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z=\frac{1}{4}x^{2}+\sin\left(\pi-x\right)\ -\frac{3}{4}y+\frac{abc}{\cos\left(x\right)}

Question

z = \frac{1}{4}x^{2} + \sin\left(\pi - x\right) - \frac{3}{4}y + \frac{abc}{\cos\left(x\right)}

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Solution

To analyze the expression given for z z :

Step 1: Break Down the Problem

The expression is composed of several parts:

  1. A quadratic term 14x2 \frac{1}{4}x^2
  2. A sine function sin(πx) \sin(\pi - x)
  3. A linear term in y y given by 34y -\frac{3}{4}y
  4. A fraction involving parameters a a , b b , c c , and a cosine function abccos(x) \frac{abc}{\cos(x)}

Step 2: Relevant Concepts

To understand the behavior of z z , we can recall:

  • The properties of sine and cosine functions, particularly their values and periodicity.
  • The structure of the quadratic function and its implications for x x .
  • The impacts of the parameters a,b,c a, b, c on z z .

Step 3: Analysis and Detail

  1. Quadratic term: 14x2 \frac{1}{4}x^2 indicates that the expression has a parabolic component opening upwards.
  2. Sine function: Using the identity sin(πx)=sin(x) \sin(\pi - x) = \sin(x) , we simplify this part: sin(πx)=sinx \sin(\pi - x) = \sin x
  3. Linear term: The term 34y -\frac{3}{4}y linearly decreases z z with increasing y y .
  4. Fractional term: The term abccos(x) \frac{abc}{\cos(x)} becomes undefined for values of x x where cos(x)=0 \cos(x) = 0 . The behavior near those points should be considered.

Step 4: Verify and Summarize

To summarize the behavior:

  • The expression for z z will vary depending on the values of x x and y y .
  • The quadratic term increases without bound as x |x| increases.
  • The overall expression must be evaluated carefully at points where cos(x)=0 \cos(x) = 0 .
  • The sine function contributes a bounded oscillation to the overall expression.

Final Answer

The final expression for z z is: z=14x2+sin(x)34y+abccos(x) z = \frac{1}{4}x^{2} + \sin(x) - \frac{3}{4}y + \frac{abc}{\cos(x)} with the understanding that care must be taken regarding the values where cos(x)=0 \cos(x) = 0 .

This problem has been solved

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