Paper 1 -- Pure Mathematics -- Full Diagnostic Exam

Time allowed: 120 minutes Total marks: 100 Topics covered: All 14 pure mathematics topics

Answer all questions. Calculators are permitted unless otherwise stated. Show all working — marks are awarded for method as well as final answer.

Given that $a^{\frac{1}{2}} + a^{-\frac{1}{2}} = 5$Find the exact value of:

$\frac{a^{\frac{3}{2}} - a^{-\frac{3}{2}}}{a^{\frac{1}{2}} - a^{-\frac{1}{2}}}$

The roots of the equation $2x^2 - 5x + 1 = 0$ are $\alpha$ and $\beta$.

Without finding the numerical values of $\alpha$ and $\beta$Find the value of:

$\frac{1}{\alpha^2 + 1} + \frac{1}{\beta^2 + 1}$

Solve the inequality:

$\frac{x^2 - 3x + 2}{x^2 + x - 6} \geq 0$

State your answer using set notation, identifying all excluded values.

Find the value of $k$ such that the line $y = kx + 5$ is tangent to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$.

Hence find the coordinates of the point of tangency.

Given $f(x) = \sqrt{2x - 1}$ and $g(x) = \frac{1}{x - 3}$:

(a) Find the domain of $f \circ g$I.e. $f(g(x))$.

(b) Find the domain of $g \circ f$I.e. $g(f(x))$.

(c) Explain why the domains of $f \circ g$ and $g \circ f$ are different.

Evaluate the sum:

$S_n = \sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)}$

Express your answer in terms of $n$And hence find $\lim_{n \to \infty} S_n$.

Find the coefficient of $x^4$ in the expansion of:

$\frac{(1 + 2x)^5}{(1 - x)^3}$

(a) Solve $\cos(3x) = \frac{1}{2}$ for $x \in [0, 2\pi)$. Find all solutions.

(b) The curve $y = \cos(3x)$ intersects the line $y = \frac{1}{2}$ at $N$ distinct points in the interval $[0, 2\pi)$. Find $N$ and the sum of all $x$-coordinates of the intersection points.

(a) Solve $e^{2x} - 5e^x + 6 = 0$Giving exact answers.

(b) Solve $e^{2x} - 5e^x + 6 = 1$ for $x \in \mathbb{R}$Giving exact answers.

(c) Explain why, when using the substitution $u = e^x$ to solve an equation of the form $e^{2x} + pe^x + q = 0$You must check that $u \gt 0$ before taking natural logarithms.

A curve has equation $y = x^4 - 4x^3 + 6x^2 - 4x + 1$.

(a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

(b) Find the coordinates of all stationary points.

(c) Classify each stationary point. A student claims that since $\frac{d^2y}{dx^2} = 0$ at the stationary point, it is a point of inflection. Explain why this reasoning is incorrect, and determine the true nature of this point.

(d) Express $y$ in a form that makes the nature of the stationary point immediately obvious.

(a) Find $\int x^3 \ln x\, dx$.

(b) A student chooses $u = x^3$ and $\frac{dv}{dx} = \ln x$ for integration by parts. Explain why this choice is problematic.

(c) Using your result from part (a), evaluate $\int_1^e x^3 \ln x\, dx$ exactly.

Line $l_1$ passes through $A(1, 2, 3)$ with direction vector $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

Line $l_2$ passes through $B(4, 1, 0)$ with direction vector $\mathbf{d}_2 = \begin{pmatrix} 1 \\ a \\ 2 \end{pmatrix}$.

(a) Find the value of $a$ for which $l_1$ and $l_2$ intersect.

(b) For $a = 3$Determine whether $l_1$ and $l_2$ are skew, parallel, or intersecting.

(c) For $a = -1$Find the shortest distance between $l_1$ and $l_2$.

For each of the following, state whether the condition is necessary, sufficient, both, or neither. Justify each answer.

(a) "$x \gt 2$" as a condition for "$x^2 \gt 4$".

(b) ”$n$ is prime” as a condition for ”$n$ is odd”.

