GSEB Class 12 Statistics Solutions Chapter 5 Differentiation

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 5 Differentiation Ex 5

Section A

Question 1. What is the formula for derivative of function f(x)?
(a) \( \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
(b) \( \lim_{h \to 0} \frac{f(x+h)+f(x)}{h} \)
(c) \( \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
(d) \( \lim_{h \to x} \frac{f(x)-f(x+h)}{h} \)
Answer: (c) \( \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
In simple words: The derivative formula helps us find how a function changes at a specific point. It uses a very small change, 'h', to calculate the slope of the function.

🎯 Exam Tip: Remember the limit definition of a derivative as it's fundamental to understanding calculus concepts.

 

Question 2. What is \( \frac{dy}{dx} \) if \( y = ax^n \), where 'a' is a constant?
(a) \( nx^{n-1} \)
(b) \( an x^{n-1} \)
(c) \( 0 \)
(d) \( anx^{n+1} \)
Answer: (b) \( an x^{n-1} \)
In simple words: If you have a constant multiplied by a variable raised to a power, to find its derivative, you multiply the constant by the power, and then decrease the power by one.

🎯 Exam Tip: This is a core rule for differentiating power functions, frequently tested in basic calculus problems.

 

Question 3. If \( y = ax + b \), where 'a' and 'b' are constants, then what will be its derivative?
(a) a
(b) b
(c) a + b
(d) 0
Answer: (a) a
In simple words: When you differentiate a simple linear function, the derivative is just the coefficient of the 'x' term. Constants on their own become zero.

🎯 Exam Tip: Understand that the derivative of a linear function represents its constant slope.

 

Question 4. What is the derivative of \( f(x) = \frac{4}{x^2} \)?
(a) \( \frac{4}{2x} \)
(b) \( - \frac{8}{x^3} \)
(c) \( \frac{8}{x^3} \)
(d) \( 0 \)
Answer: (b) \( - \frac{8}{x^3} \)
In simple words: To find the derivative of \( \frac{4}{x^2} \), first rewrite it as \( 4x^{-2} \). Then, apply the power rule for differentiation: bring down the power, multiply by the coefficient, and reduce the power by one.

🎯 Exam Tip: Always remember to rewrite fractions with variables in the denominator as terms with negative exponents before applying differentiation rules.

 

Question 5. If 'u' and 'v' are two functions of 'x', then what is the formula for the derivative of their product?
(a) \( u \cdot \frac{du}{dx} + v \cdot \frac{dv}{dx} \)
(b) \( u \cdot \frac{dv}{dx} - v \cdot \frac{du}{dx} \)
(c) \( \frac{du}{dx} \times \frac{dv}{dx} \)
(d) \( u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \)
Answer: (d) \( u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \)
In simple words: The product rule for derivatives states that you take the first function multiplied by the derivative of the second, and add it to the second function multiplied by the derivative of the first.

🎯 Exam Tip: The product rule is crucial for differentiating functions that are multiplied together; remember the "first times derivative of second plus second times derivative of first" pattern.

 

Question 6. If 'u' and 'v' are functions of 'x', then what is the formula for the derivative of \( \frac{v}{u} \)?
(a) \( \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)
(b) \( \frac{v \cdot \frac{du}{dx} + u \cdot \frac{dv}{dx}}{v^2} \)
(c) \( \frac{u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}}{u^2} \)
(d) \( \frac{u \cdot \frac{dv}{dx} - v \cdot \frac{du}{dx}}{u^2} \)
Answer: (d) \( \frac{u \cdot \frac{dv}{dx} - v \cdot \frac{du}{dx}}{u^2} \)
In simple words: The quotient rule for derivatives helps find the rate of change for a fraction of two functions. It involves the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the denominator squared.

🎯 Exam Tip: The quotient rule is often remembered as "low d high minus high d low over low squared" (where low is the denominator and high is the numerator).

 

Question 7. If the function \( f(x) \) is increasing at \( x = a \), then which is the correct option from the following?
(a) \( f'(a) < 0 \)
(b) \( f'(a) > 0 \)
(c) \( f'(a) = 0 \)
(d) \( f''(a) > 0 \)
Answer: (b) \( f'(a) > 0 \)
In simple words: A function is increasing at a point if its first derivative at that point is positive, meaning the slope of the function is going upwards.

🎯 Exam Tip: A positive first derivative signifies an increasing function, while a negative first derivative indicates a decreasing function.

 

Question 8. What are the necessary and sufficient conditions for a function to be minimum at \( x = a \)?
(a) \( f'(a) = 0, f''(a) < 0 \)
(b) \( f'(a) > 0, f''(a) > 0 \)
(c) \( f'(a) = 0, f''(a) > 0 \)
(d) \( f'(a) < 0, f'(a) > 0 \)
Answer: (c) \( f'(a) = 0, f''(a) > 0 \)
In simple words: For a function to be at a minimum point, its slope (first derivative) must be zero, and its curvature (second derivative) must be positive, indicating a concave-up shape.

🎯 Exam Tip: For finding local extrema, setting the first derivative to zero identifies critical points, and the sign of the second derivative at these points determines if it's a maximum (negative) or minimum (positive).

 

Question 9. What is the formula for elasticity of demand?
(a) \( - \frac{p}{x} \cdot \frac{dx}{dp} \)
(b) \( \frac{p}{x} \cdot \frac{dx}{dp} \)
(c) \( - \frac{x}{p} \cdot \frac{dp}{dx} \)
(d) \( \frac{p}{x} \cdot \frac{dp}{dx} \)
Answer: (a) \( - \frac{p}{x} \cdot \frac{dx}{dp} \)
In simple words: Elasticity of demand measures how much the quantity demanded changes when the price changes. It's calculated by multiplying the ratio of price to quantity by the derivative of quantity with respect to price, with a negative sign to reflect the inverse relationship.

🎯 Exam Tip: The negative sign in the elasticity of demand formula indicates the inverse relationship between price and quantity demanded; omitting it would lead to an incorrect interpretation.

 

Question 10. What are the conditions for revenue function R to be maximum?
(a) \( \frac{dR}{dx} = 0, \frac{d^2R}{dx^2} < 0 \)
(b) \( \frac{dR}{dx} = 0, \frac{d^2R}{dx^2} > 0 \)
(c) \( \frac{dR}{dx} > 0, \frac{d^2R}{dx^2} < 0 \)
(d) \( \frac{dR}{dx} > 0, \frac{d^2R}{dx^2} > 0 \)
Answer: (a) \( \frac{dR}{dx} = 0, \frac{d^2R}{dx^2} < 0 \)
In simple words: For total revenue to be at its highest point, the first derivative of the revenue function must be zero, and the second derivative must be negative, meaning the revenue curve is concave down at that point.

🎯 Exam Tip: Maximizing revenue (or profit) involves finding critical points where the first derivative is zero and confirming it's a maximum using the second derivative test.

Section B

Question 1. Define differentiation.
Answer: The mathematical method of finding the derivative of a function is known as differentiation.
In simple words: Differentiation is how we figure out how fast something is changing, like the slope of a curve at any single point.

🎯 Exam Tip: A clear, concise definition of differentiation is essential; focus on its role in finding the rate of change of a function.

 

Question 2. Find \( f'(x) \) for the function \( f(x) = 50 \).
Answer: For the function \( f(x) = 50 \), the derivative \( f'(x) = 0 \).
In simple words: If a function is just a constant number, like 50, it means it's not changing, so its rate of change (derivative) is zero.

🎯 Exam Tip: Remember that the derivative of any constant is always zero because a constant value has no rate of change.

 

Question 3. Find \( \frac{dy}{dx} \) if \( y = a \), where 'a' is constant.
Answer: If \( y = a \), where 'a' is a constant, then its derivative \( \frac{dy}{dx} = 0 \).
In simple words: When 'y' is a fixed number and does not change, its rate of change with respect to 'x' is zero.

🎯 Exam Tip: This question tests the fundamental rule that the derivative of a constant is zero.

 

Question 4. State the rule for derivative for the product of two functions of x.
Answer: If 'u' and 'v' are functions of 'x' that can be differentiated, and \( y = u \cdot v \), then the rule for finding the derivative of their product is: \( \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \).
In simple words: To find the derivative of two functions multiplied together, you take the first function, multiply it by the derivative of the second, and then add the second function multiplied by the derivative of the first.

🎯 Exam Tip: The product rule is essential for functions involving multiplication; clearly stating the formula and its components is key to scoring points.

 

Question 5. How will the first order derivative of a function be at \( x = a \) if the function is decreasing at \( x = a \)?
Answer: If a function is going down (decreasing) at a point \( x = a \), then its first derivative at that point will be a negative number, meaning \( f'(a) < 0 \).
In simple words: If a function is decreasing, its first derivative, which tells us the slope, will be less than zero.

🎯 Exam Tip: Connect the sign of the first derivative directly to the function's behavior (positive for increasing, negative for decreasing).

 

Question 6. How will the second order derivative of a function be at \( x = a \) if the function is maximum at \( x = a \)?
Answer: If a function reaches its highest point (maximum) at \( x = a \), then its second derivative at \( x = a \) will be a negative number, meaning \( f''(a) < 0 \).
In simple words: When a function hits its maximum, its second derivative will be negative, showing that the curve is bending downwards.

🎯 Exam Tip: Remember that a negative second derivative indicates a local maximum, signifying a concave-down shape at that point.

 

Question 7. What are the stationary points of a function?
Answer: The specific points on a function's graph where its value is either at its highest (maximum) or lowest (minimum) are called stationary points.
In simple words: Stationary points are where a function's slope is flat (zero), which can be a peak or a valley.

🎯 Exam Tip: Stationary points are found by setting the first derivative of a function to zero; they include local maxima, minima, and saddle points.

 

Question 8. What is marginal revenue?
Answer: Marginal revenue refers to the extra income a business gets from selling one more unit of a product, which means it's the change in total revenue when there is a small change in demand.
In simple words: Marginal revenue is the additional money earned from selling one more item.

🎯 Exam Tip: Marginal revenue is the derivative of the total revenue function with respect to the quantity sold; it's a key concept in economics for optimizing production.

 

Question 9. Define marginal cost.
Answer: Marginal cost is the additional expense a company incurs to produce one more unit of a good, representing the change in total cost due to a small change in production.
In simple words: Marginal cost is the extra money spent to make one more unit of something.

🎯 Exam Tip: Marginal cost is the derivative of the total cost function; understanding it is vital for production and pricing decisions.

 

Question 10. State the formula of elasticity of demand.
Answer: The formula for elasticity of demand is as follows:
\( \text{Elasticity of demand} = - \frac{\text{Percentage change in demand}}{\text{Percentage change in price}} \)
The demand for an item, 'x', is a function of its price, 'p' (i.e., \( x = f(p) \)). Therefore, the elasticity of demand is also given by: \( \text{Elasticity of demand} = - \frac{p}{x} \cdot \frac{dx}{dp} \). The negative sign is used because the demand for an item usually has an inverse relationship with its price.
In simple words: Elasticity of demand shows how much demand changes when price changes. It's found by dividing the percentage change in demand by the percentage change in price, and a negative sign is added because demand usually falls when prices rise.

🎯 Exam Tip: Always include the negative sign in the elasticity of demand formula, as it reflects the inverse relationship between price and quantity demanded.

 

Question 11. Find \( f'(x) \) if \( f(x) = 7x^2 - 6x + 5 \).
Answer: If \( f(x) = 7x^2 - 6x + 5 \), then the first derivative is found by differentiating each term:
\( f'(x) = 7(2x) - 6(1) + 0 \)
\( = 14x - 6 \)
In simple words: To find the derivative, multiply the power by the coefficient and subtract one from the power for each term. The derivative of a constant is zero.

🎯 Exam Tip: Practice differentiating polynomials by applying the power rule to each term and remembering that the derivative of a constant term is zero.

 

Question 12. Find \( \frac{dy}{dx} \) if \( y = 6x^3 + \frac{7}{2}x^2 + \frac{6}{5}x - 8 \).
Answer: If \( y = 6x^3 + \frac{7}{2}x^2 + \frac{6}{5}x - 8 \), then the derivative \( \frac{dy}{dx} \) is:
\( \frac{dy}{dx} = 6(3x^2) + \frac{7}{2}(2x) + \frac{6}{5}(1) - 0 \)
\( = 18x^2 + 7x + \frac{6}{5} \)
In simple words: We find the derivative by applying the power rule to each part of the equation and remembering that constants become zero.

🎯 Exam Tip: Differentiate each term of the polynomial separately. Ensure you correctly handle fractional coefficients and the constant term.

Section C

Question 1. Define derivative.
Answer: Let \( f: A \to R \) be a real-valued function where 'A' is an open interval of real numbers. If, as 'h' becomes very small (approaches 0), the limit \( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \) exists, then this limit is called the derivative of function \( f \) at point \( x = a \). This is written as \( f'(a) \).
Therefore, \( f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \).
For any value of \( x \) in the domain of \( f \), the derivative is \( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \).
If \( y = f(x) \), then \( f'(x) \) can also be written as \( \frac{dy}{dx} \).
In simple words: A derivative tells us the exact rate at which a function is changing at a specific point. It's like finding the slope of a curve at a single spot.

🎯 Exam Tip: Understanding the limit definition of the derivative is foundational; clearly state the formula and explain its components for full marks.

 

Question 2. State the division rule of derivative.
Answer: If 'u' and 'v' are functions of 'x' that can be differentiated, and \( y = \frac{u}{v} \) where \( v \neq 0 \), then the division (quotient) rule for finding the derivative is as follows:
\( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अवकलन के भागफल नियम को दर्शाता है। इसमें बताया गया है कि यदि दो फलनों 'u' और 'v' का अनुपात \( \frac{u}{v} \) है, तो उसके अवकलज को कैसे ज्ञात किया जाए। सूत्र के अनुसार, यह हर (denominator) को अंश (numerator) के अवकलज से गुणा करके, फिर अंश को हर के अवकलज से गुणा करके घटाने और पूरे को हर के वर्ग से विभाजित करने पर प्राप्त होता है।
In simple words: The division rule says that to differentiate a fraction of two functions, you multiply the bottom function by the derivative of the top, subtract the top function multiplied by the derivative of the bottom, and then divide everything by the bottom function squared.

🎯 Exam Tip: The quotient rule is vital for differentiating rational functions; precise recall of the formula, including the order of terms in the numerator and the denominator, is necessary.

 

Question 3. State necessary and sufficient conditions for a function to be maximum at \( x = a \).
Answer: Suppose \( y = f(x) \). The conditions needed for a function \( f(x) \) to reach a maximum value at \( x = a \) are as follows:

• Necessary condition: The first derivative at that point must be zero: \( f'(a) = 0 \).


