Questions and Answers (Board Exam)

Question 1 (CBSE 2019): Relations

Check whether the relation R in the set of real numbers $\mathbb{R}$ defined by $R = \{(a, b) : a \le b^3\}$ is reflexive, symmetric or transitive.

Explanation:

• Reflexive: We need to check if $a \le a^3$ for all $a \in \mathbb{R}$. This is not always true. For a counterexample, let $a=1/2$. The condition becomes $1/2 \le (1/2)^3$, which is $0.5 \le 0.125$. This is false. Therefore, R is not reflexive.

• Symmetric: We need to check if $(a,b) \in R$ implies $(b,a) \in R$. For a counterexample, let $a=1, b=2$. The pair $(1,2)$ is in R because $1 \le 2^3=8$. However, for the pair $(2,1)$, the condition is $2 \le 1^3=1$, which is false. Therefore, R is not symmetric.

• Transitive: We need to check if $(a,b) \in R$ and $(b,c) \in R$ implies $(a,c) \in R$. For a counterexample, let $a=3, b=3/2, c=6/5$.

  • $(a,b) \in R$? $3 \le (1.5)^3 = 3.375$. (True)
  • $(b,c) \in R$? $1.5 \le (1.2)^3 = 1.728$. (True)
  • Is $(a,c) \in R$? $3 \le (1.2)^3 = 1.728$. (False) Since the condition fails, R is not transitive.

Question 2 (CBSE 2011): Relations

Show that the relation R in the set A of all the books in a library, given by $R = \{(x, y) : x \text{ and } y \text{ have the same number of pages}\}$ is an equivalence relation.

Explanation:

• Reflexive: For any book $x$, it is true that book $x$ and book $x$ have the same number of pages. So, $(x,x) \in R$. R is reflexive.

• Symmetric: If $(x,y) \in R$, it means book $x$ and book $y$ have the same number of pages. This implies that book $y$ and book $x$ also have the same number of pages. So, $(y,x) \in R$. R is symmetric.

• Transitive: If $(x,y) \in R$ and $(y,z) \in R$, it means $x$ and $y$ have the same number of pages, and $y$ and $z$ have the same number of pages. It follows that $x$ and $z$ must have the same number of pages. So, $(x,z) \in R$. R is transitive. Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Question 3 (CBSE 2017): Relations

Show that the relation R on the set A = {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Explanation:

• Reflexive: For R to be reflexive on A={1,2,3}, it must contain (1,1), (2,2), and (3,3). None of these are in R. Thus, R is not reflexive.

• Symmetric: For every pair $(a,b)$ in R, the pair $(b,a)$ must also be in R. The only non-diagonal pair is (1,2). Its reverse, (2,1), is also in R. Thus, R is symmetric.

• Transitive: We have $(1,2) \in R$ and $(2,1) \in R$. For transitivity, this requires that $(1,1)$ must also be in R. Since $(1,1)$ is not in R, the condition fails. Thus, R is not transitive.

Question 4 (CBSE 2020): Relations

Let R be a relation on the set of natural numbers given by $R = \{(a, b) : a = b-2, b > 6\}$. Determine if (6,8) is in R.

Explanation: For the pair $(a,b) = (6,8)$ to be in R, it must satisfy two conditions:

1. $b > 6$: Here, $b=8$, and $8 > 6$ is true.

2. $a = b-2$: Here, $a=6$ and $b=8$. $6 = 8-2$ is true.

Since both conditions are satisfied, the pair (6,8) is in R.

Question 5 (CBSE Sample Paper): Relations

Let A = {1, 2, 3}. Write the smallest equivalence relation on A.

Explanation: An equivalence relation must be reflexive, symmetric, and transitive. To be the 'smallest', it must contain only the pairs that are absolutely necessary.

• Reflexivity: The relation must contain (1,1), (2,2), and (3,3) to be reflexive on A.