(c) "$a^2 + b^2 = 0$" (where $a, b \in \mathbb{R}$) as a condition for ”$a = 0$ and $b = 0$”.

The function $f(x) = x^3 - 2x + 2$ has a root near $x = -1.77$.

(a) Show that $f(x) = 0$ has exactly one real root.

(b) Apply the Newton-Raphson formula with initial value $x_0 = 0$. Compute $x_1$, $x_2$And $x_3$. Describe the behaviour of the iteration.

(c) Explain why the iteration fails to converge.

Key observation: The numerator $a^{3/2} - a^{-3/2}$ can be factorised using the difference of cubes identity $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$ with $x = a^{1/2}$ and $y = a^{-1/2}$:

$a^{\frac{3}{2}} - a^{-\frac{3}{2}} = \left(a^{\frac{1}{2}} - a^{-\frac{1}{2}}\right)\left(a + 1 + a^{-1}\right)$

Therefore the expression simplifies to:

$\frac{\left(a^{\frac{1}{2}} - a^{-\frac{1}{2}}\right)\left(a + 1 + a^{-1}\right)}{a^{\frac{1}{2}} - a^{-\frac{1}{2}}} = a + 1 + a^{-1}$

Provided $a^{1/2} - a^{-1/2} \neq 0$I.e. $a \neq 1$. (If $a = 1$The given condition would give $2 = 5$A contradiction, so $a \neq 1$ is guaranteed.)

Step 2: Find $a + a^{-1}$ from the given condition.

We are given $a^{1/2} + a^{-1/2} = 5$. Squaring both sides:

$a + 2 + a^{-1} = 25$

$a + a^{-1} = 23$

Step 3: Compute the final answer.

$a + 1 + a^{-1} = (a + a^{-1}) + 1 = 23 + 1 = 24$

Step 1: State Vieta”s formulas.

$\alpha + \beta = \frac{5}{2}, \quad \alpha\beta = \frac{1}{2}$

Step 2: Simplify the target expression.

$\frac{1}{\alpha^2+1} + \frac{1}{\beta^2+1} = \frac{(\beta^2+1) + (\alpha^2+1)}{(\alpha^2+1)(\beta^2+1)} = \frac{\alpha^2 + \beta^2 + 2}{\alpha^2\beta^2 + \alpha^2 + \beta^2 + 1}$

Step 3: Express $\alpha^2 + \beta^2$ using Vieta’s.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{25}{4} - 1 = \frac{21}{4}$

Step 4: Compute $\alpha^2\beta^2$.

$\alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{4}$

Step 5: Substitute into the expression.

$\frac{\frac{21}{4} + 2}{\frac{1}{4} + \frac{21}{4} + 1} = \frac{\frac{21}{4} + \frac{8}{4}}{\frac{1}{4} + \frac{21}{4} + \frac{4}{4}} = \frac{\frac{29}{4}}{\frac{26}{4}} = \frac{29}{26}$

Step 1: Factorise numerator and denominator.

Numerator: $x^2 - 3x + 2 = (x-1)(x-2)$.

Denominator: $x^2 + x - 6 = (x+3)(x-2)$.

Step 2: Identify critical values and excluded points.

Critical values: $x = -3, 1, 2$.

Excluded values: $x = -3$ (denominator zero) and $x = 2$ (denominator zero).

Step 3: Sign analysis.

Step 4: Include or exclude endpoints.

• $x = -3$: excluded (denominator zero)
• $x = 1$: included (numerator zero, expression equals zero, and $\geq 0$ is satisfied)
• $x = 2$: excluded (denominator zero)

Step 5: Assemble the solution.

The expression is non-negative when $x \lt -3$, $1 \leq x \lt 2$Or $x \gt 2$.

$x \in (-\infty, -3) \cup [1, 2) \cup (2, \infty)$

Step 1: Substitute the line into the circle equation.

$x^2 + (kx+5)^2 - 4x - 6(kx+5) + 9 = 0$ $x^2 + k^2x^2 + 10kx + 25 - 4x - 6kx - 30 + 9 = 0$ $(1+k^2)x^2 + (10k - 4 - 6k)x + 4 = 0$ $(1+k^2)x^2 + (4k - 4)x + 4 = 0$

Step 2: Set the discriminant to zero for tangency.

