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TiliK225 [7]
3 years ago
13

Assume f(x) = g(x). Which of the following two functions may be used to represent the function 3x = x – 4? *

Mathematics
1 answer:
nirvana33 [79]3 years ago
8 0

Since we know that the two functions are equal, f (x) = g (x), therefore when we equate the two equations the result should be similar to 3x = x – 4:

 

A. f(x) = 3x, g(x) = x – 4

Equating the two functions:

3x = x – 4             (correct)

 

So we now got our answer (A)

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Answer:

L.S = R.S ⇒ Proved down

Step-by-step explanation:

Let us revise some rules in trigonometry

  1. sin²α + cos²α = 1
  2. sin2α = 2 sin α cosα
  3. cscα = 1/sinα

To solve the question let us find the simplest form of the right side and the left side, then show that they are equal

∵ L.S = csc2α + 1

→ By using the 3rd rule above

∴ L.S = \frac{1}{sin2\alpha} + 1

→ Change 1 to \frac{sin2\alpha}{sin2\alpha}

∴ L.S = \frac{1}{sin2\alpha} + \frac{sin2\alpha}{sin2\alpha}

→ The denominators are equal, then add the numerators

∴ L.S = \frac{1+sin2\alpha}{sin2\alpha}

∵ R. S = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α

∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα

→ By using the 1st rule above, equate sin²α + cos²α by 1

∴ (sinα + cosα)² = 1 + 2 sinα cosα

→ By using the 2nd rule above, equate 2 sinα cosα by sin2α

∴ (sinα + cosα)² = 1 + sin2α

→ Substitute it in the R.S above

∴ R. S = \frac{1+sin2\alpha}{sin2\alpha}

∵ L.S = R.S

∴ csc 2α + 1 = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

4 0
3 years ago
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