Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.10

Question 1. If \( f(x) = x^2 - x + 1 \) then \( f(x + 1) \) is:
(c) 1
(d) \( x^2 + x + 1 \)
Answer: (d) \( x^2 + x + 1 \)
In simple words: To find \( f(x+1) \), you replace every 'x' in the original function \( f(x) \) with '\( x+1 \)'. Then you expand and simplify the expression. This substitution helps us see how the function changes when the input is slightly different.

🎯 Exam Tip: Always expand binomials carefully, like \( (x+1)^2 \), and pay attention to signs when simplifying to avoid common errors.

Question 2. If \( f(x) = \begin{cases} x^2-4x & \text{if } x \ge 2 \\ x+2 & \text{if } x < 2 \end{cases} \), then \( f(5) \) is:
(a) -1
(b) 2
(c) 5
(d) 7
Answer: (c) 5
In simple words: Since 5 is greater than or equal to 2, we use the first rule \( f(x) = x^2 - 4x \). Then we put 5 into that rule to get the answer. This kind of function has different rules for different input ranges.

🎯 Exam Tip: For piecewise functions, correctly identify which condition (e.g., \( x \ge 2 \) or \( x < 2 \)) applies to the given input value before substituting.

Question 3. If \( f(x) = \begin{cases} x^2-4x & \text{if } x \ge 2 \\ x+2 & \text{if } x < 2 \end{cases} \), then \( f(0) \) is:
(b) 5
(c) -1
(d) 0
Answer: (a) 2
In simple words: Since 0 is less than 2, we use the second rule \( f(x) = x + 2 \). Then we put 0 into that rule. Piecewise functions have different formulas depending on the input value.

🎯 Exam Tip: Always double-check the range condition for the input value (like \( x=0 \)) to ensure you apply the correct part of the piecewise function.

Question 4. If \( f(x) = \frac{1-x}{1+x} \) then \( f(-x) \) is equal to:
(a) -f(x)
(b) \( \frac{1}{f(x)} \)
(c) \( -\frac{1}{f(x)} \)
(d) f(x)
Answer: (b) \( \frac{1}{f(x)} \)
In simple words: To find \( f(-x) \), you replace every 'x' in the original function \( f(x) \) with '\( -x \)'. After simplifying the new expression, it turns out to be the inverse of the original function. Knowing how functions behave with negative inputs is important in calculus.

🎯 Exam Tip: When substituting \( -x \), be careful with signs, especially when squaring or dealing with fractions, and then try to relate the resulting expression back to the original function \( f(x) \).

Question 5. The graph of the line \( y = 3 \) is:
(a) Parallel to x-axis
(c) Passing through the origin
(d) Perpendicular to the x-axis
Answer: (a) Parallel to x-axis

In simple words: A line like \( y = 3 \) means that the 'y' value is always 3, no matter what 'x' is. This creates a flat line that runs perfectly straight across, never touching or moving away from the x-axis, just like railway tracks. It's a fundamental type of linear equation.

🎯 Exam Tip: Remember that equations of the form \( y = c \) (where \( c \) is a constant) represent horizontal lines parallel to the x-axis, and \( x = c \) represents vertical lines parallel to the y-axis.

Question 6. The graph of \( y = 2x^2 \) is passing through:
(a) (0, 0)
(b) (2, 1)
(c) (2, 0)
(d) (0, 2)
Answer: (a) (0, 0)

In simple words: To check if a point is on a graph, you put its x and y values into the equation. If both sides of the equation are equal, then the point is on the graph. For \( y = 2x^2 \), when \( x=0 \), \( y \) also becomes 0, meaning it passes through the origin. This specific function creates a parabola shape.

🎯 Exam Tip: To verify if a point lies on a curve, substitute its coordinates into the equation. If the equation holds true, the point is on the curve; otherwise, it is not.

Question 7. The graph of \( y = e^x \) intersect the y-axis at:
(a) (0, 0)
(b) (1, 0)
(c) (0, 1)
(d) (1, 1)
Answer: (c) (0, 1)
In simple words: The y-axis is where the x-value is always 0. So, to find where any graph crosses the y-axis, you just put \( x=0 \) into its equation. For \( y = e^x \), when \( x \) is 0, \( e^0 \) is 1, so the graph crosses at the point (0, 1). This point is a fixed feature for all basic exponential functions.

🎯 Exam Tip: To find the y-intercept of any function, set \( x = 0 \) and solve for \( y \). This is a standard procedure for locating where a graph crosses the vertical axis.

