Q.An organization conducted a bike race under 2 different categories — boys and girls. In all, there were 250 participants. Among all of them finally three from Category 1 and two from Category 2 were selected for the final race. Ravi forms two sets $B$ and $G$ with these participants for his college project. Let $B = \{b_1, b_2, b_3\}$, $G = \{g_1, g_2\}$ where $B$ represents the set of boys selected and $G$ the set of girls who were selected for the final race. Ravi decides to explore these sets for various types of relations and functions. On the basis of the above information, answer the following questions:

Concept and intuition: A relation from set $A$ to set $B$ is any subset of the Cartesian product $A \times B$. The number of elements in $A \times B$ is $n(A) \times n(B)$. For a set with $m$ elements, the number of subsets is $2^m$. So the total number of possible relations is $2^{n(A) \cdot n(B)}$. This is the key formula — once you know it, the problem becomes a simple exponential equation.

1. Write the formula for the number of relations.

   If $n(A) = a$ and $n(B) = b$, then the number of relations from $A$ to $B$ is $2^{a \cdot b}$.

2. Plug in the given values.

   We have $a = 2$ and the total number of relations $= 1024$. So:

3. Express 1024 as a power of 2. $1024 = 2^{10}$ (since $2^{10} = 1024$). Therefore:

4. Equate the exponents. Since the bases are equal, the exponents must be equal:

5. Solve for $b$. $b = \frac{10}{2} = 5$ …

Step 1 — The counting rule for a Cartesian product.

To build one ordered pair you choose the first coordinate in $n(A)$ ways and, independently, the second in $n(B)$ ways. By the fundamental principle of counting,

Step 2 — Set up the equation.

We are told $n(A\times B)=7$, so

Step 3 — Use the fact that 7 is PRIME.

$p$ and $q$ are cardinalities of sets, so they must be non-negative integers — you cannot have $2.5$ elements. Their product is $7$, and the only factorisations of the prime 7 into two positive integers are

$7 = 1 \times 7 \quad\text{or}\quad 7 = 7\times 1 .$ …

Concept & Intuition

The Cartesian product $A \times B$ is the set of all ordered pairs $(a,b)$ with $a \in A$ and $b \in B$. Similarly, $B \times A$ is all pairs $(b,a)$ with $b \in B$ and $a \in A$.

For a pair to be in both products, its first coordinate must belong to $A$ (to be in $A \times B$) and also belong to $B$ (to be in $B \times A$). So the first coordinate must be in $A \cap B$. By the same reasoning, the second coordinate must also be in $A \cap B$.

Thus the intersection is exactly $(A \cap B) \times (A \cap B)$. The number of elements is simply the square of the size of the intersection.

Step-by-step solution

1. Find the intersection of the two sets

   $A = \{1,2,3,4,5\}$, $B = \{2,3,6,7\}$.

   The common elements are $2$ and $3$.

   So $|A \cap B| = 2$.

2. Characterize the intersection of the Cartesian products

   A pair $(x,y)$ belongs to $(A \times B) \cap (B \times A)$ iff:

   • $x \in A$ and $y \in B$ (from $A \times B$), and
   • $x \in B$ and $y \in A$ (from $B \times A$). Together this means $x \in A \cap B$ and $y \in A \cap B$.

3. Count the number of such pairs

   The set of all such pairs is $(A \cap B) \times (A \cap B)$. …

Step 1 — The cardinality of a Cartesian product

For finite sets,

because every ordered pair pairs one of the $n(A)$ elements with one of the $n(B)$ elements.

Step 2 — Use the "non-singleton" condition

$35 = 1\times 35 = 5\times 7 = 7 \times 5 = 35 \times 1$.