$\Delta = (4k-4)^2 - 4(1+k^2)(4) = 0$ $16k^2 - 32k + 16 - 16 - 16k^2 = 0$ $-32k = 0$ $k = 0$

Step 3: Find the point of tangency.

With $k = 0$The line is $y = 5$. Substituting into the circle:

$x^2 + 25 - 4x - 30 + 9 = 0$ $x^2 - 4x + 4 = 0$ $(x-2)^2 = 0$ $x = 2$

The point of tangency is $(2, 5)$.

Geometric check: The circle $(x-2)^2 + (y-3)^2 = 4$ has centre $(2, 3)$ and radius $2$. The line $y = 5$ is at distance $|5-3| = 2$ from the centre, equal to the radius. Confirmed.

(a) $f(g(x)) = f\left(\frac{1}{x-3}\right) = \sqrt{\frac{2}{x-3} - 1} = \sqrt{\frac{2 - (x-3)}{x-3}} = \sqrt{\frac{5-x}{x-3}}$.

For this to be defined, we need:

1. $\frac{5-x}{x-3} \geq 0$.
2. $x \neq 3$.

Solving $\frac{5-x}{x-3} \geq 0$ by sign analysis:

Critical values: $x = 3$ (excluded) and $x = 5$ (included).

Domain of $f \circ g$: $(3, 5]$.

(b) $g(f(x)) = g(\sqrt{2x-1}) = \frac{1}{\sqrt{2x-1} - 3}$.

For this to be defined:

1. $2x - 1 \geq 0 \implies x \geq \frac{1}{2}$ (domain of $f$).
2. $\sqrt{2x-1} - 3 \neq 0 \implies \sqrt{2x-1} \neq 3 \implies 2x-1 \neq 9 \implies x \neq 5$.

Domain of $g \circ f$: $\left[\frac{1}{2}, 5\right) \cup (5, \infty)$.

(c) The domains differ because:

• For $f \circ g$: the input to $f$ is $g(x) = \frac{1}{x-3}$And we need $g(x) \geq \frac{1}{2}$ (since $f$ requires $2 \cdot g(x) - 1 \geq 0$). This constrains $x$ to a finite interval $(3, 5]$.
• For $g \circ f$: the input to $g$ is $f(x) = \sqrt{2x-1}$And we need $f(x) \neq 3$. Since $f(x) \geq 0$ for all $x$ in its domain, we only exclude $x = 5$. The domain is almost the entire domain of $f$.

The key difference is that $f \circ g$ requires the output of $g$ to fall within the domain of $f$ (which is $[1/2, \infty)$), while $g \circ f$ requires the output of $f$ to avoid the single excluded value of $g$ (which is $3$). The former is much more restrictive.

Step 1: Partial fraction decomposition.

We seek constants $A$, $B$, $C$ such that:

$\frac{1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}$

Multiplying through by $r(r+1)(r+2)$:

$1 = A(r+1)(r+2) + Br(r+2) + Cr(r+1)$

Substituting $r = 0$: $1 = A(1)(2) \implies A = \frac{1}{2}$.

Substituting $r = -1$: $1 = B(-1)(1) \implies B = -1$.

Substituting $r = -2$: $1 = C(-2)(-1) \implies C = \frac{1}{2}$.

Step 2: Verify.

$\frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2} = \frac{(r+1)(r+2) - 2r(r+2) + r(r+1)}{2r(r+1)(r+2)}$

Numerator: $(r^2 + 3r + 2) + (-2r^2 - 4r) + (r^2 + r) = 0 + 0 + 2 = 2$.

So the fraction is $\frac{2}{2r(r+1)(r+2)} = \frac{1}{r(r+1)(r+2)}$. Confirmed.

Step 3: Write out the telescoping sum.

