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∈ℝ, x≠1.
Hence or otherwise, solve the equation 2 sin 2θ-3-6sin 2θ-1=0 for 0≤θ≤π, θ≠π4.
Markscheme
METHOD 1
attempt to write all LHS terms with a common denominator of x-1 (M1)
2x-3-6x-1=2x(x-1)-3(x-1)-6x-1 OR (2x-3)(x-1)x-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]
consider the equation 2 sin2 2θ-5 sin 2θ-3sin 2θ-1=0 (M1)
⇒2 sin2 2θ-5 sin 2θ-3=0
EITHER
attempt to factorise in the form (2 sin 2θ+a)(sin 2θ+b) (M1)
Note: Accept any variable in place of sin 2θ.
(2 sin 2θ+1)(sin 2θ-3)=0
OR
attempt to substitute into quadratic formula (M1)
sin 2θ=5±√494
THEN
sin 2θ=-12 or sin 2θ=3 (A1)
Note: Award A1 for sin 2θ=-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]