A non-singleton set has more than one element, i.e. $n \ge 2$. That kills every factorisation containing a $1$. The only survivor is

Step 3 — Use $B \subset A$ to decide which is which

If $B$ is contained in $A$, then $B$ cannot have more elements than $A$:

$n(B) \le n(A) \;\Longrightarrow\; n(B) = 5,\quad n(A) = 7$ …

Checking one-one (injectivity)

This is positive on $(-1,1)$ and negative on $(-\infty,-1)\cup(1,\infty)$, so $f$ increases then decreases — it is not monotonic on $\mathbb{R}$, and since its two turning points $f(-1)=-\tfrac12$ and $f(1)=\tfrac12$ are different heights, values between them are attained twice.

To confirm algebraically: solving $f(x)=y$ gives $yx^2 - x + y = 0$. For $0<|y|<\tfrac12$, the discriminant $1-4y^2>0$, so there are two distinct real roots $x_1, x_2$ (with $x_1 x_2 = 1$, i.e. $x_2 = 1/x_1$). So $f$ repeats values — not one-one.

Checking onto (surjectivity)

The same equation $yx^2 - x + y = 0$ has a real solution only when the discriminant $1-4y^2 \ge 0$, i.e. $|y| \le \tfrac12$. So the range of $f$ is $[-\tfrac12, \tfrac12]$, a proper subset of the codomain $\mathbb{R}$ — not onto. …

We need to decide whether $f:\mathbb{N}\to\mathbb{Z}$ given by

is injective (one‑one) and/or surjective (onto).

The key idea: the function “splits” the natural numbers into two tracks — odds go to non‑negative integers (including 0), evens go to negative integers. If we list the outputs in order of $n$, we get $0, -1, 1, -2, 2, -3, 3, \dots$ — exactly the integers, each appearing exactly once. That suggests a bijection.

1. Check one‑one (injectivity)

   Suppose $f(a)=f(b)$. We must show $a=b$.

   • If both $a,b$ are odd: $\frac{a-1}{2}=\frac{b-1}{2} \Rightarrow a=b$.
   • If both $a,b$ are even: $-\frac{a}{2}=-\frac{b}{2} \Rightarrow a=b$.
   • If one is odd and the other even: Then one output is non‑negative (odd case) and the other is negative (even case). They cannot be equal because a non‑negative number equals a negative number only if both are 0. But the odd case gives 0 only when $n=1$; the even case gives 0 only when $n=0$, and 0 is not a natural number. So this case never happens. Hence $f$ is one‑one.

2. Check onto (surjectivity)

   We need every integer $k$ to be hit by some $n\in\mathbb{N}$.

   • If $k\ge 0$: set $n=2k+1$ (odd). Then $f(n)=\frac{(2k+1)-1}{2}=k$.
   • If $k<0$: write $k=-m$ with $m>0$. Set $n=2m$ (even). Then …

Concept & Intuition

We are asked about the function $f(A) = |A|$, where $|A|$ denotes the determinant of the $2\times2$ matrix $A$. The domain is all $2\times2$ real matrices, and the codomain is all real numbers.

• One‑one (injective) means: if $f(A)=f(B)$ then $A=B$. But many different matrices can have the same determinant — for instance, swapping rows changes the sign of the determinant, but the matrices are different. So it’s unlikely to be one‑one.
• Onto (surjective) means: every real number $r$ appears as the determinant of some $2\times2$ matrix. Since we can easily build a matrix whose determinant is any given $r$, this should be true.

Let’s check both properties carefully.

1. Check one‑one (injectivity)

   Suppose $A = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ and $B = \begin{pmatrix}2 & 0 \\ 0 & \frac12\end{pmatrix}$.

   Then $|A| = 1$ and $|B| = 2 \cdot \frac12 - 0 = 1$.

   So $f(A)=f(B)$ but $A \neq B$. Hence $f$ is not one‑one.

   (In fact, infinitely many matrices share the same determinant — any matrix with determinant 1, for example.)

2. Check onto (surjectivity)

   Take any real number $r$. We need a $2\times2$ matrix $A$ with $|A| = r$.

