1. We are asked to find the coordinates of point A where the line $y = 2x + 1$ intersects the curve $y = \frac{5}{x+1} + 2$. 2. To find the intersection, set the two equations equal: $$ 2x + 1 = \frac{5}{x+1} + 2 $$ 3. Subtract 2 from both sides: $$ 2x + 1 - 2 = \frac{5}{x+1} $$ $$ 2x -1 = \frac{5}{x+1} $$ 4. Multiply both sides by $(x+1)$ to clear the denominator: $$ (2x -1)(x+1) = 5 $$ 5. Expand the left side: $$ 2x \cdot x + 2x \cdot 1 -1 \cdot x -1 \cdot 1 = 5 $$ $$ 2x^2 + 2x - x -1 = 5 $$ $$ 2x^2 + x - 1 = 5 $$ 6. Subtract 5 from both sides: $$ 2x^2 + x - 6 = 0 $$ 7. Solve the quadratic equation $2x^2 + x - 6 = 0$. Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=2$, $b=1$, $c=-6$. 8. Calculate the discriminant: $$ \Delta = 1^2 - 4 \times 2 \times (-6) = 1 + 48 = 49 $$ 9. Substitute values: $$ x = \frac{-1 \pm 7}{2 \times 2} = \frac{-1 \pm 7}{4} $$ 10. Two possible values for $x$: - $$ x = \frac{-1 + 7}{4} = \frac{6}{4} = 1.5 $$ - $$ x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2 $$ 11. Find corresponding $y$ values for each $x$ from the line equation $y = 2x + 1$: - For $x=1.5$: $$ y = 2(1.5) + 1 = 3 + 1 = 4 $$ - For $x=-2$: $$ y = 2(-2) + 1 = -4 + 1 = -3 $$ 12. Check the curve for $x=-2$ to verify if the intersection occurs: $$ y = \frac{5}{-2 + 1} + 2 = \frac{5}{-1} + 2 = -5 + 2 = -3 $$ Matches line $y$, so point A could be $(-2, -3)$. For $x=1.5$: $$ y = \frac{5}{1.5 +1} + 2 = \frac{5}{2.5} + 2 = 2 + 2 = 4 $$ Matches line $y$ as well. Therefore, the two intersection points are $A_1 = (1.5, 4)$ and $A_2 = (-2, -3)$. But since the shaded region is between the y-axis and point A and the diagram likely shows only one intersection at positive $x$, the relevant intersection point is $A = (1.5, 4)$. 2. Solve simultaneous equations: $$ \frac{y}{x} = \frac{3}{2}, \quad \frac{y^4}{x^5} = \frac{27}{16} $$ Step 1: From the first equation: $$ y = \frac{3}{2}x $$ Step 2: Substitute into the second equation: $$ \frac{(\frac{3}{2}x)^4}{x^5} = \frac{27}{16} $$ Simplify numerator: $$ \frac{(\frac{3}{2})^4 x^4}{x^5} = \frac{27}{16} $$ Simplify powers of $x$: $$ (\frac{3}{2})^4 x^{-1} = \frac{27}{16} $$ Calculate $(\frac{3}{2})^4$: $$ (\frac{3}{2})^4 = \frac{3^4}{2^4} = \frac{81}{16} $$ So: $$ \frac{81}{16} x^{-1} = \frac{27}{16} $$ Multiply both sides by $x$: $$ \frac{81}{16} = \frac{27}{16} x $$ Multiply both sides by 16: $$ 81 = 27 x $$ Solve for $x$: $$ x = \frac{81}{27} = 3 $$ Step 3: Substitute $x=3$ into $y = \frac{3}{2} x$: $$ y = \frac{3}{2} \times 3 = \frac{9}{2} = 4.5 $$ Answer: $x = 3$, $y = 4.5$. 