@siddharth bhaiya(*Brainly teacher*) please try your best to answer!
Number of integers in the range of the given range of the function f(x).?
Other users please answer if you know else leave.
Thanks
I will be delightful to offer 100 points for the correct and appropriate answer!
X belongs to {R}-0
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Answers
Answer:
0
Step-by-step explanation:
Answer:
0
Step-by-step explanation:
Given f(x) = (x³ + 2x² + 3x + 2)/(x³ + 2x² + 2x + 1)
(i) Domain:
Equate the denominator to 0 to find the value of x.
⇒ x³ + 2x² + 2x + 1 = 0
⇒ x³ + x² + x + x² + x + 1 = 0
⇒ x(x² + x + 1) + (x² + x + 1) = 0
⇒ (x + 1)(x² + x + 1) = 0
⇒ (x + 1)(x² + x + 1 --- Cannot be considered)
⇒ x + 1 = 0
⇒ x = -1
Therefore:
Domain = (-∞,-1) ∪ (-1,∞)
In real numbers, the denominator cannot be 0.
x ≠ 0.
So, f(0) ≠ 2.
(ii) Range:
Given: f(x) = (x³ + 2x² + 3x + 2)/(x³ + 2x² + 2x + 1)
Rearrange them into groups.
= (x³ + x² + 2x + x² + x + 2)/(x³ + x² + x + x + x + 1)
= [x(x² + x + 2) + (x² + x + 2)]/[x(x² + x + 1) + (x² + x + 1)]
= [(x + 1)(x² + x + 2)]/[(x + 1)(x² + x + 1)]
= [(x² + x + 2)]/[(x² + x + 1)]
Let f(x) = y and rearrange the equation.
y = (x² + x + 2)/(x² + x + 1)
⇒ y(x² + x + 1) = x² + x + 2
⇒ x²y + xy + y = x² + x + 2
⇒ x²y + xy + y - x² - x - 2 = 0
⇒ (y - 1)x² + (y - 1)x + y - 2 = 0
We know that the Quadratic formula:
x = -b ± √b² - 4ac/2a
The solutions of x are real when b² - 4ac ≥ 0
Here, a = (y - 1), b = (y - 1), c = y - 2
Now,
Find the inequality for the set of values of y.
⇒ b² - 4ac ≥ 0
⇒ (y - 1)² - 4(y - 1)(y - 2) ≥ 0
⇒ y² + 1 - 2y - 4[y² - 2y - y + 2] ≥ 0
⇒ y² + 1 - 2y - 4y² + 8y + 4y - 8 ≥ 0
⇒ -3y² + 10y - 7 ≥ 0
⇒ -(3y² - 10y + 7) ≥ 0
⇒ -(3y² - 3y - 7y + 7) ≥ 0
⇒ -(3y(y - 1) - 7(y - 1)) ≥ 0
⇒ -(y - 1)(3y - 7) ≥ 0
⇒ (y - 1)(3y - 7) ≤ 0
⇒ y = 1, 7/3
∴ y ∈ R
Range = (1, 7/3]
Here, the integer value is 2.
But, y≠ 2.
Therefore, the number of integers = 0.
Hope it helps!
0 IS THE ANSWER
Given f(x) = (x³ + 2x² + 3x + 2)/(x³ + 2x² + 2x + 1)
(i) Domain:
Equate the denominator to 0 to find the value of x.
⇒ x³ + 2x² + 2x + 1 = 0
⇒ x³ + x² + x + x² + x + 1 = 0
⇒ x(x² + x + 1) + (x² + x + 1) = 0
⇒ (x + 1)(x² + x + 1) = 0
⇒ (x + 1)(x² + x + 1 --- Cannot be considered)
⇒ x + 1 = 0
⇒ x = -1
Therefore:
Domain = (-∞,-1) ∪ (-1,∞)
In real numbers, the denominator cannot be 0.
x ≠ 0.
So, f(0) ≠ 2.
(ii) Range:
Given: f(x) = (x³ + 2x² + 3x + 2)/(x³ + 2x² + 2x + 1)
Rearrange them into groups.
