OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (E)

S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(e)

Question 1. If the function f : N → N is defined by f (x) = \( \sqrt{x} \), then find \( \frac{f(25)}{f(16)+f(1)} \).
Answer: The function is given by \( f(x) = \sqrt{x} \).
First, we find the values of the function for the given inputs:
\( f(25) = \sqrt{25} = 5 \)
\( f(16) = \sqrt{16} = 4 \)
\( f(1) = \sqrt{1} = 1 \)
Now, we substitute these values into the expression we need to find:
\( \frac{f(25)}{f(16)+f(1)} = \frac{5}{4+1} \)
\( = \frac{5}{5} \)
\( = 1 \)
So, the value of the expression is 1. This shows how function values are combined in an expression.
In simple words: First, find the square root of 25, 16, and 1. Then, put these numbers into the fraction: the square root of 25 goes on top, and the sum of the square roots of 16 and 1 goes on the bottom. When you calculate it, the answer is 1.

🎯 Exam Tip: When evaluating functions, always substitute the input values correctly into the function definition and perform calculations step-by-step to avoid errors.

Question 2. If \( f(x) = \frac{x^3}{2}-\frac{x^2}{2}+x-16 \), find \( f(\frac { 1 }{ 2 }) \).
Answer: The given function is \( f(x) = \frac{x^3}{2}-\frac{x^2}{2}+x-16 \).
To find \( f(\frac{1}{2}) \), we substitute \( x = \frac{1}{2} \) into the function:
\( f\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^3}{2} - \frac{\left(\frac{1}{2}\right)^2}{2} + \frac{1}{2} - 16 \)
\( = \frac{\frac{1}{8}}{2} - \frac{\frac{1}{4}}{2} + \frac{1}{2} - 16 \)
\( = \frac{1}{16} - \frac{1}{8} + \frac{1}{2} - 16 \)
To combine these terms, we find a common denominator, which is 16.
\( = \frac{1}{16} - \frac{2}{16} + \frac{8}{16} - \frac{16 \times 16}{16} \)
\( = \frac{1 - 2 + 8 - 256}{16} \)
\( = \frac{9 - 258}{16} \)
\( = -\frac{249}{16} \)
The value of \( f(\frac{1}{2}) \) is \( -\frac{249}{16} \). This calculation demonstrates substituting a fractional value into a polynomial function.
In simple words: To find the value of the function when x is 1/2, replace every 'x' in the equation with '1/2'. Then, carefully do all the math, including cubing, squaring, and fractions, and finally subtract 16. The result is -249/16.

🎯 Exam Tip: When substituting fractions, be careful with powers and ensure all terms are brought to a common denominator before combining them. Always simplify fractions at the end if possible.

Question 3. If \( f (x) = 7x^4 – 2x^3 – 8x – 5 \), find \( f (- 1) \).
Answer: The given function is \( f (x) = 7x^4 – 2x^3 – 8x – 5 \).
To find \( f (- 1) \), we substitute \( x = -1 \) into the function:
\( f(-1) = 7 (-1)^4 - 2 (-1)^3 - 8 (-1) - 5 \)
First, calculate the powers of -1:
\( (-1)^4 = 1 \) (an even power makes it positive)
\( (-1)^3 = -1 \) (an odd power keeps it negative)
Now substitute these back:
\( f(-1) = 7 (1) - 2 (-1) - 8 (-1) - 5 \)
\( = 7 + 2 + 8 - 5 \)
\( = 17 - 5 \)
\( = 12 \)
So, the value of \( f(-1) \) is 12. This shows how negative inputs affect the terms in a polynomial.
In simple words: To find \( f(-1) \), put -1 in place of 'x' everywhere in the equation. Remember that a negative number raised to an even power becomes positive, and to an odd power stays negative. Then, add and subtract the numbers to get the final answer.

🎯 Exam Tip: Pay close attention to the signs when substituting negative values into a function, especially with powers, as a common mistake is sign errors.

