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kogti [31]
3 years ago
9

Which of the following trigonometric inequalities has no solution over the interval 0 ≤ x ≤ 2pi radians?

Mathematics
1 answer:
valentinak56 [21]3 years ago
6 0

Answer:

A .cos(x)<1

Step-by-step explanation:

According to the first inequality

cos(x)<1

x < arccos 1

x<0

This therefore does not have a solution within the range 0 ≤ x ≤ 2pi

x cannot be leas than 0. According to the range not value, 0≤x which is equivalent to x≥0. Thus means otvis either x = 0 or x> 0.

For the second option

.cos(x/2)<1

x/2< arccos1

x/2<0

x<0

This inequality also has solution within the range 0 ≤ x ≤ 2pi since 0 falls within the range of values.

For the inequality csc(x)<1

1/sin(x) < 1

1< sin(x)

sinx>1

x>arcsin1

x>90°

x>π/2

This inequality also has solution within the range 0 ≤ x ≤ 2pi since π/2 falls within the range of values

For the inequality csc(x/2)<1

1/sin(x/2) < 1

1< sin(x/2)

sin(x/2)> 1

x/2 > arcsin1

X/2 > 90°

x>180°

x>π

This value of x also has a solution within the range.

Therefore option A is the only inequality that does not have a solution with the range.

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3 years ago
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2 years ago
Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.
galina1969 [7]

Answer:

All potential roots are 3,3 and -\frac{5}{3}.

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

f(x)=3x^3-13x^2-3x+45

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

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q=\pm 1,\pm 3

All possible roots

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Now we check each rational root and see which are possible roots for given function.

f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0

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f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144

f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}

Similarly, we will check for all value of p/q and we get

f(-5/3)=0

Thus, All potential roots are 3,3 and -\frac{5}{3}.


5 0
3 years ago
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4 0
2 years ago
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