Question 8. The minimum value of the function \( f(x) = |x| \) is:
(a) 0
(b) -1
(c) +1
(d) \( \infty \)
Answer: (a) 0

In simple words: The absolute value of a number means its distance from zero, so it is always positive or zero. The smallest possible distance from zero is zero itself, which happens when the number is zero. Therefore, \( |x| \) can never be a negative number.

🎯 Exam Tip: Remember that the absolute value function \( |x| \) always yields a non-negative result, meaning its range is \( [0, \infty) \), with 0 being the lowest possible value.

Question 9. Which one of the following functions has the property \( f(x) = f(\frac{1}{x}) \)?
(a) \( f(x) = \frac{x^{2}-1}{x} \)
(b) \( f(x) = \frac{1-x^{2}}{x} \)
(c) \( f(x) = x \)
(d) \( f(x) = \frac{x^{2}+1}{x} \)
Answer: (d) \( f(x) = \frac{x^{2}+1}{x} \)
In simple words: We are looking for a function where if you put \( x \) into it, you get the same answer as when you put \( \frac{1}{x} \) into it. By trying each option, we find that \( f(x) = \frac{x^2+1}{x} \) fits this rule, because \( f(\frac{1}{x}) \) simplifies back to \( f(x) \). This property is specific to certain types of functions, often related to symmetry or specific algebraic structures.

🎯 Exam Tip: To check if \( f(x) = f(\frac{1}{x}) \), substitute \( \frac{1}{x} \) for \( x \) in the function and simplify. If the result is the original function, then the property holds.

Question 10. If \( f(x) = 2^x \) and \( g(x) = \frac{1}{2^x} \) then \( (fg)(x) \) is:
(a) 1
(b) 0
(c) 4
(d) \( \frac{1}{4^x} \)
Answer: (a) 1
In simple words: The notation \( (fg)(x) \) means you multiply the two functions \( f(x) \) and \( g(x) \) together. When you multiply \( 2^x \) by \( \frac{1}{2^x} \), they cancel each other out, leaving you with 1. This shows the relationship between a number and its reciprocal.

🎯 Exam Tip: Remember that \( (fg)(x) = f(x) \times g(x) \). Also, any non-zero number multiplied by its reciprocal always equals 1.

Question 11. Which of the following function is neither even nor odd?
(a) \( f(x) = x^3 + 5 \)
(b) \( f(x) = x^5 \)
(c) \( f(x) = x^{10} \)
(d) \( f(x) = x^2 \)
Answer: (a) \( f(x) = x^3 + 5 \)
In simple words: An even function is symmetric about the y-axis, meaning \( f(-x) = f(x) \). An odd function is symmetric about the origin, meaning \( f(-x) = -f(x) \). The function \( f(x) = x^3 + 5 \) is neither because its constant term means it does not satisfy either of these symmetry conditions. Functions can often be a mix of both even and odd parts, or neither.

🎯 Exam Tip: To test if a function is even or odd, substitute \( -x \) for \( x \) and simplify. If \( f(-x) = f(x) \), it's even. If \( f(-x) = -f(x) \), it's odd. If neither, it's neither.

Question 12. \( f(x) = -5 \), for all \( x \in R \) is a:
(a) an identity function
(b) modulus function
(c) exponential function
(d) constant function
Answer: (d) constant function
In simple words: A constant function is one where the output value (y-value) stays the same no matter what the input value (x-value) is. In this case, \( f(x) \) is always -5, which means it doesn't change. This type of function produces a horizontal line when graphed.

🎯 Exam Tip: A function \( f(x) = c \) (where \( c \) is any real number) is always a constant function, as its output does not depend on the input variable \( x \).

Question 13. The range of \( f(x) = |x| \), for all \( x \in R \) is:
(a) \( (0,\infty) \)
(b) \( [0, \infty) \)
(c) \( (-\infty, \infty) \)
(d) \( [1,\infty) \)
Answer: (b) \( [0, \infty) \)
In simple words: The range of a function refers to all the possible output values it can produce. For the absolute value function \( f(x) = |x| \), the output is always a positive number or zero. It can't be negative, and it can be any positive number, including zero itself. This is because absolute value represents distance from zero.

🎯 Exam Tip: Remember that the absolute value function produces only non-negative outputs. The square bracket \( [ \) indicates that 0 is included, while the parenthesis \( ) \) for infinity indicates it's unbounded.

Question 14. The graph of \( f(x) = e^x \) is identical to that of:
(a) \( f(x) = a^x, a > 1 \)
(b) \( f(x) = a^x, a < 1 \)
(c) \( f(x) = a^x, 0 < a < 1 \)
(d) \( y = ax + b, a \ne 0 \)
Answer: (a) \( f(x) = a^x, a > 1 \)
In simple words: The number 'e' is a special constant, approximately 2.718. Since this value is greater than 1, the function \( e^x \) behaves just like any other exponential function \( a^x \) where 'a' is a number greater than 1. This means the graph will always increase as \( x \) increases.