• Sufficient condition: The second derivative at that point must be negative: \( f''(a) < 0 \) (indicating a downward curve).

In simple words: For a function to have a maximum at a point, its slope must be flat (zero), and the curve must be bending downwards at that point.

🎯 Exam Tip: Clearly distinguish between necessary and sufficient conditions; both are required to confirm a local maximum accurately.

 

Question 4. Explain marginal cost and give its formula.
Answer: If the total cost of making 'x' units of an item is represented by \( C \), then \( C \) is a function of \( x \).

• Marginal cost is the change in total cost \( C \) that results from a very small change in the number of units produced, \( x \).


• Since total cost \( C \) is a function of production \( x \), the derivative of \( C \) with respect to \( x \) gives the marginal cost.


• Therefore, the formula for marginal cost is: Marginal cost \( = \frac{dC}{dx} \).

In simple words: Marginal cost is the added cost to produce one more unit. Its formula is the derivative of the total cost function.

🎯 Exam Tip: Define marginal cost as the rate of change of total cost and provide its derivative formula for full clarity.

 

Question 5. Define elasticity of demand.
Answer: Elasticity of demand is defined as the ratio of the percentage change in demand to the percentage change in price. The formula for elasticity of demand is as follows:
\( \text{Elasticity of demand} = - \frac{\text{Percentage change in demand}}{\text{Percentage change in price}} \).
The demand for an item, 'x', is a function of its price, 'p' (i.e., \( x = f(p) \)). Therefore, the elasticity of demand is also given by: \( \text{Elasticity of demand} = - \frac{p}{x} \cdot \frac{dx}{dp} \). The negative sign is included because demand and price typically have an inverse relationship.
In simple words: Elasticity of demand shows how much consumer demand changes for a product when its price changes. It's measured by comparing the percentage change in quantity demanded to the percentage change in price.

🎯 Exam Tip: Explain both the conceptual definition and the derivative-based formula for elasticity of demand, emphasizing the significance of the negative sign.

 

Question 6. What are the conditions for profit function P to be maximum?
Answer: For a profit function \( P \) to reach its highest point (maximum), the following conditions must be met:

• Necessary condition: The first derivative of the profit function must be zero: \( \frac{dP}{dx} = 0 \).


• Sufficient condition: The second derivative of the profit function must be negative: \( \frac{d^2P}{dx^2} < 0 \) (indicating a downward curve).

In simple words: To get the most profit, the rate of change of profit must be zero, and the profit curve must be bending downwards at that point.

🎯 Exam Tip: Maximum profit conditions involve both the first (critical points) and second (concavity) derivatives; ensure both are stated correctly.

 

Question 7. State the conditions for production cost function C to be minimum.
Answer: For a production cost function \( C \) to reach its lowest point (minimum), the following conditions must be met:

• Necessary condition: The first derivative of the cost function must be zero: \( \frac{dC}{dx} = 0 \).


• Sufficient condition: The second derivative of the cost function must be positive: \( \frac{d^2C}{dx^2} > 0 \) (indicating an upward curve).

In simple words: To achieve the lowest production cost, the rate of change of cost must be zero, and the cost curve must be bending upwards at that point.

🎯 Exam Tip: Minimum cost conditions require the first derivative to be zero and the second derivative to be positive for correct identification.

 

Question 8. Find \( f''(x) \) if \( f(x) = \sqrt[4]{x} \).
Answer: To find \( f''(x) \) for \( f(x) = \sqrt[4]{x} \):
First, rewrite \( f(x) \) using exponents: \( f(x) = x^{\frac{1}{4}} \).
Next, find the first derivative \( f'(x) \):
\( f'(x) = \frac{1}{4} x^{\frac{1}{4}-1} = \frac{1}{4} x^{-\frac{3}{4}} \)
Now, find the second derivative \( f''(x) \):
\( f''(x) = \frac{d}{dx} [f'(x)] = \frac{d}{dx} [\frac{1}{4} x^{-\frac{3}{4}}] \)
\( = \frac{1}{4} \left(-\frac{3}{4}\right) x^{-\frac{3}{4}-1} \)
\( = -\frac{3}{16} x^{-\frac{7}{4}} \)
\( = -\frac{3}{16x^{\frac{7}{4}}} \)
So, if \( f(x) = \sqrt[4]{x} \), then \( f''(x) = -\frac{3}{16x^{\frac{7}{4}}} \).
In simple words: We first change the root into a power. Then, we find the first derivative using the power rule, and then do it again to find the second derivative.

🎯 Exam Tip: Always convert radical expressions to exponential form before differentiating. Remember to apply the power rule carefully for both first and second derivatives.

 

Question 9. Write the chain rule of differentiation.
Answer: If \( y \) is a function that depends on \( u \), and \( u \) is a function that depends on \( x \), then the chain rule of differentiation is given as follows:
\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
In simple words: The chain rule helps us find the derivative of a function that is "inside" another function. We take the derivative of the outer function with respect to the inner one, and then multiply by the derivative of the inner function with respect to the variable.

🎯 Exam Tip: The chain rule is essential for composite functions; ensure you identify the outer and inner functions correctly before applying the rule.

 

Question 10. Find \( f''(0) \) if \( f(x) = x^4 - 4x^3 + 3x^2 + x + 1 \).
Answer: Given the function \( f(x) = x^4 - 4x^3 + 3x^2 + x + 1 \).
First, find the first derivative \( f'(x) \):
\( f'(x) = 4x^3 - 12x^2 + 6x + 1 \)
Next, find the second derivative \( f''(x) \):
\( f''(x) = \frac{d}{dx} [f'(x)] = \frac{d}{dx} [4x^3 - 12x^2 + 6x + 1] \)
\( = 12x^2 - 24x + 6 \)
Now, substitute \( x = 0 \) into \( f''(x) \):
\( f''(0) = 12(0)^2 - 24(0) + 6 = 6 \)
Therefore, for \( f(x) = x^4 - 4x^3 + 3x^2 + x + 1 \), the value of \( f''(0) \) is 6.
In simple words: We first find the derivative once, then find the derivative of that result to get the second derivative. Finally, we put \( x=0 \) into the second derivative to get the answer.

🎯 Exam Tip: Be careful with signs and powers when calculating derivatives. When substituting a value like 0, ensure all terms become zero correctly except for constants.

 

Question 11. Find marginal revenue if the revenue function is \( R = 90x - \frac{x^2}{2} \).
Answer: Given the revenue function \( R = 90x - \frac{x^2}{2} \).
To find the marginal revenue, we need to take the derivative of the revenue function with respect to \( x \):
Marginal revenue \( = \frac{dR}{dx} \)
\( = \frac{d}{dx} [90x - \frac{x^2}{2}] \)
\( = 90(1) - \frac{2x}{2} \)
\( = 90 - x \)
In simple words: We take the derivative of the given revenue formula to find the marginal revenue, which tells us how revenue changes with each extra unit sold.

🎯 Exam Tip: Marginal revenue is always the first derivative of the total revenue function; correctly applying basic differentiation rules to each term is important.

 

Question 12. What is the maximum value of a function?
Answer: For a function \( y = f(x) \), its value is considered maximum at \( x = a \) if it is the highest value in a small range around \( x = a \).
To be more precise, if we consider a very small interval (let's say of size \( h \)) around \( x = a \), and if \( f(a) \) is greater than both \( f(a + h) \) and \( f(a - h) \), then it can be concluded that the function reaches its maximum value at \( x = a \).
In simple words: The maximum value of a function at a point means that point is higher than all other points nearby.

🎯 Exam Tip: Define the maximum value in terms of a local maximum, emphasizing that the function's value at that point is greater than or equal to its values in an immediate neighborhood.

 

Question 13. When can it be said that a function is decreasing at a point?
Answer: Suppose \( y = f(x) \) is a function, and \( h \) is a very small positive number.
If, at a point \( x = a \), the value of the function \( f(a+h) \) is less than \( f(a) \), and \( f(a) \) is also less than \( f(a-h) \), then the function \( f(x) \) is considered decreasing at \( x = a \).
For such a decreasing function, its first derivative at \( x = a \) will be negative, i.e., \( f'(a) < 0 \).
In simple words: A function is decreasing at a point if its value goes down as you move slightly to the right of that point. Mathematically, its derivative at that point is negative.

🎯 Exam Tip: Clearly state the relationship between a decreasing function and its first derivative (which must be negative).

 

Question 14. Determine whether the function \( y = 12 + 4x - 7x^2 \) is increasing or decreasing at \( x = 2 \).
Answer: Given the function \( y = 12 + 4x - 7x^2 \).
First, find the first derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = 0 + 4(1) - 7(2x) \)
\( = 4 - 14x \)
Now, evaluate \( \frac{dy}{dx} \) at \( x = 2 \):
\( \frac{dy}{dx} |_{x=2} = 4 - 14(2) \)
\( = 4 - 28 \)
\( = -24 \)
Since \( \frac{dy}{dx} = -24 \), which is less than 0, the given function is decreasing at \( x = 2 \).
In simple words: We find the function's rate of change by taking its derivative, then put \( x=2 \) into that derivative. If the result is negative, the function is going down (decreasing) at that point.

🎯 Exam Tip: To determine if a function is increasing or decreasing, calculate its first derivative and check its sign at the specified point.

 

Question 15. Find the derivative of \( y = 4x^2 + 4x + 8 \). For which value of \( x \) will the derivative be zero?
Answer: Given the function \( y = 4x^2 + 4x + 8 \).
First, find the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = 4(2x) + 4(1) + 0 \)
\( = 8x + 4 \)
Now, to find the value of \( x \) for which the derivative is zero, set \( \frac{dy}{dx} = 0 \):
\( 0 = 8x + 4 \)
\( -8x = 4 \)
\( x = -\frac{4}{8} \)
\( x = -\frac{1}{2} \)
Therefore, the derivative will be zero when \( x = -\frac{1}{2} \).
In simple words: We first find how the function changes by taking its derivative. Then, we set that change to zero to find the specific 'x' value where the function stops increasing or decreasing.

🎯 Exam Tip: This question combines finding the derivative with solving an equation. Ensure accurate differentiation and algebraic manipulation to find the critical point.

 

Question 16. If \( f(x) = x^3 + 5x^2 + 3x + 7 \), prove that \( f'(2) = 35 \).
Answer: Given the function \( f(x) = x^3 + 5x^2 + 3x + 7 \).
First, find the first derivative \( f'(x) \):
\( f'(x) = 3x^2 + 5(2x) + 3(1) + 0 \)
\( = 3x^2 + 10x + 3 \)
Now, substitute \( x = 2 \) into \( f'(x) \):
\( f'(2) = 3(2)^2 + 10(2) + 3 \)
\( = 3 \times 4 + 20 + 3 \)
\( = 12 + 20 + 3 \)
\( = 35 \)
Hence, it is proven that \( f'(2) = 35 \).
In simple words: We first find the derivative of the function. Then, we replace 'x' with '2' in the derivative to show that the result is 35.

🎯 Exam Tip: Accurately calculate the derivative and substitute the given value of x; common errors include arithmetic mistakes or incorrect application of the power rule.

 

Question 17. If \( f(x) = 3x^2 + 3 \), then for which value of \( x \) is \( f'(x) = f(x) \)?
Answer: Given the function \( f(x) = 3x^2 + 3 \).
First, find the derivative \( f'(x) \):
\( f'(x) = 3(2x) + 0 \)
\( = 6x \)
Now, set \( f'(x) = f(x) \):
\( 6x = 3x^2 + 3 \)
Rearrange the equation to a standard quadratic form:
\( 3x^2 - 6x + 3 = 0 \)
Divide by 3:
\( x^2 - 2x + 1 = 0 \)
This is a perfect square trinomial:
\( (x - 1)^2 = 0 \)
Solve for \( x \):
\( x - 1 = 0 \)
\( x = 1 \)
Therefore, \( f'(x) = f(x) \) when \( x = 1 \).
In simple words: We find the derivative of the function and then set it equal to the original function. Solving this equation tells us the 'x' value where they are the same.

🎯 Exam Tip: This problem requires both differentiation and solving a quadratic equation. Ensure all algebraic steps are correct, especially factoring or using the quadratic formula.

 

Question 18. Find \( \frac{d^2y}{dx^2} \) if \( y = 2x^3 + 5x^2 - 3 + \frac{4}{x^2} - \frac{5}{x^3} \).
Answer: Given the function \( y = 2x^3 + 5x^2 - 3 + \frac{4}{x^2} - \frac{5}{x^3} \).
First, rewrite the terms with negative exponents for easier differentiation:
\( y = 2x^3 + 5x^2 - 3 + 4x^{-2} - 5x^{-3} \)
Next, find the first derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = 2(3x^2) + 5(2x) - 0 + 4(-2)x^{-2-1} - 5(-3)x^{-3-1} \)
\( = 6x^2 + 10x - 8x^{-3} + 15x^{-4} \)
Now, find the second derivative \( \frac{d^2y}{dx^2} \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx} [\frac{dy}{dx}] \)
\( = \frac{d}{dx} [6x^2 + 10x - 8x^{-3} + 15x^{-4}] \)
\( = 6(2x) + 10(1) - 8(-3)x^{-3-1} + 15(-4)x^{-4-1} \)
\( = 12x + 10 + 24x^{-4} - 60x^{-5} \)
Finally, rewrite the terms with positive exponents:
\( = 12x + 10 + \frac{24}{x^4} - \frac{60}{x^5} \)
Thus, \( \frac{d^2y}{dx^2} \) is \( 12x + 10 + \frac{24}{x^4} - \frac{60}{x^5} \).
In simple words: We take the derivative of the function once, then we take the derivative of that result again. Remember to change fractions with 'x' in the bottom to 'x' with negative powers first.

🎯 Exam Tip: Be meticulous with negative exponents and signs during differentiation. Double-check each term's derivative, especially when moving from positive to negative powers.

