosine are positive, while in Q3 both are negative. In the first quadrant, $f(x) = (\sqrt{2} - 1)(\sin + \cos)$. In the third quadrant, $f(x) = (\sqrt{2} + 1)|\sin + \cos|$, which is non-negative. I'm now determining the range of $f$ in each quadrant to find all integer values $k$ for which the equation has a solution. I'm excluding the axes from the quadrants. I'm using the identity $\sin(x) + \cos(x) = \sqrt{2}\sin(x + \pi/4)$ to find the range of $f$ across the first and third quadrants. In the first quadrant, the sum $\sin(x) + \cos(x)$ ranges from $1$ to $\sqrt{2}$. Therefore $f(x) = (\sqrt{2} - 1)(\sin + \cos)$ spans the interval $(\sqrt{2} - 1, 2 - \sqrt{2}]$, approximately $(0.414, 0.586]$. In the third quadrant, the sum $\sin(x) + \cos(x)$ ranges from $-\sqrt{2}$ to $-1$. Therefore $f(x) = (\sqrt{2} + 1)|\sin + \cos|$ spans the interval $[0, \sqrt{2})$. The maximum value of $\sqrt{2}$ is reached at $x = \pi/4$ in the first quadrant. In the third quadrant, $f(x) = (\sqrt{2} + 1)(\sin + \cos)$ spans the interval $(\sqrt{2} + 1, 2 + \sqrt{2}]$. The function therefore covers values from approximately $2.414$ up to $2 + \sqrt{2}$. The function values in the first quadrant are between approximately 0.414 and 0.586, while in the third quadrant they range from about 2.414 up to 3.414. I am now identifying which integers fall within these two intervals. The first interval contains no integers. The second interval yields only k = 3. I am now checking whether negative values are possible. The function is strictly positive in the relevant ranges, so no negative integers are possible. I am now verifying the third-quadrant expression by substituting x = 5π/4 to confirm that k = 3 is indeed achieved. I confirmed that k = 3 is the only integer solution. I checked that the value 3 falls inside the third-quadrant range and verified that a corresponding x exists in the third quadrant.

We need answer in Portuguese, solve. Need understand problem. We need determine all integer k such that equation f(x)=k has at least one real solution x in first or third quadrant of trigonometric circle. Function f: R -> R given by f(x)= sqrt(1+cos2x) + sqrt(1-cos2x) - sin x - cos x. Need integer k values achievable for x in first or third quadrant. Need analyze carefully. We have f(x). Use identities: 1+cos2x = 2 cos^2 x, 1 - cos2x = 2 sin^2 x. Since sqrt of squares = |sqrt2 cos x|? sqrt(2 cos^2 x)= sqrt2 |cos x|. Similarly sqrt(2 sin^2 x)= sqrt2 |sin x|. So f(x)= sqrt2 (|cos x| + |sin x|) - sin x - cos x. We need x in first or third quadrant of trigonometric circle. Interpret: "primeiro ou terceiro quadrante do círculo trigonométrico" means x in first quadrant (0 to π/2 maybe excluding?) or third quadrant (π to 3π/2). Usually angles in first quadrant: 0 < x < π/2 (or including axes? Quadrant typically open intervals? Need discuss. First quadrant is between 0 and π/2; third between π and 3π/2. Since "círculo trigonométrico" maybe x angle. Need include boundaries? Usually first quadrant: 0 < x < π/2, third: π < x < 3π/2 maybe. But equation may have solutions at quadrant boundaries? Let's analyze. Need determine all integer k such that exists real x in first or third quadrant satisfying f(x)=k. Let's derive expressions for f in first and third quadrants. Case 1: x in first quadrant (Q1): 0 < x < π/2 (maybe endpoints). Then sin x >=0, cos x >=0. So |cos x|=cos x, |sin x|=sin x. f(x)= sqrt2 (cos x + sin x) - sin x - cos x = (sqrt2 -1)(sin x + cos x). Since sin+cos >? positive. Range? For x in (0, π/2), s+c = sqrt2 sin(x+π/4). x+π/4 in (π/4, 3π/4). Sin ranges (sqrt2/2,1]? Actually sin max 1 at x=π/4, endpoints sqrt2/2. So s+c in (1? let's compute endpoints: at 0, s+c=1; at π/2=1; open exclude endpoints maybe range (1, sqrt2]? Since sin(pi/4)=1 -> sqrt2; endpoints 1. So range (1, sqrt2] if open, including max. Multiply