Key Takeaways & Direct Answer Briefing
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Full algebraic working causes severe time bleed; sub-90 performance relies on invariant spotting. Odd/even symmetry eliminates definite integrals over symmetric limits [-a, a] in seconds. Vieta's formulas give root sums (sum alpha = -b/a) without solving or factorising polynomials. Surd conjugate bounding replaces tedious manual decimal approximations. Testing extreme angles (0, pi/4, pi/2) prunes incorrect trig identity options immediately. A-Level Working vs ESAT Speed # School exams award partial marks for written steps. On the ESAT, method marks do not exist . You get 1 mark for the correct answer, and 0 for anything else.
Spending 2 minutes factorising a polynomial line-by-line earns the exact same mark as using Vieta's formulas in 15 seconds. To beat the 90-second clock, shift from computation to deduction .
Shortcut 1: Parity & Odd/Even Definite Integrals # For symmetric limits [ − a , a ] [-a, a] [ − a , a ]
Odd function (f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x ) ∫ − a a f odd ( x ) d x = 0 \int_{-a}^a f_{\text{odd}}(x)\,dx = 0 ∫ − a a f odd ( x ) d x = 0 Even function (f ( − x ) = f ( x ) f(-x) = f(x) f ( − x ) = f ( x ) ∫ − a a f even ( x ) d x = 2 ∫ 0 a f even ( x ) d x \int_{-a}^a f_{\text{even}}(x)\,dx = 2\int_0^a f_{\text{even}}(x)\,dx ∫ − a a f even ( x ) d x = 2 ∫ 0 a f even ( x ) d x Example : Evaluate ∫ − π / 4 π / 4 ( x 3 cos x + x 5 + sin x + x 2 ) d x \int_{-\pi/4}^{\pi/4} (x^3 \cos x + x^5 + \sin x + x^2)\,dx ∫ − π /4 π /4 ( x 3 cos x + x 5 + sin x + x 2 ) d x
School Method : Integrate term-by-term. Time: 3 minutes.
Sub-90 Shortcut : x 3 cos x x^3 \cos x x 3 cos x x 5 x^5 x 5 sin x \sin x sin x [ − π 4 , π 4 ] [-\frac{\pi}{4}, \frac{\pi}{4}] [ − 4 π , 4 π ] 0 0 0
2 ∫ 0 π / 4 x 2 d x = 2 [ x 3 3 ] 0 π / 4 = π 3 96 2\int_0^{\pi/4} x^2\,dx = 2\left[\frac{x^3}{3}\right]_0^{\pi/4} = \frac{\pi^3}{96} 2 ∫ 0 π /4 x 2 d x = 2 [ 3 x 3 ] 0 π /4 = 96 π 3 Solve time: 18 seconds .
Shortcut 2: Vieta's Formulas & Polynomial Invariants # When asked for the sum or product of roots, never factorize the polynomial:
∑ i = 1 n α i = − a n − 1 a n , ∏ i = 1 n α i = ( − 1 ) n a 0 a n \sum_{i=1}^n \alpha_i = -\frac{a_{n-1}}{a_n}, \qquad \prod_{i=1}^n \alpha_i = (-1)^n \frac{a_0}{a_n} i = 1 ∑ n α i = − a n a n − 1 , i = 1 ∏ n α i = ( − 1 ) n a n a 0 For a cubic a x 3 + b x 2 + c x + d = 0 a x^3 + b x^2 + c x + d = 0 a x 3 + b x 2 + c x + d = 0
α + β + γ = − b a \alpha + \beta + \gamma = -\frac{b}{a} α + β + γ = − a b α β + β γ + γ α = c a \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} α β + β γ + γ α = a c Sum of squares: α 2 + β 2 + γ 2 = ( − b a ) 2 − 2 ( c a ) \alpha^2 + \beta^2 + \gamma^2 = (-\frac{b}{a})^2 - 2(\frac{c}{a}) α 2 + β 2 + γ 2 = ( − a b ) 2 − 2 ( a c ) Shortcut 3: Rapid Surd Conjugate Bounding # To compare surds (e.g. which is larger: 17 − 15 \sqrt{17} - \sqrt{15} 17 − 15 11 − 9 \sqrt{11} - \sqrt{9} 11 − 9
17 − 15 = 2 17 + 15 , 11 − 9 = 2 11 + 9 \sqrt{17} - \sqrt{15} = \frac{2}{\sqrt{17} + \sqrt{15}}, \qquad \sqrt{11} - \sqrt{9} = \frac{2}{\sqrt{11} + \sqrt{9}} 17 − 15 = 17 + 15 2 , 11 − 9 = 11 + 9 2 Since 17 + 15 > 11 + 9 \sqrt{17} + \sqrt{15} > \sqrt{11} + \sqrt{9} 17 + 15 > 11 + 9
17 − 15 < 11 − 9 \sqrt{17} - \sqrt{15} < \sqrt{11} - \sqrt{9} 17 − 15 < 11 − 9 Zero decimals calculated. Solved in 10 seconds.
Shortcut 4: Trig Identity Reduction via Test Angles # In Maths 2, test specific angles against the options rather than expanding compound angle formulas:
Plug in θ = 0 \theta = 0 θ = 0 θ = π 4 \theta = \frac{\pi}{4} θ = 4 π θ = π 2 \theta = \frac{\pi}{2} θ = 2 π Any option that fails for a single test point is eliminated immediately. Practice Drilling Mathematics 1 & 2 on Sub90 # Review our 31 Maths 2 Solutions with full heuristic proofs. Train rapid mental calculation in the Speed Dojo . Practice live questions in the Exam Arena . 