 

Question 19. Find \( \frac{d^2y}{dx^2} \) if \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \).
Answer: Given the function \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \).
First, rewrite the terms using exponents:
\( y = x^{\frac{1}{2}} + x^{-\frac{1}{2}} \)
Next, find the first derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{2} x^{\frac{1}{2}-1} + \left(-\frac{1}{2}\right) x^{-\frac{1}{2}-1} \)
\( = \frac{1}{2} x^{-\frac{1}{2}} - \frac{1}{2} x^{-\frac{3}{2}} \)
Now, find the second derivative \( \frac{d^2y}{dx^2} \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx} [\frac{1}{2} x^{-\frac{1}{2}} - \frac{1}{2} x^{-\frac{3}{2}}] \)
\( = \frac{1}{2} \left(-\frac{1}{2}\right) x^{-\frac{1}{2}-1} - \frac{1}{2} \left(-\frac{3}{2}\right) x^{-\frac{3}{2}-1} \)
\( = -\frac{1}{4} x^{-\frac{3}{2}} + \frac{3}{4} x^{-\frac{5}{2}} \)
Finally, rewrite with positive exponents:
\( = -\frac{1}{4x^{\frac{3}{2}}} + \frac{3}{4x^{\frac{5}{2}}} \)
Thus, for \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \), the second derivative \( \frac{d^2y}{dx^2} = -\frac{1}{4x^{\frac{3}{2}}} + \frac{3}{4x^{\frac{5}{2}}} \).
In simple words: We convert the square roots to powers, then take the derivative twice. This helps us see how the rate of change itself is changing.

🎯 Exam Tip: Fractional and negative exponents are common sources of errors. Apply the power rule carefully for each differentiation step and simplify the final answer to its most common form.

 

Question 20. Obtain marginal cost if the production cost function is \( C = 0.0012x^2 - 0.18x + 25 \).
Answer: Given the production cost function \( C = 0.0012x^2 - 0.18x + 25 \).
To find the marginal cost, we differentiate the cost function with respect to \( x \):
Marginal cost \( = \frac{dC}{dx} \)
\( = \frac{d}{dx} [0.0012x^2 - 0.18x + 25] \)
\( = 0.0012(2x) - 0.18(1) + 0 \)
\( = 0.0024x - 0.18 \)
Therefore, if the production cost function is \( C = 0.0012x^2 - 0.18x + 25 \), the marginal cost obtained is \( 0.0024x - 0.18 \).
In simple words: To find the marginal cost, we take the derivative of the total cost function. This tells us the extra cost to produce one more item.

🎯 Exam Tip: Marginal cost is the first derivative of the total cost function. Pay attention to decimal coefficients and the constant term, which differentiates to zero.

Section D

Question 1. Find derivative of \( y = ax + b \) (a and b are constants) using definition.
Answer: Given the function \( y = ax + b \). Here, 'a' and 'b' are constants.
By definition, the derivative \( \frac{dy}{dx} \) is given by the limit:
\( \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
Substitute \( f(x) = ax + b \). Then \( f(x+h) = a(x+h) + b = ax + ah + b \).
\( \frac{dy}{dx} = \lim_{h \to 0} \frac{(ax + ah + b) - (ax + b)}{h} \)
\( = \lim_{h \to 0} \frac{ax + ah + b - ax - b}{h} \)
\( = \lim_{h \to 0} \frac{ah}{h} \)
\( = \lim_{h \to 0} a \)
\( = a \)
Thus, the derivative of \( y = ax + b \) is \( \frac{dy}{dx} = a \).
In simple words: We use the definition of a derivative to find how \( y = ax + b \) changes. We calculate the slope of the line by seeing how much \( y \) changes for a tiny change in \( x \), which turns out to be 'a'.

🎯 Exam Tip: When using the definition of a derivative, ensure all algebraic simplifications are precise, especially canceling 'h' from the numerator and denominator before taking the limit.

 

Question 2. Find derivative of \( f(x) = x^{10} \) using definition.
Answer: Given the function \( f(x) = x^{10} \).
By definition, the derivative \( f'(x) \) is given by the limit:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
Substitute \( f(x) = x^{10} \). Then \( f(x+h) = (x+h)^{10} \).
\( f'(x) = \lim_{h \to 0} \frac{(x+h)^{10}-x^{10}}{h} \)
To evaluate this limit, let \( t = x+h \). As \( h \to 0 \), \( t \to x \), and \( h = t-x \).
The limit becomes:
\( f'(x) = \lim_{t \to x} \frac{t^{10}-x^{10}}{t-x} \)
Using the formula \( \lim_{a \to x} \frac{a^n - x^n}{a-x} = nx^{n-1} \), where \( n = 10 \), we get:
\( f'(x) = 10x^{10-1} \)
\( = 10x^9 \)
Therefore, the derivative of \( f(x) = x^{10} \) is \( f'(x) = 10x^9 \).
In simple words: We use the limit definition to find the derivative of \( x^{10} \). This involves a special limit rule for powers, which helps us quickly find the answer.

🎯 Exam Tip: When using the limit definition for power functions, recognize and apply the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y-a} = na^{n-1} \) for efficient calculation.

 

Question 3. Find derivative of \( \frac{2}{3+4x} \) using definition.
Answer: Let \( f(x) = \frac{2}{3+4x} \).
Then \( f(x+h) = \frac{2}{3+4(x+h)} \).
Using the definition of the derivative:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\frac{2}{3+4(x+h)} - \frac{2}{3+4x}}{h} \)
\( = \lim_{h \to 0} \frac{2(3+4x) - 2(3+4(x+h))}{h(3+4(x+h))(3+4x)} \)
\( = \lim_{h \to 0} \frac{6+8x - 6 - 8x - 8h}{h(3+4x+4h)(3+4x)} \)
\( = \lim_{h \to 0} \frac{-8h}{h(3+4x+4h)(3+4x)} \)
Cancel \( h \) from numerator and denominator:
\( = \lim_{h \to 0} \frac{-8}{(3+4x+4h)(3+4x)} \)
Now, substitute \( h = 0 \):
\( = \frac{-8}{(3+4x+4(0))(3+4x)} \)
\( = \frac{-8}{(3+4x)(3+4x)} \)
\( = \frac{-8}{(3+4x)^2} \)
Therefore, the derivative of \( f(x) = \frac{2}{3+4x} \) is \( f'(x) = \frac{-8}{(3+4x)^2} \).
In simple words: We find the derivative using the limit definition. This means we write out the formula, combine the fractions, simplify by removing 'h', and then let 'h' become zero to get the final answer.

🎯 Exam Tip: This type of problem requires careful algebraic manipulation, especially when combining fractions. Ensure you factor out and cancel 'h' correctly before evaluating the limit.

 

Question 4. y = x³ - 3x² - 3x + 80. For which value of x, \( \frac{d y}{d x} \) = -6
Answer:The given function is \( y = x^3 - 3x^2 - 3x + 80 \). First, find the derivative: \( \frac{d y}{d x} = 3x^2 - 3(2x) - 3(1) + 0 \) \( = 3x^2 - 6x - 3 \) We are given that \( \frac{d y}{d x} = -6 \). So, \( -6 = 3x^2 - 6x - 3 \) Now, move -6 to the right side: \( 3x^2 - 6x - 3 + 6 = 0 \) \( 3x^2 - 6x + 3 = 0 \) Divide by 3: \( x^2 - 2x + 1 = 0 \) This can be written as a square: \( (x - 1)^2 = 0 \) So, \( x - 1 = 0 \) \( x = 1 \) Therefore, at \( x = 1 \), the derivative of the function is -6.
In simple words: We found the derivative of the function. Then, we set it equal to -6 and solved for 'x' to find the specific value where the derivative is -6.

🎯 Exam Tip: Remember to clearly show each step of differentiation and algebraic manipulation when solving for 'x'.

 

Question 15. Find the derivative of y = 4x² + 4x + 8. For which value of x will the derivative be zero ?
Answer:Given the function \( y = 4x^2 + 4x + 8 \). First, find the first derivative \( \frac{d y}{d x} \): \( \frac{d y}{d x} = 4(2x) + 4(1) + 0 \) \( = 8x + 4 \) To find the value of x where the derivative is zero, set \( \frac{d y}{d x} = 0 \): \( 0 = 8x + 4 \) Subtract 4 from both sides: \( -8x = 4 \) Divide by -8: \( x = -\frac{4}{8} \) \( x = -\frac{1}{2} \) So, the derivative will be zero at \( x = -\frac{1}{2} \).
In simple words: First, we calculated the derivative of the given function. Then, we set this derivative to zero and solved for 'x' to find the point where the function's slope is flat.

🎯 Exam Tip: When asked to find where a derivative is zero, always remember to set the first derivative equal to zero and solve the resulting equation for x.

 

Question 16. f(x) = x³ + 5x² + 3x + 7, prove that f'(2) = 35.
Answer:Given the function \( f(x) = x^3 + 5x^2 + 3x + 7 \). First, find the first derivative \( f'(x) \): \( f'(x) = 3x^2 + 5(2x) + 3(1) + 0 \) \( = 3x^2 + 10x + 3 \) Now, substitute \( x = 2 \) into \( f'(x) \): \( f'(2) = 3(2)^2 + 10(2) + 3 \) \( = 3 \times 4 + 20 + 3 \) \( = 12 + 20 + 3 \) \( = 35 \) Thus, it is proven that \( f'(2) = 35 \).
In simple words: We first found the formula for the derivative of the function. Then, we put the value x=2 into this derivative formula to show that the result is 35, as requested.

🎯 Exam Tip: For "prove that" questions, clearly show each step of differentiation and substitution to reach the target value.

 

Question 17. If f (x) = 3x² + 3, then for which value of x, f'(x) = f(x) ?
Answer:Given the function \( f(x) = 3x^2 + 3 \). First, find the first derivative \( f'(x) \): \( f'(x) = 3(2x) + 0 \) \( = 6x \) Now, we need to find the value of x for which \( f'(x) = f(x) \). Set the derivative equal to the original function: \( 6x = 3x^2 + 3 \) Rearrange the equation to form a quadratic equation: \( 3x^2 - 6x + 3 = 0 \) Divide the entire equation by 3: \( x^2 - 2x + 1 = 0 \) This is a perfect square trinomial: \( (x - 1)^2 = 0 \) Solve for x: \( x - 1 = 0 \) \( x = 1 \) Therefore, for \( x = 1 \), \( f'(x) = f(x) \).
In simple words: We calculated the function's derivative and then set it equal to the original function. By solving this new equation, we found the value of 'x' where both are the same.

🎯 Exam Tip: Always double-check your differentiation steps before solving the equation. Recognize perfect squares to simplify solving quadratic equations.

 

Question 18. Find \( \frac{d^{2} y}{d x^{2}} \) if \( y = 2x^3 + 5x^2 - 3 + \frac{4}{x^{2}} - \frac{5}{x^{3}} \).
Answer:Given the function \( y = 2x^3 + 5x^2 - 3 + \frac{4}{x^{2}} - \frac{5}{x^{3}} \). Rewrite the terms with negative exponents for easier differentiation: \( y = 2x^3 + 5x^2 - 3 + 4x^{-2} - 5x^{-3} \) First, find the first derivative \( \frac{d y}{d x} \): \( \frac{d y}{d x} = 2(3x^2) + 5(2x) - 0 + 4(-2)x^{-2-1} - 5(-3)x^{-3-1} \) \( = 6x^2 + 10x - 8x^{-3} + 15x^{-4} \) Now, find the second derivative \( \frac{d^{2} y}{d x^{2}} \) by differentiating \( \frac{d y}{d x} \): \( \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} [6x^2 + 10x - 8x^{-3} + 15x^{-4}] \) \( = 6(2x) + 10(1) - 8(-3)x^{-3-1} + 15(-4)x^{-4-1} \) \( = 12x + 10 + 24x^{-4} - 60x^{-5} \) Rewrite with positive exponents: \( = 12x + 10 + \frac{24}{x^{4}} - \frac{60}{x^{5}} \) Therefore, \( \frac{d^{2} y}{d x^{2}} \) is \( 12x + 10 + \frac{24}{x^{4}} - \frac{60}{x^{5}} \).
In simple words: We took the derivative of the function once to get the first derivative, then took the derivative of that result again to find the second derivative, simplifying the exponents at each step.

🎯 Exam Tip: Be careful with negative exponents and their signs during differentiation. Remember to differentiate each term separately and simplify the final expression.

 

Question 19. Find \( \frac{d^{2} y}{d x^{2}} \) if \( y = \sqrt{x}+\frac{1}{\sqrt{x}} \).
Answer:Given the function \( y = \sqrt{x}+\frac{1}{\sqrt{x}} \). Rewrite in exponential form: \( y = x^{\frac{1}{2}} + x^{-\frac{1}{2}} \) First, find the first derivative \( \frac{d y}{d x} \): \( \frac{d y}{d x} = \frac{1}{2}x^{\frac{1}{2}-1} + (-\frac{1}{2})x^{-\frac{1}{2}-1} \) \( = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}} \) Now, find the second derivative \( \frac{d^{2} y}{d x^{2}} \) by differentiating \( \frac{d y}{d x} \): \( \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} [\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}] \) \( = \frac{1}{2}(-\frac{1}{2})x^{-\frac{1}{2}-1} - \frac{1}{2}(-\frac{3}{2})x^{-\frac{3}{2}-1} \) \( = -\frac{1}{4}x^{-\frac{3}{2}} + \frac{3}{4}x^{-\frac{5}{2}} \) Rewrite with positive exponents: \( = -\frac{1}{4x^{\frac{3}{2}}} + \frac{3}{4x^{\frac{5}{2}}} \) Thus, if \( y = \sqrt{x}+\frac{1}{\sqrt{x}} \), then \( \frac{d^{2} y}{d x^{2}} = -\frac{1}{4x^{\frac{3}{2}}} + \frac{3}{4x^{\frac{5}{2}}} \).
In simple words: We converted the square roots to powers, then found the first derivative and then the second derivative of the expression, making sure to handle the negative and fractional powers correctly.

🎯 Exam Tip: Always convert radical expressions to fractional exponents before differentiating. Pay close attention to the power rule and negative signs during both first and second differentiations.

 

Question 20. Obtain marginal cost if the production cost function is C = 0.0012x² - 0.18x + 25.
Answer:The production cost function is given as \( C = 0.0012x^2 - 0.18x + 25 \). Marginal cost (MC) is found by taking the first derivative of the total cost function with respect to x. \( \text{Marginal cost} = \frac{d C}{d x} \) \( = \frac{d}{d x} [0.0012x^2 - 0.18x + 25] \) Differentiate each term: \( = 0.0012(2x) - 0.18(1) + 0 \) \( = 0.0024x - 0.18 \) Therefore, if \( C = 0.0012x^2 - 0.18x + 25 \), the marginal cost obtained is \( 0.0024x - 0.18 \).
In simple words: We found the marginal cost by taking the derivative of the total cost function, which tells us how much extra cost is added when one more unit is produced.

🎯 Exam Tip: Remember that marginal cost is always the first derivative of the total cost function. Treat constants as having a derivative of zero.

