The polynomial P(x) = x3 − 4x² + ax + 5 leaves the same remainder when divided by (x-1) and (x + 2). (a) Find the value of a. (b) Hence, find the remainder. 4. Given f(x) = (3x+1)/(x-2) (a) Find f-¹(x). (b) State the domain of f-¹.

The remainder when $P(x)$ is divided by $(x-1)$ is $P(1)$, and by $(x+2)$ is $P(-2)$.

$$P(1) = (1)^3 - 4(1)^2 + a(1) + 5 = 1 - 4 + a + 5 = a + 2$$ $$P(-2) = (-2)^3 - 4(-2)^2 + a(-2) + 5 = -8 - 4(4) - 2a + 5 = -8 - 16 - 2a + 5 = -19 - 2a$$

$$a + 2 = -19 - 2a$$ $$3a = -21$$ $$a = -7$$

The value of $a$ is $\boxed{-7}$.

b) Hence, find the remainder.

Using $P(1)$:

$$Remainder = a + 2 = -7 + 2 = -5$$

The remainder is $\boxed{-5}$.

4. Given $f(x) = \frac{3x+1}{x-2}$

a) Find $f^{-1}(x)$.

Swap $x$ and $y$ to find the inverse.

$$x = \frac{3y+1}{y-2}$$

$$x(y-2) = 3y+1$$ $$xy - 2x = 3y + 1$$ $$xy - 3y = 2x + 1$$ $$y(x-3) = 2x + 1$$ $$y = \frac{2x+1}{x-3}$$

So, $f^{-1}(x) = \boxed{\frac{2x+1}{x-3}}$.

b) State the domain of $f^{-1}$.

$$x-3 \neq 0$$ $$x \neq 3$$

The domain of $f^{-1}(x)$ is $\boxed{\{x \in \mathbb{R} \mid x \neq 3\}}$.