Question 4. If \( f (x) = \left\{\begin{array}{l} 3 x-2 \text { when } x \leq 0 \\ x+1 \text { when } x>0 \end{array}\right. \), find \( f (-1) \) and \( f (0) \).
Answer: The function \( f(x) \) is defined in two parts, depending on the value of \( x \).
The first part is \( 3x - 2 \) when \( x \leq 0 \).
The second part is \( x + 1 \) when \( x > 0 \).

To find \( f(-1) \):
Since \( -1 \leq 0 \), we use the first rule: \( f(x) = 3x - 2 \).
\( f(-1) = 3(-1) - 2 \)
\( = -3 - 2 \)
\( = -5 \)

To find \( f(0) \):
Since \( 0 \leq 0 \) (it falls into the "less than or equal to" category), we again use the first rule: \( f(x) = 3x - 2 \).
\( f(0) = 3(0) - 2 \)
\( = 0 - 2 \)
\( = -2 \)
This type of function is called a piecewise function because its definition changes based on the input value.
In simple words: This function has two rules. If the number you put in is 0 or less, use the rule \( 3x - 2 \). If the number is more than 0, use \( x + 1 \). For \( f(-1) \), since -1 is less than 0, use the first rule to get -5. For \( f(0) \), since 0 is less than or equal to 0, use the first rule to get -2.

🎯 Exam Tip: For piecewise functions, carefully identify which rule applies based on the input value's relationship to the given conditions (e.g., \( x \leq 0 \) or \( x > 0 \)).

Question 5. If \( f (x) = \log \left(\frac{1-x}{1+x}\right) \), show that \( f (a) + f (b) = f\left(\frac{a+b}{1+ab}\right) \).
Answer: The given function is \( f(x) = \log \left(\frac{1-x}{1+x}\right) \).

First, let's find \( f(a) \) and \( f(b) \):
\( f(a) = \log \left(\frac{1-a}{1+a}\right) \)
\( f(b) = \log \left(\frac{1-b}{1+b}\right) \)

Now, let's calculate the Left Hand Side (L.H.S.):
L.H.S. \( = f(a) + f(b) \)
\( = \log \left(\frac{1-a}{1+a}\right) + \log \left(\frac{1-b}{1+b}\right) \)
Using the logarithm property \( \log x + \log y = \log (xy) \):
\( = \log \left[ \left(\frac{1-a}{1+a}\right) \left(\frac{1-b}{1+b}\right) \right] \)
\( = \log \left[ \frac{(1-a)(1-b)}{(1+a)(1+b)} \right] \)
\( = \log \left[ \frac{1 - b - a + ab}{1 + b + a + ab} \right] \) (Equation 1)

Next, let's calculate the Right Hand Side (R.H.S.):
R.H.S. \( = f\left(\frac{a+b}{1+ab}\right) \)
We substitute \( x = \frac{a+b}{1+ab} \) into the function \( f(x) \):
\( = \log \left[ \frac{1 - \left(\frac{a+b}{1+ab}\right)}{1 + \left(\frac{a+b}{1+ab}\right)} \right] \)
To simplify, we find a common denominator in the numerator and denominator:
\( = \log \left[ \frac{\frac{1+ab-(a+b)}{1+ab}}{\frac{1+ab+(a+b)}{1+ab}} \right] \)
The \( (1+ab) \) terms cancel out:
\( = \log \left[ \frac{1 + ab - a - b}{1 + ab + a + b} \right] \) (Equation 2)

Comparing Equation 1 and Equation 2, we can see that L.H.S. \( = \) R.H.S.
Thus, we have shown that \( f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right) \). This result demonstrates an important property of this type of logarithmic function.
In simple words: First, write out what \( f(a) \) and \( f(b) \) mean. Then, add them together using the rule for adding logarithms, which turns into multiplying the parts inside the log. Next, calculate \( f \) of the combined term \( \frac{a+b}{1+ab} \) by putting that whole fraction into the original function. When you simplify both sides, you will see they are exactly the same, proving the statement.

