Year 13 Core - Peer Marked Homework 1

Question 1

The equation of a curve is \(y = {x^2}\ln (3x + 2)\). Find the exact value of \(\dfrac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}}\) at the point on the curve for which $x = 3$.

[8 marks]


Q1 - Mark Scheme (Page 1)

Question Marking Instructions Marks Typical Solution

1

Correct use of the product rule for their values of \(u,v,\frac{{{\rm{d}}u}}{{{\rm{d}}x}}\) and \(\frac{{{\rm{d}}v}}{{{\rm{d}}x}}\)

M1*

\[\begin{align}\dfrac{{{\rm{d}}y}}{{{\rm{d}}x}} =& \ {x^2} \cdot \dfrac{3}{{3x + 2}} + 2x \cdot \ln (3x + 2)\\\dfrac{{{\rm{d}}y}}{{{\rm{d}}x}} =& \ \dfrac{{3{x^2}}}{{3x + 2}} + 2x\ln (3x + 2)\end{align}\]


Obtains $2x\ln (3x + 2)$ or $\frac{{3{x^2}}}{{3x + 3}}$

A1

Obtains correct first derivative (may be unsimplified).

A1

Attempts to use product rule

M1*dep

\[\begin{array}{l} \dfrac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} = \dfrac{{(3x + 2) \cdot 6x - 3{x^2} \cdot 3}}{{{{(3x + 2)}^2}}} + 2x \cdot \dfrac{3}{{3x + 2}} + 2\ln (3x + 2)\\ \dfrac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} = \dfrac{{9{x^2} + 12x}}{{{{(3x + 2)}^2}}} + \dfrac{{6x}}{{3x + 2}} + 2\ln (3x + 2) \end{array}\]

Attempts use of quotient rule.

M1*dep

Obtains correct second derivative (may be unsimplified).

A1

Question continues on page below...

Q1 - Mark Scheme (Page 2)


Question Marking Instructions Marks Typical Solution

1

cont...

Substitutes $x = 3$ into their expression for \(\dfrac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}}\)

Note: Must have optained all previous method (M1) marks


M1*dep

\[\begin{align} {\left. {\dfrac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}}} \right|_{x = 3}} =& \ \dfrac{{9{{(3)}^2} + 12(3)}}{{{{(3(3) + 2)}^2}}} + \dfrac{{6(3)}}{{3(3) + 2}} + 2\ln (3(3) + 2)\\ =&\ \dfrac{{315}}{{121}} + 2\ln 11 \end{align}\]

Obtains correct final answer (must be exact value).

A1

8 Marks



Slide 2