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Date	November 2020	Marks available	4	Reference code	20N.2.SL.TZ0.S_10
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	S_10	Adapted from	N/A

Question

Consider a function $f (x)$ , for $x \geq 0$ . The derivative of $f$ is given by $f' (x) = \frac{6 x}{x^{2} + 4}$ .

The graph of $f$ is concave-down when $x > n$ .

Show that $f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ .

[4]

Find the least value of $n$ .

[2]

Find $\int \frac{6 x}{x^{2} + 4} d x$ .

[3]

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = 1$ and $x = 3$ . The area of $R$ is $19.6$ , correct to three significant figures.

Find $f (x)$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

evidence of choosing the quotient rule (M1)

eg $\frac{v u' - u v'}{v^{2}}$

derivative of $6 x$ is $6$ (must be seen in rule) (A1)

derivative of $x^{2} + 4$ is $2 x$ (must be seen in rule) (A1)

correct substitution into the quotient rule A1

eg $\frac{6 (x^{2} + 4) - (6 x) (2 x)}{{(x^{2} + 4)}^{2}}$

$f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ AG N0

METHOD 2

evidence of choosing the product rule (M1)

eg $v u' + u v'$

derivative of $6 x$ is $6$ (must be seen in rule) (A1)

derivative of ${(x^{2} + 4)}^{- 1}$ is $- 2 x {(x^{2} + 4)}^{- 2}$ (must be seen in rule) (A1)

correct substitution into the product rule A1

eg $6 {(x^{2} + 4)}^{- 1} + (- 1) (6 x) (2 x) {(x^{2} + 4)}^{- 2}$

$f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ AG N0

[4 marks]

METHOD 1 (2^nd derivative) (M1)

valid approach

eg $f'' < 0, 24 - 6 x^{2} < 0, n = \pm 2, x = 2$

$n = 2$ (exact) A1 N2

METHOD 2 (1^st derivative)

valid attempt to find local maximum on $f'$ (M1)

eg sketch with max indicated, $(2, 1.5), x = 2$

$n = 2$ (exact) A1 N2

[2 marks]

evidence of valid approach using substitution or inspection (M1)

eg $\int 3 (2 x) \frac{1}{u} d x, u = x^{2} + 4, d u = 2 x d x, \int 3 \times (\frac{1}{u}) d u$

$\int \frac{6 x}{(x^{2} + 4)} d x = 3 \ln (x^{2} + 4) + c$ A2 N3

[3 marks]

recognizing that area $= \int_{1}^{3} f (x) d x$ (seen anywhere) (M1)

recognizing that their answer to (c) is their $f (x)$ (accept absence of $c$ ) (M1)

eg $f (x) = 3 \ln (x^{2} + 4) + c, f (x) = 3 \ln (x^{2} + 4)$

correct value for $\int_{1}^{3} 3 \ln (x^{2} + 4) d x$ (seen anywhere) (A1)

eg $12.4859$

correct integration for $\int_{1}^{3} c d x$ (seen anywhere) (A1)

${[c x]}_{1}^{3}, 2 c$

adding their integrated expressions and equating to $19.6$ (do not accept an expression which involves an integral) (M1)

eg $12.4859 + 2 c = 19.6, 2 c = 7.114$

$c = 3.55700$ (A1)

$f (x) = 3 \ln (x^{2} + 4) + 3.56$ A1 N4

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » AHL 5.9—Differentiating standard functions and derivative rules

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