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Date November 2021 Marks available 5 Reference code 21N.1.SL.TZ0.6
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Solve and Hence or otherwise Question number 6 Adapted from N/A

Question

Show that 2x-3-6x-1=2x2-5x-3x-1, x, x1.

[2]
a.

Hence or otherwise, solve the equation  2sin2θ-3-6sin2θ-1=0  for  0θπ, θπ4.

[5]
b.

Markscheme

METHOD 1

attempt to write all LHS terms with a common denominator of x-1                 (M1)

2x-3-6x-1=2xx-1-3x-1-6x-1   OR   2x-3x-1x-1-6x-1

=2x2-2x-3x+3-6x-1   OR   2x2-5x+3x-1-6x-1                 A1

=2x2-5x-3x-1                 AG

 

METHOD 2

attempt to use algebraic division on RHS                 (M1)

correctly obtains quotient of 2x-3 and remainder -6                 A1

=2x-3-6x-1 as required.                 AG

 

[2 marks]

a.

consider the equation 2sin22θ-5sin2θ-3sin2θ-1=0                 (M1)

2sin22θ-5sin2θ-3=0


EITHER

attempt to factorise in the form 2sin2θ+asin2θ+b                 (M1)


Note:
Accept any variable in place of sin2θ.


2sin2θ+1sin2θ-3=0


OR

attempt to substitute into quadratic formula                 (M1)

sin2θ=5±494


THEN

sin2θ=-12  or  sin2θ=3                 (A1)


Note:
Award A1 for sin2θ=-12 only.


one of 7π6  OR  11π6   (accept 210 or 330)                 (A1)

θ=7π12,11π12  (must be in radians)                 A1


Note:
Award A0 if additional answers given.

 

[5 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1—Number and algebra » SL 1.6—Simple proof
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Topic 3— Geometry and trigonometry » SL 3.8—Solving trig equations
Topic 1—Number and algebra
Topic 3— Geometry and trigonometry

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