June 2022 Paper 1 Q4

OCR ACurrent spec8 marksQuadratics Trigonometry

(a) Write \(2x^2 + 6x + 7\) in the form \(p(x + q)^2 + r\), where \(p\), \(q\) and \(r\) are constants. [3]

(b) State the coordinates of the minimum point on the graph of \(y = 2x^2 + 6x + 7\). [2]

the minimum value of \(2\tan^2\theta + 6\tan\theta + 7\),
the smallest positive value of \(\theta\), in degrees, for which the minimum value occurs. [3]

Mark scheme (a)

Scheme	Marks	AO
\(2(x + 1.5)^2 + 2.5\) \(p = 2\)	B1	1.1a
\(q = 1.5\)	B1	1.1a
\(r = 2.5\)	B1FT	1.1a
	[3]

Notes

B1: Could be implied by \(2(x + q)^2 + r\)

B1: Could be implied by \(p(x + 1.5)^2 + r\)

B1FT: FT on their \(p\) and \(q\) ie \(7 - pq^2\)

Mark scheme (b)

Scheme	Marks	AO
\((-1.5, 2.5)\) Correct \(x\)-coordinate	B1FT	1.1
Correct \(y\)-coordinate	B1FT	1.1
	[2]

Notes

B1FT: FT on their (a)
Could come from differentiation

B1FT: FT on their (a)
No FT on incorrect \(x\)-value from differentiation

Mark scheme (c)

Scheme	Marks	AO
minimum value of the function \(= 2.5\)	B1FT	3.1a
\(\tan\theta = -1.5\) \(\theta = -56.3^\circ\)	M1	3.1a
\(\theta = 124^\circ\)	A1	1.1
	[3]

Notes

B1FT: FT on their minimum value
Allow BOD if different answers in (a) and (b)
2.5 must be stated as, or clearly intended to be, the minimum value
Just \((\ldots, 2.5)\) is insufficient

M1: Attempt to solve \(\tan\theta =\) their \((-1.5)\)
To obtain numerical value for \(\theta\)
Allow an angle in radians (expect \(-0.983\) rad)
Allow BOD if different answers in (a) and (b)

A1: Obtain \(124^\circ\), or better
A0 if additional solutions

Condone approaches other than ‘hence’ eg
B1 – attempt to solve \(\tan\theta = -1.5\), from correct derivative (expect \(4\tan\theta\sec^2\theta + 6\sec^2\theta = 0\))
B1 – obtain \(\theta = 124^\circ\)
B1 – obtain min value of 2.5 (no FT)