🎯 Exam Tip: Recall that \( e \) is Euler's number, an irrational constant approximately 2.718. Thus, \( e^x \) is a specific type of exponential function \( a^x \) where the base \( a \) is greater than 1.

Question 15. If \( f(x) = x^2 \) and \( g(x) = 2x + 1 \) then \( (fg)(0) \) is:
(a) 0
(b) 2
(c) 1
(d) 4
Answer: (a) 0
In simple words: First, we find the value of \( g(0) \) by putting 0 into \( g(x) \). Then, we take that result and put it into \( f(x) \). Since \( g(0) = 1 \) and \( f(1) = 1^2 = 1 \), there was a calculation mistake in my analysis. Let's re-evaluate the hint. Hint: \( (fg)(0) = f(0)g(0) = 0^2(2(0)+1) = 0(1) = 0 \). Ah, \( (fg)(x) \) means the product \( f(x) \cdot g(x) \), not composition \( f(g(x)) \). So \( (fg)(0) \) is \( f(0) \times g(0) \). \( f(0) = 0^2 = 0 \). \( g(0) = 2(0) + 1 = 1 \). So \( f(0) \times g(0) = 0 \times 1 = 0 \). The answer is correct. This function operation means multiplying the two functions. First, evaluate each function at \( x=0 \). Then, multiply their results. The fact that \( f(0) \) is 0 makes the final product also 0. This is a basic rule of multiplying by zero.

🎯 Exam Tip: Distinguish between the product of functions \( (fg)(x) = f(x) \cdot g(x) \) and the composition of functions \( (f \circ g)(x) = f(g(x)) \). Each operation requires a different approach.

Question 16. \( \lim_{x \rightarrow 0} \frac{\tan \theta}{\theta} = \)
(a) 1
(b) \( \infty \)
(c) \( -\infty \)
(d) \( \theta \)
Answer: (a) 1 (By formula)
In simple words: This is a special limit that you should remember. As \( \theta \) gets closer and closer to zero, the value of \( \frac{\tan \theta}{\theta} \) gets closer and closer to 1. This limit is often used in calculus when dealing with trigonometric functions and their derivatives.

🎯 Exam Tip: Memorize fundamental limits involving trigonometric functions, such as \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \), as they are frequently used in derivations and problem-solving.

Question 17. \( \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}= \)
(a) e
(b) \( nx^{n-1} \)
(c) 1
(d) 0
Answer: (c) 1 (By formula)
In simple words: This is another important limit in calculus. As \( x \) gets very close to 0, the expression \( \frac{e^x - 1}{x} \) approaches the value of 1. This limit is crucial for understanding the derivative of the exponential function \( e^x \).

🎯 Exam Tip: This is a standard limit related to the definition of the derivative of \( e^x \) at \( x=0 \). Recognizing such fundamental limits can save time in exams.

Question 18. For what value of \( x \), \( f(x) = \frac{x+2}{x-1} \) is not continuous?
(a) -2
(b) 1
(c) 2
(d) -1
Answer: (b) 1
In simple words: A function is not continuous when its denominator becomes zero, because you can't divide by zero. If \( x-1 = 0 \), then \( x \) must be 1. At this point, the function has a break or a jump, which means it is not continuous. This is a common discontinuity type for rational functions.

🎯 Exam Tip: Rational functions \( f(x) = \frac{P(x)}{Q(x)} \) are discontinuous at values of \( x \) where the denominator \( Q(x) \) equals zero.

Question 19. A function \( f(x) \) is continuous at \( x = a \lim _{x \rightarrow a} f(x) \) is equal to:
(a) \( f(-a) \)
(b) \( f(\frac{1}{a}) \)
(c) \( 2f(a) \)
(d) \( f(a) \)

🎯 Exam Tip: The definition of continuity at a point \( x=a \) states that \( \lim_{x \to a} f(x) = f(a) \). This means the limit exists and equals the function's value at that point.

Question 20. \( \frac{d}{d x}\left(\frac{1}{x}\right) \) is equal to:
(a) \( -\frac{1}{x^{2}} \)
(b) \( -\frac{1}{x} \)
(c) log x
(d) \( \frac{1}{x^{2}} \)
Answer: (a) \( -\frac{1}{x^{2}} \)
In simple words: This question asks for the derivative of \( \frac{1}{x} \). You can rewrite \( \frac{1}{x} \) as \( x^{-1} \). Using the power rule for derivatives, you bring the exponent down and subtract 1 from the exponent, which gives \( -1x^{-2} \), or \( -\frac{1}{x^2} \). The derivative tells us the rate of change of the function.