by sqrt2 -1 >0. Range: (sqrt2-1)*1 = sqrt2 -1? Wait lower open 1, so range (sqrt2-1, (sqrt2-1)*sqrt2 = 2 - sqrt2]? Compute (sqrt2-1)*sqrt2 = 2 - sqrt2 ≈0.586. sqrt2-1≈0.414. So range (0.414..., 0.586...] if open. Integer k possible? Only k=0? But range positive? minimum >0.4, so no integer? k=0 impossible since positive. But maybe k=1? range max 0.586 <1, so none. But wait maybe x at boundary? If include endpoints, range [sqrt2-1, 2-sqrt2] ≈ [0.414,0.586] no integer. So first quadrant yields no integer k? Hmm maybe k=0? no positive. Need check third quadrant too. Case 2: x in third quadrant: π < x < 3π/2. There sin x <=0, cos x <=0 (both negative except axes). So |cos x| = -cos x, |sin x| = -sin x. f(x)= sqrt2 ( -cos x - sin x) - sin x - cos x = -sqrt2(sin x + cos x) - (sin x + cos x) = -(sqrt2 +1)(sin x + cos x). Now in Q3, sin x + cos x? Let's determine range. Let t = x - π? Or x in (π, 3π/2). sin x + cos x = sqrt2 sin(x+π/4). x+π/4 in (5π/4, 7π/4). Sin in that interval? From 5π/4 to 7π/4: sin at 5π/4 = -√2/2, decreases? Wait sin on (π,2π) is negative, from 0 to -1 (at 3π/2) then back; here interval (5π/4, 7π/4) contains 3π/2, so sin ranges? At 5π/4 sin=-√2/2; at 3π/2 sin=-1; at 7π/4 sin=-√2/2. Continuous, range [-1, -√2/2) or (-√2/2? endpoints? open). Actually sin(3π/2)=-1 max? Since negative, closer to 0 is max -√2/2. On interval open endpoints, range [-1, -√2/2) ? At 3π/2 included? If open interval, includes 3π/2? x in (π,3π/2) open, so 3π/2 not included, so range [-? wait sin at 3π/2 = -1 not attained if x < 3π/2 (strict), so range (-1, -√2/2)? Let's check: interval (5π/4, 7π/4) open endpoints, includes 3π/2, so sin attains -1. Minimum -1 attained at 3π/2. Maximum? Actually at endpoints sin=-√2/2; near endpoints less? Function on [5π/4,3π/2] decreases from -√2/2 to -1; on [3π/2,7π/4] increases from -1 to -√2/2. So range [-1, -√2/2]? At endpoints sin=-√2/2 included? But interval open at 5π/4, but 3π/2 included, so maximum -√2/2 is limit not attained? Since 5π/4 not included, but value -√2/2 not attained because x+π/4=5π/4 -> x=π excluded (also endpoint). Similarly 7π/4 -> x=3π/2 excluded? Wait x=3π/2 gives x+π/4=7π/4, which is endpoint, excluded if x<3π/2 strict. But x=3π/2 gives sin=-1, not -√2/2. So -√2/2 not attained; values > -√2/2? Actually between -√2/2 and -1? At x just greater than π, sin x+cos x? x near π: sin~0, cos~-1 => sum~-1, which is less than -√2/2? -1 < -0.707? Yes -1 < -0.707. So near π, sum near -1. At x=5π/4? sum=-√2≈-1.414? Wait hold on! I confused: sin(x+π/4). Let's compute directly sin x + cos x = √2 sin(x+π/4). At x=π, sinπ+cosπ = -1. At x=5π/4, sin+cos = -√2/2 -√2/2 = -√2 ≈ -1.414. At x=3π/2, -1. At x=7π/4, -√2/2? Actually sin(7π/4)+cos(7π/4)= -√2/2 + √2/2 =0? Wait cos(7π/4)=√2/2 positive? Wait 7π/4 is fourth quadrant, cos positive, sin negative. But x in third quadrant is (π,3π/2), not 7π/4? Wait I made mistake: x+π/4 in (5π/4, 7π/4). Yes sin at 7π/4 = -√2/2, cos? But sin(x+π/4) at 7π/4 = -√2/2. But sin x + cos x = √2 sin(x+π/4). At x=π? x+π/4=5π/4, sin=-√2/2 => sum=-1. At x=3π/2, x+π/4=7π/4, sin=-√2/2 => sum=-1? Wait sin(7π/4)=-√2/2, times √2 = -1. Hmm direct sin(3π/2)+cos(3π/2)= -1 + 0 = -1. So range maybe [-√2? Wait at x=5π/4 (not included) sin+cos=-√2. But x=5π/4 not in (π,3π/2) if interval? 5π/4 = 1.25π = 225°, which is less than 3π/2=270°? Actually yes 5π/4 (225°) lies in third quadrant (180°-270°). But interval x in (π,3π/2) includes 5π/4. Wait π to 3π/2 is 180° to 270°, includes 225°=5π/4 yes. I mistakenly thought 5π/4 is not in? It is. Domain (π,3π/2) includes all angles between 180 and 270, including 225. So x+π/4 in (5π/4,7π/4) includes 5π/4 (not endpoint? Wait if x=π, x+π/4=5π/4; if x>π, then >5π/4; if x<3π/2, then <7π/4. So interval (5π/4,7π/4) open endpoints. It includes values up to? At x=3π/2, x+π/4=7π/4; excluded if x<3π/2. So range of sin on (5π/4,7π/4): from -√2/2 (at left limit, not attained) down to -1 (attained at 3π/2? Wait 3π/2 corresponds to 7π/4? Actually x+π/4=7π/4 corresponds x=3π/2, yes attained if x=3π/2 included? If open, not included. But sin at 3π/2 = -1; so included? x=3π/2 is endpoint of Q3? Third quadrant maybe includes negative y-axis? Typically "terceiro quadrante" maybe x in (π, 3π/2), not including axes? Hmm trigonometric circle quadrants usually open except axes? Actually axes belong to no quadrant? Some conventions include axes? Usually first quadrant: 0 < θ < π/2, etc. So endpoints not included. Then x=3π/2 excluded. Thus sin range (-1? Let's see: x in (π,3π/2), x+π/4 in (5π/4,7π/4). At x→π+, angle→5π/4+, sin→-√2/2 from? sin near 5π/4: derivative cos=-√2/2 negative, as angle increases, sin decreases? At angle just >5π/4, sin < -√2/2? Actually sin(225+ε) ~ -√2/2 - √2/2 ε, so decreases. So near left endpoint, values just less than -√2/2. At angle=3π/2 (x=3π/2? actually angle=7π/4 when x=3π/2) sin=-√2/2? Wait hold: x+π/4 = 3π/2 corresponds x=5π/4, which is in middle. At x=3π/2, angle=7π/4 = 315°, sin=-√2/2. So maximum? Let's list critical: angle 5π/4 sin=-√2/2; then 3π/2 (x=5π/4) sin=-1; then 7π/4 sin=-√2/2. But interval (5π/4,7π/4) includes 3π/2? Yes 3π/2 = 270° between 225° and 315°, so includes. So range [-1, -√2/2) maybe since -√2/2 not attained at endpoints; -1 attained at angle=3π/2 (x=5π/4? Wait angle=3π/2 corresponds x=5π/4, which is inside? Actually x=5π/4 => angle=5π/4+π/4=6π/4=3π/2. Yes x=5π/4 is inside (π,3π/2). So sin=-1 attained. So range includes -1. Values closer to -√2/2 at endpoints not attained. So range [-1, -√2/2) maybe because at x→π+, sum→-1; at x→3π/2-, sum→-1? Wait direct at x=3π/2-, sin~-1, cos~0 => sum~-1. So range not below -1? Let's check direct: x in (π,3π/2): sin negative, cos negative; sum = sin+cos. At x=π, sum=-1; at x=5π/4, sum=-√2≈-1.414; at x=3π/2, sum=-1. So min -√2 at x=5π/4. So range [-√2, -1]? Actually at x=5π/4, sum=-√2 (minimum). At endpoints, sum→-1. So range [-√2, -1] with -1 included? At x=π, not included but limit; at x=3π/2 not included but limit; so -1 not attained. Minimum -√2 attained. So range [-√2, -1). Wait since near endpoints, sum > -1? e.g x=π+0.1: sin≈-0.1, cos≈-0.995 => -1.095 < -1; so values between -√2 and -1. At x=3π/2-0.1: sin≈-0.995, cos≈-0.1 => -1.095 < -1. So yes range [-√2, -1)? But -1 at x? x=π? no, x=? solve sin+cos=-1. sin+cos = √2 sin(x+π/4) = -1 => sin(x+π/4)= -1/√2. In interval (5π/4,7π/4), solutions? x+π/4 = 5π/4? gives -√2/2? Wait -1/√2 = -√2/2. Angles in (5π/4,7π/4) where sin=-√2/2: only endpoints? Actually sin=-√2/2 at 5π/4 and 7π/4, but open interval, so no. So -1 not attained. Good. Thus sin+cos ∈ [-√2, -1) for x∈(π,3π/2). Multiply by -(√2+1): f(x)= -(√2+1)(sin+cos)? Wait f = - (√2+1)(sin+cos). Since sin+cos negative, f positive. Range: since sin+cos ∈ [-√2, -1), f ∈ ( (√2+1)*1, (√2+1)*√2 ]? Need careful with negative. Let s = sin+cos ∈ [-√2, -1). Then - (√2+1) s ∈? s negative, multiply negative -> positive. f = -(√2+1)s = (√2+1)(-s). -s ∈ (1, √2]? Since s ∈ [-√2, -1) => -s ∈ (1, √2] (because s=-1 gives -s=1 but not attained; s=-√2 gives √2). Thus f ∈ (√2+1)*1, (√2+1)*√2] = (√2+1, 2+√2]? Compute (√2+1)*√2 = 2 + √2 ≈3.414. Lower ≈2.414. Values in (2.414..., 3.414...]. Integers k possible: k=3? 3 lies in (2.414,3.414] yes. k=? k=2? 2 <2.414 no. k=4? 4>3.414 no. So only k=3. But wait check if f can equal 3 exactly. Need solve f(x)=3 for x in third quadrant. f(x)= (√2+1)(-sin x - cos x)? Actually f = - (√2+1)(sin+cos) = (√2+1)(-sin-cos). Let t = -sin x - cos x = -(sin+cos). t ∈ [1, √2). Need t = 3/(√2+1) =? 