 

Section D

Question 1. Find derivative of y = ax + b (a and b are constants) using definition.
Answer:Given the function \( y = ax + b \). According to the definition of a derivative, \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). Here, \( f(x) = ax + b \). So, \( f(x+h) = a(x+h) + b = ax + ah + b \). Now, substitute these into the definition: \( \frac{d y}{d x} = \lim_{h \to 0} \frac{(ax + ah + b) - (ax + b)}{h} \) \( = \lim_{h \to 0} \frac{ax + ah + b - ax - b}{h} \) \( = \lim_{h \to 0} \frac{ah}{h} \) Cancel 'h' from the numerator and denominator: \( = \lim_{h \to 0} a \) Since 'a' is a constant, the limit is 'a'. \( = a \) Hence, if \( y = ax + b \), then \( \frac{d y}{d x} = a \).
In simple words: We used the limit definition of a derivative. We plugged the function into the formula and simplified it by subtracting terms and cancelling 'h', which showed the derivative is 'a'.

🎯 Exam Tip: When using the limit definition, ensure to expand \( f(x+h) \) correctly and simplify the numerator before taking the limit as \( h \to 0 \).

 

Question 2. Find derivative of f (x) = x¹⁰ using definition.
Answer:Given the function \( f(x) = x^{10} \). Using the definition of a derivative: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). Here, \( f(x) = x^{10} \), so \( f(x+h) = (x+h)^{10} \). Substituting into the definition: \( f'(x) = \lim_{h \to 0} \frac{(x+h)^{10} - x^{10}}{h} \) Let \( t = x+h \). As \( h \to 0 \), \( t \to x \). Also, \( h = t-x \). So, the limit becomes: \( f'(x) = \lim_{t \to x} \frac{t^{10} - x^{10}}{t-x} \) This is a standard limit form: \( \lim_{a \to b} \frac{a^n - b^n}{a-b} = nb^{n-1} \). Applying this rule: \( f'(x) = 10x^{10-1} \) \( = 10x^9 \) Hence, if \( f(x) = x^{10} \), then \( f'(x) = 10x^9 \).
In simple words: We used the limit definition of a derivative and changed the variable to simplify the expression. We then applied a known limit formula to quickly find that the derivative of \( x^{10} \) is \( 10x^9 \).

🎯 Exam Tip: Recognizing the standard limit form \( \lim_{t \to x} \frac{t^n - x^n}{t-x} = nx^{n-1} \) is crucial for efficiently solving derivatives using the definition for power functions.

 

Question 3. Find derivative of \( \frac{2}{3+4 x} \) using definition.
Answer:Given the function \( f(x) = \frac{2}{3+4x} \). Using the definition of a derivative: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). Here, \( f(x+h) = \frac{2}{3+4(x+h)} \). Substitute into the definition: \( f'(x) = \lim_{h \to 0} \frac{\frac{2}{3+4(x+h)} - \frac{2}{3+4x}}{h} \) Combine the fractions in the numerator: \( = \lim_{h \to 0} \frac{\frac{2(3+4x) - 2(3+4(x+h))}{(3+4(x+h))(3+4x)}}{h} \) \( = \lim_{h \to 0} \frac{2(3+4x) - 2(3+4x+4h)}{h(3+4(x+h))(3+4x)} \) Expand the numerator: \( = \lim_{h \to 0} \frac{6+8x - 6-8x-8h}{h(3+4(x+h))(3+4x)} \) Simplify the numerator: \( = \lim_{h \to 0} \frac{-8h}{h(3+4(x+h))(3+4x)} \) Cancel 'h': \( = \lim_{h \to 0} \frac{-8}{(3+4(x+h))(3+4x)} \) Now, substitute \( h=0 \) into the expression: \( = \frac{-8}{(3+4(x+0))(3+4x)} \) \( = \frac{-8}{(3+4x)(3+4x)} \) \( = \frac{-8}{(3+4x)^2} \) Hence, if \( f(x) = \frac{2}{3+4x} \), then \( f'(x) = \frac{-8}{(3+4x)^2} \).
In simple words: We used the definition of derivative, which involves a limit. We combined the fractions in the numerator, simplified by canceling terms, and then took the limit as 'h' approached zero to get the final derivative.

🎯 Exam Tip: When dealing with rational functions and the limit definition, remember to find a common denominator in the numerator and carefully simplify before canceling 'h'.

 

Question 4. y = x³ - 3x² - 3x + 80. For which value of x, \( \frac{d y}{d x} \) = -6
Answer:Given the function \( y = x^3 - 3x^2 - 3x + 80 \). First, find the derivative \( \frac{d y}{d x} \): \( \frac{d y}{d x} = 3x^2 - 3(2x) - 3(1) + 0 \) \( = 3x^2 - 6x - 3 \) We are given that \( \frac{d y}{d x} = -6 \). Set the derivative equal to -6: \( 3x^2 - 6x - 3 = -6 \) Move -6 to the left side: \( 3x^2 - 6x - 3 + 6 = 0 \) \( 3x^2 - 6x + 3 = 0 \) Divide the entire equation by 3: \( x^2 - 2x + 1 = 0 \) This is a perfect square: \( (x - 1)^2 = 0 \) Therefore, \( x - 1 = 0 \), which means \( x = 1 \). So, at \( x = 1 \), the derivative \( \frac{d y}{d x} = -6 \).
In simple words: We calculated the derivative of the given function. Then, we set this derivative to -6 and solved the resulting equation to find the value of 'x' where this condition holds true.

🎯 Exam Tip: After differentiating, form an equation with the given derivative value. Factorizing or using the quadratic formula will help solve for x.

 

Question 5. Find f '(2) if \( f(x) = \frac{4 x^{5}+3 x^{3}+2 x^{2}+24}{x^{2}} \).
Answer:Given the function \( f(x) = \frac{4 x^{5}+3 x^{3}+2 x^{2}+24}{x^{2}} \). First, simplify the function by dividing each term in the numerator by \( x^2 \): \( f(x) = \frac{4x^5}{x^2} + \frac{3x^3}{x^2} + \frac{2x^2}{x^2} + \frac{24}{x^2} \) \( f(x) = 4x^3 + 3x + 2 + 24x^{-2} \) Now, find the first derivative \( f'(x) \): \( f'(x) = 4(3x^2) + 3(1) + 0 + 24(-2)x^{-2-1} \) \( = 12x^2 + 3 - 48x^{-3} \) Rewrite with a positive exponent: \( = 12x^2 + 3 - \frac{48}{x^3} \) Next, substitute \( x = 2 \) into \( f'(x) \): \( f'(2) = 12(2)^2 + 3 - \frac{48}{(2)^3} \) \( = 12(4) + 3 - \frac{48}{8} \) \( = 48 + 3 - 6 \) \( = 51 - 6 \) \( = 45 \) Hence, if \( f(x) = \frac{4 x^{5}+3 x^{3}+2 x^{2}+24}{x^{2}} \), then \( f'(2) = 45 \).
In simple words: We first simplified the given fraction by dividing each part. Then, we calculated the derivative of the simplified function. Finally, we put x=2 into the derivative to get the numerical answer.

🎯 Exam Tip: Always simplify rational functions before differentiating, if possible. This makes the differentiation process much simpler and reduces calculation errors.

 

Question 6. Find the derivative of \( y = (3x^2 + 4x - 2)(3x + 2) \) with respect to x.
Answer:Given the function \( y = (3x^2 + 4x - 2)(3x + 2) \). We can use the product rule for differentiation, which states that if \( y = u \cdot v \), then \( \frac{d y}{d x} = u \cdot \frac{d v}{d x} + v \cdot \frac{d u}{d x} \). Let \( u = 3x^2 + 4x - 2 \) and \( v = 3x + 2 \). Find the derivative of u with respect to x: \( \frac{d u}{d x} = \frac{d}{d x} (3x^2 + 4x - 2) = 3(2x) + 4(1) - 0 = 6x + 4 \) Find the derivative of v with respect to x: \( \frac{d v}{d x} = \frac{d}{d x} (3x + 2) = 3(1) + 0 = 3 \) Now, apply the product rule: \( \frac{d y}{d x} = (3x^2 + 4x - 2)(3) + (3x + 2)(6x + 4) \) Expand and simplify: \( = (9x^2 + 12x - 6) + (18x^2 + 12x + 12x + 8) \) \( = 9x^2 + 12x - 6 + 18x^2 + 24x + 8 \) Combine like terms: \( = (9x^2 + 18x^2) + (12x + 24x) + (-6 + 8) \) \( = 27x^2 + 36x + 2 \) Hence, for \( y = (3x^2 + 4x - 2)(3x + 2) \), \( \frac{d y}{d x} = 27x^2 + 36x + 2 \).
In simple words: We used the product rule for differentiation, which involves taking the derivative of each part of the multiplication and combining them in a specific way. After that, we expanded and simplified the expression to get the final derivative.

🎯 Exam Tip: Clearly identify 'u' and 'v' and their derivatives before applying the product rule. Be careful with algebraic expansion and combining like terms.

 

Question 7. Find \( \frac{d y}{d x} \) if \( y = \frac{a x+b}{b x+a} \) (a and b are constants).
Answer:Given the function \( y = \frac{ax+b}{bx+a} \). We use the quotient rule for differentiation, which states that if \( y = \frac{u}{v} \), then \( \frac{d y}{d x} = \frac{v \cdot \frac{d u}{d x} - u \cdot \frac{d v}{d x}}{v^2} \). Let \( u = ax+b \) and \( v = bx+a \). Find the derivative of u with respect to x: \( \frac{d u}{d x} = \frac{d}{d x} (ax+b) = a(1) + 0 = a \) Find the derivative of v with respect to x: \( \frac{d v}{d x} = \frac{d}{d x} (bx+a) = b(1) + 0 = b \) Now, apply the quotient rule: \( \frac{d y}{d x} = \frac{(bx+a)(a) - (ax+b)(b)}{(bx+a)^2} \) Expand the numerator: \( = \frac{abx + a^2 - (abx + b^2)}{(bx+a)^2} \) \( = \frac{abx + a^2 - abx - b^2}{(bx+a)^2} \) Simplify the numerator by canceling \( abx \): \( = \frac{a^2 - b^2}{(bx+a)^2} \) Hence, if \( y = \frac{ax+b}{bx+a} \), then \( \frac{d y}{d x} = \frac{a^2 - b^2}{(bx+a)^2} \).
In simple words: We used the quotient rule to find the derivative of the fraction. We differentiated the top and bottom parts separately, then put them into the rule's formula, and finally simplified the expression.

🎯 Exam Tip: Be meticulous with the order of terms in the quotient rule's numerator and the signs. A common error is mixing up \( v \frac{du}{dx} \) and \( u \frac{dv}{dx} \).

 

Question 8. Find the derivative of \( y = 1 + \frac{1}{1+\frac{1}{x}} \) with respect to x.
Answer:Given the function \( y = 1 + \frac{1}{1+\frac{1}{x}} \). First, simplify the expression for y: Start with the inner fraction: \( 1+\frac{1}{x} = \frac{x+1}{x} \) Substitute this back into y: \( y = 1 + \frac{1}{\frac{x+1}{x}} \) \( y = 1 + \frac{x}{x+1} \) Combine the terms: \( y = \frac{x+1}{x+1} + \frac{x}{x+1} \) \( y = \frac{x+1+x}{x+1} \) \( y = \frac{2x+1}{x+1} \) Now, differentiate \( y = \frac{2x+1}{x+1} \) using the quotient rule. Let \( u = 2x+1 \) and \( v = x+1 \). Find the derivative of u with respect to x: \( \frac{d u}{d x} = \frac{d}{d x} (2x+1) = 2(1) + 0 = 2 \) Find the derivative of v with respect to x: \( \frac{d v}{d x} = \frac{d}{d x} (x+1) = 1(1) + 0 = 1 \) Apply the quotient rule: \( \frac{d y}{d x} = \frac{v \cdot \frac{d u}{d x} - u \cdot \frac{d v}{d x}}{v^2} \) \( \frac{d y}{d x} = \frac{(x+1)(2) - (2x+1)(1)}{(x+1)^2} \) Expand the numerator: \( = \frac{2x+2 - (2x+1)}{(x+1)^2} \) \( = \frac{2x+2 - 2x - 1}{(x+1)^2} \) Simplify the numerator: \( = \frac{1}{(x+1)^2} \) Hence, if \( y = 1 + \frac{1}{1+\frac{1}{x}} \), then \( \frac{d y}{d x} = \frac{1}{(x+1)^2} \).
In simple words: First, we simplified the complex fraction into a simpler one. Then, we used the quotient rule to find the derivative of this simplified fraction, which gave us the final answer.

🎯 Exam Tip: Always simplify complex algebraic expressions before attempting to differentiate. This reduces the complexity of differentiation rules and minimizes errors.

 

Question 9. Find \( \frac{d y}{d x} \) if \( (2x + 3) (y + 2) = 15 \).
Answer:Given the equation \( (2x + 3) (y + 2) = 15 \). First, express y explicitly as a function of x: Divide both sides by \( (2x+3) \): \( y + 2 = \frac{15}{2x+3} \) Subtract 2 from both sides: \( y = \frac{15}{2x+3} - 2 \) Combine the terms to get a single fraction: \( y = \frac{15 - 2(2x+3)}{2x+3} \) \( y = \frac{15 - 4x - 6}{2x+3} \) \( y = \frac{9 - 4x}{2x+3} \) Now, differentiate \( y = \frac{9-4x}{2x+3} \) using the quotient rule. Let \( u = 9-4x \) and \( v = 2x+3 \). Find the derivative of u with respect to x: \( \frac{d u}{d x} = \frac{d}{d x} (9-4x) = 0 - 4(1) = -4 \) Find the derivative of v with respect to x: \( \frac{d v}{d x} = \frac{d}{d x} (2x+3) = 2(1) + 0 = 2 \) Apply the quotient rule: \( \frac{d y}{d x} = \frac{v \cdot \frac{d u}{d x} - u \cdot \frac{d v}{d x}}{v^2} \) \( \frac{d y}{d x} = \frac{(2x+3)(-4) - (9-4x)(2)}{(2x+3)^2} \) Expand the numerator: \( = \frac{-8x - 12 - (18 - 8x)}{(2x+3)^2} \) \( = \frac{-8x - 12 - 18 + 8x}{(2x+3)^2} \) Simplify the numerator: \( = \frac{-30}{(2x+3)^2} \) Hence, for \( (2x + 3) (y + 2) = 15 \), \( \frac{d y}{d x} = \frac{-30}{(2x+3)^2} \).
In simple words: First, we rearranged the given equation to isolate 'y' as a function of 'x'. Then, we used the quotient rule to find the derivative of this expression.

🎯 Exam Tip: Always solve for 'y' explicitly before differentiating implicitly or using a direct differentiation rule. Careful algebraic simplification is key to avoid errors.

