Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Hi all im not sure how to do question 10. couldnt get the right answer. Your help is greatly appreciated! Tge answer is y = ln(2x - 4)/(x - 1) + 2
Date Posted: 3 years ago
Views: 423
You'll need to split the fraction into partial fractions first.
You can do this in the usual way by letting 1/ [(x - 1)(x - 2)] = A/(x - 1) + B/(x - 2), then solve for A and B.
Or, simply rewrite it like this :
1 / [(x - 1)(x - 2)]
= (x - 1 - (x - 2)) / [(x - 1)(x - 2)]
= (x - 1) / [(x - 1)(x - 2)] - (x - 2) / [(x - 1)(x - 2)]
= 1/(x - 2) - 1/(x - 1)
You can do this in the usual way by letting 1/ [(x - 1)(x - 2)] = A/(x - 1) + B/(x - 2), then solve for A and B.
Or, simply rewrite it like this :
1 / [(x - 1)(x - 2)]
= (x - 1 - (x - 2)) / [(x - 1)(x - 2)]
= (x - 1) / [(x - 1)(x - 2)] - (x - 2) / [(x - 1)(x - 2)]
= 1/(x - 2) - 1/(x - 1)
Then,
dy/dx = 1/(x - 2) - 1/(x - 1)
y = ∫ ( 1/(x - 2) - 1/(x - 1) ) dx
y = ln(x - 2) - ln(x - 1) + c
where c is an arbitrary constant.
y = ln [(x-2)/(x-1)] + c
(Logarithm laws, subtraction)
Sub x = 3, y = 2,
2 = ln [(3-2)/(3-1)] + c
2 = ln½ + c
c = 2 - ln½
c = 2 - ln(2)⁻¹
c = 2 - (-ln2) = 2 + ln2
So,
y = ln[(x - 2)/(x - 1)] + 2 + ln2
y = ln[(x - 2)/(x - 1) × 2] + 2
(Logarithm laws, addition)
y = ln[(2x - 4)/(x - 1)] + 2
dy/dx = 1/(x - 2) - 1/(x - 1)
y = ∫ ( 1/(x - 2) - 1/(x - 1) ) dx
y = ln(x - 2) - ln(x - 1) + c
where c is an arbitrary constant.
y = ln [(x-2)/(x-1)] + c
(Logarithm laws, subtraction)
Sub x = 3, y = 2,
2 = ln [(3-2)/(3-1)] + c
2 = ln½ + c
c = 2 - ln½
c = 2 - ln(2)⁻¹
c = 2 - (-ln2) = 2 + ln2
So,
y = ln[(x - 2)/(x - 1)] + 2 + ln2
y = ln[(x - 2)/(x - 1) × 2] + 2
(Logarithm laws, addition)
y = ln[(2x - 4)/(x - 1)] + 2
Thank you so much! I got it!
See 1 Answer
Answered in the question's comments section.
360 Tutoring Program