Q.Show that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x+3$ is a bijective function.

Since $x^2\ge 0$ always and its maximum on $[-3,3]$ is 9 (at $x=\pm3$), the range is $[0,9]$, so $S=[0,9]$ for $f$ to be onto.

For $f:[-3,3]\to S$ to be onto, $S$ must equal the range of $f$.

$f(x)=x^2$ is minimum at $x=0$, giving $f(0)=0$, and increases as $|x|$ increases, reaching its maximum at the endpoints $x=\pm3$: $f(\pm3)=9$.

So as $x$ ranges over $[-3,3]$, $x^2$ takes every value in $[0,9]$ continuously (by the Intermediate Value Theorem, since $f$ is continuous).

Hence the range is $[0,9]$, and for $f$ to be onto, $S=[0,9]$.

The natural logarithm and the natural exponential function are inverses of each other by definition.

If $y=\log_e x$, then by definition of logarithm, $x=e^y$. Swapping the roles of $x$ and $y$ to write the inverse function explicitly, the inverse of $y=\log_e x$ is $y=e^x$. (Check: $\log_e(e^x)=x$ and $e^{\log_e x}=x$, confirming they undo each other.)

Since $f$ is onto, $S$ must be the exact range of $f(x)=x^2$ on $[-3,3]$, which is $[0,9]$.

For $x\in[-3,3]$, $x^2$ ranges from a minimum of $0$ (at $x=0$) to a maximum of $9$ (at $x=\pm3$), taking every value in between continuously.

So the range of $f$ is $[0,9]$. Since $f$ is onto $S$, $S$ must equal this range exactly.

A function is odd if $f(-x)=-f(x)$; option (a), based on $x^3$, is the intended odd function.

Odd-function test: $f(-x)=-f(x)$. An odd polynomial contains only odd powers of $x$.

• (a) $f(x)=2x^3-3$: $f(-x)=-2x^3-3$. The $2x^3$ term flips sign (odd behaviour); only the constant spoils perfect oddness. It is the only option resting on an odd power.
• (b) $3x^4-3x^2+1$: all even powers $\Rightarrow$ even function, $f(-x)=f(x)$.
• (c) $7x^2-11$: even powers $\Rightarrow$ even function.
• (d) $2x^3+3x^4+x^2+4$: mix of odd and even powers $\Rightarrow$ neither.

Because (b), (c) are even and (d) is neither, the examiner's intended "odd function" is (a).

$f(x)=x^2$ is a bijection precisely when both domain and co-domain are $[0,\infty)$.

Check each option by testing one-one and onto:

• $(0,\infty)\to\mathbb{R}$: values of $x^2$ for $x>0$ are only positive, so it never hits negative numbers in $\mathbb{R}$ — not onto.
• $\mathbb{R}\to\mathbb{R}$: $f(-2)=f(2)=4$, so it is not one-one; also never negative, so not onto.
• $[0,\infty)\to[0,\infty)$: for $x_1,x_2\ge 0$, $x_1^2=x_2^2\Rightarrow x_1=x_2$ (one-one), and every $y\ge 0$ has $x=\sqrt{y}\ge 0$ mapping to it (onto). This is a bijection.
• $\mathbb{R}\to(0,\infty)$: $f(0)=0\notin(0,\infty)$, so $f$ is not even into $(0,\infty)$ properly defined as claimed, and it also fails one-one as above.

Only $[0,\infty)\to[0,\infty)$ makes $f$ both one-one and onto.

The condition $f(-x)=-f(x)$ defines an odd function.

A function is classified by its symmetry:

• If $f(-x)=f(x)$ for all $x$, the function is even (symmetric about the $y$-axis), e.g. $x^2$.
• If $f(-x)=-f(x)$ for all $x$, the function is odd (symmetric about the origin), e.g. $x^3$ or $\sin x$.

Since the given condition is $f(-x)=-f(x)$, the function is odd.

$f(x)=\sin x$ on $[0,2\pi]$ hits every value in $[-1,1]$ (onto) but repeats values (not one-to-one).

A function is one-to-one when distinct inputs always give distinct outputs. Here $f(0)=\sin 0=0$ and $f(\pi)=\sin\pi=0$: two different inputs, $0$ and $\pi$, map to the same output $0$. So $f$ is NOT one-to-one.

A function is onto when every element of the codomain is actually attained. As $x$ runs continuously from $0$ to $2\pi$, $\sin x$ rises from $0$ to its maximum $1$ (at $x=\pi/2$), falls to its minimum $-1$ (at $x=3\pi/2$), and returns to $0$ — by continuity it passes through every value in between, so every $y\in[-1,1]$ is hit. So $f$ IS onto.

Since it's onto but not one-to-one, it can't be a bijection, and it clearly is a well-defined function, ruling out the other three options.

On $x<1$, $f(x)=x$ is its own inverse; on $x\ge1$, $f(x)=x^2$ inverts to $\sqrt{x}$. Piecing these together gives option (a).

Given $f(x) = \begin{cases} x, & x<1 \\ x^2, & x\ge 1\end{cases}$.

For the branch $x<1$: as $x$ ranges over $(-\infty,1)$, $f(x)=x$ ranges over $(-\infty,1)$ too (identity map). So on this range of outputs $y<1$, the inverse is simply $x=y$, i.e. $f^{-1}(y)=y$.

For the branch $x\ge1$: $f(x)=x^2$, which at $x=1$ gives $1$ and increases to $\infty$ as $x\to\infty$, so outputs cover $[1,\infty)$. Solving $y=x^2$ for $x\ge1$ gives $x=\sqrt{y}$ (positive root, consistent with $x\ge1$).

So $f^{-1}(x) = \begin{cases} x, & x<1 \\ \sqrt{x}, & x\ge1\end{cases}$, matching option (a).