   A simple choice: $A = \begin{pmatrix}r & 0 \\ 0 & 1\end{pmatrix}$. Then $|A| = r \cdot 1 - 0 = r$. …

We are given

We need to decide whether $f$ is injective (one-to-one), surjective (onto), both, or neither.

Concept and intuition

The absolute value in the numerator makes the function behave differently for $x\ge 0$ and $x<0$.

• For $x\ge 0$, $|x|=x$, so the numerator becomes $e^{x}-e^{-x}$, which is $2\sinh x$. The denominator is $e^{x}+e^{-x}=2\cosh x$. So for $x\ge 0$, $f(x)=\tanh x$.
• For $x<0$, $|x|=-x$, so the numerator becomes $e^{-x}-e^{-x}=0$. Hence for $x<0$, $f(x)=0$.

Thus the function is constant ($0$) on all negative numbers, and equals $\tanh x$ on $[0,\infty)$.

This immediately suggests:

• Not injective: many different $x<0$ give the same output $0$, and also $f(0)=0$ as well.
• Not surjective: $\tanh x$ only takes values in $[0,1)$ for $x\ge 0$, and $0$ for $x<0$, so the range is $[0,1)$, not all of $\mathbb{R}$.

Step-by-step reasoning

1. Simplify the expression piecewise For $x\ge 0$: $|x|=x$, so

For $x<0$: $|x|=-x$, so

2. Check injectivity

   A function is injective if $f(a)=f(b)$ implies $a=b$.

   Take $a=-1$ and $b=-2$: both are $<0$, so $f(-1)=0$ and $f(-2)=0$, but $-1\neq -2$.

   Also $f(0)=\tanh 0 = 0$, so $f(0)=f(-1)$ yet $0\neq -1$.

   Hence $f$ is not injective.

3. Check surjectivity

   A function is surjective if every real number appears as an output. …

Since $f'(x)=e^x>0$ everywhere, $f$ is strictly increasing, so distinct inputs give distinct outputs — it is injective (one-one). However, $e^x>0$ for all $x$, so the range is $(0,\infty)$, which is a proper subset of the codomain $\mathbb R$ (e …

Step 1 — Identify set $A$.

Step 2 — Identify set $B$.

Both are integers, so they qualify for $x\in\mathbb{Z}$:

Step 3 — Count all functions $A\to B$.

Each of the $5$ elements of $A$ can be sent independently to either of the $2$ elements of $B$:

Step 4 — Remove the ones that are not onto. …

1. Find the identity element $e$. It must satisfy $a * e = a$ for all positive rationals $a$:

(Check: $e*a = \frac{2ea}{5} = a$ too, so $*$ is commutative and $e=\tfrac52$ is the two-sided identity.)

2. Find $3^{-1}$, the inverse of $3$ under $*$. It satisfies $3 * 3^{-1} = e$:

3. Solve $2 * x = 3^{-1}$. $\frac{2\cdot 2\cdot x}{5} = \frac{25}{12} \;\Longrightarrow\; \frac{4x}{5} = \frac{25}{12} \;\Longrightarrow\; x = \frac{5}{4}\cdot\frac{25}{12} = \frac{125}{48}$ …

"$P(n)$ is true for all $n \in \mathbb{N}$" means the divisibility must hold for every natural number, so $k$ must divide $2^{2n}-1$ for all $n$. We find such a $k$ by testing small cases and then confirming in general.

1. Test $n=1$.

   $2^{2(1)} - 1 = 4 - 1 = 3$. So $k$ must divide $3$; the only option that does is $k=3$. (Immediately, $k=2$, $6$ and $7$ fail, since none divides $3$.)

2. Test $n=2$ as a check.

   $2^{2(2)} - 1 = 16 - 1 = 15$, and $3 \mid 15$. So $k=3$ still works.

3. Confirm for all $n$.

   Since $2^{2n} = 4^n$ and $4 \equiv 1 \pmod 3$, we have $4^n \equiv 1^n = 1 \pmod 3$, so

for every $n$. Thus $3$ divides $2^{2n}-1$ for all natural numbers $n$. …