3. Find values of $k$ for which equation $$ 4x^2 - k = 4kx - 2 $$ has no real roots. Step 1: Rearrange into standard quadratic form: $$ 4x^2 - 4kx + ( -k + 2 ) = 0 $$ Step 2: Coefficients are: - $a = 4$ - $b = -4k$ - $c = -k + 2$ Step 3: For no real roots, the discriminant must be less than zero: $$ \Delta = b^2 - 4ac < 0 $$ Calculate discriminant: $$ (-4k)^2 - 4 \times 4 \times (-k + 2) < 0 $$ $$ 16k^2 - 16(-k + 2) < 0 $$ $$ 16k^2 + 16k - 32 < 0 $$ Divide by 16: $$ k^2 + k - 2 < 0 $$ Step 4: Factor: $$ (k + 2)(k - 1) < 0 $$ Step 5: This inequality is true where $k$ is between $-2$ and $1$. Answer: $$ -2 < k < 1 $$ 4. Given quadratic: $$ kx^2 + (2k - 1)x + (k + 1) = 0 $$ Find values of $k$ for no real roots. Step 1: Coefficients: - $a = k$ - $b = 2k - 1$ - $c = k + 1$ Step 2: Discriminant: $$ \Delta = b^2 - 4ac < 0 $$ Calculate: $$ (2k - 1)^2 - 4 k (k + 1) < 0 $$ $$ 4k^2 - 4k + 1 - 4k^2 - 4k < 0 $$ $$ 1 - 8k < 0 $$ Step 3: Solve inequality: $$ 1 < 8k $$ $$ k > \frac{1}{8} $$ Step 4: Also, since $a = k$ is the leading coefficient, to have a quadratic equation ($a \neq 0$), $k \neq 0$. So the solution is: $$ k > \frac{1}{8} $$ 5. Curve: $$ y = x^2 + kx + (4k - 15) $$ Find $k$ such that curve is above $x$-axis (i.e., $y > 0$ for all $x$). Step 1: For curve to be above $x$-axis always, quadratic must have no real roots and open upwards. Step 2: Coefficients: - $a=1 > 0$ (opens upwards) Step 3: Discriminant must be less than 0: $$ \Delta = b^2 - 4ac < 0 $$ Here: $$ b = k $$ $$ c = 4k - 15 $$ Calculate: $$ k^2 - 4(1)(4k - 15) < 0 $$ $$ k^2 - 16k + 60 < 0 $$ Step 4: Solve quadratic inequality: Factorize or use quadratic formula: $$ k = \frac{16 \pm \sqrt{256 - 240}}{2} = \frac{16 \pm 4}{2} $$ Roots: $$ k = \frac{16 - 4}{2} = 6 $$ $$ k = \frac{16 + 4}{2} = 10 $$ Step 5: Inequality $k^2 -16k +60 <0$ is true between roots: $$ 6 < k < 10 $$ Answer: Curve is above $x$-axis for $$ 6 < k < 10 $$ 6. Find non-zero $k$ such that line $$ y = -2x - 6k - 1 $$ is tangent to curve $$ y = x(x + 2k) = x^2 + 2kx $$ Step 1: Equate line and curve: $$ x^2 + 2kx = -2x - 6k -1 $$ Step 2: Bring all terms to one side: $$ x^2 + 2kx + 2x + 6k + 1 = 0 $$ Group terms: $$ x^2 + (2k + 2)x + (6k + 1) = 0 $$ Step 3: For tangency, discriminant is zero: $$ \Delta = b^2 - 4ac = 0 $$ Coefficients: $$ a=1, b=2k+2, c=6k +1 $$ Step 4: Calculate discriminant: $$ (2k+2)^2 - 4(1)(6k+1) = 0 $$ $$ 4(k+1)^2 - 24k - 4 = 0 $$ $$ 4(k^2 + 2k + 1) - 24k - 4 = 0 $$ $$ 4k^2 + 8k + 4 - 24k -4 = 0 $$ $$ 4k^2 - 16k = 0 $$ Step 5: Factorize: $$ 4k(k - 4) = 0 $$ Step 6: Solutions: $$ k=0 $$ or $$ k=4 $$ Since $k \neq 0$, solution is: $$ k = 4 $$ 7. Find exact values of $k$ such that line $$ y = 1 - k - x $$ is tangent to curve $$ y = kx^2 + x + 2k $$ Step 1: Set equal: $$ 1 - k - x = kx^2 + x + 2k $$ Step 2: Rearrange: $$ kx^2 + x + 2k + x + k - 1 = 0 $$ $$ kx^2 + 2x + 3k - 1 = 0 $$ Step 3: For tangency, discriminant $=0$. Coefficients: $$ a = k, b = 2, c = 3k -1 $$ Step 4: Discriminant: $$ 2^2 - 4k(3k -1) = 0 $$ $$ 4 - 12k^2 + 4k = 0 $$ Rearranged: $$ -12 k^2 + 4 k + 4 = 0 $$ Multiply both sides by $-1$: $$ 12 k^2 - 4k - 4 = 0 $$ Step 5: Solve quadratic for $k$: $$ k = \frac{4 \pm \sqrt{( -4)^2 - 4 \times 12 \times (-4)}}{2 \times 12} = \frac{4 \pm \sqrt{16 + 192}}{24} = \frac{4 \pm \sqrt{208}}{24} $$ Simplify $\sqrt{208} = \sqrt{16 \times 13} = 4 \sqrt{13}$: $$ k = \frac{4 \pm 4 \sqrt{13}}{24} = \frac{1 \pm \sqrt{13}}{6} $$ Answer: $$ k = \frac{1 + \sqrt{13}}{6}, \quad k = \frac{1 - \sqrt{13}}{6} $$ 8. Solve simultaneous equations: $$ x + 5y = -4 $$ $$ 3y - xy = 6 $$ Step 1: From first equation: $$ x = -4 - 5y $$ Step 2: Substitute into second equation: $$ 3y - (-4 - 5y)y = 6 $$ $$ 3y + 4y + 5y^2 = 6 $$ $$ 7y + 5 y^2 = 6 $$ Step 3: Rearranged: $$ 5 y^2 + 7 y - 6 = 0 $$ Step 4: Solve quadratic for $y$ using formula: $$ y = \frac{-7 \pm \sqrt{7^2 - 4 \times 5 \times (-6)}}{2 \times 5} = \frac{-7 \pm \sqrt{49 + 120}}{10} = \frac{-7 \pm \sqrt{169}}{10} $$ $$ y = \frac{-7 \pm 13}{10} $$ Solutions: - $$ y = \frac{-7 + 13}{10} = \frac{6}{10} = 0.6 $$ - $$ y = \frac{-7 - 13}{10} = \frac{-20}{10} = -2 $$ Step 5: Find $x$ for each $y$: - For $y=0.6$: $$ x = -4 - 5(0.6) = -4 - 3 = -7 $$ - For $y=-2$: $$ x = -4 - 5(-2) = -4 + 10 = 6 $$ Solutions: $$ (x, y) = (-7, 0.6), (6, -2) $$ 9. Solve simultaneously: $$ x + 3y = 11 $$ $$ x - \sqrt{7} y = 7 $$ Step 1: Subtract second equation from first: $$ (x + 3y) - (x - \sqrt{7} y) = 11 - 7 $$ $$ x + 3y - x + \sqrt{7} y = 4 $$ $$ (3 + \sqrt{7}) y = 4 $$ Step 2: Solve for $y$: $$ y = \frac{4}{3 + \sqrt{7}} $$ Rationalize denominator: $$ y = \frac{4}{3 + \sqrt{7}} \times \frac{3 - \sqrt{7}}{3 - \sqrt{7}} = \frac{4(3 - \sqrt{7})}{9 - 7} = \frac{4(3 - \sqrt{7})}{2} = 2 (3 - \sqrt{7}) = 6 - 2 \sqrt{7} $$ Step 3: Substitute $y$ into first equation: $$ x + 3(6 - 2 \sqrt{7}) = 11 $$ $$ x + 18 - 6 \sqrt{7} = 11 $$ $$ x = 11 - 18 + 6 \sqrt{7} = -7 + 6 \sqrt{7} $$ Answer: $$ x = -7 + 6 \sqrt{7}, \quad y = 6 - 2 \sqrt{7} $$ 10. (a) Solve inequality: $$ (5x - 1)(6 - x) < 0 $$ Step 1: Find roots: $$ 5x - 1 = 0 \Rightarrow x = \frac{1}{5} $$ $$ 6 - x = 0 \Rightarrow x=6 $$ Step 2: Test intervals: - For $x < \frac{1}{5}$, e.g. $0$, $(5(0) -1)(6-0) = -1 \times 6 < 0$ TRUE - For $\frac{1}{5} < x < 6$, e.g. $1$, $(5(1)-1)(6 -1) = 4 \times 5 > 0$ FALSE - For $x > 6$, e.g. $7$, $(5(7) -1)(6 - 7) = 34 \times (-1) < 0$ TRUE Step 3: Final solution: $$ \boxed{x < \frac{1}{5} \quad \text{or} \quad x > 6} $$ (b) Show equation $$ (2k + 1) x^2 - 4k x + 2k - 1 = 0 $$ has distinct real roots for $k \neq \frac{1}{2}$. Step 1: Coefficients: $$ a = 2k + 1, \quad b = -4k, \quad c = 2k - 1 $$ Step 2: Compute discriminant: $$ \Delta = b^2 - 4ac = (-4k)^2 - 4 (2k + 1)(2k - 1) = 16k^2 - 4 (4k^2 - 1) $$ $$ = 16k^2 - 16k^2 + 4 = 4 $$ Step 3: Since $\Delta = 4 > 0$ for all $k$, roots are always distinct and real for any $k$. Step 4: Check $a \neq 0$, i.e., $2k + 1 \neq 0$ $\Rightarrow k \neq -\frac{1}{2}$. Since problem states $k \neq \frac{1}{2}$, it is safe. Conclusion: The equation has distinct real roots for all $k \neq -\frac{1}{2}$ including $k \neq \frac{1}{2}$. 11. (a) Write $$ 19 - 12x - 3x^2 $$ in form $$ a (x + b)^2 + c $$ where $a,b,c$ are integers. Step 1: Rewrite as $$ -3x^2 -12x + 19 $$ Factor out $-3$: $$ -3(x^2 + 4x) + 19 $$ Step 2: Complete the square inside parentheses: $$ x^2 + 4x = (x + 2)^2 - 4 $$ So: $$ -3[(x + 2)^2 - 4] + 19 = -3(x + 2)^2 + 12 + 19 = -3(x + 2)^2 + 31 $$ Answer: $$ a = -3, b = 2, c = 31 $$ (b) Hence find maximum value and where it occurs. Since $a < 0$, parabola opens downward, so maximum at vertex $x = -b = -2$. Maximum value: $$ y = -3(0)^2 + 31 = 31 $$ Answer: Maximum value = 31 at $x = -2$. 12. Curve $$ y = x^2 + 2x - 3 $$ (a) Find stationary point by completing the square. Step 1: $$ y = (x^2 + 2x) - 3 $$ $$ = (x + 1)^2 - 1 - 3 = (x + 1)^2 - 4 $$ Stationary point at vertex $x = -1$, $$ y = -4 $$ Coordinates: $(-1, -4)$ (b) Intercepts: - $y$-intercept when $x=0$: $$ y = 0 + 0 - 3 = -3 $$ - $x$-intercepts when $y=0$: Solve $$ x^2 + 2x - 3 = 0 $$ Factorize: $$ (x + 3)(x - 1) = 0 $$ So $$ x = -3, x = 1 $$ 13. Solve inequality: $$ (2x + 3)(x -4) > (3x + 4)(x -1) $$ Step 1: Expand both sides: $$ (2x + 3)(x - 4) = 2x^2 - 8x + 3x -12 = 2x^2 - 5x -12 $$ $$ (3x +4)(x -1) = 3x^2 - 3x + 4x - 4 = 3x^2 + x -4 $$ Step 2: Set inequality: $$ 2x^2 - 5x -12 > 3x^2 + x -4 $$ Step 3: Bring all terms to left side: $$ 2x^2 - 5x -12 - 3x^2 - x +4 > 0 $$ $$ -x^2 - 6x - 8 > 0 $$ Multiply inequality by $-1$ (flip inequality): $$ x^2 + 6x + 8 < 0 $$ Step 4: Solve quadratic inequality. Discriminant: $$ \Delta = 6^2 - 4 \times 1 \times 8 = 36 - 32 = 4 $$ Roots: $$ x = \frac{-6 \pm 2}{2} $$ $$ x = -2, -4 $$ Step 5: $x^2 + 6x + 8 < 0$ between roots: $$ -4 < x < -2 $$ Answer: $$ \boxed{-4 < x < -2} $$