= (x³ + x² + 2x + x² + x + 2)/(x³ + x² + x + x + x + 1)
= [x(x² + x + 2) + (x² + x + 2)]/[x(x² + x + 1) + (x² + x + 1)]
= [(x + 1)(x² + x + 2)]/[(x + 1)(x² + x + 1)]
= [(x² + x + 2)]/[(x² + x + 1)]
Let f(x) = y and rearrange the equation.
y = (x² + x + 2)/(x² + x + 1)
⇒ y(x² + x + 1) = x² + x + 2
⇒ x²y + xy + y = x² + x + 2
⇒ x²y + xy + y - x² - x - 2 = 0
⇒ (y - 1)x² + (y - 1)x + y - 2 = 0
We know that the Quadratic formula:
x = -b ± √b² - 4ac/2a
The solutions of x are real when b² - 4ac ≥ 0
Here, a = (y - 1), b = (y - 1), c = y - 2
Now,
Find the inequality for the set of values of y.
⇒ b² - 4ac ≥ 0
⇒ (y - 1)² - 4(y - 1)(y - 2) ≥ 0
⇒ y² + 1 - 2y - 4[y² - 2y - y + 2] ≥ 0
⇒ y² + 1 - 2y - 4y² + 8y + 4y - 8 ≥ 0
⇒ -3y² + 10y - 7 ≥ 0
⇒ -(3y² - 10y + 7) ≥ 0
⇒ -(3y² - 3y - 7y + 7) ≥ 0
⇒ -(3y(y - 1) - 7(y - 1)) ≥ 0
⇒ -(y - 1)(3y - 7) ≥ 0
⇒ (y - 1)(3y - 7) ≤ 0
⇒ y = 1, 7/3
∴ y ∈ R
Range = (1, 7/3]
Here, the integer value is 2.
But, y≠ 2.
Therefore, the number of integers = 0.
Given f(x) = (x³ + 2x² + 3x + 2)/(x³ + 2x² + 2x + 1)
(i) Domain:
Equate the denominator to 0 to find the value of x.
⇒ x³ + 2x² + 2x + 1 = 0
⇒ x³ + x² + x + x² + x + 1 = 0
⇒ x(x² + x + 1) + (x² + x + 1) = 0
⇒ (x + 1)(x² + x + 1) = 0
⇒ (x + 1)(x² + x + 1 --- Cannot be considered)
⇒ x + 1 = 0
⇒ x = -1
Therefore:
Domain = (-∞,-1) ∪ (-1,∞)
In real numbers, the denominator cannot be 0.
x ≠ 0.
So, f(0) ≠ 2.
(ii) Range:
Given: f(x) = (x³ + 2x² + 3x + 2)/(x³ + 2x² + 2x + 1)
Rearrange them into groups.
= (x³ + x² + 2x + x² + x + 2)/(x³ + x² + x + x + x + 1)
= [x(x² + x + 2) + (x² + x + 2)]/[x(x² + x + 1) + (x² + x + 1)]
= [(x + 1)(x² + x + 2)]/[(x + 1)(x² + x + 1)]
= [(x² + x + 2)]/[(x² + x + 1)]
Let f(x) = y and rearrange the equation.
y = (x² + x + 2)/(x² + x + 1)
⇒ y(x² + x + 1) = x² + x + 2
⇒ x²y + xy + y = x² + x + 2
⇒ x²y + xy + y - x² - x - 2 = 0
⇒ (y - 1)x² + (y - 1)x + y - 2 = 0
We know that the Quadratic formula:
x = -b ± √b² - 4ac/2a
The solutions of x are real when b² - 4ac ≥ 0
Here, a = (y - 1), b = (y - 1), c = y - 2
Now,
Find the inequality for the set of values of y.
⇒ b² - 4ac ≥ 0
⇒ (y - 1)² - 4(y - 1)(y - 2) ≥ 0
⇒ y² + 1 - 2y - 4[y² - 2y - y + 2] ≥ 0
⇒ y² + 1 - 2y - 4y² + 8y + 4y - 8 ≥ 0
⇒ -3y² + 10y - 7 ≥ 0
⇒ -(3y² - 10y + 7) ≥ 0
⇒ -(3y² - 3y - 7y + 7) ≥ 0
⇒ -(3y(y - 1) - 7(y - 1)) ≥ 0
⇒ -(y - 1)(3y - 7) ≥ 0
⇒ (y - 1)(3y - 7) ≤ 0
⇒ y = 1, 7/3
∴ y ∈ R
Range = (1, 7/3]
Here, the integer value is 2.
But, y≠ 2.
Therefore, the number of integers = 0.