Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.1

 

Question 1. Determine whether the following functions are odd or even?
(i) \( f(x) = \left(\frac{a^{x}-1}{a^{x}+1}\right) \)
(ii) \( f(x) = \log(x^{2} + \sqrt{x^{2}+1}) \)
(iii) \( f(x) = \sin x + \cos x \)
(iv) \( f(x) = x^{2} - |x| \)
(v) \( f(x) = x + x^{2} \)
Answer:
(i) Given \( f(x) = \left(\frac{a^{x}-1}{a^{x}+1}\right) \)
To check if it's odd or even, we find \( f(-x) \):
\( f(-x) = \frac{a^{-x}-1}{a^{-x}+1} = \frac{\frac{1}{a^{x}}-1}{\frac{1}{a^{x}}+1} = \frac{\frac{1-a^{x}}{a^{x}}}{\frac{1+a^{x}}{a^{x}}} = \frac{1-a^{x}}{1+a^{x}} \)
We can also write this as \( -\left(\frac{a^{x}-1}{a^{x}+1}\right) = -f(x) \).
Since \( f(-x) = -f(x) \), the function \( f(x) \) is an odd function. Odd functions are symmetric with respect to the origin.
(ii) Given \( f(x) = \log(x^{2} + \sqrt{x^{2}+1}) \)
To check if it's odd or even, we find \( f(-x) \):
\( f(-x) = \log((-x)^{2} + \sqrt{(-x)^{2}+1}) \)
\( = \log(x^{2} + \sqrt{x^{2}+1}) \)
Since \( f(-x) = f(x) \), the function \( f(x) \) is an even function. Even functions are symmetric about the y-axis.
(iii) Given \( f(x) = \sin x + \cos x \)
To check if it's odd or even, we find \( f(-x) \):
\( f(-x) = \sin(-x) + \cos(-x) \)
\( = -\sin x + \cos x \)
If \( f(x) \) were odd, \( f(-x) \) would be \( -(\sin x + \cos x) = -\sin x - \cos x \). This is not the case.
If \( f(x) \) were even, \( f(-x) \) would be \( \sin x + \cos x \). This is also not the case.
Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), the function \( f(x) \) is neither an odd nor an even function. This means it doesn't have the special symmetry of odd or even functions.
(iv) Given \( f(x) = x^{2} - |x| \)
To check if it's odd or even, we find \( f(-x) \):
\( f(-x) = (-x)^{2} - |-x| \)
\( = x^{2} - |x| \)
Since \( f(-x) = f(x) \), the function \( f(x) \) is an even function. The absolute value function \( |x| \) is itself an even function.
(v) Given \( f(x) = x + x^{2} \)
To check if it's odd or even, we find \( f(-x) \):
\( f(-x) = (-x) + (-x)^{2} \)
\( = -x + x^{2} \)
If \( f(x) \) were odd, \( f(-x) \) would be \( -(x + x^{2}) = -x - x^{2} \). This is not the case.
If \( f(x) \) were even, \( f(-x) \) would be \( x + x^{2} \). This is also not the case.
Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), the function \( f(x) \) is neither an odd nor an even function. It combines parts that behave differently under \( x \to -x \).
In simple words: To tell if a function is odd or even, replace \( x \) with \( -x \). If the new function is the same as the original, it's even. If it's the negative of the original, it's odd. If it's neither, then it has no special symmetry.

🎯 Exam Tip: Remember that \( \sin(-x) = -\sin x \) (odd) and \( \cos(-x) = \cos x \) (even). These basic properties are key for determining the symmetry of trigonometric functions.

 

Question 2. Let f be defined by \( f(x) = x^{3} - kx^{2} + 2x, x \in R \). Find k, if 'f' is an odd function.
Answer: For a polynomial function to be an odd function, every term in the polynomial must have an odd power of \( x \). This means there should be no terms with even powers of \( x \). In the given function \( f(x) = x^{3} - kx^{2} + 2x \), the term \( -kx^{2} \) has an even power of \( x \) (which is 2). For the function to be odd, this term must be eliminated. The only way to remove \( -kx^{2} \) is if its coefficient, \( k \), is zero. This ensures that only odd-powered terms remain, making the function symmetric about the origin.
\( \implies k = 0 \)
In simple words: For a function to be "odd," all the powers of \( x \) in its equation must be odd numbers (like 1, 3, 5...). If there's an \( x^{2} \) term, its coefficient must be zero to make the function odd. So, \( k \) must be 0.

