We are given the following:
To find the value of \( x \) (the number of items) that will give a profit, we need to calculate the profit function. The profit function is the difference between revenue and cost:
\[ \text{Profit} = R(x) - C(x) \]
Substituting the given functions:
\[ \text{Profit} = (60x + 2000) - (20x + 4000) \]
Simplifying:
\[ \text{Profit} = 60x + 2000 - 20x - 4000 \]
\[ \text{Profit} = 40x - 2000 \]
To earn a profit, the profit must be greater than 0:
\[ 40x - 2000 > 0 \]
Solving for \( x \):
\[ 40x > 2000 \]
\[ x > \frac{2000}{40} \]
\[ x > 50 \]
Therefore, the value of \( x \) to earn a profit is greater than 50.
The correct answer is (A): \( x > 50 \)
We are given the cost function \( C(x) \) and the revenue function \( R(x) \) for a product, where x is the number of items produced and sold:
The Profit function \( P(x) \) is calculated as the difference between the revenue and the cost:
\[ P(x) = R(x) - C(x) \]
Substitute the given functions into the profit formula:
\[ P(x) = (60x + 2000) - (20x + 4000) \]
Simplify the expression:
\[ P(x) = 60x + 2000 - 20x - 4000 \]
\[ P(x) = (60x - 20x) + (2000 - 4000) \]
\[ P(x) = 40x - 2000 \]
To earn a profit, the profit \( P(x) \) must be greater than zero:
\[ P(x) > 0 \]
Substitute the expression for \( P(x) \):
\[ 40x - 2000 > 0 \]
Now, solve this inequality for x:
Add 2000 to both sides:
\[ 40x > 2000 \]
Divide both sides by 40:
\[ x > \frac{2000}{40} \]
\[ x > \frac{200}{4} \]
\[ x > 50 \]
Therefore, a profit is earned when the number of items produced and sold (x) is greater than 50.
Comparing this result with the given options:
The correct option is >50.
Let A be the set of 30 students of class XII in a school. Let f : A -> N, N is a set of natural numbers such that function f(x) = Roll Number of student x.
Give reasons to support your answer to (i).
Find the domain of the function \( f(x) = \cos^{-1}(x^2 - 4) \).