JEE Exam > JEE Questions > If x and y are strictly positive such that x ...
Start Learning for Free
If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y is
- a)log 2
- b)- log 2
- c)2 log 2
- d)0
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
If x and y are strictly positive such that x + y = 1, then the minimum...
Explanation:
Given:
x and y are strictly positive such that x + y = 1
To Find:
Minimum value of x log x + y log y
Approach:
To find the minimum value of x log x + y log y, we can use the concept of weighted arithmetic mean and weighted geometric mean.
Weighted Arithmetic Mean (WAM):
Given two positive numbers a and b with weights p and q (p + q = 1), the weighted arithmetic mean is given by:
WAM = pa + qb
Weighted Geometric Mean (WGM):
Given two positive numbers a and b with weights p and q (p + q = 1), the weighted geometric mean is given by:
WGM = (a^p)(b^q)
Solution:
Given x + y = 1, we can consider x as 'a' and y as 'b' with weights x and y respectively.
Using WAM and WGM, we have:
x log x + y log y >= (x log x)(y log y)
= log[(x^x)(y^y)]
Using the given condition x + y = 1, we can rewrite the expression as:
log[(x^x)(y^y)] = log[x^x * (1 - x)^y]
Now, to find the minimum value, we can take the derivative of the expression with respect to x and set it to zero. Solving the derivative, we get the minimum value at x = 1/2.
Substitute x = 1/2 in the expression, we get:
x log x + y log y >= -(1/2) log (1/2) = -log 2
Therefore, the minimum value of x log x + y log y is -log 2, which corresponds to option B.
Given:
x and y are strictly positive such that x + y = 1
To Find:
Minimum value of x log x + y log y
Approach:
To find the minimum value of x log x + y log y, we can use the concept of weighted arithmetic mean and weighted geometric mean.
Weighted Arithmetic Mean (WAM):
Given two positive numbers a and b with weights p and q (p + q = 1), the weighted arithmetic mean is given by:
WAM = pa + qb
Weighted Geometric Mean (WGM):
Given two positive numbers a and b with weights p and q (p + q = 1), the weighted geometric mean is given by:
WGM = (a^p)(b^q)
Solution:
Given x + y = 1, we can consider x as 'a' and y as 'b' with weights x and y respectively.
Using WAM and WGM, we have:
x log x + y log y >= (x log x)(y log y)
= log[(x^x)(y^y)]
Using the given condition x + y = 1, we can rewrite the expression as:
log[(x^x)(y^y)] = log[x^x * (1 - x)^y]
Now, to find the minimum value, we can take the derivative of the expression with respect to x and set it to zero. Solving the derivative, we get the minimum value at x = 1/2.
Substitute x = 1/2 in the expression, we get:
x log x + y log y >= -(1/2) log (1/2) = -log 2
Therefore, the minimum value of x log x + y log y is -log 2, which corresponds to option B.
Free Test
FREE
| Start Free Test |
Community Answer
If x and y are strictly positive such that x + y = 1, then the minimum...
Explanation:
Given:
x + y = 1
x, y > 0
Objective:
Find the minimum value of x log x + y log y
Strategy:
To find the minimum value of the given expression, we can use the concept of weighted AM-GM inequality for logarithms.
Weighted AM-GM Inequality:
For strictly positive numbers a and b with weights p and q such that p + q = 1, the weighted arithmetic mean is greater than or equal to the weighted geometric mean, i.e.,
p * log a + q * log b >= log(a^p * b^q)
Application:
Given x + y = 1, we can rewrite it as x * 1/2 + y * 1/2 = 1/2.
Now, using the weighted AM-GM inequality with p = 1/2, q = 1/2, a = x, b = y, we have:
1/2 * log x + 1/2 * log y >= log(x^(1/2) * y^(1/2))
=> 1/2 * log x + 1/2 * log y >= log(sqrt(xy))
=> 1/2 * log x + 1/2 * log y >= log(sqrt(xy))
Further Simplification:
Since x + y = 1, we have sqrt(xy) <= 1/2="" (from="" am-gm="">
=> log(sqrt(xy)) <=>
=> log(sqrt(xy)) <= -log="">
=> 1/2 * log x + 1/2 * log y >= -log 2
=> x log x + y log y >= -2 log 2
Final Answer:
The minimum value of x log x + y log y is -log 2, which corresponds to option B.
Given:
x + y = 1
x, y > 0
Objective:
Find the minimum value of x log x + y log y
Strategy:
To find the minimum value of the given expression, we can use the concept of weighted AM-GM inequality for logarithms.
Weighted AM-GM Inequality:
For strictly positive numbers a and b with weights p and q such that p + q = 1, the weighted arithmetic mean is greater than or equal to the weighted geometric mean, i.e.,
p * log a + q * log b >= log(a^p * b^q)
Application:
Given x + y = 1, we can rewrite it as x * 1/2 + y * 1/2 = 1/2.
Now, using the weighted AM-GM inequality with p = 1/2, q = 1/2, a = x, b = y, we have:
1/2 * log x + 1/2 * log y >= log(x^(1/2) * y^(1/2))
=> 1/2 * log x + 1/2 * log y >= log(sqrt(xy))
=> 1/2 * log x + 1/2 * log y >= log(sqrt(xy))
Further Simplification:
Since x + y = 1, we have sqrt(xy) <= 1/2="" (from="" am-gm="">
=> log(sqrt(xy)) <=>
=> log(sqrt(xy)) <= -log="">
=> 1/2 * log x + 1/2 * log y >= -log 2
=> x log x + y log y >= -2 log 2
Final Answer:
The minimum value of x log x + y log y is -log 2, which corresponds to option B.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Question Description
If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer?.
If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer?.
Solutions for If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer?, a detailed solution for If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice If x and y are strictly positive such that x + y = 1, then the minimum value of x log x + y log y isa)log 2b)- log 2c)2 log 2d)0Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup