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Date May 2013 Marks available 4 Reference code 13M.1.sl.TZ1.3
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 3 Adapted from N/A

Question

Consider \(f(x) = {x^2}\sin x\) .

Find \(f'(x)\) .

[4]
a.

Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .

[3]
b.

Markscheme

evidence of choosing product rule     (M1)

eg   \(uv' + vu'\)

correct derivatives (must be seen in the product rule) \(\cos x\) , \(2x\)     (A1)(A1)

\(f'(x) = {x^2}\cos x + 2x\sin x\)     A1 N4

[4 marks]

a.

substituting \(\frac{\pi }{2}\) into their \(f'(x)\)     (M1)

eg   \(f'\left( {\frac{\pi }{2}} \right)\) , \({\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)\)

correct values for both \(\sin \frac{\pi }{2}\) and \(\cos \frac{\pi }{2}\) seen in \(f'(x)\)     (A1)

eg   \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\) 

\(f'\left( {\frac{\pi }{2}} \right) = \pi \)     A1 N2

[3 marks]

b.

Examiners report

Many candidates correctly applied the product rule for the derivative, although a common error was to answer \(f'(x) = 2x\cos x\) .

a.

Candidates generally understood that the gradient of the curve uses the derivative, although in some cases the substitution was made in the original function. Some candidates did not know the values of sine and cosine at \({\frac{\pi }{2}}\) .

b.

Syllabus sections

Topic 6 - Calculus » 6.2 » Derivative of \({x^n}\left( {n \in \mathbb{Q}} \right)\) , \(\sin x\) , \(\cos x\) , \(\tan x\) , \({{\text{e}}^x}\) and \(\ln x\) .
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