🎯 Exam Tip: An odd function means \( f(-x) = -f(x) \). For polynomials, this simplifies to ensuring all powers of \( x \) are odd. If an even power term exists, its coefficient must be zero.

 

Question 3. If \( f(x) = x^{3}-\frac{1}{x^{3}} \), then show that \( f(x) + f\left(\frac{1}{x}\right) = 0 \).
Answer: Given the function \( f(x) = x^{3}-\frac{1}{x^{3}} \) ........ (1)
First, we need to find \( f\left(\frac{1}{x}\right) \). We replace \( x \) with \( \frac{1}{x} \) in the function:
\( f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^{3} - \frac{1}{\left(\frac{1}{x}\right)^{3}} \)
\( = \frac{1}{x^{3}} - x^{3} \) ........ (2)
Now, we add equation (1) and equation (2):
\( f(x) + f\left(\frac{1}{x}\right) = \left(x^{3} - \frac{1}{x^{3}}\right) + \left(\frac{1}{x^{3}} - x^{3}\right) \)
When we combine the terms, the \( x^{3} \) and \( -x^{3} \) cancel out, and the \( -\frac{1}{x^{3}} \) and \( +\frac{1}{x^{3}} \) also cancel out.
\( \implies f(x) + f\left(\frac{1}{x}\right) = 0 \)
Hence proved. This shows a specific property of this type of reciprocal function.
In simple words: We have a function \( f(x) \). We find what \( f \) is when \( x \) is replaced by \( \frac{1}{x} \). Then, we add the two results together. All the parts cancel each other out, leaving us with 0.

🎯 Exam Tip: When proving an identity, always start with one side (or combine both if needed) and algebraically manipulate it to reach the other side. Clearly show each step.

 

Question 4. If \( f(x) = \frac{x+1}{x-1} \), then prove that \( f(f(x)) = x \).
Answer: Given the function \( f(x) = \frac{x+1}{x-1} \).
We need to find \( f(f(x)) \). This means we substitute the entire expression for \( f(x) \) into \( f(x) \) itself, wherever \( x \) appears.
\( f(f(x)) = f\left(\frac{x+1}{x-1}\right) \)
So, we replace \( x \) in \( f(x) \) with \( \left(\frac{x+1}{x-1}\right) \):
\( = \frac{\left(\frac{x+1}{x-1}\right)+1}{\left(\frac{x+1}{x-1}\right)-1} \)
Now, we find a common denominator for the numerator and the denominator of this complex fraction:
For the numerator: \( \frac{x+1}{x-1} + 1 = \frac{x+1 + (x-1)}{x-1} = \frac{x+1+x-1}{x-1} = \frac{2x}{x-1} \)
For the denominator: \( \frac{x+1}{x-1} - 1 = \frac{x+1 - (x-1)}{x-1} = \frac{x+1-x+1}{x-1} = \frac{2}{x-1} \)
Now we put these back into the main expression:
\( f(f(x)) = \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \)
We can cancel out the common denominator \( (x-1) \) from both the numerator and the denominator.
\( \implies f(f(x)) = \frac{2x}{2} \)
\( \implies f(f(x)) = x \)
Hence proved. This type of function is its own inverse, meaning applying it twice returns the original input.
In simple words: We start with a function \( f(x) \). We then apply the function to its own output, which is like putting the function inside itself. After doing all the math, the result is simply \( x \), showing that applying the function twice cancels itself out.

🎯 Exam Tip: When evaluating composite functions like \( f(f(x)) \), perform the inner function first, then substitute that result into the outer function. Simplify carefully, especially with fractions.

