Let's solve each part step by step. --- ## Q2 (a): Concavity ### (i) \( y = \frac{x^2}{x^2 + 1} \) **Step 1:** Find the first and second derivatives. Let \( y = \frac{x^2}{x^2 + 1} \). First derivative using quotient rule: \[ y' = \frac{(x^2 + 1)(2x) - x^2(2x)}{(x^2 + 1)^2} = \frac{2x(x^2 + 1 - x^2)}{(x^2 + 1)^2} = \frac{2x}{(x^2 + 1)^2} \] Second derivative: Use quotient rule again on \( y' \): \[ y'' = \frac{(x^2+1)^2 \cdot 2 - 2x \cdot 2(x^2 + 1) \cdot 2x}{(x^2 + 1)^4} = \frac{2(x^2 + 1)^2 - 8x^2(x^2 + 1)}{(x^2 + 1)^4} \] Expand numerator: \[ 2(x^2 + 1)^2 - 8x^2(x^2 + 1) = 2(x^4 + 2x^2 + 1) - 8x^4 - 8x^2 = 2x^4 + 4x^2 + 2 - 8x^4 - 8x^2 \] \[ = (2x^4 - 8x^4) + (4x^2 - 8x^2) + 2 = -6x^4 - 4x^2 + 2 \] So, \[ y'' = \frac{-6x^4 - 4x^2 + 2}{(x^2 + 1)^4} \] **Step 2:** Determine where \( y'' = 0 \) (possible inflection points): Set numerator to 0: \[ -6x^4 - 4x^2 + 2 = 0 \implies 6x^4 + 4x^2 - 2 = 0 \] Divide by 2: \[ 3x^4 + 2x^2 - 1 = 0 \] Let \( u = x^2 \): \[ 3u^2 + 2u - 1 = 0 \] Solve quadratic: \[ u = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm 4}{6} \] So: \[ u_1 = \frac{2}{6} = \frac{1}{3}, \quad u_2 = \frac{-6}{6} = -1 \] But \( x^2 = -1 \) has no real solutions, so \( x^2 = \frac{1}{3} \), \( x = \pm \frac{1}{\sqrt{3}} \). **Step 3:** Test intervals around these points to determine sign of \( y'' \): - For \( x < -\frac{1}{\sqrt{3}} \) - For \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \) - For \( x > \frac{1}{\sqrt{3}} \) Pick test points, say \( x = -1, 0, 1 \): - \( x = 0 \): \( y'' = \frac{2}{1^4} = 2 > 0 \) (concave up) - \( x = 1 \): \( y'' = \frac{-6 - 4 + 2}{16} = \frac{-8}{16} = -0.5 < 0 \) (concave down) - \( x = -1 \): Same as above, also \( -0.5 \) (concave down) So: - **Concave upward:** \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \) - **Concave downward:** \( x < -\frac{1}{\sqrt{3}} \) and \( x > \frac{1}{\sqrt{3}} \) --- ### (ii) \( y = \frac{x^2 - 1}{2x - 1} \) **Step 1:** First derivative (quotient rule): \[ y' = \frac{(2x-1) \cdot 2x - (x^2-1) \cdot 2}{(2x-1)^2} = \frac{4x^2 - 2x - 2x^2 + 2}{(2x-1)^2} = \frac{2x^2 - 2x + 2}{(2x-1)^2} \] **Step 2:** Second derivative: Differentiate numerator and denominator: Let \( u = 2x^2 - 2x + 2 \), \( v = (2x - 1)^2 \): \[ y'' = \frac{u'v - uv'}{v^2} \] Where \( u' = 4x - 2 \), \( v' = 2(2x-1) \cdot 2 = 8x - 4 \): So, \[ y'' = \frac{(4x - 2)(2x-1)^2 - (2x^2 - 2x + 2)(8x - 4)}{(2x-1)^4} \] Now, expand numerator: First term: \( (4x-2)(2x-1)^2 \) \[ (2x-1)^2 = 4x^2 - 4x + 1 \] So, \[ (4x-2)(4x^2 - 4x + 1) = 16x^3 - 16x^2 + 4x - 8x^2 + 8x - 2 = 16x^3 - 24x^2 + 12x - 2 \] Second term: \( (2x^2 - 2x + 2)(8x-4) \) \[ = (2x^2)(8x-4) - (2x)(8x-4) + 2(8x-4) = 16x^3 - 8x^2 - 16x^2 + 8x + 16x - 8 = 16x^3 - 8x^2 - 16x^2 + 8x + 16x - 8 = 16x^3 - 24x^2 + 24x - 8 \] Now, subtract second term from first: \[ \text{Numerator} = [16x^3 - 24x^2 + 4x - 2] - [16x^3 - 24x^2 + 24x - 8] = 16x^3 - 24x^2 + 4x - 2 - 16x^3 + 24x^2 - 24x + 8 = (0) + (0) + (4x - 24x) + (-2 + 8) = -20x + 6 \] So, \[ y'' = \frac{-20x + 6}{(2x-1)^4} \] **Step 3:** Set \( y'' = 0 \): \[ -20x + 6 = 0 \implies x = \frac{3}{10} \] **Step 4:** Test