Transformations and Modulus Question Help

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#1
Hi, I need help with part d of this question but ive attached all of the questions below too.
I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.
My working/thought process:
I'm thinking after plotting the graphs they intersect at (2, 0) when a is 0. There is only one point of intersection. Therefore, I set the modulus side to 0 as y = 0 and then multiplied by e to get 0 = -e^(1/2x - a). I then set 1/2x - a to 0, making x = 2a. after subbing in (2,0) i know that when a is 0, x is 2. Then i get y = 2a + 2 but im not sure if thats right or if my method to get there is correct.

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.
a) Give full details of the translation and stretch involved
b)Sketch y = ln(1/2x - a)
c) Sketch y = |ln(1/2x - a)|
d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)
Last edited by Alex159; 1 month ago
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1 month ago
#2
(Original post by Alex159)
Hi, I need help with part 4 of this question
I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.
a) Give full details of the translation and stretch involved
b)Sketch y = ln(1/2x - a)
c) Sketch y = |ln(1/2x - a)|
d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)
Please post your working 0
#3
(Original post by Muttley79)
Please post your working edited the post with info
0
1 month ago
#4
(Original post by Alex159)
edited the post with info
You really need to upload the whole question [see the forum rules] ... I'll try to check later but have a lot of stuff to do this evening [which is why it's useful to see working so I don't have to do the question from the beginning]
0
1 month ago
#5
(Original post by Alex159)
Hi, I need help with part d of this question but ive attached all of the questions below too.
I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.
My working/thought process:
I'm thinking after plotting the graphs they intersect at (2, 0) when a is 0. There is only one point of intersection. Therefore, I set the modulus side to 0 as y = 0 and then multiplied by e to get 0 = -e^(1/2x - a). I then set 1/2x - a to 0, making x = 2a. after subbing in (2,0) i know that when a is 0, x is 2. Then i get y = 2a + 2 but im not sure if thats right or if my method to get there is correct.

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.
a) Give full details of the translation and stretch involved
b)Sketch y = ln(1/2x - a)
c) Sketch y = |ln(1/2x - a)|
d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)
For part d, the modulus of the ln function will only equal the negative of the ln function if the ln function itself equals 0 (since 0 = -0) or is negative (since a double negative is always positive, just as the modulus of a function is always positive). So you can solve the equation ln((1/2)x- a) <= 0 to receive your set of solutions. Just remember to include x > 2a in the result, as you can't take the ln of anything less than or equal to 0.
I imagine using your graph sketches as you have done is a valid alternative route to take with this question instead btw, but I personally prefer the algebraic approach.
Reply here if you need help solving the equation. Last edited by BlueBazooka; 1 month ago
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