1.1 Los op vir x, in elk van die volgende: 1.1.1 2x² - 7x = 0 1.1.2 4x⁴ + 11 = 0 ; x ≠ 0 1.1.3 (2x - 1)(x - 3) > 0 1.1.4 3³.3²¹ = 27²ˣ 1.2 Los geliktydig op vir x en y in die volgende vergelykings: 3 + y = 2x 4x² + y² = 2xy + 7 1.3 Gegee: F(x) = -3x⁴ - 2x² + 20 = (x + 2)(x² - 6x + 10) Bewys dat f(x) slegs een reële wortel het. - NSC Mathematics - Question 1 - 2016 - Paper 1

Since $4x^4 + 11 = 0$ has no real solutions, we can see that $4x^4 = -11$ However, for any real value of $x$, $4x^4$ cannot equal a negative number. Therefore, there are no real solutions for $x$.

To solve the inequality, we first find the critical points by setting each factor to zero: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ $x - 3 = 0 \Rightarrow x = 3$ Creating a sign chart or testing intervals:

1. For $x < \frac{1}{2}$, the product is positive.
2. For $\frac{1}{2} < x < 3$, the product is negative.
3. For $x > 3$, the product is positive. Hence, the solution is $x < \frac{1}{2}$ or $x > 3$.

Converting to exponent form: $3^{3 + 21x} = 27^{2x}$ Since $27 = 3^3$, we rewrite it as: $3^{3 + 21x} = (3^3)^{2x} = 3^{6x}$ Thus, setting the exponents equal: $3 + 21x = 6x\Rightarrow 15x = -3\Rightarrow x = -\frac{1}{5}$

To solve for $x$ and $y$, we can first express $y$ from the first equation: $y = 2x - 3$ Substituting $y$ in the second equation: $4x^2 + (2x - 3)^2 = 2x(2x - 3) + 7$ Expanding and simplifying yields: $4x^2 + (4x^2 - 12x + 9) = 4x^2 - 6x + 7\Rightarrow 4x^2 - 12x + 9 = -6x + 7\Rightarrow 4x^2 - 6x + 2 = 0$ The solutions for $x$ can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4(4)(2)}}{2(4)}$

To show that $F(x)$ has only one real root, we need to investigate the discriminant of the polynomial $x² - 6x + 10$: $D = b^2 - 4ac = (-6)^2 - 4(1)(10) = 36 - 40 = -4$ Since the discriminant is negative, this indicates that there are no real roots for this quadratic, thus $F(x)$ has only one real root at $x = -2$.