 

Question 10. Find \( \frac{d y}{d x} \) if \( y = 5 + \frac{6}{7 x+8} \).
Answer:Given the function \( y = 5 + \frac{6}{7x+8} \). First, combine the terms to form a single fraction: \( y = \frac{5(7x+8)}{7x+8} + \frac{6}{7x+8} \) \( y = \frac{35x+40+6}{7x+8} \) \( y = \frac{35x+46}{7x+8} \) Now, differentiate \( y = \frac{35x+46}{7x+8} \) using the quotient rule. Let \( u = 35x+46 \) and \( v = 7x+8 \). Find the derivative of u with respect to x: \( \frac{d u}{d x} = \frac{d}{d x} (35x+46) = 35(1) + 0 = 35 \) Find the derivative of v with respect to x: \( \frac{d v}{d x} = \frac{d}{d x} (7x+8) = 7(1) + 0 = 7 \) Apply the quotient rule: \( \frac{d y}{d x} = \frac{v \cdot \frac{d u}{d x} - u \cdot \frac{d v}{d x}}{v^2} \) \( \frac{d y}{d x} = \frac{(7x+8)(35) - (35x+46)(7)}{(7x+8)^2} \) Expand the numerator: \( = \frac{245x + 280 - (245x + 322)}{(7x+8)^2} \) \( = \frac{245x + 280 - 245x - 322}{(7x+8)^2} \) Simplify the numerator: \( = \frac{-42}{(7x+8)^2} \) Hence, if \( y = 5 + \frac{6}{7x+8} \), then \( \frac{d y}{d x} = \frac{-42}{(7x+8)^2} \).
In simple words: We first combined the constant and the fraction into one single fraction. Then, we applied the quotient rule to this simplified fraction to calculate its derivative.

🎯 Exam Tip: Combining terms into a single fraction before applying the quotient rule often simplifies the differentiation process, especially when a constant is involved.

 

Question 11. Find f '(x) if \( f(x) = \sqrt{x^{2}+5} \).
Answer:Given the function \( f(x) = \sqrt{x^2+5} \). We can use the chain rule for differentiation. Let \( u = x^2+5 \). Then \( f(x) = \sqrt{u} = u^{\frac{1}{2}} \). The chain rule states: \( \frac{d f}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x} \). First, find \( \frac{d f}{d u} \): \( \frac{d f}{d u} = \frac{d}{d u} (u^{\frac{1}{2}}) = \frac{1}{2}u^{\frac{1}{2}-1} = \frac{1}{2}u^{-\frac{1}{2}} \) Next, find \( \frac{d u}{d x} \): \( \frac{d u}{d x} = \frac{d}{d x} (x^2+5) = 2x + 0 = 2x \) Now, multiply these two derivatives: \( f'(x) = (\frac{1}{2}u^{-\frac{1}{2}}) \cdot (2x) \) Substitute back \( u = x^2+5 \): \( f'(x) = \frac{1}{2}(x^2+5)^{-\frac{1}{2}} \cdot (2x) \) Simplify the expression: \( f'(x) = x(x^2+5)^{-\frac{1}{2}} \) Rewrite with a positive exponent and radical form: \( f'(x) = \frac{x}{\sqrt{x^2+5}} \) Hence, if \( f(x) = \sqrt{x^2+5} \), then \( f'(x) = \frac{x}{\sqrt{x^2+5}} \).
In simple words: We used the chain rule to find the derivative. We broke the function into an outer part (square root) and an inner part (\( x^2+5 \)), differentiated each, and multiplied the results, then simplified.

🎯 Exam Tip: When dealing with composite functions like square roots, apply the chain rule correctly. Remember to differentiate the outer function first, then multiply by the derivative of the inner function.

 

Question 12. Find the derivative of \( \left(3 x^{3}-2 x^{2}+1\right)^{\frac{5}{2}} \) with respect to x.
Answer:Given the function \( y = (3x^3 - 2x^2 + 1)^{\frac{5}{2}} \). We will use the chain rule. Let \( u = 3x^3 - 2x^2 + 1 \). Then \( y = u^{\frac{5}{2}} \). The chain rule states: \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \). First, find \( \frac{d y}{d u} \): \( \frac{d y}{d u} = \frac{d}{d u} (u^{\frac{5}{2}}) = \frac{5}{2}u^{\frac{5}{2}-1} = \frac{5}{2}u^{\frac{3}{2}} \) Next, find \( \frac{d u}{d x} \): \( \frac{d u}{d x} = \frac{d}{d x} (3x^3 - 2x^2 + 1) = 3(3x^2) - 2(2x) + 0 = 9x^2 - 4x \) Now, multiply these two derivatives: \( \frac{d y}{d x} = (\frac{5}{2}u^{\frac{3}{2}}) \cdot (9x^2 - 4x) \) Substitute back \( u = 3x^3 - 2x^2 + 1 \): \( \frac{d y}{d x} = \frac{5}{2}(3x^3 - 2x^2 + 1)^{\frac{3}{2}}(9x^2 - 4x) \) Hence, for \( y = (3x^3 - 2x^2 + 1)^{\frac{5}{2}} \), the derivative obtained is \( \frac{5}{2}(3x^3 - 2x^2 + 1)^{\frac{3}{2}}(9x^2 - 4x) \).
In simple words: We applied the chain rule. We treated the entire expression inside the power as one unit, differentiated the power first, then multiplied it by the derivative of the inner expression.

🎯 Exam Tip: For complex powers of functions, the chain rule is essential. Break down the differentiation into 'outer' and 'inner' parts and combine them correctly.

 

Question 13. Find f'(x) if \( f(x) = (x^2 + 3x + 4)^7 \).
Answer:Given the function \( f(x) = (x^2 + 3x + 4)^7 \). We will use the chain rule. Let \( u = x^2 + 3x + 4 \). Then \( f(x) = u^7 \). The chain rule states: \( \frac{d f}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x} \). First, find \( \frac{d f}{d u} \): \( \frac{d f}{d u} = \frac{d}{d u} (u^7) = 7u^{7-1} = 7u^6 \) Next, find \( \frac{d u}{d x} \): \( \frac{d u}{d x} = \frac{d}{d x} (x^2 + 3x + 4) = 2x + 3(1) + 0 = 2x + 3 \) Now, multiply these two derivatives: \( f'(x) = (7u^6) \cdot (2x+3) \) Substitute back \( u = x^2 + 3x + 4 \): \( f'(x) = 7(x^2 + 3x + 4)^6(2x+3) \) Hence, for \( f(x) = (x^2 + 3x + 4)^7 \), \( f'(x) = 7(x^2 + 3x + 4)^6(2x+3) \).
In simple words: We used the chain rule, treating the entire base of the power as 'u'. We first differentiated the power term and then multiplied it by the derivative of the base itself.

🎯 Exam Tip: Pay attention to the exponent when applying the power rule as the first step of the chain rule. Ensure the derivative of the inner function is correctly calculated and multiplied.

 

Question 14. If f(x) = 3x² + 4x + 5, then for which value of x, f '(x) = f "(x)?
Answer:Given the function \( f(x) = 3x^2 + 4x + 5 \). First, find the first derivative \( f'(x) \): \( f'(x) = 3(2x) + 4(1) + 0 \) \( = 6x + 4 \) Next, find the second derivative \( f''(x) \): \( f''(x) = \frac{d}{d x} (6x+4) \) \( = 6(1) + 0 \) \( = 6 \) Now, we need to find the value of x for which \( f'(x) = f''(x) \). Set the first derivative equal to the second derivative: \( 6x + 4 = 6 \) Subtract 4 from both sides: \( 6x = 6 - 4 \) \( 6x = 2 \) Divide by 6: \( x = \frac{2}{6} \) \( x = \frac{1}{3} \) Hence, for \( f(x) = 3x^2 + 4x + 5 \), when \( x = \frac{1}{3} \), \( f'(x) = f''(x) \).
In simple words: We calculated the first derivative and the second derivative of the function. Then, we set these two derivatives equal to each other and solved for 'x' to find where they are the same.

🎯 Exam Tip: Differentiate carefully to find both the first and second derivatives. Solving the equation \( f'(x) = f''(x) \) often leads to a simple linear equation.

 

Question 15. Find marginal revenue if demand function is \( P = \frac{2500-x^{2}}{100} \).
Answer:Given the demand function: \( P = \frac{2500-x^{2}}{100} \). Revenue (R) is calculated as the product of price (P) and quantity (x): \( R = x \cdot P \). Substitute the demand function into the revenue function: \( R = x \left(\frac{2500-x^{2}}{100}\right) \) \( R = \frac{2500x - x^3}{100} \) To find marginal revenue (MR), differentiate the total revenue function with respect to x: \( \text{Marginal revenue} = \frac{d R}{d x} \) \( = \frac{d}{d x} \left[\frac{2500x - x^3}{100}\right] \) We can factor out the constant \( \frac{1}{100} \): \( = \frac{1}{100} \frac{d}{d x} [2500x - x^3] \) Differentiate the terms inside the bracket: \( = \frac{1}{100} [2500(1) - 3x^2] \) \( = \frac{1}{100} (2500 - 3x^2) \) Distribute \( \frac{1}{100} \): \( = \frac{2500}{100} - \frac{3x^2}{100} \) \( = 25 - \frac{3x^2}{100} \) Hence, the marginal revenue obtained is \( 25 - \frac{3x^2}{100} \).
In simple words: We first found the total revenue by multiplying the demand function by 'x'. Then, we took the derivative of this total revenue function to get the marginal revenue, which shows the change in revenue from selling one more unit.

🎯 Exam Tip: Always derive the total revenue function (R=Px) first before calculating marginal revenue. Remember to differentiate each term carefully.

 

Question 16. Determine whether the function y = 3x² - x + 7 is increasing or decreasing at x = 1 and x = 2.
Answer:Given the function \( y = 3x^2 - x + 7 \). To determine if the function is increasing or decreasing, we need to find its first derivative, \( \frac{d y}{d x} \). \( \frac{d y}{d x} = \frac{d}{d x} (3x^2 - x + 7) \) \( = 3(2x) - 1(1) + 0 \) \( = 6x - 1 \) Now, evaluate the derivative at the given points: At \( x = 1 \): \( \frac{d y}{d x} = 6(1) - 1 = 6 - 1 = 5 \) Since \( \frac{d y}{d x} = 5 > 0 \), the function is increasing at \( x = 1 \). At \( x = 2 \): \( \frac{d y}{d x} = 6(2) - 1 = 12 - 1 = 11 \) Since \( \frac{d y}{d x} = 11 > 0 \), the function is increasing at \( x = 2 \).
In simple words: We calculated the derivative of the function, which tells us its slope. Then, we checked the sign of this slope at x=1 and x=2. A positive slope means the function is going up (increasing).

🎯 Exam Tip: A positive first derivative means the function is increasing, while a negative first derivative means it is decreasing. If the derivative is zero, it's a stationary point.

 

Question 17. Determine whether the function y = 2x³ - 7x² - 11x + 5 is increasing or decreasing at x = \( \frac{1}{2} \) and x = 3.
Answer:Given the function \( y = 2x^3 - 7x^2 - 11x + 5 \). To determine if the function is increasing or decreasing, we need to find its first derivative, \( \frac{d y}{d x} \). \( \frac{d y}{d x} = \frac{d}{d x} (2x^3 - 7x^2 - 11x + 5) \) \( = 2(3x^2) - 7(2x) - 11(1) + 0 \) \( = 6x^2 - 14x - 11 \) Now, evaluate the derivative at the given points: At \( x = \frac{1}{2} \): \( \frac{d y}{d x} = 6\left(\frac{1}{2}\right)^2 - 14\left(\frac{1}{2}\right) - 11 \) \( = 6\left(\frac{1}{4}\right) - 7 - 11 \) \( = \frac{6}{4} - 18 \) \( = \frac{3}{2} - 18 \) \( = \frac{3 - 36}{2} = -\frac{33}{2} \) Since \( \frac{d y}{d x} = -\frac{33}{2} < 0 \), the function is decreasing at \( x = \frac{1}{2} \). At \( x = 3 \): \( \frac{d y}{d x} = 6(3)^2 - 14(3) - 11 \) \( = 6(9) - 42 - 11 \) \( = 54 - 42 - 11 \) \( = 12 - 11 \) \( = 1 \) Since \( \frac{d y}{d x} = 1 > 0 \), the function is increasing at \( x = 3 \).
In simple words: We found the derivative of the function to see its rate of change. Then, we checked the derivative's value at two different 'x' points. If the value was negative, the function was decreasing; if positive, it was increasing.

🎯 Exam Tip: Carefully substitute fractional values into the derivative. Pay close attention to fraction arithmetic and the rules of signs to correctly determine if the function is increasing or decreasing.

 

Question 18. Determine whether the function y = 3 + 2x - 7x² is increasing or decreasing at x = -4 and x = 4.
Answer:Given the function \( y = 3 + 2x - 7x^2 \). To determine if the function is increasing or decreasing, we need to find its first derivative, \( \frac{d y}{d x} \). \( \frac{d y}{d x} = \frac{d}{d x} (3 + 2x - 7x^2) \) \( = 0 + 2(1) - 7(2x) \) \( = 2 - 14x \) Now, evaluate the derivative at the given points: At \( x = -4 \): \( \frac{d y}{d x} = 2 - 14(-4) \) \( = 2 + 56 \) \( = 58 \) Since \( \frac{d y}{d x} = 58 > 0 \), the function is increasing at \( x = -4 \). At \( x = 4 \): \( \frac{d y}{d x} = 2 - 14(4) \) \( = 2 - 56 \) \( = -54 \) Since \( \frac{d y}{d x} = -54 < 0 \), the function is decreasing at \( x = 4 \).
In simple words: We calculated the derivative to see how the function changes. By checking if the derivative was positive or negative at x=-4 and x=4, we found out if the function was going up or down at those points.

🎯 Exam Tip: Be cautious with negative numbers in substitutions, as a missed sign can lead to an incorrect determination of increasing/decreasing behavior.

 

Question 19. Production cost of a factory producing sugar is C = \( \frac{x^{2}}{10} \) + 5x + 200. Find the marginal cost if the production is 100 units and interpret it.
Answer:Given the production cost function: \( C = \frac{x^2}{10} + 5x + 200 \). Marginal cost (MC) is the first derivative of the total cost function with respect to x: \( \text{Marginal cost} = \frac{d C}{d x} \) \( = \frac{d}{d x} \left[\frac{x^2}{10} + 5x + 200\right] \) Differentiate each term: \( = \frac{1}{10}(2x) + 5(1) + 0 \) \( = \frac{2x}{10} + 5 \) \( = \frac{x}{5} + 5 \) Now, find the marginal cost when production \( x = 100 \) units: \( \text{Marginal cost} = \frac{100}{5} + 5 \) \( = 20 + 5 \) \( = 25 \) Rs.
In simple words: We calculated the marginal cost by taking the derivative of the total cost function. Then, we found that producing the 101st unit of sugar would cost an additional Rs. 25, meaning the cost for an extra unit is Rs. 25.

