Q.$f(x) = \begin{cases}-x+4 & -\infty<x\leq -3\\ x+4 & -3<x<-2\\ x^2-x & -2\leq x<1\\ x-x^2 & 1\leq x<7\\ 0 & \text{otherwise}\end{cases}$ Write the values of $f$ at $-4, 1, -2, 7, 0$. OR If $\theta$ is an acute angle, then find $\sin\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)$ when $\sin\theta = \dfrac{1}{25}$.

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 $x=\pm3$: $f(\pm3)=9$.

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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 $y=e^x$. (Che …

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.

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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. …

$f(x)=x^2$ 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. …

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. $x^2$. …

$f(x)=\sin x$ 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.

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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. $f^{-1}(y)=y$.

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