$S_n = \sum_{r=1}^{n}\left(\frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2}\right)$

$= \frac{1}{2}\sum_{r=1}^{n}\frac{1}{r} - \sum_{r=1}^{n}\frac{1}{r+1} + \frac{1}{2}\sum_{r=1}^{n}\frac{1}{r+2}$

Re-indexing:

$= \frac{1}{2}\sum_{r=1}^{n}\frac{1}{r} - \sum_{r=2}^{n+1}\frac{1}{r} + \frac{1}{2}\sum_{r=3}^{n+2}\frac{1}{r}$

The $\sum_{r=3}^{n}\frac{1}{r}$ terms cancel (coefficient: $\frac{1}{2} - 1 + \frac{1}{2} = 0$).

Remaining terms:

$\frac{1}{2} + \frac{1}{4} - \frac{1}{2} - \frac{1}{n+1} + \frac{1}{2(n+1)} + \frac{1}{2(n+2)}$

$= \frac{1}{4} + \frac{1}{2(n+2)}$

Step 4: Find the limit.

$\lim_{n \to \infty} S_n = \frac{1}{4} + 0 = \frac{1}{4}$

Step 1: Expand the numerator.

$(1 + 2x)^5 = \sum_{r=0}^{5} \binom{5}{r} (2x)^r = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5$

Step 2: Expand the denominator using the general binomial theorem.

$(1 - x)^{-3} = \sum_{s=0}^{\infty} \binom{-3}{s}(-x)^s$

$\binom{-3}{s} = \frac{(-1)^s (s+2)!}{2! \cdot s!} = (-1)^s \binom{s+2}{2}$

Therefore:

$(1-x)^{-3} = \sum_{s=0}^{\infty} (-1)^s \binom{s+2}{2} (-x)^s = \sum_{s=0}^{\infty} \binom{s+2}{2} x^s$

$= 1 + 3x + 6x^2 + 10x^3 + 15x^4 + 21x^5 + \cdots$

Step 3: Multiply the series and extract the $x^4$ coefficient.

We need all pairs $(r, s)$ with $r + s = 4$ where $0 \leq r \leq 5$ and $s \geq 0$:

Coefficient of $x^4$: $15 + 100 + 240 + 240 + 80 = 675$.

(a) $\cos(3x) = \frac{1}{2}$.

Let $\theta = 3x$. Since $x \in [0, 2\pi)$We have $\theta \in [0, 6\pi)$.

$\cos\theta = \frac{1}{2}$ gives $\theta = \frac{\pi}{3} + 2n\pi$ or $\theta = \frac{5\pi}{3} + 2n\pi$ for $n \in \mathbb{Z}$.

Systematically listing all $\theta \in [0, 6\pi)$:

All six values are in $[0, 6\pi)$ since $\frac{17\pi}{3} = 5\frac{2}{3}\pi \lt 6\pi$.

Dividing by 3:

$x = \frac{\pi}{9},\quad \frac{5\pi}{9},\quad \frac{7\pi}{9},\quad \frac{11\pi}{9},\quad \frac{13\pi}{9},\quad \frac{17\pi}{9}$

(b) $N = 6$ (from part (a)).

The sum of all $x$-coordinates:

$S = \frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} + \frac{11\pi}{9} + \frac{13\pi}{9} + \frac{17\pi}{9} = \frac{60\pi}{9} = \frac{20\pi}{3}$

(a) Let $u = e^x$. Since $e^x \gt 0$ for all $x \in \mathbb{R}$We require $u \gt 0$.

$u^2 - 5u + 6 = 0 \implies (u-2)(u-3) = 0 \implies u = 2 \text{ or } u = 3$

Both satisfy $u \gt 0$.

$e^x = 2 \implies x = \ln 2$

$e^x = 3 \implies x = \ln 3$

Solutions: $x = \ln 2$ and $x = \ln 3$.

(b) $e^{2x} - 5e^x + 6 = 1 \implies e^{2x} - 5e^x + 5 = 0$.

Let $u = e^x$ ($u \gt 0$):

$u^2 - 5u + 5 = 0 \implies u = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$

Both roots are positive: $\frac{5 - \sqrt{5}}{2} \approx 1.382 \gt 0$ and $\frac{5 + \sqrt{5}}{2} \approx 3.618 \gt 0$.