3/(√2+1) = 3(√2-1)/(2-1)=3√2 - 3 ≈1.2426. This is between 1 and √2≈1.414, so attainable. Need find x in third quadrant? Domain open, but t range (1, √2]? Wait t ∈ [1? Actually s ∈ [-√2, -1); -s ∈ (1, √2]. Because s can equal -√2 => t=√2 attained; s never equals -1, so t=1 not attained but values tend to 1 from above. So t ∈ (1, √2]. Yes t=1.2426 ∈ (1,√2], so exists x. So k=3. But wait first quadrant maybe k? Range f ∈ [√2-1, 2-√2]? Let's recalc. If x in first quadrant open, sin+cos ∈ (1, √2]? Actually at x=π/4, s=√2; at endpoints s→1. So s ∈ (1,√2] (including √2? x=π/4 included yes; excluding 1? endpoints excluded). f=(√2-1)s ∈ (√2-1, 2-√2]? Since lower s>1 -> f>√2-1 (open), upper s≤√2 -> f≤2-√2 (included at x=π/4). Integer none. But need consider if x in first or third quadrant of trigonometric circle maybe includes angles rotating? "x pertencente ao primeiro ou ao terceiro quadrante do círculo trigonométrico." Since f defined on R, x real; quadrant determined by angle x modulo 2π. But "primeiro quadrante do círculo trigonométrico" usually means after positioning angle, its terminal side lies in first quadrant. So x could be any real with mod 2π in (0,π/2) or (π,3π/2). So our analysis applies because sin/cos periodic? f uses sin/cos and cos2x also periodic, f periodic? sin, cos period 2π, cos2x period π. But |cos x| and |sin x| periodic with? f may be periodic 2π? Let's check f(x+2π)=f(x). But if x in Q1 shifted by 2π? Actually quadrant is modulo 2π, so if x in first quadrant modulo 2π, x can be in any interval (2πn, 2πn + π/2). But sin/cos same as in (0,π/2) with signs positive, so f same. Similarly third quadrant. So our expressions valid for any representative. Good. But wait: what about x in first quadrant but not necessarily 0<x<π/2? Quadrant of circle trig: first quadrant also includes angles after full rotations, e.g x=2π+0.5. But sin/cos signs same as 0.5, so f same. So no issue. Need be careful with "primeiro ou terceiro quadrante" could mean x itself in interval (0,π/2) ∪ (π,3π/2) considering principal? In trigonometria, often "x pertence ao primeiro quadrante" if 0 < x < π/2 (for x in [0,2π) perhaps). Since x real, but if not restricted, "quadrante" refers to terminal side. Our result same for each period. But let's verify possible k=0? f positive? Let's inspect original function maybe f can be 0? sqrt terms are nonnegative; -sin -cos maybe positive/negative. In Q1, sqrt(1+cos2x)=√2 cos x, sqrt(1-cos2x)=√2 sin x, so f=(√2-1)(sin+cos)>0. In Q3, sqrt terms = √2(-cos)+√2(-sin)= -√2(sin+cos), subtract sin+cos => (-√2-1)(sin+cos)= (√2+1)(-sin-cos)>0. So f positive always in Q1 and Q3. So k=0 no. Integers 1,2? First quadrant max 0.586 <1; third quadrant range >2.414, so no 1,2. k=3 yes. k=4? third max 3.414 <4; first max 0.586. So no. But wait what about k maybe negative? f positive so no. So only k=3? But problem asks "Determine all integers k such that equation f(x)=k has at least one real solution x belonging to first or third quadrant." Could be just {3}. But let's double-check. However there may be nuance: f(x) = sqrt(1+cos2x)+sqrt(1-cos2x) - sin x - cos x. Using sqrt(2 cos^2 x)=√2 |cos x|. But sqrt(1+cos 2x)=sqrt(2 cos^2 x)=√2 |cos x| yes. Then f = √2(|cos|+|sin|) - sin - cos. For first quadrant: positive; for third: negative both, so f = √2(-cos -sin) - sin - cos = -(√2+1)(sin+cos). Since sin+cos negative, f positive. Yes. But wait, "primeiro ou terceiro quadrante do círculo trigonométrico" maybe means the angle x is in standard position with terminal side in first or third quadrant, but x could be any real number. If x in third quadrant, both sin and cos negative. We handled. Let's test numerically at x=π/4 (Q1 boundary? yes inside first quadrant? π/4=45°, first quadrant): f=√2(|cos|+|sin|)-sin-cos = √2(√2/2+√2/2) - √2/2 - √2/2 = 2 - √2 ≈0.586. Integer? no. At x=5π/4 (third quadrant? 225°): sin=cos=-√2/2. f=√2(√2/2+√2/2) - (-√2/2) - (-√2/2)? Wait sqrt(1+cos2x)=sqrt(1+cos(5π/2))=sqrt(1+0)=1? Actually cos(2x)=cos(5π/2? 