• Symmetry & Transitivity: The set $R = \{(1,1), (2,2), (3,3)\}$ (the identity relation) has no pairs that violate symmetry or transitivity. For example, for symmetry, there are no $(a,b)$ pairs where $a \ne b$, so the condition is vacuously true. The same applies to transitivity.

Therefore, the identity relation itself is the smallest possible equivalence relation on any set.

Answer: $R = \{(1,1), (2,2), (3,3)\}$.

Question 6 (CBSE 2018): Relations

Show that the relation R on the set of points in a plane, given by $R = \{(P,Q) : \text{distance of P from the origin is the same as the distance of Q from the origin}\}$ is an equivalence relation.

Explanation: Let OP denote the distance of a point P from the origin.

• Reflexive: For any point P, OP = OP. Thus, $(P,P) \in R$. R is reflexive.

• Symmetric: If $(P,Q) \in R$, then OP = OQ. This equality implies OQ = OP. Thus, $(Q,P) \in R$. R is symmetric.

• Transitive: If $(P,Q) \in R$ and $(Q,S) \in R$, then OP = OQ and OQ = OS. By the transitive property of equality, this implies OP = OS. Thus, $(P,S) \in R$. R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation. (Geometrically, this relation groups all points that lie on the same circle centered at the origin).

Question 7 (CBSE 2019): Relations

Let A = {1, 2, 3, …, 9} and R be the relation on $A \times A$ defined by $(a,b)R(c,d)$ if $a+d=b+c$. Prove that R is an equivalence relation.

Explanation: The condition $a+d=b+c$ can be rearranged to $a-b=c-d$.

• Reflexive: We check if $(a,b)R(a,b)$. This requires $a+b=b+a$, which is true. R is reflexive.

• Symmetric: If $(a,b)R(c,d)$, then $a+d=b+c$. We check if $(c,d)R(a,b)$, which requires $c+b=d+a$. Since these are the same equation, the property holds. R is symmetric.

• Transitive: If $(a,b)R(c,d)$ and $(c,d)R(e,f)$, then (1) $a+d=b+c$ and (2) $c+f=d+e$. We want to prove $a+f=b+e$. Adding the two equations: $$(a+d)+(c+f)=(b+c)+(d+e)$$

  Cancelling $d$ and $c$ from both sides gives $a+f=b+e$. R is transitive.

Therefore, R is an equivalence relation.

Question 8 (CBSE 2012): Functions

Show that the function $f: \mathbb{N} \to \mathbb{N}$, given by $f(x) = 2x$, is one-one but not onto.

Explanation:

• One-one (Injective): Let $f(x_1)=f(x_2)$ for some $x_1, x_2 \in \mathbb{N}$. This means $2x_1=2x_2$. Dividing by 2, we get $x_1=x_2$. Since $f(x_1)=f(x_2)$ implies $x_1=x_2$, the function is one-one.

• Onto (Surjective): For a function to be onto, its range must be equal to its codomain.

  • The codomain is $\mathbb{N} = \{1, 2, 3, 4, ...\}$.

  • The range is the set of all outputs: $f(1)=2, f(2)=4, f(3)=6, ...$. The range is the set of all even natural numbers.

  Since the range is not equal to the codomain (for example, the odd number 3 is in the codomain but has no pre-image), the function is not onto.

Question 9 (CBSE 2017): Functions

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^4$. Show that f is neither one-one nor onto.

Explanation:

• One-one: To show it's not one-one, we need a counterexample. Let $x_1 = -1$ and $x_2=1$. Here $x_1 \ne x_2$. But $f(-1) = (-1)^4 = 1$ and $f(1) = 1^4 = 1$. Since different inputs produce the same output, f is not one-one (it is many-one).

• Onto: The codomain is $\mathbb{R}$. The range of $f(x)=x^4$ consists of only non-negative numbers, since any real number raised to the fourth power is non-negative. The range is $[0, \infty)$. Since the range is not equal to the codomain (e.g., -1 has no pre-image), f is not onto.

Question 10 (CBSE 2016): Functions

State whether the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x)=x^2$ is one-one and onto.