$x = \ln\!\left(\frac{5 + \sqrt{5}}{2}\right) \quad \text{or} \quad x = \ln\!\left(\frac{5 - \sqrt{5}}{2}\right)$

(c) The check is necessary because if a root of the quadratic in $u$ were negative or zero, taking $\ln u$ would be undefined. For example, if the equation were $e^{2x} - 3e^x - 4 = 0$Then $u = -1$ or $u = 4$And $u = -1$ would give $e^x = -1$Which has no real solution. The substitution $u = e^x$ implicitly constrains $u \gt 0$And students who forget this constraint accept spurious solutions.

(a) $y = x^4 - 4x^3 + 6x^2 - 4x + 1$

$\frac{dy}{dx} = 4x^3 - 12x^2 + 12x - 4$

$\frac{d^2y}{dx^2} = 12x^2 - 24x + 12 = 12(x^2 - 2x + 1) = 12(x-1)^2$

(b) $\frac{dy}{dx} = 4x^3 - 12x^2 + 12x - 4 = 4(x^3 - 3x^2 + 3x - 1) = 4(x-1)^3 = 0$.

So $x = 1$ is the only stationary point.

$y(1) = 1 - 4 + 6 - 4 + 1 = 0$.

Stationary point: $(1, 0)$.

(c) $\frac{d^2y}{dx^2}\big\rvert_{x=1} = 12(0)^2 = 0$.

The student claims this is a point of inflection. This reasoning is incorrect because $\frac{d^2y}{dx^2} = 0$ is necessary but not sufficient for a point of inflection. We must examine the sign change of $\frac{dy}{dx}$ either side of $x = 1$.

$\frac{dy}{dx} = 4(x-1)^3$.

For $x \lt 1$ (e.g. $x = 0$): $\frac{dy}{dx} = 4(-1)^3 = -4 \lt 0$.

For $x \gt 1$ (e.g. $x = 2$): $\frac{dy}{dx} = 4(1)^3 = 4 \gt 0$.

The gradient changes from negative to positive, so $x = 1$ is a local minimum, not a point of inflection.

The second derivative test is inconclusive when $\frac{d^2y}{dx^2} = 0$; the first derivative test (sign change analysis) is the definitive method.

(d) $y = x^4 - 4x^3 + 6x^2 - 4x + 1 = (x-1)^4$.

This is immediately obvious because $(x-1)^4 \geq 0$ for all $x$With equality only at $x = 1$. So $(1, 0)$ is a global (and local) minimum.

(a) By LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential), $\ln x$ is prioritised for $u$.

Set $u = \ln x$, $\frac{dv}{dx} = x^3$.

$du = \frac{1}{x}\, dx$, $v = \frac{x^4}{4}$.

$\int x^3 \ln x\, dx = \frac{x^4}{4}\ln x - \int \frac{x^4}{4} \cdot \frac{1}{x}\, dx$

$= \frac{x^4}{4}\ln x - \frac{1}{4}\int x^3\, dx$

$= \frac{x^4}{4}\ln x - \frac{x^4}{16} + C$

$= \frac{x^4}{16}(4\ln x - 1) + C$

(b) If the student chooses $u = x^3$ and $\frac{dv}{dx} = \ln x$:

$du = 3x^2\, dx$But $v = \int \ln x\, dx = x\ln x - x$ (which itself requires integration by parts to find).

This produces an equation involving the original integral on both sides, which eventually works but requires more steps and an additional integration by parts just to find $v$. The LIATE choice is more efficient.

(c) $\int_1^e x^3\ln x\, dx = \left[\frac{x^4}{16}(4\ln x - 1)\right]_1^e$

At $x = e$: $\frac{e^4}{16}(4 - 1) = \frac{3e^4}{16}$

At $x = 1$: $\frac{1}{16}(0 - 1) = -\frac{1}{16}$

$= \frac{3e^4}{16} - \left(-\frac{1}{16}\right) = \frac{3e^4 + 1}{16}$

(a) $\mathbf{d}_1$ and $\mathbf{d}_2$ are not proportional for any value of $a$So the lines are never parallel.