sign of \( y'' \): For \( x < 0.3 \), say \( x = 0 \): \( y'' = \frac{6}{1} > 0 \) For \( x > 0.3 \), say \( x = 1 \): \( y'' = \frac{-20 + 6}{1^4} = -14 < 0 \) So: - **Concave upward:** \( x < 0.3 \) - **Concave downward:** \( x > 0.3 \) - Note: \( x \neq 0.5 \), as denominator is zero there (vertical asymptote). --- ### **Final Answers:** #### (i) \( y = \frac{x^2}{x^2 + 1} \) - **Concave upward:** \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \) - **Concave downward:** \( x < -\frac{1}{\sqrt{3}} \) and \( x > \frac{1}{\sqrt{3}} \) #### (ii) \( y = \frac{x^2 - 1}{2x - 1} \) - **Concave upward:** \( x < 0.3 \) (excluding \( x = 0.5 \)) - **Concave downward:** \( x > 0.3 \) (excluding \( x = 0.5 \))

Certainly! Let's proceed step-by-step with detailed calculations for each part of the question, ensuring clarity at each stage. --- ## Q2 (a): Concavity Analysis ### (i) \( y = \frac{x^2}{x^2 + 1} \) #### Step 1: Find the first derivative \( y' \) Given: \[ y = \frac{x^2}{x^2 + 1} \] Use the quotient rule: \[ y' = \frac{(x^2 + 1) \cdot 2x - x^2 \cdot 2x}{(x^2 + 1)^2} \] Calculate numerator: \[ (2x)(x^2 + 1) - 2x \cdot x^2 = 2x^3 + 2x - 2x^3 = 2x \] Thus: \[ \boxed{ y' = \frac{2x}{(x^2 + 1)^2} } \] --- #### Step 2: Find the second derivative \( y'' \) Differentiate \( y' \) using quotient rule: \[ y'' = \frac{d}{dx} \left( \frac{2x}{(x^2 + 1)^2} \right) \] Set numerator \( u = 2x \), denominator \( v = (x^2 + 1)^2 \). Compute derivatives: \[ u' = 2, \quad v' = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1) \] Apply quotient rule: \[ y'' = \frac{u'v - uv'}{v^2} = \frac{2 \cdot (x^2 + 1)^2 - 2x \cdot 4x(x^2 + 1)}{(x^2 + 1)^4} \] Simplify numerator: \[ 2(x^2 + 1)^2 - 8x^2 (x^2 + 1) \] Factor numerator: \[ 2(x^2 + 1)^2 - 8x^2 (x^2 + 1) = (x^2 + 1) \left[ 2(x^2 + 1) - 8x^2 \right] \] Calculate inside brackets: \[ 2x^2 + 2 - 8x^2 = -6x^2 + 2 \] Thus, numerator: \[ (x^2 + 1)(-6x^2 + 2) \] Therefore: \[ y'' = \frac{(x^2 + 1)(-6x^2 + 2)}{(x^2 + 1)^4} = \frac{-6x^2 + 2}{(x^2 + 1)^3} \] --- #### Step 3: Determine concavity by analyzing \( y'' \) Set numerator to zero: \[ -6x^2 + 2 = 0 \implies 6x^2 = 2 \implies x^2 = \frac{1}{3} \] \[ x = \pm \frac{1}{\sqrt{3}} \] The denominator \( (x^2 + 1)^3 \) is always positive, so the sign of \( y'' \) depends on numerator: - For \( x^2 < \frac{1}{3} \): \[ -6x^2 + 2 > 0 \] since \( -6x^2 \) less than 2, numerator positive. - For \( x^2 > \frac{1}{3} \): \[ -6x^2 + 2 < 0 \] numerator negative. Translate into intervals: - When \( |x| < \frac{1}{\sqrt{3}} \): \[ y'' > 0 \Rightarrow \text{graph is concave upward} \] - When \( |x| > \frac{1}{\sqrt{3}} \): \[ y'' < 0 \Rightarrow \text{graph is concave downward} \] --- ### **Final answer for (i):** | Interval | Concavity | Explanation | |------------|--------------|--------------| | \( -\infty < x < -\frac{1}{\sqrt{3}} \) | Concave downward | \( y'' < 0 \) | | \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \) | Concave upward | \( y'' > 0 \) | | \( \frac{1}{\sqrt{3}} < x < \infty \) | Concave downward | \( y'' < 0 \) | --- ### (ii) \( y = \frac{x^2 - 1}{2x - 1} \) #### Step 1: Find \( y' \) Apply