Explanation:

• One-one: Let $f(x_1)=f(x_2)$ for $x_1, x_2 \in \mathbb{N}$.

  Then $x_1^2=x_2^2$. Since the domain is natural numbers (which are positive), we can take the positive square root of both sides, which implies $x_1=x_2$. Therefore, the function is one-one.

• Onto: The codomain is $\mathbb{N} = \{1,2,3,...\}$. The range is the set of outputs: $f(1)=1, f(2)=4, f(3)=9, ...$.

  The range is the set of perfect squares. Since the range is not equal to the codomain (e.g., 2 has no pre-image), the function is not onto.

Question 11 (CBSE 2018): Functions

Show that the function $f(x)=|x|$ from $\mathbb{R}$ to $\mathbb{R}$ is neither one-one nor onto.

Explanation:

• One-one: To show it's not one-one, we find a counterexample. Let $x_1=-1$ and $x_2=1$. Then $f(-1)=|-1|=1$ and $f(1)=|1|=1$.

  Since different inputs give the same output, f is not one-one.

• Onto: The codomain is $\mathbb{R}$. The range of the modulus function is the set of all non-negative numbers, $[0, \infty)$.

  Since the range is not equal to the codomain (e.g., any negative number like -1 has no pre-image), f is not onto.

Question 12 (CBSE 2011): Functions

Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Consider the function $f: A \to B$ defined by $f(x) = \frac{x-2}{x-3}$. Is f one-one and onto?

Explanation:

• One-one: Let $f(x_1)=f(x_2)$. Then $$\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$$

  Cross-multiplying gives $$(x_1-2)(x_2-3)=(x_2-2)(x_1-3),$$ which simplifies to $$x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6$$

  This further simplifies to $-x_1=-x_2$, or $x_1=x_2$. So, f is one-one.

• Onto: Let $y$ be an arbitrary element in the codomain B. Set $y = \frac{x-2}{x-3}$ and solve for x. $$y(x-3)=x-2$$ $$\implies xy-3y=x-2$$ $$\implies x(y-1)=3y-2$$ $$\implies x = \frac{3y-2}{y-1}$$

  Since $y \in B$, we know $y \ne 1$, so a valid pre-image $x$ can always be found. We must also check that this $x$ is never 3. If $x=3$, then $$3 = \frac{3y-2}{y-1} \implies 3y-3=3y-2 \implies -3=-2$$ which is a contradiction. Thus, the pre-image $x$ is never 3, so it is always in the domain A. f is onto.

Conclusion: Yes, the function is one-one and onto (bijective).

Question 13 (CBSE 2014): Composition and Inverses

Let $f(x) = x+7$ and $g(x) = x-7$, $x \in \mathbb{R}$. Find $(f \circ g)(x)$.

Explanation: The composition $(f \circ g)(x)$ is defined as $f(g(x))$.

1. Substitute g(x): Start with the outer function $f$ and replace its variable with the entire function $g(x)$. $$f(g(x)) = f(x-7)$$

2. Apply f: Now apply the rule for $f$, which is 'add 7' to the input. $$f(x-7) = (x-7)+7 = x$$

Answer: $(f \circ g)(x) = x$.

Question 14 (CBSE 2015): Composition and Inverses

If $f(x) = 8x^3$ and $g(x) = x^{1/3}$, then find $(g \circ f)(x)$.

Explanation: The composition $(g \circ f)(x)$ is defined as $g(f(x))$.

1. Substitute f(x):

   Start with the outer function $g$ and replace its variable with the entire function $f(x)$.

   $g(f(x)) = g(8x^3)$.

2. Apply g:

   Now apply the rule for $g$, which is 'take the cube root' of the input.

   $g(8x^3) = (8x^3)^{1/3} = 8^{1/3} \cdot (x^3)^{1/3} = 2x$.

Answer: $(g \circ f)(x) = 2x$.

Question 15 (CBSE 2018): Composition and Inverses

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 3-4x$. Find its inverse.

