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Question

Question: Find the range of \({\log _e}(\sin x)\) is A)\(( - \infty ,\infty )\) B)\(( - \infty ,1)\) C)\...

Find the range of loge(sinx){\log _e}(\sin x) is
A)(,)( - \infty ,\infty )
B)(,1)( - \infty ,1)
C)(,0]( - \infty ,0]
D)(,0)( - \infty ,0)

Explanation

Solution

Hint: Here we will proceed with the solution as we know the range of sinx=[1,1]\sin x = [1, - 1] which is required to solve this problem.
Here we need to find the range of given value that is loge(sinx){\log _e}(\sin x)
As we know that the domain of logarithmic functions are of positive value only
Since we know the range of sinx\sin x is [1,1][ - 1,1]
Then the domain value of above log function would be (0,1](0,1]
So, now to get the range of loge(sinx){\log _e}(\sin x)
Let us substitute x=0x = 0 in the given function loge(sinx){\log _e}(\sin x)
i.e.
For x0logex=x \to 0 \Rightarrow {\log _e}x = - \infty
Now let us substitute x=1x = 1 in the given function loge(sinx){\log _e}(\sin x)
i.e.
Forx1logex=0x \to 1 \Rightarrow {\log _e}x = 0
Hence from this we can say that the range of given function loge(sinx){\log _e}(\sin x)=(,0]( - \infty ,0]

NOTE: In this particular problem we know the range of sinx\sin x is [1,1][ - 1,1] and domain of log function is (0,1](0,1] so by substituting the domain values (xx values) in the given function i.e. loge(sinx){\log _e}(\sin x) . We will get the range of the given function.