🎯 Exam Tip: Remember to calculate the derivative first to find the marginal cost function. The interpretation of marginal cost means the approximate additional cost of producing one more unit (e.g., the (x+1)th unit).

 

Question 20. The cost of producing x units of a commodity is C = 50 + 2x + \( \sqrt{x} \). Find the marginal cost if the production is 100 units and interpret it.
Answer:Given the cost function: \( C = 50 + 2x + \sqrt{x} \). Rewrite \( \sqrt{x} \) as \( x^{\frac{1}{2}} \): \( C = 50 + 2x + x^{\frac{1}{2}} \) Marginal cost (MC) is the first derivative of the total cost function with respect to x: \( \text{Marginal cost} = \frac{d C}{d x} \) \( = \frac{d}{d x} [50 + 2x + x^{\frac{1}{2}}] \) Differentiate each term: \( = 0 + 2(1) + \frac{1}{2}x^{\frac{1}{2}-1} \) \( = 2 + \frac{1}{2}x^{-\frac{1}{2}} \) Rewrite with a positive exponent and radical form: \( = 2 + \frac{1}{2\sqrt{x}} \) Now, find the marginal cost when production \( x = 100 \) units: \( \text{Marginal cost} = 2 + \frac{1}{2\sqrt{100}} \) \( = 2 + \frac{1}{2 \times 10} \) \( = 2 + \frac{1}{20} \) \( = 2 + 0.05 \) \( = 2.05 \) Rs.
In simple words: We found the marginal cost by taking the derivative of the total cost function, including the square root term. Then, we calculated its value when 100 units are produced, which tells us the additional cost to produce the 101st unit is Rs. 2.05.

🎯 Exam Tip: Remember to convert square roots to fractional exponents for differentiation. Accurate calculation of marginal cost at a specific production level is essential for interpretation.

 

Question 21. State the method of obtaining maximum or minimum value of a function.
Answer:To find the maximum or minimum value of a function \( y = f(x) \), follow these steps:
(i) Find the first derivative, \( \frac{d y}{d x} = f'(x) \), for the given function.
(ii) Set the first derivative equal to zero, \( \frac{d y}{d x} = 0 \), and solve the equation to find the values of x. These values are called stationary points or critical points of the function.
(iii) Find the second derivative, \( \frac{d^{2} y}{d x^{2}} = f''(x) \).
(iv) Evaluate the second derivative at each stationary value of x obtained in step (ii).
(v) If at a stationary value of x, \( \frac{d^{2} y}{d x^{2}} > 0 \) (Positive), then that value of x gives the minimum value of the function. Substitute this x-value into the original function \( f(x) \) to find the minimum value.
(vi) If at a stationary value of x, \( \frac{d^{2} y}{d x^{2}} < 0 \) (Negative), then that value of x gives the maximum value of the function. Substitute this x-value into the original function \( f(x) \) to find the maximum value.
In simple words: To find the highest or lowest points of a function, we first find where its slope is zero by taking the first derivative. Then, we use the second derivative to check if these points are peaks (maximum) or valleys (minimum).

🎯 Exam Tip: Clearly differentiate between the conditions for maximum (\( f''(x) < 0 \)) and minimum (\( f''(x) > 0 \)). Remember to substitute the 'x' values back into the *original* function for the actual max/min *values*.

 

Section E

Question 1. Give working rules for differentiation.
Answer:The main rules for differentiation, assuming u and v are differentiable functions of x, are:
(i) **Rule of Addition and Subtraction:** If \( y = u \pm v \), then \( \frac{d y}{d x} = \frac{d u}{d x} \pm \frac{d v}{d x} \).
(ii) **Rule of Multiplication (Product Rule):** If \( y = u \cdot v \), then \( \frac{d y}{d x} = u \cdot \frac{d v}{d x} + v \cdot \frac{d u}{d x} \).
(iii) **Rule of Division (Quotient Rule):** If \( y = \frac{u}{v} \) and \( v \neq 0 \), then \( \frac{d y}{d x} = \frac{v \cdot \frac{d u}{d x} - u \cdot \frac{d v}{d x}}{v^2} \).
(iv) **Chain Rule:** If y is a function of u, and u is a function of x, then \( \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} \).
In simple words: These are the basic rules for finding derivatives. They tell us how to differentiate when functions are added or subtracted, multiplied, divided, or when one function is inside another.

🎯 Exam Tip: Memorize these fundamental rules of differentiation thoroughly. Understanding their application is crucial for solving any differentiation problem.

 

Question 2. How can it be decided using derivative that the function is increasing or decreasing at a point?
Answer:Suppose we have a function \( y = f(x) \). To determine if \( f(x) \) is increasing or decreasing at a specific point \( x = a \), we use its first derivative, \( f'(x) \).
(i) First, calculate the first derivative of the function, \( f'(x) \).
(ii) Evaluate \( f'(x) \) at the point \( x = a \), which gives \( f'(a) \).
(iii) If \( f'(a) > 0 \) (a positive value), it means the function is increasing at \( x = a \). This implies that if h is a very small positive number, then \( f(a+h) > f(a) \) and \( f(a) > f(a-h) \).
(iv) If \( f'(a) < 0 \) (a negative value), it means the function is decreasing at \( x = a \). This implies that if h is a very small positive number, then \( f(a+h) < f(a) \) and \( f(a) < f(a-h) \).
In simple words: To know if a function is going up or down at a point, we find its derivative at that point. If the derivative is positive, the function is going up; if it's negative, the function is going down.

🎯 Exam Tip: The sign of the first derivative (positive or negative) directly indicates whether a function is increasing or decreasing at a particular point. This is a core concept in calculus applications.

 

Question 3. What is maximum value of a function ? State the conditions for maximum value.
Answer:The maximum value of a function \( y = f(x) \) at a point \( x = a \) is the highest value the function reaches in a small interval around \( x = a \). This means that for a very small positive number h, \( f(a) > f(a+h) \) and \( f(a) > f(a-h) \). The conditions required for a function \( f(x) \) to have a maximum value at \( x = a \) are:
(i) **Necessary condition:** The first derivative of the function at \( x = a \) must be zero. \( f'(a) = 0 \)
(ii) **Sufficient condition:** The second derivative of the function at \( x = a \) must be negative. \( f''(a) < 0 \) (Negative)
In simple words: The maximum value is the highest point a function reaches. To find it, the function's slope must be zero, and its second derivative must be negative, meaning it's curving downwards at that point.

🎯 Exam Tip: Remember both the necessary (first derivative zero) and sufficient (second derivative negative) conditions for a maximum. Applying only one can lead to incorrect conclusions.

 

Question 4. What is minimum value of a function ? State the conditions for minimum value.
Answer:The minimum value of a function \( y = f(x) \) at a point \( x = a \) is the lowest value the function reaches in a small interval around \( x = a \). This means that for a very small positive number h, \( f(a) < f(a+h) \) and \( f(a) < f(a-h) \). The conditions required for a function \( f(x) \) to have a minimum value at \( x = a \) are:
(i) **Necessary condition:** The first derivative of the function at \( x = a \) must be zero. \( f'(a) = 0 \)
(ii) **Sufficient condition:** The second derivative of the function at \( x = a \) must be positive. \( f''(a) > 0 \) (Positive)
In simple words: The minimum value is the lowest point a function reaches. To find it, the function's slope must be zero, and its second derivative must be positive, meaning it's curving upwards at that point.

🎯 Exam Tip: A key difference from maximum conditions is that for a minimum, the second derivative must be *positive*. Confusion between positive and negative second derivative conditions is a common mistake.

 

Question 5. In a factory, production cost per hundred tons of steel is \( \frac{1}{10}x^3 - 4x^2 + 50x + 300 \). Determine the production for minimum cost.
Answer:Given the production cost function: \( C = \frac{1}{10}x^3 - 4x^2 + 50x + 300 \). To find the production level for minimum cost, we use the conditions for a minimum:
(i) **Necessary condition:** Set the first derivative to zero. First, find the first derivative \( \frac{d C}{d x} \): \( \frac{d C}{d x} = \frac{d}{d x} [\frac{1}{10}x^3 - 4x^2 + 50x + 300] \) \( = \frac{1}{10}(3x^2) - 4(2x) + 50(1) + 0 \) \( = \frac{3x^2}{10} - 8x + 50 \) Set \( \frac{d C}{d x} = 0 \): \( \frac{3x^2}{10} - 8x + 50 = 0 \) Multiply by 10 to clear the fraction: \( 3x^2 - 80x + 500 = 0 \) Factorize the quadratic equation: We need two numbers that multiply to \( 3 \times 500 = 1500 \) and add up to -80. These numbers are -30 and -50. \( 3x^2 - 30x - 50x + 500 = 0 \) Factor by grouping: \( 3x(x - 10) - 50(x - 10) = 0 \) \( (x - 10)(3x - 50) = 0 \) This gives two possible values for x: \( x - 10 = 0 \implies x = 10 \) or \( 3x - 50 = 0 \implies 3x = 50 \implies x = \frac{50}{3} \)
(ii) **Sufficient condition:** Check the sign of the second derivative. Find the second derivative \( \frac{d^{2} C}{d x^{2}} \): \( \frac{d^{2} C}{d x^{2}} = \frac{d}{d x} [\frac{3x^2}{10} - 8x + 50] \) \( = \frac{3}{10}(2x) - 8(1) + 0 \) \( = \frac{6x}{10} - 8 \) \( = \frac{3x}{5} - 8 \) Now, evaluate \( \frac{d^{2} C}{d x^{2}} \) at each value of x: At \( x = 10 \): \( \frac{d^{2} C}{d x^{2}} = \frac{3(10)}{5} - 8 = \frac{30}{5} - 8 = 6 - 8 = -2 \) Since \( \frac{d^{2} C}{d x^{2}} = -2 < 0 \), this point corresponds to a maximum cost. At \( x = \frac{50}{3} \): \( \frac{d^{2} C}{d x^{2}} = \frac{3(\frac{50}{3})}{5} - 8 = \frac{50}{5} - 8 = 10 - 8 = 2 \) Since \( \frac{d^{2} C}{d x^{2}} = 2 > 0 \), this point corresponds to a minimum cost. Therefore, the production for minimum cost is \( x = \frac{50}{3} \) hundred tons.
In simple words: To find the production level for the lowest cost, we first found the point where the cost's rate of change is zero. Then, using the second derivative, we confirmed that this point is indeed a minimum, indicating the optimal production amount.

🎯 Exam Tip: Always verify both stationary points using the second derivative test to correctly distinguish between maximum and minimum costs. Factorization of quadratic equations is a critical skill here.

Question 6.
The cost of producing x units of an item is \(C = 1000 + 8x + \frac{5000}{x}\). What should be the production for minimum cost? Also find the minimum cost.
Answer:
The cost of making \(x\) units is given by the function \(C = 1000 + 8x + \frac{5000}{x}\).
We can write this as \(C = 1000 + 8x + 5000x^{-1}\).
To find the minimum cost, we first calculate the first derivative of \(C\) with respect to \(x\):
\( \frac{dC}{dx} = \frac{d}{dx}(1000 + 8x + 5000x^{-1}) \)
\( \frac{dC}{dx} = 0 + 8 + (-1)5000x^{-1-1} \)
\( \frac{dC}{dx} = 8 - 5000x^{-2} \)
To find the points where the cost might be minimum, we set the first derivative to zero (Necessary condition):
\( \frac{dC}{dx} = 0 \)
\( 8 - 5000x^{-2} = 0 \)
\( 8 = 5000x^{-2} \)
\( 8 = \frac{5000}{x^2} \)

\( 8x^2 = 5000 \)
\( x^2 = \frac{5000}{8} \)
\( x^2 = 625 \)
\( x = \pm\sqrt{625} \)
\( x = 25 \) OR \( x = -25 \)
Since production units cannot be negative, we take \(x = 25\).
Next, we check the second derivative to ensure it's a minimum (Sufficient condition):
\( \frac{d^2C}{dx^2} = \frac{d}{dx}(8 - 5000x^{-2}) \)
\( \frac{d^2C}{dx^2} = 0 - (-2)5000x^{-2-1} \)
\( \frac{d^2C}{dx^2} = 10000x^{-3} \)
\( \frac{d^2C}{dx^2} = \frac{10000}{x^3} \)
Now, substitute \(x = 25\) into the second derivative:
\( \frac{d^2C}{dx^2} \text{ at } x=25 = \frac{10000}{(25)^3} = \frac{10000}{15625} \)
Since \( \frac{10000}{15625} > 0 \), the cost is indeed minimum at \(x = 25\) units.
To find the minimum cost, substitute \(x = 25\) into the original cost function:
\( C_{min.} = 1000 + 8(25) + \frac{5000}{25} \)
\( C_{min.} = 1000 + 200 + 200 \)
\( C_{min.} = 1400 \)
So, the production should be 25 units for minimum cost, and the minimum cost is Rs. 1400.
In simple words: To find the lowest cost, we first find when the rate of cost change is zero. This tells us the number of items to make. Then, we check if this number actually gives the lowest cost, and finally, we calculate that lowest cost.

🎯 Exam Tip: Remember to always check both the first and second derivative conditions when finding maximum or minimum values. Incorrectly identifying a maximum as a minimum (or vice versa) is a common error.