Explanation:

1. Replace f(x) with y: $y=3-4x$.

2. Swap x and y: $x=3-4y$.

3. Solve for y: $x-3 = -4y \implies 4y = 3-x \implies y=\frac{3-x}{4}$.

4. State the inverse: $f^{-1}(x)=\frac{3-x}{4}$.

Answer: $f^{-1}(x)=\frac{3-x}{4}$.

Question 16 (CBSE 2017): Composition and Inverses

Let $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x) = (3-x^3)^{1/3}$. Find $(f \circ f)(x)$.

Explanation: We need to compute $f(f(x))$.

1. Substitute f(x) into itself: $$(f \circ f)(x) = f(f(x)) = (3 - (f(x))^3)^{1/3}$$

2. Replace f(x) with its definition: $$(f \circ f)(x) = (3 - ((3-x^3)^{1/3})^3)^{1/3}$$

3. Simplify: The cube and the cube root cancel each other out. $$(f \circ f)(x) = (3 - (3-x^3))^{1/3} = (3 - 3 + x^3)^{1/3} = (x^3)^{1/3} = x$$

Answer: $(f \circ f)(x) = x$.

Question 17 (CBSE 2014): Composition and Inverses

Let $f(x) = \frac{4x+3}{6x-4}, x \ne 2/3$. Show that $f(f(x))=x$ and find the inverse of f.

Explanation: Part 1: Show that $f(f(x))=x$.

1. Compose the function with itself: $$f(f(x)) = \frac{4(f(x))+3}{6(f(x))-4} = \frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4}$$

2. Simplify: Multiply the numerator and the denominator by $(6x-4)$ to clear the complex fraction. $$f(f(x)) = \frac{4(4x+3)+3(6x-4)}{6(4x+3)-4(6x-4)} = \frac{16x+12+18x-12}{24x+18-24x+16} = \frac{34x}{34} = x$$

Part 2: Find the inverse of f.

Since we have shown that $f(f(x))=x$, this means that applying the function twice returns the original input. This is the definition of a function that is its own inverse.

Therefore, $f^{-1}(x) = f(x)$.

Answer: $f^{-1}(x) = \frac{4x+3}{6x-4}$.

Question 18 (CBSE 2011): Composition and Inverses

Consider $f: \mathbb{R}_+ \to [4, \infty)$ given by $f(x)=x^2+4$. Find the inverse of f.

Explanation: Here, $\mathbb{R}_+$ represents the set of non-negative real numbers, i.e., $[0, \infty)$.

1. Replace f(x) with y: $y=x^2+4$.

2. Swap x and y: $x=y^2+4$.

3. Solve for y: $y^2=x-4 \implies y=\pm\sqrt{x-4}$.

4. Choose the Correct Root: The range of the inverse function $f^{-1}$ must be the domain of the original function $f$, which is $[0, \infty)$. To ensure our output $y$ is non-negative, we must choose the positive root.

   Therefore, $f^{-1}(x) = \sqrt{x-4}$.

Question 19 (CBSE 2019): Composition and Inverses

If $f(x)=|x|$ and $g(x)=|5x-2|$, find $g \circ f$.

Explanation: We need to find $(g \circ f)(x) = g(f(x))$.

1. Substitute f(x) into g(x): $$g(f(x)) = g(|x|)$$

2. Apply the function g: The rule for $g$ is to take 5 times the input, subtract 2, and take the absolute value. $$g(|x|) = |5|x|-2|$$

Answer: $(g \circ f)(x) = |5|x|-2|$.

Question 20 (CBSE Sample Paper): Composition and Inverses

If $f(x)=x^2-1$, find $(f \circ f)(x)$.

Explanation: We need to compute $f(f(x))$.

1. Substitute f(x) into itself: $$(f \circ f)(x) = f(f(x)) = f(x^2-1)$$

2. Apply the function f: The rule for $f$ is to square the input and subtract 1. $$f(x^2-1) = (x^2-1)^2-1$$

3. Expand and Simplify: $$(x^2-1)^2-1 = (x^4 - 2x^2 + 1) - 1 = x^4-2x^2$$

Answer: $(f \circ f)(x) = x^4-2x^2$.

