IJMB 2025 A LEVEL MATHEMATICS PAPER 1
QUESTIONS BELOW
IJMB 2025 A LEVEL MATHEMATICS PAPER 1
NUMBER 1
NUMBER 2
NUMBER 3
NUMBER 4
NUMBER 5
NUMBER 7
NUMBER 8
NUMBER 9
NUMBER 10
*NUMBER 11a*
sin(A+B)
tan(A+B)=————–
cos(A+B)
sinAcosB+sinBcosA
“””=———————————
cosAcosB–sinAsinB
*Divide through by cosAcosB*
sinAcosB+sinBcosA
——————————–
cosAcosB
“””=———————————
cosAcosB–sinAsinB
————————————
cosAcosB
tanA+tanB
“””=——————-
1–tanAtanB
11B
*To express 12sinx–5cosx in terms of Rsin(x–α)*
Rsin(x–α)=R(sinxcosα–sinαcosx)
”’=Rsinxcosα–Rsinαcosx
Comparing we have
Rsinxcosα=12sinx……(1)
Rsinαcosx=5cosx…….(2)
*From equ(1)*
R ~sinx~ cosα=12 ~sinx~
Rcosα=12………(3)
*From equ(2)*
Rsinα ~cosx~=5 ~cosx~
Rsinα=5……….(4)
*Divide equ(4) by (3)*
tanα=⁵/₁₂
α=tan⁻¹(⁵/₁₂)
α≈22.62
To find R we have
R=√12²+5²
R=√144+25
R=√169
R=13
*Hence we have*
13sin(x–22.62)
To solve 12sinx–5cosx=13
13sin(x–22.62)=13
sin(x–22.62)=1
(x–22.62)=sin⁻¹1
(x–22.62)=90°
Using sin function
x=θ+180n°,n=0,1,2……
*When n=0 we have*
(x–22.62)=90+180•0
x–22.62=90
x=90+22.62
x=112.62°
*When n=1*
(x–22.62)=90+180•1
x–22.62=90+180
x–22.62=270
x=270+22.62
x=292.62°
*Hence x=(112.62°,292.62°)*