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Question: Let $x$ be the solution of equation $2[\alpha + 32] = 3[\alpha - 64]$, where $[\alpha]$ is the great...

Let xx be the solution of equation 2[α+32]=3[α−64]2[\alpha + 32] = 3[\alpha - 64], where [α][\alpha] is the greatest integer less than or equal to α\alpha and let y=∏j=19sin⁡(2j−118π)y = \prod_{j=1}^{9} \sin (\frac{2j-1}{18} \pi), then which of the following is/are incorrect?

A

[x]=[y][x] = [y]

B

x=20518x = \frac{2051}{8}

C

[x]=[1y]=1[x] = [\frac{1}{y}] = 1

D

[1x]+[1y]=210[\frac{1}{x}] + [\frac{1}{y}] = 2^{10}

Answer

A, C, D

Explanation

Solution

The problem requires us to find the values of xx and yy and then check the given options.

Part 1: Solving for xx

The given equation is 2[α+32]=3[α−64]2[\alpha + 32] = 3[\alpha - 64]. We use the property of the greatest integer function: [n+k]=[n]+k[n+k] = [n] + k for any integer kk. Let [α]=n[\alpha] = n, where nn is an integer. Then the equation becomes: 2([α]+32)=3([α]−64)2([\alpha] + 32) = 3([\alpha] - 64) 2(n+32)=3(n−64)2(n + 32) = 3(n - 64) 2n+64=3n−1922n + 64 = 3n - 192 64+192=3n−2n64 + 192 = 3n - 2n 256=n256 = n So, [α]=256[\alpha] = 256. This implies 256≤α<257256 \le \alpha < 257. The problem states "xx be the solution of equation...", and option B gives a specific value for xx. Let's check if this value satisfies the condition. From option B, x=20518x = \frac{2051}{8}. x=256.375x = 256.375. This value falls within the interval [256,257)[256, 257), so [x]=[256.375]=256[x] = [256.375] = 256. Let's verify this value in the original equation: LHS = 2[x+32]=2[256.375+32]=2[288.375]=2×288=5762[x + 32] = 2[256.375 + 32] = 2[288.375] = 2 \times 288 = 576. RHS = 3[x−64]=3[256.375−64]=3[192.375]=3×192=5763[x - 64] = 3[256.375 - 64] = 3[192.375] = 3 \times 192 = 576. Since LHS = RHS, x=20518x = \frac{2051}{8} is a valid solution for the equation. Thus, we take x=256.375x = 256.375.

Part 2: Calculating yy

The expression for yy is y=∏j=19sin⁡(2j−118π)y = \prod_{j=1}^{9} \sin (\frac{2j-1}{18} \pi). Let's list the terms in the product: For j=1:sin⁡(118π)=sin⁡(10∘)j=1: \sin(\frac{1}{18}\pi) = \sin(10^\circ) For j=2:sin⁡(318π)=sin⁡(30∘)j=2: \sin(\frac{3}{18}\pi) = \sin(30^\circ) For j=3:sin⁡(518π)=sin⁡(50∘)j=3: \sin(\frac{5}{18}\pi) = \sin(50^\circ) For j=4:sin⁡(718π)=sin⁡(70∘)j=4: \sin(\frac{7}{18}\pi) = \sin(70^\circ) For j=5:sin⁡(918π)=sin⁡(90∘)j=5: \sin(\frac{9}{18}\pi) = \sin(90^\circ) For j=6:sin⁡(1118π)=sin⁡(110∘)=sin⁡(180∘−70∘)=sin⁡(70∘)j=6: \sin(\frac{11}{18}\pi) = \sin(110^\circ) = \sin(180^\circ - 70^\circ) = \sin(70^\circ) For j=7:sin⁡(1318π)=sin⁡(130∘)=sin⁡(180∘−50∘)=sin⁡(50∘)j=7: \sin(\frac{13}{18}\pi) = \sin(130^\circ) = \sin(180^\circ - 50^\circ) = \sin(50^\circ) For j=8:sin⁡(1518π)=sin⁡(150∘)=sin⁡(180∘−30∘)=sin⁡(30∘)j=8: \sin(\frac{15}{18}\pi) = \sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ) For j=9:sin⁡(1718π)=sin⁡(170∘)=sin⁡(180∘−10∘)=sin⁡(10∘)j=9: \sin(\frac{17}{18}\pi) = \sin(170^\circ) = \sin(180^\circ - 10^\circ) = \sin(10^\circ)