Question 21 (CBSE 2019): Composition and Inverses

Show that the function $f$ in $A = \mathbb{R} - \{2/3\}$ defined as $f(x) = \frac{4x+3}{6x-4}$ is one-one and onto. Hence, find $f^{-1}$.

Explanation: This question is related to Example 17.

Method 1: Using Composition In Example 17, we showed that $f(f(x))=x$. Let's analyze this result.

• Since $f(f(x)) = x$, applying the function $f$ to any output of $f$ returns the original input. This means $f$ is its own inverse. $f^{-1}(x) = f(x)$.

• A function that has an inverse must be bijective (one-one and onto).

• Therefore, the function is one-one and onto, and its inverse is $f^{-1}(x) = \frac{4x+3}{6x-4}$.

Method 2: Direct Proof

• One-one: We would show $f(x_1)=f(x_2) \implies x_1=x_2$ through cross-multiplication, which holds true.

• Onto: We would solve for $x$ in terms of $y$ to get $x = \frac{4y+3}{6y-4}$. This shows that a pre-image exists for every $y$ except $y=2/3$. Since the codomain is given as $A = \mathbb{R} - \{2/3\}$, the function is onto.

Question 22 (CBSE 2020): Composition and Inverses

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 3x+2$. Show that f is invertible.

Explanation: A function is invertible if it is bijective (both one-one and onto).

• One-one:

  Let $f(x_1)=f(x_2)$ $$\implies 3x_1+2=3x_2+2 \implies 3x_1=3x_2 \implies x_1=x_2$$

  The function is injective.

• Onto: Let $y \in \mathbb{R}$. We need to find an $x$ such that $f(x)=y$. Set $y=3x+2$. Solving for $x$ gives $x=(y-2)/3$. For every real number $y$, this formula gives a valid real number $x$. Thus, the function is surjective.

Since the function is both one-one and onto, it is bijective and therefore invertible.

Question 23 (CBSE 2012): Composition and Inverses

If a function $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 2x+5$, find $f^{-1}(x)$.

Explanation:

1. Replace f(x) with y: $y=2x+5$.

2. Swap x and y: $x=2y+5$.

3. Solve for y: $x-5 = 2y \implies y=(x-5)/2$.

4. State the inverse: $f^{-1}(x)=(x-5)/2$.

Answer: $f^{-1}(x) = \frac{x-5}{2}$.

Question 24 (CBSE 2015): Functions

Let $f, g: \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=x^2$ and $g(x)=x+1$. Find $(f+g)(2)$ and $(fg)(1)$.

Explanation: This question is about the algebra of functions.

• Sum of functions, $(f+g)(x)$: This is defined as $f(x)+g(x)$.

  $$(f+g)(x) = x^2+x+1$$

  To find $(f+g)(2)$, we substitute $x=2$: $$(f+g)(2) = 2^2+2+1 = 4+2+1 = 7$$

• Product of functions, $(fg)(x)$: This is defined as $f(x)g(x)$. $$(fg)(x) = x^2(x+1)$$

  To find $(fg)(1)$, we substitute $x=1$: $$(fg)(1) = 1^2(1+1) = 1(2) = 2$$

Answer: $(f+g)(2) = 7$ and $(fg)(1) = 2$.

Question 25 (CBSE 2013): Composition and Inverses

If $f(x) = x/(x-1)$, find $(f \circ f)(x)$.

Explanation: We need to compute $f(f(x))$.

$$(f \circ f)(x) = f(f(x)) = \frac{f(x)}{f(x)-1} = \frac{x/(x-1)}{x/(x-1)-1}$$

To simplify the complex fraction, multiply the numerator and denominator by $(x-1)$: $$(f \circ f)(x) = \frac{x}{x - 1(x-1)} = \frac{x}{x-x+1} = \frac{x}{1} = x$$

Answer: $(f \circ f)(x) = x$.

