19M.1.AHL.TZ1.H_10

pestleMathematicsAIHLPaper 119M· sl-4-8-binomial-distributionsource ↗

The function $p\left( x \right)$ is defined by $p\left( x \right) = {x^3} + 3{x^2} + 8x - 24$ where $x \in \mathbb{R}$ .

Find the remainder when $p\left( x \right)$ is divided by $\left( {x - 2} \right)$ .

[2]

a.i.

Find the remainder when $p\left( x \right)$ is divided by $\left( {x - 3} \right)$ .

[1]

a.ii.

Prove that $p\left( x \right)$ has only one real zero.

[4]

Write down the transformation that will transform the graph of $y = p\left( x \right)$ onto the graph of $y = 8{x^3} + 12{x^2} + 16x - 24$ .

[2]

The random variable $X$ follows a Poisson distribution with a mean of $\lambda$ and $6{\text{P}}\left( {X = 3} \right) = 3{\text{P}}\left( {X = 2} \right) - 2{\text{P}}\left( {X = 1} \right) + 3{\text{P}}\left( {X = 0} \right)$ .

Find the value of $\lambda$ .

[6]

Markscheme / solution

$p\left( 2 \right) = 8 - 12 + 16 - 24$ (M1)

Note: Award M1 for a valid attempt at remainder theorem or polynomial division.

= −12 A1

remainder = −12

[2 marks]

a.i.

$p\left( 3 \right) = 27 - 27 + 24 - 24$ = 0 A1

remainder = 0

[1 mark]

a.ii.

$x = 3$ (is a zero) A1

Note: Can be seen anywhere.

EITHER

factorise to get $\left( {x - 3} \right)\left( {{x^2} + 8} \right)$ (M1)A1

${x^2} + 8 \ne 0$ (for $x \in \mathbb{R}$ ) (or equivalent statement) R1

Note: Award R1 if correct two complex roots are given.

$p'\left( x \right) = 3{x^2} - 6x + 8$ A1

attempting to show $p'\left( x \right) \ne 0$ M1

eg discriminant = 36 – 96 < 0, completing the square

no turning points R1

THEN

only one real zero (as the curve is continuous) AG

[4 marks]

new graph is $y = p\left( {2x} \right)$ (M1)

stretch parallel to the $x$ -axis (with $x = 0$ invariant), scale factor 0.5 A1

Note: Accept “horizontal” instead of “parallel to the $x$ -axis”.

[2 marks]

$\frac{{6{\lambda ^3}{e^{ - \lambda }}}}{6} = \frac{{3{\lambda ^2}{e^{ - \lambda }}}}{2} - 2\lambda {e^{ - \lambda }} + 3{e^{ - \lambda }}$ M1A1

Note: Allow factorials in the denominator for A1.

$2{\lambda ^3} - 3{\lambda ^2} + 4\lambda - 6 = 0$ A1

Note: Accept any correct cubic equation without factorials and ${{e^{ - \lambda }}}$ .

EITHER

$4\left( {2{\lambda ^3} - 3{\lambda ^2} + 4\lambda - 6} \right) = 8{\lambda ^3} - 12{\lambda ^2} + 16\lambda - 24 = 0$ (M1)

$2\lambda = 3$ (A1)

$\left( {2\lambda - 3} \right)\left( {{\lambda ^2} + 2} \right) = 0$ (M1)(A1)

THEN

$\lambda$ = 1.5 A1

[6 marks]

Examiners’ report

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a.i.

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a.ii.

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