So, y=sin⁡(10∘)sin⁡(30∘)sin⁡(50∘)sin⁡(70∘)sin⁡(90∘)sin⁡(70∘)sin⁡(50∘)sin⁡(30∘)sin⁡(10∘)y = \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \sin(90^\circ) \sin(70^\circ) \sin(50^\circ) \sin(30^\circ) \sin(10^\circ). This can be written as: y=(sin⁡(10∘)sin⁡(30∘)sin⁡(50∘)sin⁡(70∘))2sin⁡(90∘)y = (\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ))^2 \sin(90^\circ). We know sin⁡(90∘)=1\sin(90^\circ) = 1 and sin⁡(30∘)=12\sin(30^\circ) = \frac{1}{2}. Now, consider the product sin⁡(10∘)sin⁡(50∘)sin⁡(70∘)\sin(10^\circ) \sin(50^\circ) \sin(70^\circ). This is of the form sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)\sin A \sin(60^\circ - A) \sin(60^\circ + A) with A=10∘A=10^\circ. Using the identity sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=14sin⁡(3A)\sin A \sin(60^\circ - A) \sin(60^\circ + A) = \frac{1}{4} \sin(3A): sin⁡(10∘)sin⁡(50∘)sin⁡(70∘)=14sin⁡(3×10∘)=14sin⁡(30∘)=14×12=18\sin(10^\circ) \sin(50^\circ) \sin(70^\circ) = \frac{1}{4} \sin(3 \times 10^\circ) = \frac{1}{4} \sin(30^\circ) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}. Substitute this value back into the expression for yy: y=(18×sin⁡(30∘))2×1y = \left(\frac{1}{8} \times \sin(30^\circ)\right)^2 \times 1 y=(18×12)2=(116)2=1256y = \left(\frac{1}{8} \times \frac{1}{2}\right)^2 = \left(\frac{1}{16}\right)^2 = \frac{1}{256}.

So, we have x=256.375x = 256.375 and y=1256=0.00390625y = \frac{1}{256} = 0.00390625.

Part 3: Checking the options

A. [x]=[y][x] = [y] [x]=[256.375]=256[x] = [256.375] = 256. [y]=[1256]=[0.00390625]=0[y] = [\frac{1}{256}] = [0.00390625] = 0. Since 256≠0256 \ne 0, option A is incorrect.

B. x=20518x = \frac{2051}{8} As verified in Part 1, this statement is correct.

C. [x]=[1y]=1[x] = [\frac{1}{y}] = 1 [x]=256[x] = 256. 1y=11/256=256\frac{1}{y} = \frac{1}{1/256} = 256. [1y]=[256]=256[\frac{1}{y}] = [256] = 256. So, [x]=[1y]=256[x] = [\frac{1}{y}] = 256. The option states that this value is 1, which is incorrect. Option C is incorrect.

D. [1x]+[1y]=210[\frac{1}{x}] + [\frac{1}{y}] = 2^{10} 1x=12051/8=82051\frac{1}{x} = \frac{1}{2051/8} = \frac{8}{2051}. Since 0<82051<10 < \frac{8}{2051} < 1, [1x]=0[\frac{1}{x}] = 0. [1y]=256[\frac{1}{y}] = 256 (from option C calculation). So, [1x]+[1y]=0+256=256[\frac{1}{x}] + [\frac{1}{y}] = 0 + 256 = 256. 210=10242^{10} = 1024. Since 256≠1024256 \ne 1024, option D is incorrect.

The question asks for the incorrect options. Based on our analysis, options A, C, and D are incorrect.