functions

[shadow veldt]

and 2 and 3 here

[patent hamlet]

With q3, note that solutions will only be valid

within the domain where $-1 \leq 4x^4+ x^2 \leq 1$.

Within that domain, you have $4x^4 + x^2 = \sin\frac{\pi}{6}$

which is a quadratic equation in $x^2$.

[hearty vessel]

Jay

[patent hamlet]

With q2, anti-differentiate $f'(x)$ to find a primitive function with an arbitrary constant (that is, an indefinite integral) $f(x)$, then use the given point to evaluate the constant and find the unique $f(x)$ solution.

[hearty vessel]

Jay

[shadow veldt]

can u do 6 2?

jus

the last part

[patent hamlet]

(i) the curves meet where the equations are the same,

so rearrange $y = x^2 - x + 3 = 3x + a$

(ii) two solutions means the discriminant is positive...

use the solution $x = -1$ to find $a$, then solve...

or, note that the sum of the solutions is 4

(Vieta's formulae) and deduce the second solution.

(iii) one solution for the tangent case, so the

discriminant of the quadratic from (i) is zero.

[hearty vessel]

Jay

[shadow veldt]

tysm