Question

If \[{\text{x}}\] and \[{\text{y}}\] are positive and \[{\text{x + y = 1}}\] then the minimum value of \[{\text{x logx + y logy}}\] is
A. \[{\text{log}}2\]
B. \[{\text{ - log}}2\]
C. \[{\text{2log}}2\]
D. \[0\]

Answer 1

Hint: In the question they have given some values we want to use these values to solve the problem
First, we want to rewrite the given equation as we needed
Then we need to apply them on the given values
And we need to differentiate those values
After that thing, we want to simplify those values
By using these steps, you will get the answer which we needed.

Formula used: The formulas which we are needed to solve the problems are
\[{\text{uv = u}}{{\text{v}}^{\text{'}}}{\text{ + v}}{{\text{u}}^{\text{'}}}\]

Complete step-by-step answer:
In the question they have given that,
\[{\text{x + y = 1}}\]
While rewriting the above values we will get as
\[{\text{y = 1 - x}}\]
Let us assume that the given value \[{\text{x logx + y logy}}\] be \[{\text{z}}\]
\[{\text{z = }}\]\[{\text{x logx + y logy}}\]
As we already know the value of \[{\text{y}}\], we are going to apply that on the above. So, we will get
\[{\text{z = }}\]\[{\text{x logx + (1 - x) log(1 - x) }}\]
On differentiate above equation, we will get that
\[\dfrac{{{\text{dz}}}}{{{\text{dx}}}}{\text{ = x}}{\text{.}}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + logx + (1 - x)}}\dfrac{{\text{1}}}{{{\text{(1 - x)}}}}{\text{ + log(1 - x)( - 1) }}\]\[ = 0\] [since, \[{\text{uv = u}}{{\text{v}}^{\text{'}}}{\text{ + v}}{{\text{u}}^{\text{'}}}\]]
By Simplifying the above values, we will get that
\[{\text{log x = log(1 - x)}}\]
As log is common term on both sides, we are cancelling log on both sides, then we will get,
\[{\text{x = 1 - x}}\]
By bringing the variables on one side we will get
\[ \Rightarrow \]\[{\text{2x = 1}}\]
Here we are bringing the numbers on one side
\[\therefore {\text{x = }}\dfrac{1}{2}\]
Hence, we got the \[{\text{x}}\] value.
[\{\text{x = 1 - x}}\]……………. (A)
We know that \[{\text{y = 1 - x}}\]
Substitute \[{\text{y}}\] in equation (A) we get,
Hence,\[{\text{x = y}}\]
So, \[{\text{x}}\]and \[{\text{y}}\] values are equal]
\[{\text{Z}}\] Minimum\[{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ log}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ log}}\dfrac{{\text{1}}}{{\text{2}}}\]
Taking out the term \[\dfrac{1}{2}\] which is the common
\[{\text{ = }}\]\[\dfrac{1}{2}\left( {2\log \dfrac{1}{2}} \right)\]
On Simplification we will get
\[{\text{Z}}\] Minimum\[{\text{ = }}\]\[\log \dfrac{1}{2}\]
\[{\text{Z}}\] Minimum\[{\text{ = }}\]\[{\text{ - log2}}\]
Therefore, finally we got that the minimum value of \[{\text{x logx + y logy}}\] is \[{\text{ - log}}2\]

Option B is the correct answer.

Note: Logarithmic functions are inverse of exponential functions. Differentiable means that the derivative exists and it must exist for every value in the function’s domain. In these kinds of sum, differentiation is more important. Students were making mistakes while differentiating.