🎯 Exam Tip: Remember key logarithm properties like \( \log x + \log y = \log(xy) \) and \( \log x - \log y = \log(\frac{x}{y}) \). Algebraic manipulation, especially with fractions within fractions, requires careful attention to detail.

Question 6. If \( f (x) = 2x\sqrt{1-x^2} \), then show that \( f \left(\sin \frac{x}{2}\right) = \sin x \).
Answer: The given function is \( f (x) = 2x\sqrt{1-x^2} \).
To show that \( f\left(\sin \frac{x}{2}\right) = \sin x \), we substitute \( x = \sin \frac{x}{2} \) into the function definition.
\( f\left(\sin \frac{x}{2}\right) = 2\left(\sin \frac{x}{2}\right)\sqrt{1-\left(\sin \frac{x}{2}\right)^2} \)
We know the trigonometric identity \( 1 - \sin^2 \theta = \cos^2 \theta \). Applying this:
\( = 2\sin \frac{x}{2}\sqrt{\cos^2 \frac{x}{2}} \)
\( = 2\sin \frac{x}{2} \cos \frac{x}{2} \)
This expression is a standard double-angle identity: \( \sin(2\theta) = 2\sin\theta\cos\theta \).
Here, \( \theta = \frac{x}{2} \), so \( 2\theta = 2 \times \frac{x}{2} = x \).
\( = \sin x \)
Therefore, we have shown that \( f\left(\sin \frac{x}{2}\right) = \sin x \). This demonstrates the connection between functions and trigonometric identities.
In simple words: Replace 'x' in the function with \( \sin \frac{x}{2} \). Then, use the trig rule that \( 1 - \sin^2 \theta \) is the same as \( \cos^2 \theta \). After that, remember the double angle formula for sine, which is \( 2\sin\theta\cos\theta = \sin(2\theta) \). This will simplify the expression to \( \sin x \).

🎯 Exam Tip: Keep common trigonometric identities, especially Pythagorean identities (\( \sin^2 \theta + \cos^2 \theta = 1 \)) and double-angle formulas (\( \sin 2\theta = 2\sin\theta\cos\theta \)), readily in mind for such problems.

Question 7. If \( f(x) = \cos (\log x) \), then prove that \( f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)- \frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right] = 0 \).
Answer: The given function is \( f(x) = \cos (\log x) \).
We need to prove that \( f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)- \frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right] = 0 \).

Let's evaluate the terms on the Left Hand Side (L.H.S.):
\( f\left(\frac{1}{x}\right) = \cos\left(\log\frac{1}{x}\right) = \cos(-\log x) = \cos(\log x) \) (since \( \cos(-\theta) = \cos\theta \))
Similarly, \( f\left(\frac{1}{y}\right) = \cos\left(\log\frac{1}{y}\right) = \cos(-\log y) = \cos(\log y) \)

\( f\left(\frac{x}{y}\right) = \cos\left(\log\frac{x}{y}\right) = \cos(\log x - \log y) \)
\( f(xy) = \cos(\log(xy)) = \cos(\log x + \log y) \)

Now substitute these into the L.H.S. expression:
L.H.S. \( = \cos(\log x) \cos(\log y) - \frac{1}{2} [\cos(\log x - \log y) + \cos(\log x + \log y)] \)

We use the cosine sum and difference formulas:
\( \cos(A - B) = \cos A \cos B + \sin A \sin B \)
\( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
Let \( A = \log x \) and \( B = \log y \).