2*5π/4=5π/2? cos(5π/2)=0? yes, sqrt(1+0)=1, not √2. Our formula √2|cos x| = √2*√2/2=1. yes. sqrt(1-cos2x)=sqrt(1-0)=1. Then f=1+1 - (-√2/2)-(-√2/2)=2+√2≈3.414. Good. At x=π, sqrt1+cos2π=sqrt2, sqrt1-cos2π=0, sin=0, cos=-1 => f=√2 -0 -(-1)=√2+1≈2.414. x in third quadrant? π is boundary between second and third? Actually quadrant III begins just after π (180°), but π itself is negative x-axis, usually not in any quadrant. Our range near π f→√2+1 ≈2.414 from above? At x=π+0.1, f ~? yes >2.414. At x=3π/2, f=√2? Wait x=3π/2=270°, sin=-1, cos=0, cos2x=cos3π=-1? sqrt(1+cos)=0, sqrt(1-cos)=√2, f=0+√2 -(-1)-0 = √2+1≈2.414? Wait that's 2.414, but formula range third quadrant up to 3.414. At x=5π/4 f=3.414. At x→3π/2-, f→? f=-(√2+1)(sin+cos); sin+cos=-1? at 3π/2 sin+cos=-1, so f=√2+1=2.414. So range [2.414?, 3.414]? Wait at x=5π/4 f=3.414 max; at endpoints f→2.414. So range (2.414, 3.414]? Actually at x=π? excluded, but limit 2.414; at x=3π/2? excluded, limit 2.414; but at x=5π/4 included max 3.414. So range (2.414..., 3.414]? It attains all values between? Since continuous and min? Actually f has maximum at 5π/4, minima as limits at endpoints. It attains values approaching 2.414 but not exactly? Let's solve f=2.5: possible? At x near π+? f? Let's analyze f= (√2+1)(-sin-cos). Let u=-sin-cos. For x in (π,3π/2), u = -(sin+cos). Since sin+cos continuous negative. u has range? At x=π+ε, sin+cos≈ - (1 - ε^2/2?) + (-1+ε^2/2?) Actually sin≈ -ε, cos≈ -1 + ε^2/2, sum≈ -1 - ε +..., so u≈1+ε? Wait u=-sin-cos = -(-ε -1 +ε^2/2)=1+ε -ε^2/2. So as ε→0+, u→1+. At x=3π/2 - ε, sin≈ -1 + ε^2/2, cos≈ -ε? Actually cos(3π/2 - ε)= -sin? Let's compute cos(270° - ε)= -sin ε? Actually cos(3π/2 - ε)= cos3π/2 cosε + sin3π/2 sinε = 0 + (-1)*sinε = -sinε ≈ -ε. sin(3π/2 - ε)= -cosε ≈ -1 + ε^2/2. Sum≈ -1 - ε + ε^2/2. u≈1+ε. So u>1, tends 1. Max u=√2 at x=5π/4 (sin+cos=-√2). So u ∈ (1, √2]? Actually at x=5π/4 u=√2; at endpoints u tends 1, not attained; but all values between attained? Since u continuous, max √2, sup 1 not attained? Wait domain open endpoints, continuous on open interval. The image of open interval under continuous function is interval (maybe not closed). Since limits at endpoints are 1, but function never equals 1? Solve u=1 => -sin-cos=1 => sin+cos=-1; but in Q3, sin+cos ≥ -√2, equal -1 at endpoints? Actually sin+cos=-1 at x=π and x=3π/2? Check x=π: -1; x=3π/2: -1. But endpoints not in domain. Solve sin+cos=-1: √2 sin(x+π/4)=-1 => sin(x+π/4)=-1/√2. In (π,3π/2), x+π/4 ∈ (5π/4,7π/4). Solutions sin=-√2/2 are at 5π/4 and 7π/4, endpoints, not interior. So no. Thus u>1. So range (1, √2]. Yes. Then f=(√2+1)u ∈ (√2+1, 2+√2]. Correct. Integer 3 lies in (2.414,3.414]. So k=3. But wait, problem statement: "Determine todos os números inteiros k tais que a equação f(x)=k possua pelo menos uma solução real x pertencente ao primeiro ou ao terceiro quadrante do círculo trigonométrico." Could there be solutions in first/third quadrant but with x not necessarily principal? Already considered. Could k be 4? f max=2+√2≈3.414 <4. So no. Could k=2? f first max 0.586 <2, third min >2.414 >2. So no. k=1? third min >2.414, first max 0.586 <1. no. k=0, negative no. So only 3. But let's ensure no other quadrants? Problem says first or third only. If x in third quadrant but maybe x in (π,3π/2) principal? Yes. Need write full solution in Portuguese. But let's see if there is any catch: "primeiro ou terceiro quadrante do círculo trigonométrico" might mean x belongs to first or third quadrant, where quadrant includes unbounded intervals? In trig, first quadrant angles: 2kπ < x < 2kπ + π/2 for some integer k. Third: 2kπ+π < x < 2kπ+3π/2. Then f periodic 2π? Check f(x+2π) = same because sin/cos and |sin|/|cos|? Actually sqrt(1+cos2x) period π? cos2(x+2π)=cos(2x+4π)=cos2x; sin/cos period 2π. |sin|? In first, shifted by 2π same signs; in third same. So yes. But could x belong to first quadrant