For intersection, there exist $s, t$ such that:

$1 + 2s = 4 + t, \quad 2 - s = 1 + at, \quad 3 + s = 0 + 2t$

From the first equation: $t = 2s - 3$.

From the third equation: $3 + s = 2(2s - 3) = 4s - 6 \implies 3s = 9 \implies s = 3$.

Then $t = 2(3) - 3 = 3$.

Substituting into the second equation: $2 - 3 = 1 + 3a \implies -1 = 1 + 3a \implies a = -\frac{2}{3}$.

The lines intersect when $a = -\frac{2}{3}$At the point $\begin{pmatrix} 7 \\ -1 \\ 6 \end{pmatrix}$.

(b) For $a = 3$: the lines are not parallel. Check for intersection:

From the first and third equations: $s = 3$, $t = 3$.

Second equation: $2 - 3 = 1 + 3(3) = 10$Giving $-1 = 10$Which is false.

The system is inconsistent, so the lines are skew.

(c) For $a = -1$: the lines are skew (checking gives $-1 = 1 + (-1)(3) = -2$False).

The shortest distance between two skew lines is:

$d = \frac{\lvert(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)\rvert}{\lvert\mathbf{d}_1 \times \mathbf{d}_2\rvert}$

$\mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 \\ -1 \\ -3 \end{pmatrix}$.

$\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \begin{pmatrix} -3 \\ -3 \\ -1 \end{pmatrix}$

$\lvert\mathbf{d}_1 \times \mathbf{d}_2\rvert = \sqrt{9+9+1} = \sqrt{19}$

$(\mathbf{b}-\mathbf{a}) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) = -9 + 3 + 3 = -3$

$d = \frac{\lvert -3 \rvert}{\sqrt{19}} = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}$

(a) "$x \gt 2$" implies "$x^2 \gt 4$": if $x \gt 2$ then $x^2 \gt 4$. However, "$x \gt 2$" is not necessary: $x = -3$ gives $x^2 = 9 \gt 4$But $x \lt 2$.

Answer: sufficient but not necessary.

(b) If $n$ is prime and $n \neq 2$Then $n$ is odd. But $n = 2$ is prime and even. Also, “odd” does not imply “prime” (counterexample: 9).

Answer: neither necessary nor sufficient.

(c) Since $a^2 \geq 0$ and $b^2 \geq 0$The sum $a^2 + b^2 = 0$ only when both are zero. So $a^2 + b^2 = 0 \iff a = 0 \text{ and } b = 0$.

Answer: both necessary and sufficient.

(a) $f'(x) = 3x^2 - 2 = 0$ gives $x = \pm\sqrt{2/3} \approx \pm 0.816$.

$f\!\left(-\sqrt{2/3}\right) \approx 3.09$ (local maximum).

$f\!\left(\sqrt{2/3}\right) \approx 0.91$ (local minimum).

Since the local minimum is positive ($\approx 0.91$), the graph crosses the $x$-axis only once (to the left of the local maximum). So there is exactly one real root.

(b) $f'(x) = 3x^2 - 2$. Newton-Raphson: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

$x_0 = 0$: $f(0) = 2$, $f'(0) = -2$.

$x_1 = 0 - \frac{2}{-2} = 1$

$x_1 = 1$: $f(1) = 1$, $f'(1) = 1$.

$x_2 = 1 - \frac{1}{1} = 0$

$x_2 = 0$: $f(0) = 2$, $f'(0) = -2$.

$x_3 = 0 - \frac{2}{-2} = 1$

The iteration cycles: $0, 1, 0, 1, 0, 1, \ldots$

(c) The iteration fails because at $x_0 = 0$The tangent to the curve has gradient $f'(0) = -2$Which points towards $x = 1$ rather than towards the root at $x \approx -1.77$. The Newton-Raphson method overshoots because the tangent at $x = 0$ intersects the $x$-axis at $x = 1$Which is on the opposite side of the local minimum at $x \approx 0.816$ where $f \approx 0.91 \gt 0$Creating a barrier that the iteration cannot cross. The iteration gets trapped in a 2-cycle between $x = 0$ and $x = 1$.

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