So, \( \cos(\log x - \log y) = \cos(\log x)\cos(\log y) + \sin(\log x)\sin(\log y) \)
And \( \cos(\log x + \log y) = \cos(\log x)\cos(\log y) - \sin(\log x)\sin(\log y) \)

Adding these two expressions:
\( \cos(\log x - \log y) + \cos(\log x + \log y) = 2\cos(\log x)\cos(\log y) \)

Substitute this back into the L.H.S.:
L.H.S. \( = \cos(\log x) \cos(\log y) - \frac{1}{2} [2\cos(\log x)\cos(\log y)] \)
\( = \cos(\log x) \cos(\log y) - \cos(\log x)\cos(\log y) \)
\( = 0 \)
Since L.H.S. \( = 0 \), and the Right Hand Side is 0, we have proven the identity. This proof relies on the properties of logarithms and trigonometric identities.
In simple words: First, find the values of \( f \) for \( \frac{1}{x}, \frac{1}{y}, \frac{x}{y} \), and \( xy \), remembering that \( \log \frac{1}{x} = -\log x \) and \( \cos(-\theta) = \cos\theta \). Then, put these into the main equation. Use the rules for adding and subtracting logarithms, and the cosine addition formulas \( \cos(A+B) \) and \( \cos(A-B) \). When you combine everything, the terms will cancel out, showing the whole expression equals zero.

🎯 Exam Tip: This problem requires a solid understanding of both logarithmic properties (\( \log(\frac{1}{x}) = -\log x \), \( \log(\frac{x}{y}) = \log x - \log y \), \( \log(xy) = \log x + \log y \)) and trigonometric identities like \( \cos(-\theta) = \cos\theta \) and the sum/difference formulas for cosine.

Question 8. If \( y = f (x) = \frac{5 x+3}{4 x-5} \), then show that \( f (y) = x \).
Answer: The given function is \( y = f (x) = \frac{5 x+3}{4 x-5} \).
To show that \( f(y) = x \), we need to substitute \( y \) into the function definition. However, the function \( f \) is defined in terms of \( x \). So, a common approach for showing \( f(f(x)) = x \) is to find the inverse function.
Let's rearrange the given equation to express \( x \) in terms of \( y \):
\( y = \frac{5x+3}{4x-5} \)
Multiply both sides by \( (4x-5) \):
\( y(4x-5) = 5x+3 \)
\( 4xy - 5y = 5x+3 \)
Now, we want to isolate \( x \). Move all terms with \( x \) to one side and terms without \( x \) to the other side:
\( 4xy - 5x = 5y+3 \)
Factor out \( x \) from the left side:
\( x(4y-5) = 5y+3 \)
Divide by \( (4y-5) \) to solve for \( x \):
\( x = \frac{5y+3}{4y-5} \)
This expression for \( x \) in terms of \( y \) is exactly in the form of \( f(y) \).
Therefore, we can say that \( f(y) = x \). This type of function is its own inverse, which is a special property.
In simple words: We have \( y \) in terms of \( x \). To show \( f(y) = x \), we need to rewrite the equation so that \( x \) is on one side and everything else (including \( y \)) is on the other. Start by multiplying \( y \) by the bottom part of the fraction. Then, move all terms with \( x \) to one side and all terms with \( y \) to the other side. Factor out \( x \) and then divide to solve for \( x \). You will find that \( x \) is equal to the same expression as the original function, but with \( y \) instead of \( x \).

🎯 Exam Tip: This problem asks to show \( f(y) = x \), which is equivalent to finding the inverse function. The steps involve algebraic manipulation to express \( x \) in terms of \( y \). Be careful with distributing and collecting terms.

Question 9. If \( f (x) = x^2 + kx + 1 \), for all \( x \) and if it is an even function, find \( k \).
Answer: The given function is \( f (x) = x^2 + kx + 1 \).
A function is defined as an even function if \( f(-x) = f(x) \) for all values of \( x \).

First, let's find \( f(-x) \):
\( f(-x) = (-x)^2 + k(-x) + 1 \)
\( = x^2 - kx + 1 \)

Now, we set \( f(-x) \) equal to \( f(x) \) since the function is even:
\( x^2 - kx + 1 = x^2 + kx + 1 \)

Subtract \( x^2 \) from both sides:
\( -kx + 1 = kx + 1 \)

Subtract 1 from both sides:
\( -kx = kx \)

Add \( kx \) to both sides:
\( 0 = 2kx \)

For this equation to be true for all values of \( x \), the coefficient of \( x \) must be zero.
So, \( 2k = 0 \)
\( \implies k = 0 \)
Therefore, for the function to be even, the value of \( k \) must be 0. This is because odd power terms vanish in an even function's expression.
In simple words: An even function is one where \( f(-x) \) is the same as \( f(x) \). So, replace 'x' with '-x' in the given equation and then set this new equation equal to the original equation. When you simplify, all the terms that have \( x^2 \) and the number 1 will cancel out, leaving you with \( -kx = kx \). This means \( 2kx = 0 \), and for this to be true for any 'x', \( k \) must be 0.

