Here are the solutions to questions 17, 18, and 19 based on the image provided: 17. Find the value of $x$ if $\begin{vmatrix} x & 3 \\ 4 & 7 \end{vmatrix} = 9$ Step 1: Calculate the determinant of the $2 \times 2$ matrix. The determinant of a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $ad - bc$. For the given matrix $\begin{bmatrix} x & 3 \\ 4 & 7 \end{bmatrix}$, the determinant is $(x)(7) - (3)(4)$. $$7x - 12$$ Step 2: Set the determinant equal to 9 as given in the problem. $$7x - 12 = 9$$ Step 3: Solve the linear equation for $x$. Add 12 to both sides: $$7x = 9 + 12$$ $$7x = 21$$ Divide by 7: $$x = \frac{21}{7}$$ $$x = 3$$ The correct answer is (b) 3. $$\boxed{3}$$ 18. Rationalise $\frac{2\sqrt{3} - \sqrt{5}}{\sqrt{5} - \sqrt{3}}$ Step 1: Identify the conjugate of the denominator. The denominator is $\sqrt{5} - \sqrt{3}$. Its conjugate is $\sqrt{5} + \sqrt{3}$. Step 2: Multiply the numerator and the denominator by the conjugate. $$\frac{2\sqrt{3} - \sqrt{5}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$$ Step 3: Expand the numerator: $(2\sqrt{3} - \sqrt{5})(\sqrt{5} + \sqrt{3})$. $$= (2\sqrt{3})(\sqrt{5}) + (2\sqrt{3})(\sqrt{3}) - (\sqrt{5})(\sqrt{5}) - (\sqrt{5})(\sqrt{3})$$ $$= 2\sqrt{15} + 2(3) - 5 - \sqrt{15}$$ $$= 2\sqrt{15} + 6 - 5 - \sqrt{15}$$ $$= (2\sqrt{15} - \sqrt{15}) + (6 - 5)$$ $$= \sqrt{15} + 1$$ Step 4: Expand the denominator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})$. $$= (\sqrt{5})^2 - (\sqrt{3})^2$$ $$= 5 - 3$$ $$= 2$$ Step 5: Combine the simplified numerator and denominator. $$\frac{\sqrt{15} + 1}{2}$$ Upon reviewing the provided options, none of them match the calculated answer $\frac{\sqrt{15} + 1}{2}$. There might be a typo in the question or the options. $$\boxed{\frac{\sqrt{15} + 1}{2}}$$ 19. If $y=x$, find $dy/dx$. Step 1: The given function is $y=x$. Step 2: Differentiate $y$ with respect to $x$. Using the power rule for differentiation, if $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$. For $y=x$, we have $n=1$. $$\frac{dy}{dx} = 1 \cdot x^{1-1}$$ $$\frac{dy}{dx} = 1 \cdot x^0$$ Since any non-zero number raised to the power of 0 is 1 ($x^0=1$ for $x \neq 0$), $$\frac{dy}{dx} = 1 \cdot 1$$ $$\frac{dy}{dx} = 1$$ The correct answer is (a) 1. $$\boxed{1}$$