Question 7.
Production cost function of a commodity is \(C = 1500 + 0.05x - 2\sqrt{x}\). Prove that production is minimum when 400 units are produced.
Answer:
The production cost function is given by \(C = 1500 + 0.05x - 2\sqrt{x}\).
We can rewrite \(\sqrt{x}\) as \(x^{\frac{1}{2}}\). So, \(C = 1500 + 0.05x - 2x^{\frac{1}{2}}\).
To find the minimum cost, we first find the first derivative of \(C\) with respect to \(x\):
\( \frac{dC}{dx} = \frac{d}{dx}(1500 + 0.05x - 2x^{\frac{1}{2}}) \)
\( \frac{dC}{dx} = 0 + 0.05(1) - 2(\frac{1}{2})x^{\frac{1}{2}-1} \)
\( \frac{dC}{dx} = 0.05 - x^{-\frac{1}{2}} \)
To find the production level for minimum cost, we set the first derivative to zero (Necessary condition):
\( \frac{dC}{dx} = 0 \)
\( 0.05 - x^{-\frac{1}{2}} = 0 \)
\( 0.05 = x^{-\frac{1}{2}} \)
\( 0.05 = \frac{1}{\sqrt{x}} \)
To solve for \(x\), we square both sides:
\( (0.05)^2 = (\frac{1}{\sqrt{x}})^2 \)
\( 0.0025 = \frac{1}{x} \)
\( x = \frac{1}{0.0025} \)
\( x = 400 \)
So, 400 units should be produced.
Next, we verify this is a minimum by checking the second derivative (Sufficient condition):
\( \frac{d^2C}{dx^2} = \frac{d}{dx}(0.05 - x^{-\frac{1}{2}}) \)
\( \frac{d^2C}{dx^2} = 0 - (-\frac{1}{2})x^{-\frac{1}{2}-1} \)
\( \frac{d^2C}{dx^2} = \frac{1}{2}x^{-\frac{3}{2}} \)
\( \frac{d^2C}{dx^2} = \frac{1}{2x^{\frac{3}{2}}} \)
Now, substitute \(x = 400\) into the second derivative:
\( \frac{d^2C}{dx^2} \text{ at } x=400 = \frac{1}{2(400)^{\frac{3}{2}}} \)
Since \(400 = 20^2\), \(400^{\frac{3}{2}} = (20^2)^{\frac{3}{2}} = 20^3 = 8000\).
\( \frac{d^2C}{dx^2} \text{ at } x=400 = \frac{1}{2(8000)} = \frac{1}{16000} \)
Since \( \frac{1}{16000} > 0 \), the production cost is minimum when 400 units are produced. This proves the statement.
In simple words: We found the production level where the cost stops changing, which is 400 units. Then, we checked a second time to make sure this point gives the absolute lowest cost, confirming that making 400 units is indeed the best for minimum cost.

🎯 Exam Tip: When proving a minimum, make sure to explicitly state both the necessary condition (first derivative equals zero) and the sufficient condition (second derivative is positive) and show the calculations for both at the specific production level.

Question 8.
The demand function of an item is \(p = 30 - \frac{x^{2}}{10}\). Find the demand and price for maximum revenue.
Answer:
The demand function is \(p = 30 - \frac{x^{2}}{10}\).
Revenue (\(R\)) is calculated as price (\(p\)) multiplied by demand (\(x\)): \(R = x \cdot p\).
Substitute the demand function into the revenue function:
\( R = x(30 - \frac{x^{2}}{10}) \)
\( R = 30x - \frac{x^{3}}{10} \)
To find the maximum revenue, we first calculate the first derivative of \(R\) with respect to \(x\):
\( \frac{dR}{dx} = \frac{d}{dx}(30x - \frac{x^{3}}{10}) \)
\( \frac{dR}{dx} = 30(1) - \frac{1}{10}(3x^2) \)
\( \frac{dR}{dx} = 30 - \frac{3x^2}{10} \)
To find the demand for maximum revenue, we set the first derivative to zero (Necessary condition):
\( \frac{dR}{dx} = 0 \)
\( 30 - \frac{3x^2}{10} = 0 \)
\( 30 = \frac{3x^2}{10} \)
\( 300 = 3x^2 \)
\( x^2 = \frac{300}{3} \)
\( x^2 = 100 \)
\( x = \pm\sqrt{100} \)
\( x = 10 \) OR \( x = -10 \)
Since demand cannot be negative, we take \(x = 10\) units.
Next, we check the second derivative to ensure it's a maximum (Sufficient condition):
\( \frac{d^2R}{dx^2} = \frac{d}{dx}(30 - \frac{3x^2}{10}) \)
\( \frac{d^2R}{dx^2} = 0 - \frac{3}{10}(2x) \)
\( \frac{d^2R}{dx^2} = -\frac{6x}{10} = -\frac{3x}{5} \)
Now, substitute \(x = 10\) into the second derivative:
\( \frac{d^2R}{dx^2} \text{ at } x=10 = -\frac{3(10)}{5} = -\frac{30}{5} = -6 \)
Since \( -6 < 0 \), the revenue is maximum when the demand \(x = 10\) units.
To find the price for maximum revenue, substitute \(x = 10\) into the demand function:
\( p = 30 - \frac{x^{2}}{10} \)
\( p = 30 - \frac{(10)^{2}}{10} \)
\( p = 30 - \frac{100}{10} \)
\( p = 30 - 10 \)
\( p = 20 \)
Thus, for maximum revenue, the demand is 10 units and the price is Rs. 20.
The maximum revenue will be:
\( R_{max.} = 30(10) - \frac{(10)^3}{10} = 300 - \frac{1000}{10} = 300 - 100 = 200 \)
In simple words: We found the number of items to sell (demand) that makes the total money earned (revenue) the highest. We did this by finding when the rate of revenue change is zero, checking it's a peak, and then finding the best selling price for that demand.

🎯 Exam Tip: Always remember to find both the demand (\(x\)) and the price (\(p\)) when asked for maximum revenue, as the price is dependent on the demand function.

Question 9.
In a market, demand law of rice is \(x = 3(60 - p)\). Find the demand for maximum revenue. Also find the price and revenue for that demand.
Answer:
The demand law for rice is \(x = 3(60 - p)\).
First, we need to express price (\(p\)) in terms of demand (\(x\)).
\( \frac{x}{3} = 60 - p \)

\( p = 60 - \frac{x}{3} \)
Revenue (\(R\)) is price (\(p\)) multiplied by quantity demanded (\(x\)): \(R = x \cdot p\).
Substitute the expression for \(p\) into the revenue function:
\( R = x(60 - \frac{x}{3}) \)
\( R = 60x - \frac{x^2}{3} \)
To find the demand for maximum revenue, we calculate the first derivative of \(R\) with respect to \(x\):
\( \frac{dR}{dx} = \frac{d}{dx}(60x - \frac{x^2}{3}) \)
\( \frac{dR}{dx} = 60(1) - \frac{1}{3}(2x) \)
\( \frac{dR}{dx} = 60 - \frac{2x}{3} \)
Set the first derivative to zero to find critical points (Necessary condition):
\( \frac{dR}{dx} = 0 \)
\( 60 - \frac{2x}{3} = 0 \)
\( 60 = \frac{2x}{3} \)
\( 180 = 2x \)
\( x = 90 \)
So, the demand for maximum revenue is 90 units.
Next, we check the second derivative to confirm it's a maximum (Sufficient condition):
\( \frac{d^2R}{dx^2} = \frac{d}{dx}(60 - \frac{2x}{3}) \)
\( \frac{d^2R}{dx^2} = 0 - \frac{2}{3}(1) \)
\( \frac{d^2R}{dx^2} = -\frac{2}{3} \)
Since \( -\frac{2}{3} < 0 \), the revenue is maximum when demand \(x = 90\) units.
Now, find the price corresponding to this demand:
\( p = 60 - \frac{x}{3} \)
\( p = 60 - \frac{90}{3} \)
\( p = 60 - 30 \)
\( p = 30 \)
The price for maximum revenue is Rs. 30.
Finally, calculate the maximum revenue:
\( R_{max.} = x \cdot p \)
\( R_{max.} = 90 \cdot 30 = 2700 \)
So, when demand is 90 units and price is Rs. 30, the maximum revenue obtained is Rs. 2700.
In simple words: We changed the demand rule to find the price based on items sold. Then, we calculated the total money earned and found the number of items that would make this money highest. We also found the price for that quantity and the maximum money earned.

🎯 Exam Tip: When the demand function is given in terms of price (\(x = f(p)\)), always rearrange it to express price in terms of quantity (\(p = f(x)\)) before forming the revenue function (\(R = x \cdot p\)).

Question 10.
If the demand function is \(p = 75 - \frac{x^{2}}{2500}\), then at which demand is revenue maximum? Also find the price for maximum revenue.
Answer:
The demand function is \(p = 75 - \frac{x^{2}}{2500}\).
Revenue (\(R\)) is given by \(R = x \cdot p\).
Substitute the demand function into the revenue function:
\( R = x(75 - \frac{x^{2}}{2500}) \)
\( R = 75x - \frac{x^{3}}{2500} \)
To find the demand for maximum revenue, we calculate the first derivative of \(R\) with respect to \(x\):
\( \frac{dR}{dx} = \frac{d}{dx}(75x - \frac{x^{3}}{2500}) \)
\( \frac{dR}{dx} = 75(1) - \frac{1}{2500}(3x^2) \)
\( \frac{dR}{dx} = 75 - \frac{3x^2}{2500} \)
Set the first derivative to zero (Necessary condition):
\( \frac{dR}{dx} = 0 \)
\( 75 - \frac{3x^2}{2500} = 0 \)
\( 75 = \frac{3x^2}{2500} \)
\( 75 \cdot 2500 = 3x^2 \)
\( 187500 = 3x^2 \)
\( x^2 = \frac{187500}{3} \)
\( x^2 = 62500 \)
\( x = \pm\sqrt{62500} \)
\( x = 250 \) OR \( x = -250 \)
Since demand cannot be negative, we choose \(x = 250\) units.
Next, we verify this is a maximum by checking the second derivative (Sufficient condition):
\( \frac{d^2R}{dx^2} = \frac{d}{dx}(75 - \frac{3x^2}{2500}) \)
\( \frac{d^2R}{dx^2} = 0 - \frac{3}{2500}(2x) \)
\( \frac{d^2R}{dx^2} = -\frac{6x}{2500} \)
Now, substitute \(x = 250\) into the second derivative:
\( \frac{d^2R}{dx^2} \text{ at } x=250 = -\frac{6(250)}{2500} = -\frac{1500}{2500} = -\frac{6}{10} < 0 \)
Since the second derivative is negative, the revenue is maximum at \(x = 250\) units.
To find the price for maximum revenue, substitute \(x = 250\) into the demand function:
\( p = 75 - \frac{x^{2}}{2500} \)
\( p = 75 - \frac{(250)^{2}}{2500} \)
\( p = 75 - \frac{62500}{2500} \)
\( p = 75 - 25 \)
\( p = 50 \)
Thus, when demand is 250 units and the price is Rs. 50, the maximum revenue is achieved.
In simple words: We used the given demand function to build a revenue function. By checking when the revenue stops changing and then checking again to ensure it's a peak, we found that selling 250 units at Rs. 50 per unit gives the highest total earnings.

🎯 Exam Tip: Always clearly state the optimal demand and price separately after calculations, as the question specifically asks for both. Showing the work for the second derivative is crucial for full marks.

Question 11.
The profit function of a producer is \(P = 40x + 10000 - 0.1x^2\). At what production is the profit maximum? Also find this maximum profit.
Answer:
The profit function is given as \(P = 40x + 10000 - 0.1x^2\).
To find the production level for maximum profit, we first calculate the first derivative of \(P\) with respect to \(x\):
\( \frac{dP}{dx} = \frac{d}{dx}(40x + 10000 - 0.1x^2) \)
\( \frac{dP}{dx} = 40(1) + 0 - 0.1(2x) \)
\( \frac{dP}{dx} = 40 - 0.2x \)
Set the first derivative to zero (Necessary condition):
\( \frac{dP}{dx} = 0 \)
\( 40 - 0.2x = 0 \)
\( 40 = 0.2x \)
\( x = \frac{40}{0.2} \)
\( x = 200 \)
So, the production for maximum profit is 200 units.
Next, we check the second derivative to confirm it's a maximum (Sufficient condition):
\( \frac{d^2P}{dx^2} = \frac{d}{dx}(40 - 0.2x) \)
\( \frac{d^2P}{dx^2} = 0 - 0.2(1) \)
\( \frac{d^2P}{dx^2} = -0.2 \)
Since \( -0.2 < 0 \), the profit is indeed maximum at \(x = 200\) units.
To find the maximum profit, substitute \(x = 200\) into the original profit function:
\( P_{max.} = 40(200) + 10000 - 0.1(200)^2 \)
\( P_{max.} = 8000 + 10000 - 0.1(40000) \)
\( P_{max.} = 18000 - 4000 \)
\( P_{max.} = 14000 \)
Therefore, when 200 units are produced, the maximum profit obtained is Rs. 14000.
In simple words: We first found the production level where the profit stops changing, which is 200 units. After confirming this gives the highest profit using a second check, we plugged this number back into the profit formula to find the largest possible profit amount.

🎯 Exam Tip: Always remember to calculate both the optimal production level and the resulting maximum profit, as typically both are requested in such questions. Double-check your arithmetic, especially with decimals and squares.

Question 12.
The profit function of a merchant is \(P = 5x - 100 - 0.01x^2\). How many units should be produced for maximum profit?
Answer:
The profit function is given by \(P = 5x - 100 - 0.01x^2\).
To find the number of units for maximum profit, we first calculate the first derivative of \(P\) with respect to \(x\):
\( \frac{dP}{dx} = \frac{d}{dx}(5x - 100 - 0.01x^2) \)
\( \frac{dP}{dx} = 5(1) - 0 - 0.01(2x) \)
\( \frac{dP}{dx} = 5 - 0.02x \)
Set the first derivative to zero (Necessary condition):
\( \frac{dP}{dx} = 0 \)
\( 5 - 0.02x = 0 \)
\( 5 = 0.02x \)
\( x = \frac{5}{0.02} \)
\( x = 250 \)
So, 250 units should be produced for potential maximum profit.
Next, we check the second derivative to confirm it's a maximum (Sufficient condition):
\( \frac{d^2P}{dx^2} = \frac{d}{dx}(5 - 0.02x) \)
\( \frac{d^2P}{dx^2} = 0 - 0.02(1) \)
\( \frac{d^2P}{dx^2} = -0.02 \)
Since \( -0.02 < 0 \), the profit is maximum when 250 units are produced.
In simple words: To find the number of items that generate the most profit, we first calculated when the profit growth stops. This told us to produce 250 units. We then verified that this quantity indeed leads to the highest possible profit.

🎯 Exam Tip: For profit maximization problems, ensure you correctly differentiate the profit function and remember that a negative second derivative confirms a maximum point.