 

Question 5. For \( f(x) = \frac{x-1}{3x+1} \), write the expressions of \( f\left(\frac{1}{x}\right) \) and \( \frac{1}{f(x)} \).
Answer: Given the function \( f(x) = \frac{x-1}{3x+1} \).
First, we find the expression for \( f\left(\frac{1}{x}\right) \). We replace every \( x \) in \( f(x) \) with \( \frac{1}{x} \):
\( f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}-1}{3\left(\frac{1}{x}\right)+1} \)
To simplify this complex fraction, we find a common denominator for the numerator and the denominator.
Numerator: \( \frac{1}{x}-1 = \frac{1-x}{x} \)
Denominator: \( \frac{3}{x}+1 = \frac{3+x}{x} \)
So, \( f\left(\frac{1}{x}\right) = \frac{\frac{1-x}{x}}{\frac{3+x}{x}} \)
We can cancel out the common denominator \( x \) from both the numerator and denominator.
\( \implies f\left(\frac{1}{x}\right) = \frac{1-x}{3+x} \)
Next, we find the expression for \( \frac{1}{f(x)} \). This means taking the reciprocal of the given function \( f(x) \):
\( \frac{1}{f(x)} = \frac{1}{\left(\frac{x-1}{3x+1}\right)} \)
To find the reciprocal of a fraction, we simply flip the numerator and the denominator.
\( \implies \frac{1}{f(x)} = \frac{3x+1}{x-1} \)
These calculations help us understand how the function changes with different inputs or when inverted. This is a common operation when analyzing functions.
In simple words: We are asked to find two new versions of the function. For \( f\left(\frac{1}{x}\right) \), we swap every \( x \) with \( \frac{1}{x} \) and then simplify. For \( \frac{1}{f(x)} \), we simply flip the top and bottom parts of the original function.

🎯 Exam Tip: Be careful with signs when simplifying fractions. Remember that \( \frac{A/B}{C/D} = \frac{A}{B} \cdot \frac{D}{C} \).

 

Question 6. If \( f(x) = e^{x} \) and \( g(x) = \log_{e} x \) then find
(i) \( (f + g) (1) \)
(ii) \( (fg) (1) \)
(iii) \( (3f) (1) \)
(iv) \( (5g) (1) \)
Answer: Given \( f(x) = e^{x} \) and \( g(x) = \log_{e} x \). We need to evaluate these expressions at \( x = 1 \).
(i) \( (f+g)(1) \)
This means we find \( f(1) \) and \( g(1) \) separately, then add them.
\( f(1) = e^{1} = e \)
\( g(1) = \log_{e} 1 \)
Remember that the logarithm of 1 to any base is always 0.
So, \( g(1) = 0 \)
\( \implies (f+g)(1) = f(1) + g(1) = e + 0 = e \)
(ii) \( (fg)(1) \)
This means we find \( f(1) \) and \( g(1) \) separately, then multiply them.
\( f(1) = e^{1} = e \)
\( g(1) = \log_{e} 1 = 0 \)
\( \implies (fg)(1) = f(1) \cdot g(1) = e \cdot 0 = 0 \)
(iii) \( (3f)(1) \)
This means we find \( f(1) \) and then multiply it by 3.
\( f(1) = e^{1} = e \)
\( \implies (3f)(1) = 3 \cdot f(1) = 3e \)
(iv) \( (5g)(1) \)
This means we find \( g(1) \) and then multiply it by 5.
\( g(1) = \log_{e} 1 = 0 \)
\( \implies (5g)(1) = 5 \cdot g(1) = 5 \cdot 0 = 0 \)
These evaluations demonstrate how standard function operations apply at a specific point.
In simple words: We are given two special functions, \( e^{x} \) and \( \log_{e} x \). We need to calculate what happens when we combine them (add, multiply, or multiply by a number) and then put the number 1 into the result. We find that \( e^1 \) is \( e \), and \( \log_e 1 \) is always 0. Then we do the math for each part.

🎯 Exam Tip: Always recall the basic properties of exponential and logarithmic functions: \( e^0=1 \), \( e^1=e \), \( \log_e 1 = 0 \), and \( \log_e e = 1 \).

 

Question 7. Draw the graph of the following functions:
(i) \( f(x) = 16 - x^{2} \)
(ii) \( f(x) = |x - 2| \)
(iii) \( f(x) = x|x| \)
(iv) \( f(x) = e^{2x} \)
(v) \( f(x) = e^{-2x} \)
(vi) \( f(x) = \frac{|x|}{x} \)
Answer:
(i) \( f(x) = 16 - x^{2} \)
Let \( y = f(x) = 16 - x^{2} \). This is the equation of a parabola opening downwards, with its vertex at \( (0, 16) \). To graph it, we choose some values for \( x \) and find the corresponding \( y \) values.

