Mixing
Definite integration inequality
$begingroup$
I tried doing it by area concept. Since the function varies from 1/2 to 1 in 0 to 1 it's area should also vary from 1/2 to 1 in 0 to 1 and similarly for 1 to 2 as well. When I add the answer should be from 1/2 to 3/2 but it is not in the option and the answer given is C. Where I am going wrong.
definite-integrals
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
$endgroup$
add a comment |
$begingroup$
I tried doing it by area concept. Since the function varies from 1/2 to 1 in 0 to 1 it's area should also vary from 1/2 to 1 in 0 to 1 and similarly for 1 to 2 as well. When I add the answer should be from 1/2 to 3/2 but it is not in the option and the answer given is C. Where I am going wrong.
definite-integrals
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
$endgroup$
add a comment |
$begingroup$
I tried doing it by area concept. Since the function varies from 1/2 to 1 in 0 to 1 it's area should also vary from 1/2 to 1 in 0 to 1 and similarly for 1 to 2 as well. When I add the answer should be from 1/2 to 3/2 but it is not in the option and the answer given is C. Where I am going wrong.
definite-integrals
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
$endgroup$
I tried doing it by area concept. Since the function varies from 1/2 to 1 in 0 to 1 it's area should also vary from 1/2 to 1 in 0 to 1 and similarly for 1 to 2 as well. When I add the answer should be from 1/2 to 3/2 but it is not in the option and the answer given is C. Where I am going wrong.
definite-integrals
definite-integrals
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
asked Jan 30 at 4:26
Akash GautamaAkash Gautama
955
955
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're not going wrong - it's just that they're being tricky with the answers, and not providing the best possible range. The interval $[frac12,frac32]$ is contained in $[0,2)$, so that inequality is true. We just don't have the full range of possibilities inside that.
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
You are right and the answer given is wrong , c can not be even a possible range to have a subset of it as the actual answer.
Don't worry and good luck.
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
oldest
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$begingroup$
You're not going wrong - it's just that they're being tricky with the answers, and not providing the best possible range. The interval $[frac12,frac32]$ is contained in $[0,2)$, so that inequality is true. We just don't have the full range of possibilities inside that.
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
You're not going wrong - it's just that they're being tricky with the answers, and not providing the best possible range. The interval $[frac12,frac32]$ is contained in $[0,2)$, so that inequality is true. We just don't have the full range of possibilities inside that.
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
You're not going wrong - it's just that they're being tricky with the answers, and not providing the best possible range. The interval $[frac12,frac32]$ is contained in $[0,2)$, so that inequality is true. We just don't have the full range of possibilities inside that.
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
$endgroup$
You're not going wrong - it's just that they're being tricky with the answers, and not providing the best possible range. The interval $[frac12,frac32]$ is contained in $[0,2)$, so that inequality is true. We just don't have the full range of possibilities inside that.
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
answered Jan 30 at 4:44
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
$begingroup$
You are right and the answer given is wrong , c can not be even a possible range to have a subset of it as the actual answer.
Don't worry and good luck.
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
$endgroup$
add a comment |
$begingroup$
You are right and the answer given is wrong , c can not be even a possible range to have a subset of it as the actual answer.
Don't worry and good luck.
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
$endgroup$
add a comment |
$begingroup$
You are right and the answer given is wrong , c can not be even a possible range to have a subset of it as the actual answer.
Don't worry and good luck.
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
$endgroup$
You are right and the answer given is wrong , c can not be even a possible range to have a subset of it as the actual answer.
Don't worry and good luck.
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
answered Jan 30 at 4:55
Adarsha AmanAdarsha Aman
113
113
add a comment |
add a comment |
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