after multiple rotations, e.g x=2π + π/4: sin/cos positive same, cos2x=cos(4π+π/2)=0, sqrt terms same. So f same. Good. Let's craft solution: 1. Simplify f using identities: sqrt(1+cos2x) = sqrt(2cos²x)=√2|cos x|; sqrt(1−cos2x)=√2|sin x|. Thus f(x)=√2(|cos x|+|sin x|) - sin x - cos x. 2. Analyze first quadrant: x in QI => sin x>0, cos x>0. So f(x)=√2(sin x+cos x) - (sin x+cos x) = (√2−1)(sin x+cos x). But sin x+cos x = √2 sin(x+π/4). For x in QI (0,π/2), x+π/4 ∈ (π/4,3π/4), so sin(x+π/4) ∈ (√2/2,1], hence sin+cos ∈ (1,√2]. Then f ∈ (√2−1, 2−√2] (or [√2−1? open). Since √2−1≈0.41, 2−√2≈0.59, no integer k since interval contains no integer. Actually note range (0.414,0.586] contains no integer. So first quadrant gives none. 3. Analyze third quadrant: x in QIII => sin x<0, cos x<0. So |sin|=-sin, |cos|=-cos. Then f(x)=√2(-sin x - cos x) - sin x - cos x = -(√2+1)(sin x+cos x). Let s= sin x+cos x. In QIII, s = √2 sin(x+π/4). For x∈(π,3π/2), x+π/4∈(5π/4,7π/4). The range of sin on this interval is [-1, -√2/2)? Wait need precise: sin ranges from -√2/2 at endpoints to -1 at 3π/2 (x=5π/4). Since interval open at endpoints, maximum (closest to 0) is -√2/2 not attained? Actually at x=π+? sin? Let's recompute: x+π/4 ∈ (5π/4, 7π/4). At x→π+, angle→5π/4+, sin→ -√2/2 from below? Wait sin(5π/4+ε)= -√2/2 cos ε - √2/2 sin ε? Actually sin(a+ε)=sin a cos ε + cos a sin ε = (-√2/2)cosε + (-√2/2) sinε ≈ -√2/2 - √2/2 ε < -√2/2. Similarly at x→3π/2-, angle→7π/4-, sin→? sin(7π/4 - ε)= sin7π/4 cosε - cos7π/4 sinε = (-√2/2)cosε - (√2/2) sinε? Wait cos7π/4=√2/2, so = -√2/2 cosε - √2/2 sinε ≈ -√2/2 - √2/2 ε < -√2/2. So values near endpoints are less than -√2/2. At x=5π/4, angle=3π/2, sin=-1. So range includes -1 and values between -1 and -√2/2? Actually max? Since sin is negative, maximum (closest to 0) is -√2/2 but not attained; minimum -1 attained. So range [-1, -√2/2) maybe? But the interval includes angle 3π/2 at x=5π/4, so -1 included. It doesn't include -√2/2. So s ∈ [-1, -√2/2). Wait check direct: At x=π, s=-1; at x=5π/4, s=-√2≈-1.414; at x=3π/2, s=-1. So the minimum is -√2 (at x=5π/4), maximum? The function s(x) has maximum -1? Since near endpoints s→-1, but at x=π? not included; but at x=3π/2? not included; but can s equal -1? Solve sin+cos=-1; in interval, no. So s ∈ [-√2, -1). Wait yes s min -√2, max -1 not attained. So s ∈ [-√2, -1). Let's verify: At x=π+0.1, s≈sin0.1? Wait sin(π+0.1)=-sin0.1≈-0.0998; cos(π+0.1)=-cos0.1≈-0.995; sum≈-1.0948 < -1. So values can be less than -1. At x=3π/2-0.1: sin≈-0.995, cos≈-0.0998? Actually cos(3π/2-0.1)=cos(270°-5.7°)=? cos(3π/2 - a)= -sin a ≈ -0.0998; sum≈-1.0948 < -1. So s < -1 except at endpoints x=π (s=-1) and x=3π/2 (s=-1). So range [-√2, -1). Good. Then f= -(√2+1)s. Since s ∈ [-√2, -1), -s ∈ (1, √2]? Wait if s=-1? no; s=-√2 gives -s=√2; s→ -1 from below => -s→1 from above. So -s ∈ (1, √2]. Then f ∈ (√2+1, 2+√2]. This interval contains integer 3 (since 2.414 <3 ≤3.414). Need also ensure every value in interval attained? Since s continuous and range [-√2, -1), f continuous, image is (√2+1, 2+√2]? Actually because s range not closed at -1 but includes -√2. Since continuous image of connected interval is interval; endpoints limits. All values between are attained? The function s assumes all values between -√2 and -1? Since s continuous on open interval (π,3π/2), image is connected interval (maybe not closed). It contains min -√2 and approaches -1 but not attains. The image of a continuous function on an interval is an interval; it will include all values between min and sup? Yes by intermediate value: for any y between -1 and -√2? Since it attains -√2 and can approach -1 but maybe not attain -1; it will attain all values between -1 and -√2? Let's prove: pick value c in (-√2, -1). Since f continuous on connected domain (open interval), image is connected subset