🎯 Exam Tip: The key definition for an even function is \( f(-x) = f(x) \). Remember that terms with odd powers of \( x \) (like \( kx \)) are responsible for a function being odd, and for an even function, these terms must have a coefficient of zero.

Question 10. If \( f(x) = x^3 – (k – 2) x^2 + 2x \), for all \( x \) and if it is an odd function, find \( k \).
Answer: The given function is \( f(x) = x^3 – (k – 2) x^2 + 2x \).
A function is defined as an odd function if \( f(-x) = -f(x) \) for all values of \( x \).

First, let's find \( f(-x) \):
\( f(-x) = (-x)^3 – (k – 2) (-x)^2 + 2(-x) \)
\( = -x^3 – (k – 2) (x^2) - 2x \)
\( = -x^3 – (k – 2) x^2 - 2x \)

Next, let's find \( -f(x) \):
\( -f(x) = -[x^3 – (k – 2) x^2 + 2x] \)
\( = -x^3 + (k – 2) x^2 - 2x \)

Now, we set \( f(-x) \) equal to \( -f(x) \) since the function is odd:
\( -x^3 – (k – 2) x^2 - 2x = -x^3 + (k – 2) x^2 - 2x \)

Add \( x^3 \) to both sides:
\( – (k – 2) x^2 - 2x = (k – 2) x^2 - 2x \)

Add \( 2x \) to both sides:
\( – (k – 2) x^2 = (k – 2) x^2 \)

Move all terms to one side:
\( 0 = (k – 2) x^2 + (k – 2) x^2 \)
\( 0 = 2(k – 2) x^2 \)

For this equation to be true for all values of \( x \), the coefficient of \( x^2 \) must be zero.
So, \( 2(k – 2) = 0 \)
\( \implies k – 2 = 0 \)
\( \implies k = 2 \)
Therefore, for the function to be odd, the value of \( k \) must be 2. This makes the even power term vanish, which is required for an odd function.
In simple words: An odd function is one where \( f(-x) \) is the same as \( -f(x) \). So, first replace 'x' with '-x' in the given equation. Then, take the negative of the original equation. Set these two results equal to each other. When you simplify, all the terms with \( x^3 \) and \( x \) will cancel, leaving you with \( -(k-2)x^2 = (k-2)x^2 \). This means \( 2(k-2)x^2 = 0 \), so \( k-2 \) must be 0, which means \( k=2 \).

🎯 Exam Tip: For an odd function, \( f(-x) = -f(x) \). An odd function must only contain terms with odd powers of \( x \), or terms that cancel out to zero when tested for oddness. Therefore, any even power terms (like \( x^2 \)) must have a coefficient of zero.

Question 11. Is there a function \( f \) which is both even and odd ?
Answer: Yes, there is a function that can be both even and odd.
Let's assume a function \( f(x) \) is both an even function and an odd function.
If \( f(x) \) is an even function, by definition, we have:
\( f(-x) = f(x) \) (Equation 1)

If \( f(x) \) is an odd function, by definition, we have:
\( f(-x) = -f(x) \) (Equation 2)

Since both statements are true for the same function \( f(x) \), we can equate the right-hand sides of Equation 1 and Equation 2:
\( f(x) = -f(x) \)

Now, we solve for \( f(x) \):
Add \( f(x) \) to both sides:
\( f(x) + f(x) = 0 \)
\( 2f(x) = 0 \)
Divide by 2:
\( f(x) = 0 \)
This means that the only function which is both even and odd is the zero function, i.e., \( f(x) = 0 \) for all \( x \). The zero function satisfies both conditions simultaneously.
In simple words: Yes, only one function can be both even and odd. If a function is even, \( f(-x) \) is the same as \( f(x) \). If it's odd, \( f(-x) \) is the same as \( -f(x) \). If both are true, then \( f(x) \) must be equal to \( -f(x) \). This can only happen if \( f(x) \) is always 0. So, the function \( f(x) = 0 \) is the only one that fits both definitions.