Question 26 (CBSE 2018): Relations

Let R be a relation on $\mathbb{Z}$ defined by $aRb$ if $|a-b|$ is divisible by 4. Show it is an equivalence relation.

Explanation:

• Reflexive: For any integer $a$, $|a-a|=0$. Since $0 = 4 \times 0$, 0 is divisible by 4. R is reflexive.

• Symmetric: If $aRb$, then $|a-b|$ is divisible by 4. Since $|b-a|=|-(a-b)|=|a-b|$, $|b-a|$ is also divisible by 4. Thus, $bRa$. R is symmetric.

• Transitive: If $aRb$ and $bRc$, then $|a-b|$ and $|b-c|$ are divisible by 4. This means $a-b = 4k_1$ and $b-c = 4k_2$ for some integers $k_1, k_2$. (Note: if $|x|$ is a multiple of 4, then $x$ must also be a multiple of 4). Adding the two equations: $$(a-b)+(b-c) = 4k_1+4k_2 \implies a-c = 4(k_1+k_2)$$

  This means $a-c$ is divisible by 4, and so is $|a-c|$. Thus, $aRc$. R is transitive.

Conclusion: It is an equivalence relation.

Question 27 (CBSE 2019): Functions

Let $f: \mathbb{R} - \{0\} \to \mathbb{R}$ be defined by $f(x)=1/x$. Is f one-one and onto?

Explanation:

• One-one: Let $f(x_1)=f(x_2)$. Then $1/x_1=1/x_2$. Taking the reciprocal of both sides gives $x_1=x_2$. The function is one-one.

• Onto: The codomain is $\mathbb{R}$. The range of $f(x)=1/x$ is the set of all non-zero real numbers, $\mathbb{R}-\{0\}$. Since the range is not equal to the codomain (the value 0 has no pre-image), the function is not onto.

Question 28 (CBSE 2020): Relations

Let $A=\{1,2,3\}$. Let $R=\{(1,1),(2,2),(3,3),(1,3)\}$. Is R transitive?

Explanation: The condition for transitivity is: if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c)$ must also be in R. We only need to check cases where the 'if' part is true.

• Let's check the chain $(1,1) \in R$ and $(1,3) \in R$. This requires $(1,3)$ to be in R, which it is. This case holds.
• Let's check $(3,3) \in R$. There are no pairs of the form $(3,c)$ in R to continue a chain.

There are no other chains of the form $(a,b)$ and $(b,c)$ where $a \ne b$ and $b \ne c$.

Since we cannot find any instance where the 'if' condition is met but the 'then' condition fails, the relation is transitive by vacuous truth.

Question 29 (CBSE 2016): Composition and Inverses

Let $f(x)=x^2$ and $g(x)=x+1$. Find $(f \circ g)(2)$.

Explanation: We need to calculate $(f \circ g)(2) = f(g(2))$. We work from the inside out.

1. Evaluate the inner function g(2): $g(2) = 2+1 = 3$.

2. Evaluate the outer function f(3): $f(g(2)) = f(3) = 3^2 = 9$.

Answer: 9.

Question 30 (CBSE 2012): Relations

If a relation R is defined on the set A={1,2,3,4,5,6} as $R=\{(x,y) : y \text{ is divisible by } x\}$. Is R an equivalence relation?

Explanation:

• Reflexive: Is $x$ divisible by $x$? Yes, any non-zero integer is divisible by itself. R is reflexive.

• Symmetric: If $y$ is divisible by $x$, is $x$ divisible by $y$? Not always. For a counterexample, take the pair (2,4). Since 4 is divisible by 2, $(2,4) \in R$. However, 2 is not divisible by 4, so $(4,2) \notin R$. R is not symmetric.

• Transitive: If $y$ is divisible by $x$ and $z$ is divisible by $y$, is $z$ divisible by $x$? Yes. If $y=kx$ and $z=jy$, then $z=j(kx)=(jk)x$. R is transitive.

Conclusion: Since the relation is not symmetric, it is not an equivalence relation.