of reals, so interval. Since it contains -√2 and has limit -1, it contains all intermediate values? Actually connected subset of R is interval, so if it contains -√2 and arbitrarily close to -1 (but not -1), the interval must contain (-√2, -1)? Because if it missed some value between, interval would have a gap, not connected. Since image is interval and contains points arbitrarily close to -1 and the minimum -√2, it contains all values between? More rigor: The image is interval [min, sup?) or [min, sup)? Since min attained, sup not; image is [−√2, L) where L is limit? Actually because image is connected and bounded below by -√2, it may be [−√2, a) or [−√2, a] etc. Since it can't attain -1 but sup is -1, the image is [−√2, -1) or [−√2, -1)? It must be [−√2, -1) because it contains -√2 and not -1, and is connected; any interval subset of R with min -√2 and sup -1 not included is [−√2, -1). Yes. So f ∈ (√2+1, 2+√2]. Contains 3. Then determine k integer: only k=3. Need maybe check if k=2? First quadrant max <1, third min >2. So no. So final answer: {3} or k=3. Need write in Portuguese, item by item? There are no items, one question. Need include development. Use "Convenções" maybe not needed. Final line: "Resposta final: k=3" or "Resposta final: {3}". The prompt says "Resposta final: <o resultado de cada item, de forma curta>" if multiple; here single. We'll write "Resposta final: k=3". Let's craft final answer with enough detail but concise. Potential pitfall: The equation f(x)=k has at least one real solution x belonging to first or third quadrant. What if k can be non-integer? They ask integers only. We found 3. Need ensure no integer negative etc. f positive. Good. Let's consider if "primeiro ou terceiro quadrante" might include boundary axes? Usually not, but our answer would maybe include? If include endpoints, first quadrant includes axes: x=0,π/2. f at x=0: sqrt(1+1)+sqrt(1-1)-0-1 =√2 -1≈0.414; at x=π/2 same. Still no integer. Third quadrant boundaries: x=π,3π/2. f at π=√2+1≈2.414; at 3π/2? sqrt(1+cos3π)+sqrt(1-cos3π)-sin-cos =0+√2 -(-1)-0=√2+1≈2.414. Still no integer? Wait 2.414 <3 <3.414, but if endpoints included? 3 still within (2.414,3.414] yes. No new integer. So same. But wait if third quadrant includes x=π? f=√2+1≈2.414, which is not integer 2? Actually 2.414 >2, so not 2. k=3 remains. If includes 3π/2, f=√2+1 same. No change. But what about f could equal 4 at x=? max 3.414 <4. So no. Let's double-check simplification for third quadrant: sqrt(1-cos2x) = sqrt(2 sin^2 x) = √2 |sin x| = -√2 sin x since sin negative. sqrt(1+cos2x)=√2 |cos x| = -√2 cos x. Thus f= -√2(sin x + cos x) - sin x - cos x = -(√2+1)(sin x+cos x). Yes. Maybe there is another approach: Let y = sin x + cos x. For first quadrant y>0, f=(√2-1)y. For third y<0, f=-(√2+1)y? Actually negative times negative positive. But if y in third? y=sin+cos <0, f= -(√2+1)y = (√2+1)(-y). Since -y positive. y range third: [-√2,-1). Then -y ∈ (1,√2]. etc. Need maybe show integer k=3 is attained: "Como (√2+1, 2+√2] contém 3, e pelo teorema do valor intermediário, existe x no terceiro quadrante com f(x)=3." Also mention first quadrant no integers. Let's ensure final answer no extra "all integers k": perhaps "k=3" or "O único inteiro é k=3". The final line format examples: "Resposta final: a) 3/2; b) 5√2". If single answer, "Resposta final: k=3". Good. Let's produce final in Portuguese.