🎯 Exam Tip: This is a conceptual question. To prove it, use the definitions of both even and odd functions simultaneously. Equating the results for \( f(-x) \) will directly lead to the conclusion that \( f(x) \) must be zero.

Question 12. The function \( f (x) = \log (x + \sqrt{x^2+1}) \), is
(a) an even function
(b) an odd function
(c) a periodic function
(d) Neither an even nor an odd function.
Answer: (b) an odd function
In simple words: To check if a function is even or odd, 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. For this function, after replacing 'x' with '-x' and doing some math, you will find it becomes the negative of the original function, so it is an odd function.

🎯 Exam Tip: To determine if a function is even or odd, calculate \( f(-x) \). If \( f(-x) = f(x) \), it's even. If \( f(-x) = -f(x) \), it's odd. If neither applies, it's neither.

Question 13. Prove that \( f (x) = \frac{1}{x}\log\left(x+\sqrt{x^2+1}\right) \) is an even function.
Answer: The given function is \( f(x) = \frac{1}{x}\log\left(x+\sqrt{x^2+1}\right) \).
To prove it is an even function, we need to show that \( f(-x) = f(x) \).

First, let's find \( f(-x) \):
\( f(-x) = \frac{1}{-x}\log\left(-x+\sqrt{(-x)^2+1}\right) \)
\( = -\frac{1}{x}\log\left(-x+\sqrt{x^2+1}\right) \)

Now, we want to manipulate \( \log\left(-x+\sqrt{x^2+1}\right) \) to relate it back to \( \log\left(x+\sqrt{x^2+1}\right) \).
We can multiply the argument of the logarithm by a conjugate:
\( -x+\sqrt{x^2+1} = \left(-x+\sqrt{x^2+1}\right) \times \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}} \)
\( = \frac{(\sqrt{x^2+1})^2 - x^2}{x+\sqrt{x^2+1}} \)
\( = \frac{x^2+1-x^2}{x+\sqrt{x^2+1}} \)
\( = \frac{1}{x+\sqrt{x^2+1}} \)

Now, substitute this back into \( f(-x) \):
\( f(-x) = -\frac{1}{x}\log\left(\frac{1}{x+\sqrt{x^2+1}}\right) \)
Using the logarithm property \( \log\left(\frac{1}{A}\right) = -\log A \):
\( = -\frac{1}{x}\left(-\log\left(x+\sqrt{x^2+1}\right)\right) \)
\( = \frac{1}{x}\log\left(x+\sqrt{x^2+1}\right) \)

This is exactly \( f(x) \).
Thus, \( f(-x) = f(x) \), which proves that the function \( f(x) \) is an even function. This involves using the conjugate to simplify the logarithmic argument.
In simple words: To prove the function is even, we need to show that if you replace 'x' with '-x', the function stays the same. So, first replace 'x' with '-x'. Then, in the logarithm part, multiply the inside term \( (-x+\sqrt{x^2+1}) \) by \( \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}} \) (its conjugate pair). This will simplify the inside of the log to \( \frac{1}{x+\sqrt{x^2+1}} \). Use the log rule that \( \log(\frac{1}{A}) = -\log A \). This will cancel out the initial negative sign, bringing you back to the original function, proving it is even.

🎯 Exam Tip: When dealing with logarithmic functions involving \( x+\sqrt{x^2+1} \) or similar forms, remember to use the conjugate \( \sqrt{x^2+1}-x \) or \( \sqrt{x^2+1}+x \) to simplify expressions, especially when checking for even or odd properties.