Q.Prove that the Greatest Integer Function $f: \mathbf{R} \to \mathbf{R}$, given by $f(x) = [x]$, is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.

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 $f(n)=-\frac{2m}{2}=-m=k$. Every integer appears, so $f$ is onto.

3. Conclusion

   Since $f$ is both one‑one and onto, it is a bijection from $\mathbb{N}$ to $\mathbb{Z}$.

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 $|A| = r \cdot 1 - 0 = r$.

   This works for every $r \in \mathbb{R}$. So every real number is attained — $f$ is onto.

3. Conclusion

   Since $f$ is onto but not one‑one, the correct classification is (C).

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.

   For $x\ge 0$, $f(x)=\tanh x$ ranges from $0$ (at $x=0$) to $1$ (as $x\to\infty$, $\tanh x\to 1^-$). So outputs from $[0,1)$ are covered.

   For $x<0$, output is exactly $0$.

   Thus the range of $f$ is $[0,1)$.

   Numbers like $-1$, $2$, or $100$ are never attained.

   Hence $f$ is not surjective.

4. Conclusion

   Since $f$ is neither injective nor surjective, the correct choice is (C).

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.g. $-1$ is never attained). Thus $f$ is not surjective (onto).

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.

A function is onto (surjective) if every element of $B$ is the image of at least one element of $A$. A function $A\to B$ with $|B|=2$ fails to be onto exactly when it misses one of the two elements — i.e. when it is constant. There are exactly $2$ such constant functions (everything $\mapsto 2$, and everything $\mapsto 3$). Hence

(This is the inclusion–exclusion formula $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(n-k)^{m}$ with $n=2$, $m=5$: $2^5-2\cdot1^5=30$.)

ANSWER: C

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 $2 * x = 3^{-1}$.

4. Why not $\tfrac{5}{12}$? That is what you get by mis-reading $3^{-1}$ as the ordinary reciprocal $\tfrac13$ ($\tfrac{4x}{5}=\tfrac13 \Rightarrow x=\tfrac{5}{12}$). In the binary-operation context the notation $3^{-1}$ denotes the inverse with respect to $*$, which is why $\tfrac{125}{48}$ is offered — and it is the consistent reading.

ANSWER: D

"$P(n)$ 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 $n$.