🎯 Exam Tip: Remember the power rule for differentiation: \( \frac{d}{dx}(x^n) = nx^{n-1} \). Convert fractions like \( \frac{1}{x} \) to negative exponents \( x^{-1} \) before applying the rule.

Question 21. \( \frac{d}{d x} (5e^x – 2 \log x) \) is equal to:
(a) \( 5e^x - \frac{2}{x} \)
(b) \( 5e^x - 2x \)
(c) \( 5e^x - \frac{1}{x} \)
(d) 2 log x
Answer: (a) \( 5e^x - \frac{2}{x} \)
In simple words: To find the derivative of a sum or difference of terms, you find the derivative of each term separately. The derivative of \( 5e^x \) is \( 5e^x \), and the derivative of \( -2 \log x \) is \( -2 \times \frac{1}{x} \). Combining these gives the final answer. Understanding how to differentiate common functions is key to solving these types of problems.

🎯 Exam Tip: Remember the basic differentiation rules: \( \frac{d}{dx}(ce^x) = ce^x \) and \( \frac{d}{dx}(c \log x) = \frac{c}{x} \), where \( c \) is a constant.

Question 22. If \( y = x \) and \( z = \frac{1}{x} \) then \( \frac{d y}{d z} = \)
(a) \( x^2 \)
(b) 1
(c) \( -x^2 \)
(d) \( \frac{-1}{x^{2}} \)
Answer: (c) \( -x^2 \)
In simple words: We need to find the derivative of \( y \) with respect to \( z \). We can use the chain rule, which says \( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \). First, find the derivative of \( y \) with respect to \( x \), which is 1. Then, find the derivative of \( z \) with respect to \( x \), which is \( -\frac{1}{x^2} \). Finally, divide the two derivatives to get the result. This method is useful when direct differentiation is hard.

🎯 Exam Tip: When \( y \) and \( z \) are both functions of a third variable \( x \), use the chain rule \( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \) to find \( \frac{dy}{dz} \).

Question 23. If \( y = e^{2x} \) then \( \frac{d^{2} y}{d x^{2}} \) at \( x = 0 \) is:
(a) 4
(b) 9
(c) 2
(d) 0
Answer: (a) 4
In simple words: This asks for the second derivative of the function \( y = e^{2x} \), evaluated at \( x=0 \). First, find the first derivative \( \frac{dy}{dx} \). Then, differentiate that result again to get \( \frac{d^2y}{dx^2} \). Finally, substitute \( x=0 \) into the second derivative. Each derivative step uses the chain rule for exponential functions.

🎯 Exam Tip: To find the second derivative \( \frac{d^2y}{dx^2} \), differentiate the function twice. Remember to apply the chain rule correctly when differentiating composite functions like \( e^{2x} \).

Question 24. If \( y = \log x \) then \( y_2 = \)
(a) \( \frac{1}{x} \)
(b) \( -\frac{1}{x^{2}} \)
(c) \( -\frac{2}{x^{2}} \)
(d) \( e^2 \)
Answer: (b) \( -\frac{1}{x^{2}} \)
In simple words: The notation \( y_2 \) means the second derivative of \( y \) with respect to \( x \). First, find the first derivative of \( y = \log x \), which is \( \frac{1}{x} \). Then, find the derivative of \( \frac{1}{x} \) (or \( x^{-1} \)), which is \( -\frac{1}{x^2} \). Each step shows how the rate of change itself changes.

🎯 Exam Tip: Always remember the basic derivative: \( \frac{d}{dx}(\log x) = \frac{1}{x} \). For the second derivative, differentiate this result using the power rule by writing \( \frac{1}{x} \) as \( x^{-1} \).

Question 25. \( \frac{d}{dx}\left(a^{x}\right)= \)
(a) \( \frac{1}{x \log _{e} a} \)
(b) \( a^a \)
(c) \( x \log_e a \)
(d) \( a^x \log_e a \)
Answer: (d) \( a^x \log_e a \) (by formula)
In simple words: This is a standard differentiation formula for an exponential function with any base 'a'. The derivative of \( a^x \) is \( a^x \) multiplied by the natural logarithm of the base, \( \log_e a \). This formula helps us calculate the rate of change for any exponential growth or decay.

🎯 Exam Tip: It is crucial to memorize the differentiation formula for exponential functions with an arbitrary base: \( \frac{d}{dx}(a^x) = a^x \ln(a) \). Note that \( \ln(a) \) is the same as \( \log_e a \).