CAIE A-Level Mathematics A2 6.3 Continuous Random Variables Questions

$\int_{0}^{\sqrt{2}}\left(a x-x^{3}\right) \mathrm{d} x=1$$$\left[a \frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{\sqrt{2}}=1$$$a-\frac{4}{4}=1$a=2

$\int_{0}^{m}\left(2 x-x^{3}\right) d x=\frac{1}{2}$$m^{2}-\frac{m^{4}}{4}=\frac{1}{2}$$m^{4}-4 m^{2}+2=0 \Rightarrow m^{2}=\frac{4 \pm \sqrt{16-8}}{2} \quad[=2 \pm \sqrt{2}]$$m=\sqrt{2-\sqrt{2}}$ or 0.765 (3 sf)

$\int_{0}^{\sqrt{2}}\left(2 x^{2}-x^{4}\right) \mathrm{d} x$$\left[\frac{2 x^{3}}{3}-\frac{x^{5}}{5}\right]_{0}^{\sqrt{2}}$$\left[=\frac{4 \sqrt{2}}{3}-\frac{4 \sqrt{2}}{5}\right]=\frac{8}{15} \sqrt{2}$

$k \int_{0}^{\pi}(1+\cos x) \mathrm{d} x=1$
Attempt integrate f(x) with correct limits and equate to 1.

$k[x+\sin x]_{0}^{\pi}=1$
Correct integration.

[e.g. $k(\pi+\sin \pi-(0+0))=1$ ], $k \pi=1, k=\frac{1}{\pi}$
AG

Some evidence of substitution of limits, i.e. at least one interim step (e.g. $k \pi=1$ ) as minimum requirement.

Convincingly obtained;
no errors seen.

Complete correct answer:
Using h(m)=m+sin(m), h(0.83)=1.57 and h(0.84)=1.58 (or 1.59), while pi/2 lies between these values. Hence 0.83 < median < 0.84. Equivalently, g(m)=m+sin(m)-pi/2 gives g(0.83)=-0.00286... and g(0.84)=0.0138..., so the median lies between 0.83 and 0.84.

Marking guidance:
M1 for substituting 0.83 and 0.84 into a valid integral or into g(m)=m+sin(m)-pi/2 or h(m)=m+sin(m). A1 for both correct values, e.g. h(0.83)=1.57 and h(0.84)=1.58/1.59, or g(0.83)=-0.00286/7 and g(0.84)=0.0138/9. A1FT for concluding that 0.83 < median < 0.84.

$\frac{1}{\pi} \int_{0}^{\pi}(x+x \cos x) \mathrm{d} x$

M1*

Attempt integrate x f(x). Ignore limits.

$=\frac{1}{\pi}\left(\left[\frac{x^{2}}{2}\right]_{0}^{\pi}+[x \sin x]_{0}^{\pi}-\int_{0}^{\pi}(\sin x \mathrm{~d} x)\right)$
OE Attempt to integrate (using 'parts') with correct limits, reaching an expression of the form $a x^{2}+u v-\int v \mathrm{~d} u$.

OR using parts to integrate $x(1+\cos x)$ reaching an expression of the form $u v-\int v \mathrm{~d} u$ i.e. $\left.\frac{1}{\pi}\left(x^{2}+x \sin x-\int(x+\sin x) d x\right)\right)$$$=\frac{1}{\pi}\left(\frac{x^{2}}{2}+x \sin x+\cos x\right)$$

or e.g. $\frac{\pi}{2}+\frac{1}{\pi}\left(0-[-\cos x]_{0}^{\pi}\right)$

or e.g. $\frac{\pi}{2}+\frac{1}{\pi}((-1-1))$
Integration fully correct.

$=\frac{\pi}{2}-\frac{2}{\pi}$
OE

ISW after correct exact value seen.

SC1: Unsupported answer of $\frac{\pi}{2}-\frac{2}{\pi}$ scores B3.

SC2: Unsupported answer of 0.934 scores B2.

1/4 a^2=1
OE

Attempt to set area =1.

a=2
AG

Correct equation and correctly rearranged to a=

No errors seen.

x^2+y^2=(2)^2
Or x^2+y^2=a^2.

[y^2=4/-x^2]

Must see at least one intermediate step

f(x)=4/-x^2
AG

Convincingly rearranged to reach given answer.

No errors seen.

_0^2 x 4/-x^2 d x
Correct expression for E(X)= x f(x) d x with correct limits (accept
limits 0 and a ).

-13[(4/-x^2)^3/2]_0^2}
Integrate x f(x) with any limits or none.

Must reach expression of form: any constant × (4/-x^2)^3/2.
Wholly correct integration and limits.

83 ^3
AG

Correctly obtained with no errors seen.