**Simplifying the Expression** I'm currently working on simplifying the given function. The focus is on resolving the square roots involving trigonometric identities to get a cleaner form of $f(x)$, which will then allow me to determine the possible values of $k$. **Analyzing Quadrant Constraints** I'm clarifying the definition of the first and third quadrants to accurately apply the absolute value functions to $\cos(x)$ and $\operatorname{sen}(x)$ within the simplified expression for $f(x)$. This step is crucial for the subsequent analysis of the equation. **Investigating Quadrant Definitions** I'm now exploring how the inclusion or exclusion of quadrant boundaries affects the function's image. I'm calculating the range of $f(x)$ for both open and closed intervals within the first quadrant to see if this distinction alters the integer values for $k$. **Evaluating Third Quadrant** I'm now analyzing the third quadrant ($x \in (\pi, 3\pi/2)$). Given $\cos(x) < 0$ and $\operatorname{sen}(x) < 0$, I've derived $f(x) = -(\sqrt{2}+1)\cos(x) - (\sqrt{2}+1)\operatorname{sen}(x)$ for this region. **Analyzing Third Quadrant Range** I've determined that in the third quadrant, $f(x)$ falls within the interval $(\sqrt{2}+1, 2+\sqrt{2}]$. This means $k=3$ is a possible integer value. **Confirming Quadrant Results** I've confirmed that the calculated range for the third quadrant holds whether or not the quadrant boundaries are included. This means $k=3$ remains the only possible integer value, a satisfyingly elegant outcome! **Verifying Function Simplification** I've successfully verified the simplified forms of $f(x)$ across various points. The tests confirm the accuracy of the absolute value treatments for $\cos(x)$ and $\operatorname{sen}(x)$ in each quadrant, bolstering confidence in the derived range calculations. **Confirming Third Quadrant Solutions** I've confirmed that the equation $f(x)=3$ has two valid solutions strictly within the third quadrant. This reaffirms that $k=3$ is indeed achievable, completing the analysis for this region. **Confirming Quadrant Definitions** I've confirmed that whether or not quadrant boundaries are included, the integer value of $k=3$ remains unique. This is because the function's values at the axes are not integers, so the range is unaffected. **Exploring First Quadrant** I'm now analyzing the first quadrant where $\operatorname{sen}(x)>0$ and $\cos(x)>0$. I've derived $f(x) = (\sqrt{2}-1)(\operatorname{sen}(x)+\cos(x))$, leading to a range of $(\sqrt{2}-1, 2-\sqrt{2}]$, which contains no integers. **Analyzing Third Quadrant Calculations** I've completed the detailed analysis for the third quadrant. My work shows the function $f(x)$ lies within $(\sqrt{2}+1, 2+\sqrt{2}]$, confirming that $k=3$ is the sole integer in this interval, even when considering boundary points.

**Analyzing Function's Behavior** I'm currently dissecting the ITA math problem, specifically focusing on the function $f(x) = \sqrt{1 + \cos(2x)} + \sqrt{1 - \cos(2x)} - \text{sen}(x)$. My immediate goal is to simplify this expression by leveraging trigonometric identities and understanding the behavior of the square roots. **Simplifying Trigonometric Expression** I'm working through the function $f(x)$, and I've successfully used identities to simplify the square roots. My current focus is on analyzing how these simplifications behave across different quadrants, especially considering whether axes are included. **Determining Function Range** I've successfully simplified the expression for the first quadrant, establishing a specific range. Now, I'm extending this analysis to other quadrants, carefully considering the behavior of the absolute values and how they impact the final range of $f(x)$. **Confirming Integer Absence** I've confirmed that the range of $f(x)$ in the first quadrant, regardless of interval endpoints, falls entirely between approximately 0.414 and 0.586. This interval demonstrably contains no integers. My next step involves analyzing the third quadrant. **Investigating Third Quadrant** I've calculated the range of the function in the third quadrant to be approximately $(2.414, 3.414]$. This interval contains the integer 3. My next step is to confirm if this integer value is actually attained by the function.