2014 H2 Mathematics Paper 1 Question 7

Definite Integrals: Areas and Volumes

Answers

(i)

α=1.885{\alpha = 1.885}
β=3{\beta = \sqrt{3}}

(ii)

∫βαf(x)  dx=−0.597{\displaystyle \int_\beta^\alpha f(x) \; \mathrm{d}x = -0.597}

(iii)

54353 units2{\frac{54}{35} \sqrt{3} \textrm{ units}^2}

(iv)

The six roots of f(x)=0{f(x)=0} consists of two real roots, x=α{x=\alpha} and x=−α{x=-\alpha} and four complex roots which made up of two complex conjugate pairs of the form x=a+bi,{x=a+b\mathrm{i}, } x=a−bi,{x=a-b\mathrm{i}, } x=−a−bi,{x=-a-b\mathrm{i}, } and x=−a+bi{x=-a+b\mathrm{i} } where a,b∈R{a,b \in \mathbb{R}}

Full solutions

(i)

Using a GC,
α=1.8852=1.885 (3 dp)  ■\begin{align*} \alpha &= 1.8852 \\ &= 1.885 \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}
For β,{\beta,}
x6−3x4−7=−7x6−3x4=0x4(x−3)=0x=0 or x=±3\begin{gather*} x^6 - 3 x^4 - 7 = -7 \\ x^6 - 3x^4 = 0 \\ x^4 (x^ - 3 ) = 0 \\ x = 0 \textrm{ or } x = \pm \sqrt{3} \end{gather*}
Since β>0,{\beta > 0,}
β=3  ■\beta = \sqrt{3} \; \blacksquare

(ii)

∫βαf(x)  dx=−0.597 (3 dp)  ■\int_\beta^\alpha f(x) \; \mathrm{d}x = -0.597 \textrm{ (3 dp)} \; \blacksquare

(iii)

Area of region=∫03−7−(x6−3x4−7)  dx=∫03(−x6+3x4)  dx=[−17x7+35x5]03=−17(3)7+35(3)5=−17(273)+35(93)=54353 units2  ■\begin{align*} & \textrm{Area of region} \\ & = \int_0^{\sqrt{3}} -7 - (x^6 - 3x^4 - 7) \; \mathrm{d}x \\ & = \int_0^{\sqrt{3}} ( - x^6 + 3x^4 ) \; \mathrm{d}x \\ & = \left[ -\frac{1}{7} x^7 + \frac{3}{5} x^5 \right]_0^{\sqrt{3}} \\ & = -\frac{1}{7} \left( \sqrt{3} \right)^7 + \frac{3}{5} \left( \sqrt{3} \right)^5 \\ & = - \frac{1}{7} (27 \sqrt{3}) + \frac{3}{5} (9 \sqrt{3}) \\ & = \frac{54}{35} \sqrt{3} \textrm{ units}^2 \; \blacksquare \end{align*}

(iv)

RHS=f(−x)=(−x)6−3(−x)4−7=x6−3x4−7=f(x)=LHS  ■\begin{align*} \textrm{RHS} &= f(-x) \\ &= (-x)^6 - 3(-x)^4 - 7 \\ &= x^6 - 3 x^4 - 7 \\ & = f(x) \\ &= \textrm{LHS} \; \blacksquare \end{align*}
From the graph, the positive real root to the equation f(x)=0{f(x)=0} is x=α.{x = \alpha.}
Since f(x)=f(−x),{f(x)=f(-x), } x=−α{x=-\alpha} is also a root.
The remaining four roots of the equation are complex.
Let x=a+bi{x = a+b\mathrm{i}} be a complex root of f(x)=0{f(x)=0} where a,b∈R{a,b \in \mathbb{R}}
Since f(x)=f(−x),{f(x)=f(-x), } x=−a−bi{x = -a - b\mathrm{i}} is also a root.
Since all the coefficients are real, by the conjugate root theorem, the complex conjugates (a+bi)∗=a−bi{(a+b\mathrm{i})^* = a-b\mathrm{i}} and (−a−bi)∗=−a+bi{(-a-b\mathrm{i})^* = -a+b\mathrm{i}} are also roots
Hence the six roots of f(x)=0{f(x)=0} consists of two real roots, x=α{x=\alpha} and x=−α{x=-\alpha} and four complex roots which made up of two complex conjugate pairs of the form x=a+bi,{x=a+b\mathrm{i}, } x=a−bi,{x=a-b\mathrm{i}, } x=−a−bi,{x=-a-b\mathrm{i}, } and x=−a+bi.{x=-a+b\mathrm{i}. }