The constant \(a\) satisfies \(a \gt 1\). Show that \[\int_{1}^{a}\left(3x^{2} - 4x + 1\right)\mathrm{d}x = a(a - 1)^{2}.\] (3) Given that the value of this...

This is a "show that" chain in which a definite integral produces a cubic, and the last part asks not merely for a root but for a proof that it is the only real one. That final demand is what distinguishes the question: finding \(a = 3\) is easy, and the marks are for closing off the other possibilities.

(a) Showing that \(\displaystyle\int_{1}^{a}\left(3x^{2} - 4x + 1\right)\mathrm{d}x = a\left(a - 1\right)^{2}\) [3 marks]

Integrate term by term, raising each index by one and dividing by the new index:

\[\int\left(3x^{2} - 4x + 1\right)\mathrm{d}x = \frac{3x^{3}}{3} - \frac{4x^{2}}{2} + x = x^{3} - 2x^{2} + x \qquad \textbf{M1 A1}\]

The M1 is for the integration process and survives a coefficient slip; the A1 requires the antiderivative to be exactly right. In a definite integral the constant of integration cancels between the limits, so it is omitted. Now substitute the limits, upper minus lower:

\[\left[x^{3} - 2x^{2} + x\right]_{1}^{a} = \left(a^{3} - 2a^{2} + a\right) - \left(1 - 2 + 1\right) = a^{3} - 2a^{2} + a\]

The lower limit contributes zero, which is a small gift but must still be shown. Since the target is a product, factorise rather than expand:

\[a^{3} - 2a^{2} + a = a\left(a^{2} - 2a + 1\right) = a\left(a - 1\right)^{2} \qquad \textbf{A1}\]

as required. Because the answer is printed, the marks are for the derivation: taking out the common factor \(a\) and recognising \(a^{2} - 2a + 1\) as the perfect square \(\left(a - 1\right)^{2}\). A candidate who stops at \(a^{3} - 2a^{2} + a\) has done the calculus but not the algebra the part asks for, and loses the final accuracy mark. Check the identity at a convenient value: with \(a = 2\), the left side is \(8 - 8 + 2 = 2\) and the right side is \(2\left(1\right)^{2} = 2\).

(b) Showing that \(a^{3} - 2a^{2} + a - 12 = 0\) [1 mark]

Set the result of part (a) equal to the stated value of the integral and expand:

\[a\left(a - 1\right)^{2} = 12 \quad \Longrightarrow \quad a^{3} - 2a^{2} + a = 12 \quad \Longrightarrow \quad a^{3} - 2a^{2} + a - 12 = 0 \qquad \textbf{[1]}\]

The single mark is for reversing the factorisation and moving the \(12\) across, so it is genuinely one line of work. Leaving the answer as \(a\left(a - 1\right)^{2} = 12\) does not score, because the requested form is the expanded cubic set equal to zero.

(c) Finding \(a\) and proving it is the only real value [2 marks]

A cubic with integer coefficients and leading coefficient \(1\) can only have an integer root that divides the constant term, so the candidates are the factors of \(12\). Testing \(a = 3\):

\[27 - 18 + 3 - 12 = 0,\]

so \(a = 3\) is a root and, by the factor theorem, \(\left(a - 3\right)\) is a factor. Dividing out:

\[a^{3} - 2a^{2} + a - 12 = \left(a - 3\right)\left(a^{2} + a + 4\right) = 0 \qquad \textbf{[1]}\]

The factorisation can be checked by expanding: \(a^{3} + a^{2} + 4a - 3a^{2} - 3a - 12 = a^{3} - 2a^{2} + a - 12\). It is worth noting the middle coefficient is \(+1\), not \(-1\), which is a common slip when dividing a cubic whose \(a^{2}\) coefficient is negative.

Now the uniqueness argument. The product is zero only if one factor is zero, so any further real root must satisfy \(a^{2} + a + 4 = 0\). Its discriminant is

\[b^{2} - 4ac = 1^{2} - 4\left(1\right)\left(4\right) = 1 - 16 = -15,\]

which is negative, so that quadratic has no real roots. Hence \(a = 3\) is the only real value satisfying the equation [1].

The discriminant is the point of this part. A candidate who simply asserts that \(a^{2} + a + 4\) "cannot be factorised" has not proved anything, since an unfactorisable quadratic may still have irrational real roots; the negative discriminant is what settles it. Completing the square gives the same conclusion in a different language, because \(a^{2} + a + 4 = \left(a + \tfrac{1}{2}\right)^{2} + \tfrac{15}{4}\) is a sum of a square and a positive number and so is always positive.

Finally, the value is consistent with the condition \(a \gt 1\) stated at the top of the question, and it checks against part (a): \(3\left(3 - 1\right)^{2} = 3 \times 4 = 12\), which is the given value of the integral.

Examination takeaway. When a "show that" asks for a factorised form, factorise rather than expand, and always take out a common factor before hunting for a perfect square. To prove that a cubic has exactly one real root, divide out the known linear factor and use the discriminant of the remaining quadratic; a claim that it "does not factorise" is not a proof.