Section F

Question 1.
Find the values of \(x\) which maximize or minimize \(y = 2x^3 - 15x^2 + 36x + 12\). Also find the maximum and minimum values of \(y\).
Answer:
The given function is \(y = 2x^3 - 15x^2 + 36x + 12\).
First, find the first derivative of \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 12) \)
\( \frac{dy}{dx} = 2(3x^2) - 15(2x) + 36(1) + 0 \)
\( \frac{dy}{dx} = 6x^2 - 30x + 36 \)
Next, find the second derivative of \(y\) with respect to \(x\):
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(6x^2 - 30x + 36) \)
\( \frac{d^2y}{dx^2} = 6(2x) - 30(1) + 0 \)
\( \frac{d^2y}{dx^2} = 12x - 30 \)
For maximum or minimum values of \(y\), set the first derivative to zero:
\( \frac{dy}{dx} = 0 \)
\( 6x^2 - 30x + 36 = 0 \)
Divide by 6:
\( x^2 - 5x + 6 = 0 \)
Factor the quadratic equation:
\( x^2 - 3x - 2x + 6 = 0 \)
\( x(x - 3) - 2(x - 3) = 0 \)
\( (x - 3)(x - 2) = 0 \)
This gives two critical points: \(x = 3\) OR \(x = 2\).
Now, use the second derivative to determine if these points are maximum or minimum:
At \(x = 3\):
\( \frac{d^2y}{dx^2} \text{ at } x=3 = 12(3) - 30 = 36 - 30 = 6 \)
Since \(6 > 0\), \(y\) will be minimum at \(x = 3\).
At \(x = 2\):
\( \frac{d^2y}{dx^2} \text{ at } x=2 = 12(2) - 30 = 24 - 30 = -6 \)
Since \( -6 < 0 \), \(y\) will be maximum at \(x = 2\).
Calculate the maximum value of \(y\) (at \(x = 2\)):
\( y_{max.} = 2(2^3) - 15(2^2) + 36(2) + 12 \)
\( y_{max.} = 2(8) - 15(4) + 72 + 12 \)
\( y_{max.} = 16 - 60 + 72 + 12 \)
\( y_{max.} = 40 \)
Calculate the minimum value of \(y\) (at \(x = 3\)):
\( y_{min.} = 2(3^3) - 15(3^2) + 36(3) + 12 \)
\( y_{min.} = 2(27) - 15(9) + 108 + 12 \)
\( y_{min.} = 54 - 135 + 108 + 12 \)
\( y_{min.} = 39 \)
So, the maximum value of \(y\) is 40 at \(x = 2\), and the minimum value of \(y\) is 39 at \(x = 3\).
In simple words: We found specific points where the function's slope is flat. By checking how the slope changes at these points, we determined that at \(x=2\) the function reaches its highest point (40), and at \(x=3\) it reaches its lowest point (39).

🎯 Exam Tip: Clearly show all steps for both first and second derivatives. Factoring the quadratic correctly is essential to finding the critical points. Always remember to substitute the \(x\) values back into the *original* function to find the maximum and minimum values of \(y\).

Question 2.
Find the values of \(x\) which maximize or minimize \(f(x) = 2x^3 + 3x^2 - 36x + 10\). Also find the maximum and minimum values of \(f(x)\).
Answer:
The given function is \(f(x) = 2x^3 + 3x^2 - 36x + 10\).
First, find the first derivative of \(f(x)\):
\( f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 36x + 10) \)
\( f'(x) = 2(3x^2) + 3(2x) - 36(1) + 0 \)
\( f'(x) = 6x^2 + 6x - 36 \)
Next, find the second derivative of \(f(x)\):
\( f''(x) = \frac{d}{dx}(6x^2 + 6x - 36) \)
\( f''(x) = 6(2x) + 6(1) - 0 \)
\( f''(x) = 12x + 6 \)
For maximum or minimum values, set the first derivative to zero:
\( f'(x) = 0 \)
\( 6x^2 + 6x - 36 = 0 \)
Divide by 6:
\( x^2 + x - 6 = 0 \)
Factor the quadratic equation:
\( x^2 + 3x - 2x - 6 = 0 \)
\( x(x + 3) - 2(x + 3) = 0 \)
\( (x + 3)(x - 2) = 0 \)
This gives two critical points: \(x = -3\) OR \(x = 2\).
Now, use the second derivative to determine if these points are maximum or minimum:
At \(x = -3\):
\( f''(-3) = 12(-3) + 6 = -36 + 6 = -30 \)
Since \( -30 < 0 \), \(f(x)\) is maximum at \(x = -3\).
At \(x = 2\):
\( f''(2) = 12(2) + 6 = 24 + 6 = 30 \)
Since \( 30 > 0 \), \(f(x)\) is minimum at \(x = 2\).
Calculate the maximum value of \(f(x)\) (at \(x = -3\)):
\( f_{max.}(-3) = 2(-3)^3 + 3(-3)^2 - 36(-3) + 10 \)
\( f_{max.}(-3) = 2(-27) + 3(9) + 108 + 10 \)
\( f_{max.}(-3) = -54 + 27 + 108 + 10 \)
\( f_{max.}(-3) = 91 \)
Calculate the minimum value of \(f(x)\) (at \(x = 2\)):
\( f_{min.}(2) = 2(2)^3 + 3(2)^2 - 36(2) + 10 \)
\( f_{min.}(2) = 2(8) + 3(4) - 72 + 10 \)
\( f_{min.}(2) = 16 + 12 - 72 + 10 \)
\( f_{min.}(2) = -34 \)
Therefore, the maximum value of \(f(x)\) is 91 at \(x = -3\), and the minimum value of \(f(x)\) is -34 at \(x = 2\).
In simple words: We found where the function's slope is flat, identifying two points. By checking the second derivative, we learned that at \(x = -3\) the function reaches its peak value of 91, and at \(x = 2\) it hits its lowest point of -34.

🎯 Exam Tip: When dealing with cubic functions, you'll often find two critical points, one for a local maximum and one for a local minimum. Always clearly label and calculate both. Pay close attention to negative signs during substitution and calculation.

Question 3.
Find the maximum and minimum values of \(f(x) = x^3 - x^2 - x + 2\).
Answer:
The given function is \(f(x) = x^3 - x^2 - x + 2\).
First, find the first derivative of \(f(x)\):
\( f'(x) = \frac{d}{dx}(x^3 - x^2 - x + 2) \)
\( f'(x) = 3x^2 - 2x - 1 \)
Next, find the second derivative of \(f(x)\):
\( f''(x) = \frac{d}{dx}(3x^2 - 2x - 1) \)
\( f''(x) = 6x - 2 \)
For maximum or minimum values, set the first derivative to zero:
\( f'(x) = 0 \)
\( 3x^2 - 2x - 1 = 0 \)
Factor the quadratic equation:
\( 3x^2 - 3x + x - 1 = 0 \)
\( 3x(x - 1) + 1(x - 1) = 0 \)
\( (x - 1)(3x + 1) = 0 \)
This gives two critical points: \(x = 1\) OR \(x = -\frac{1}{3}\).
Now, use the second derivative to determine if these points are maximum or minimum:
At \(x = 1\):
\( f''(1) = 6(1) - 2 = 4 \)
Since \(4 > 0\), \(f(x)\) is minimum at \(x = 1\).
At \(x = -\frac{1}{3}\):
\( f''(-\frac{1}{3}) = 6(-\frac{1}{3}) - 2 = -2 - 2 = -4 \)
Since \( -4 < 0 \), \(f(x)\) is maximum at \(x = -\frac{1}{3}\).
Calculate the maximum value of \(f(x)\) (at \(x = -\frac{1}{3}\)):
\( f_{max.} = (-\frac{1}{3})^3 - (-\frac{1}{3})^2 - (-\frac{1}{3}) + 2 \)
\( f_{max.} = -\frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 2 \)
To add these fractions, find a common denominator, which is 27:
\( f_{max.} = -\frac{1}{27} - \frac{3}{27} + \frac{9}{27} + \frac{54}{27} \)
\( f_{max.} = \frac{-1 - 3 + 9 + 54}{27} = \frac{59}{27} \)
Calculate the minimum value of \(f(x)\) (at \(x = 1\)):
\( f_{min.}(1) = (1)^3 - (1)^2 - 1 + 2 \)
\( f_{min.}(1) = 1 - 1 - 1 + 2 \)
\( f_{min.}(1) = 1 \)
Therefore, the maximum value of \(f(x)\) is \(\frac{59}{27}\) at \(x = -\frac{1}{3}\), and the minimum value of \(f(x)\) is 1 at \(x = 1\).
In simple words: We found the points where the function's slope is flat, which are \(x=1\) and \(x=-1/3\). By using the second derivative, we determined that \(x=-1/3\) gives a maximum value of \(59/27\), and \(x=1\) gives a minimum value of 1.

🎯 Exam Tip: Be careful with fraction arithmetic when substituting values into the original function. Showing the steps for finding a common denominator can prevent errors. Always verify critical points with the second derivative test.

Question 4.
A producer produces \(x\) units at cost \(200x + 15x^2\). The demand function is \(p = 1200 - 10x\). Find the profit function and how many units should be produced for maximum profit?
Answer:
The production cost function is \(C = 200x + 15x^2\).
The demand function (price) is \(p = 1200 - 10x\).
Revenue (\(R\)) is calculated as price (\(p\)) multiplied by quantity (\(x\)): \(R = x \cdot p\).
Substitute the demand function into the revenue function:
\( R = x(1200 - 10x) \)
\( R = 1200x - 10x^2 \)
Profit (\(P\)) is Revenue minus Cost: \(P = R - C\).
Substitute the expressions for \(R\) and \(C\) into the profit function:
\( P = (1200x - 10x^2) - (200x + 15x^2) \)
\( P = 1200x - 10x^2 - 200x - 15x^2 \)
\( P = 1000x - 25x^2 \)
This is the profit function.
To find the number of units for maximum profit, we calculate the first derivative of \(P\) with respect to \(x\):
\( \frac{dP}{dx} = \frac{d}{dx}(1000x - 25x^2) \)
\( \frac{dP}{dx} = 1000 - 50x \)
Set the first derivative to zero (Necessary condition):
\( \frac{dP}{dx} = 0 \)
\( 1000 - 50x = 0 \)
\( 1000 = 50x \)
\( x = \frac{1000}{50} \)
\( x = 20 \)
So, 20 units should be produced for potential maximum profit.
Next, we check the second derivative to confirm it's a maximum (Sufficient condition):
\( \frac{d^2P}{dx^2} = \frac{d}{dx}(1000 - 50x) \)
\( \frac{d^2P}{dx^2} = -50 \)
Since \( -50 < 0 \), the profit is maximum when 20 units are produced.
In simple words: First, we created a profit formula by subtracting costs from the money earned. Then, we found out how many units to make so that the profit stops increasing. By doing a quick double-check, we confirmed that making 20 units gives the most profit.

🎯 Exam Tip: Always clearly define the Revenue, Cost, and Profit functions before differentiating. Ensure correct subtraction when forming the profit function, as common errors occur here.

Question 5.
The selling price of a refrigerator as determined by the company is Rs. 10000. The total cost of the production for \(x\) refrigerator is \(C = 0.1x^2 + 9000x + 100\) rupees. How many refrigerators should be manufactured for maximum profit?
Answer:
The selling price of one refrigerator is Rs. 10000.
So, the revenue (\(R\)) from selling \(x\) refrigerators is \(R = 10000x\).
The total cost function is \(C = 0.1x^2 + 9000x + 100\).
Profit (\(P\)) is Revenue minus Cost: \(P = R - C\).
Substitute the expressions for \(R\) and \(C\) into the profit function:
\( P = 10000x - (0.1x^2 + 9000x + 100) \)
\( P = 10000x - 0.1x^2 - 9000x - 100 \)
\( P = 1000x - 0.1x^2 - 100 \)
To find the number of refrigerators for maximum profit, we calculate the first derivative of \(P\) with respect to \(x\):
\( \frac{dP}{dx} = \frac{d}{dx}(1000x - 0.1x^2 - 100) \)
\( \frac{dP}{dx} = 1000 - 0.1(2x) - 0 \)
\( \frac{dP}{dx} = 1000 - 0.2x \)
Set the first derivative to zero (Necessary condition):
\( \frac{dP}{dx} = 0 \)
\( 1000 - 0.2x = 0 \)
\( 1000 = 0.2x \)
\( x = \frac{1000}{0.2} \)
\( x = 5000 \)
So, 5000 refrigerators should be produced for potential maximum profit.
Next, we check the second derivative to confirm it's a maximum (Sufficient condition):
\( \frac{d^2P}{dx^2} = \frac{d}{dx}(1000 - 0.2x) \)
\( \frac{d^2P}{dx^2} = -0.2 \)
Since \( -0.2 < 0 \), the profit is maximum when 5000 refrigerators are manufactured.
In simple words: We created a profit formula by subtracting the production cost from the total money earned. Then, we found the production level where profit stops growing, which is 5000 refrigerators. A quick check confirmed that this amount indeed gives the highest profit.

🎯 Exam Tip: Remember that selling price per unit directly contributes to total revenue. Be careful with decimal calculations during differentiation and solving for \(x\). Verify the maximum with the second derivative test.

Question 6.
A toy is sold at Rs. 20. Total cost of producing \(x\) such toys is \(C = 1000 + 16.5x + 0.001x^2\) rupees. How many toys should be produced for maximum profit?
Answer:
The selling price of one toy is Rs. 20.
So, the revenue (\(R\)) from selling \(x\) toys is \(R = 20x\).
The total cost of producing \(x\) toys is \(C = 1000 + 16.5x + 0.001x^2\).
Profit (\(P\)) is Revenue minus Cost: \(P = R - C\).
Substitute the expressions for \(R\) and \(C\) into the profit function:
\( P = 20x - (1000 + 16.5x + 0.001x^2) \)
\( P = 20x - 1000 - 16.5x - 0.001x^2 \)
\( P = 3.5x - 1000 - 0.001x^2 \)
To find the number of toys for maximum profit, we calculate the first derivative of \(P\) with respect to \(x\):
\( \frac{dP}{dx} = \frac{d}{dx}(3.5x - 1000 - 0.001x^2) \)
\( \frac{dP}{dx} = 3.5 - 0 - 0.001(2x) \)
\( \frac{dP}{dx} = 3.5 - 0.002x \)
Set the first derivative to zero (Necessary condition):
\( \frac{dP}{dx} = 0 \)
\( 3.5 - 0.002x = 0 \)
\( 3.5 = 0.002x \)
\( x = \frac{3.5}{0.002} \)
\( x = 1750 \)
So, 1750 toys should be produced for potential maximum profit.
Next, we check the second derivative to confirm it's a maximum (Sufficient condition):
\( \frac{d^2P}{dx^2} = \frac{d}{dx}(3.5 - 0.002x) \)
\( \frac{d^2P}{dx^2} = -0.002 \)
Since \( -0.002 < 0 \), the profit is maximum when 1750 toys are produced.
In simple words: We calculated the profit by subtracting production costs from sales revenue. By finding when the profit stops changing and then confirming it's a peak, we determined that making 1750 toys will give the highest possible profit.

🎯 Exam Tip: When given a fixed selling price per unit, remember that total revenue is simply that price multiplied by the number of units. Be meticulous with decimal operations, especially in profit functions involving small coefficients.

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