Plot these points on a coordinate plane and connect them with a smooth curve to form a parabola that opens downwards, symmetric about the y-axis, and passing through \( (0, 16) \).
(ii) \( f(x) = |x - 2| \)
This is an absolute value function, which forms a V-shape graph. It can be defined piecewise:
If \( x - 2 \ge 0 \implies x \ge 2 \), then \( f(x) = x - 2 \).
If \( x - 2 < 0 \implies x < 2 \), then \( f(x) = -(x - 2) = -x + 2 \).
So, we have two linear parts. The vertex of the V-shape is where \( x - 2 = 0 \), which is at \( x=2 \). At \( x=2 \), \( f(x) = |2-2| = 0 \), so the vertex is \( (2, 0) \).


Plot these points and draw a line segment for each part, forming a V-shape with its lowest point at \( (2,0) \).
(iii) \( f(x) = x|x| \)
This function can also be defined piecewise:
If \( x \ge 0 \), then \( |x| = x \), so \( f(x) = x \cdot x = x^{2} \).
If \( x < 0 \), then \( |x| = -x \), so \( f(x) = x \cdot (-x) = -x^{2} \).
Thus, the function is \( f(x) = \begin{cases} x^{2} & \text{if } x \ge 0 \\ -x^{2} & \text{if } x < 0 \end{cases} \). This graph looks like a parabola for \( x \ge 0 \) and an inverted parabola for \( x < 0 \), creating a curve similar to a cubic function. The graph passes through the origin \( (0,0) \).


Plot these points and draw a smooth curve that passes through the origin. It rises on the right side (like \( x^2 \)) and falls on the left side (like \( -x^2 \)).
(iv) \( f(x) = e^{2x} \)
This is an exponential growth function. For \( x=0 \), \( f(x) = e^{2(0)} = e^0 = 1 \), so the curve cuts the y-axis at \( y=1 \). As \( x \) increases, \( f(x) \) increases rapidly. As \( x \) decreases, \( f(x) \) approaches 0 but never reaches it, meaning the x-axis is a horizontal asymptote. The graph always stays above the x-axis. This shows how quickly exponential functions can grow.
(v) \( f(x) = e^{-2x} \)
This is an exponential decay function. For \( x=0 \), \( f(x) = e^{-2(0)} = e^0 = 1 \), so the curve cuts the y-axis at \( y=1 \). As \( x \) increases, \( f(x) \) decreases rapidly and approaches 0, but never reaches it. As \( x \) decreases, \( f(x) \) increases rapidly. The x-axis is a horizontal asymptote. This function is a mirror image of \( e^{2x} \) across the y-axis.
(vi) \( f(x) = \frac{|x|}{x} \)
This function can be defined piecewise, but only for \( x \neq 0 \) since division by zero is not allowed.
If \( x > 0 \), then \( |x| = x \), so \( f(x) = \frac{x}{x} = 1 \).
If \( x < 0 \), then \( |x| = -x \), so \( f(x) = \frac{-x}{x} = -1 \).
So, the function is \( f(x) = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \end{cases} \). The function is undefined at \( x=0 \). The graph consists of two horizontal lines: one at \( y=1 \) for all positive \( x \), and another at \( y=-1 \) for all negative \( x \). There is a jump discontinuity at \( x=0 \). The domain of the function is \( R \setminus \{0\} \), and the range is \( \{-1, 1\} \).
In simple words: To graph these, we look at what kind of function each is. For parabolas and absolute values, we pick points or find the special 'vertex' to draw the curve. For exponential functions, we know they either grow or shrink fast and cross the y-axis at a specific point. For the last one, it's like a switch: if \( x \) is positive, the answer is 1; if \( x \) is negative, the answer is -1, and it's undefined at 0.

🎯 Exam Tip: When drawing graphs, always label your axes, indicate the scale, and mark any key points like intercepts or vertices. For piecewise functions, pay close attention to the points where the definition changes.