quotient rule: \[ y' = \frac{(2x - 1) \cdot 2x - (x^2 - 1) \cdot 2}{(2x - 1)^2} \] Calculate numerator: \[ (2x - 1) \cdot 2x = 4x^2 - 2x \] \[ (x^2 - 1) \cdot 2 = 2x^2 - 2 \] Numerator: \[ 4x^2 - 2x - (2x^2 - 2) = 4x^2 - 2x - 2x^2 + 2 = (4x^2 - 2x^2) - 2x + 2 = 2x^2 - 2x + 2 \] So: \[ \boxed{ y' = \frac{2x^2 - 2x + 2}{(2x - 1)^2} } \] --- #### Step 2: Find \( y'' \) Let numerator \( u = 2x^2 - 2x + 2 \), denominator \( v = (2x - 1)^2 \). Compute derivatives: \[ u' = 4x - 2 \] \[ v' = 2(2x - 1) \cdot 2 = 8x - 4 \] Apply quotient rule: \[ y'' = \frac{u'v - uv'}{v^2} \] Calculate numerator: \[ u'v = (4x - 2)(2x - 1)^2 \] \[ uv' = (2x^2 - 2x + 2)(8x - 4) \] --- #### Step 3: Expand numerator parts First, expand \( u'v \): \[ (4x - 2)(2x - 1)^2 \] Recall: \[ (2x - 1)^2 = 4x^2 - 4x + 1 \] So: \[ (4x - 2)(4x^2 - 4x + 1) \] Distribute: \[ 4x \cdot 4x^2 = 16x^3 \] \[ 4x \cdot (-4x) = -16x^2 \] \[ 4x \cdot 1 = 4x \] \[ -2 \cdot 4x^2 = -8x^2 \] \[ -2 \cdot (-4x) = 8x \] \[ -2 \cdot 1 = -2 \] Sum all: \[ 16x^3 + (-16x^2 - 8x^2) + (4x + 8x) - 2 = 16x^3 - 24x^2 + 12x - 2 \] Next, expand \( uv' \): \[ (2x^2 - 2x + 2)(8x - 4) \] Distribute: \[ 2x^2 \cdot 8x = 16x^3 \] \[ 2x^2 \cdot (-4) = -8x^2 \] \[ -2x \cdot 8x = -16x^2 \] \[ -2x \cdot (-4) = 8x \] \[ 2 \cdot 8x = 16x \] \[ 2 \cdot (-4) = -8 \] Sum: \[ 16x^3 - 8x^2 - 16x^2 + 8x + 16x - 8 = 16x^3 - 24x^2 + 24x -8 \] --- #### Step 4: Compute numerator of \( y'' \) \[ \text{Numerator} = u'v - uv' = (16x^3 - 24x^2 + 12x - 2) - (16x^3 - 24x^2 + 24x - 8) \] Subtract: \[ (16x^3 - 16x^3) + (-24x^2 + 24x^2) + (12x - 24x) + (-2 + 8) \] Simplify: \[ 0 + 0 - 12x + 6 = -12x + 6 \] So: \[ \boxed{ y'' = \frac{-12x + 6}{(2x - 1)^4} } \] --- #### Step 5: Find inflection points where \( y'' = 0 \): \[ -12x + 6 = 0 \implies 12x = 6 \implies x = \frac{6}{12} = \frac{1}{2} \] But note that at \( x = \frac{1}{2} \), the denominator \( (2x - 1)^4 = (1 - 1)^4 = 0 \), indicating a vertical asymptote. Since the second derivative involves division by zero, there is **no** inflection point at \( x = 0.5 \). Instead, the change in concavity occurs around this vertical asymptote. **Note:** The sign of \( y'' \) depends on numerator \( -12x + 6 \): - For \( x < 0.5 \): \[ -12x + 6 > 0 \text{ (since } x < 0.5 \implies -12x + 6 > 0) \] - For \( x > 0.5 \): \[ -12x + 6 < 0 \] --- ### **Final conclusion for (ii):** - \( y'' > 0 \) when \( x < 0.5 \) (graph is concave upward) - \( y'' < 0 \) when \( x > 0.5 \) (graph is concave downward) **Note:** At \( x = 0.5 \), the second derivative is undefined due to vertical asymptote, and the concavity changes across this point. --- ## **Summary of Part (a):** | Function | Concave Upward | Concave Downward | Inflection Points | |----------|------------------|---------------------|---------------------| | \( y = \frac{x^2}{x^2 + 1} \) | \( |x| < \frac{1}{\sqrt{3}} \) | \( |x| > \frac{1}{\sqrt{3}} \) | at \( x = \pm \frac{1}{\sqrt{3}} \) | | \( y = \frac{x^2 - 1}{2x - 1} \) | \( x < 0.5 \) | \( x > 0.5 \) | at \( x = 0.5 \) (vertical asymptote, change in concavity) | --- If you'd like, I can proceed with the remaining parts of the question!