Problem 127 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=\sin ^{-1}(\log [x])+\log \left(\sin ^{-1}[x]\right)\), where [] denotes the greatest integer function. Then, (A) domain of \(f\) is \([1,2)\) (B) domain of \(f\) is \([1,3)\) (C) range of \(f\) is \(\left\\{\log \frac{\pi}{2}\right\\}\) (D) range of \(f\) is \(\\{0\\}\)

Step-by-Step Solution

Answer

The domain of \(f\) is \([1, 2)\) and its range is \(\{\log \frac{\pi}{2}\}\).

Step 1: Understanding the Function

Given the function \( f(x) = \sin^{-1}(\log[x]) + \log(\sin^{-1}[x]) \), we need to analyze both components: \( \sin^{-1}(\log[x]) \) and \( \log(\sin^{-1}[x]) \). Here, \([x]\) denotes the greatest integer less than or equal to \(x\). This function is only defined for particular values of \(x\) due to the constraints of the inverse sine and logarithm functions.

Step 2: Analyzing \( \sin^{-1}(\log[x]) \)

The component \( \sin^{-1}(\log[x]) \) is defined when \( \log[x] \) is in the interval \([-1, 1]\), because the range of \( \sin^{-1}(x) \) is defined only for \(x\) in this interval. Hence, \(\log[x] \geq -1\) implies \([x] \geq \frac{1}{e}\) and \(\log[x] \leq 1\) implies \([x] \leq e\). Therefore, \([x]\) must be in \(\{1, 2\}\).

Step 3: Analyzing \( \log(\sin^{-1}[x]) \)

The component \( \log(\sin^{-1}[x]) \) is defined when \( \sin^{-1}[x] > 0 \) since the logarithm requires positive inputs. \(\sin^{-1}[x] > 0\) implies \([x] \geq 1\), as the sine inverse is positive for positive integers.

Step 4: Establishing Domain of \(f\)

Combining these conditions, both components restrict \([x]\) to \(\{1, 2\}\). Thus, the domain of the function, considering \([x] = 1\) and \([x] = 2\) within \(x\), is \([1, 2)\).

Step 5: Finding the Range of \(f\)

For \([x] = 1\), \(f(x) = \sin^{-1}(\log[1]) + \log(\sin^{-1}1) = \sin^{-1}(0) + \log(\frac{\pi}{2}) = 0 + \log(\frac{\pi}{2}) = \log(\frac{\pi}{2})\). Thus, the range of the function is \(\{\log(\frac{\pi}{2})\}\) only.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the reverse operations of trigonometric functions. They are used to find angles when the value of a trigonometric ratio is known. The most common inverse trigonometric functions are the inverse sine \(\sin^{-1}(x)\), inverse cosine \(\cos^{-1}(x)\), and inverse tangent \(\tan^{-1}(x)\).

These functions have specific domains and ranges. For instance, the domain of \(\sin^{-1}(x)\) is \([-1, 1]\), and its range is \[\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\]. This means that \(\sin^{-1}(x)\) can only accept values of \(x\) between -1 and 1, inclusive. The function returns an angle between \(-\frac{\pi}{2}\) and \(+\frac{\pi}{2}\).

In our given function \(f(x) = \sin^{-1}(\log[x]) + \log(\sin^{-1}[x])\), care must be taken that the expression inside \(\sin^{-1}(\log[x])\) satisfies the domain conditions, which is why \([-1, 1]\) is important to determine where the function is defined.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. The general form is \(\log_b(a)\), where \(b\) is the base and \(a\) is the argument. This implies asking, "What power must we raise the base to in order to get \(a\)?"

In the context of the natural logarithm, the base is \(e\), an irrational number approximately equal to 2.718. Hence, when using \(\log\) with no base specified, it often implies natural logarithms which are denoted as \(\ln(x)\).

The function \(\log(x)\) is defined only for \(x > 0\) because you cannot take the logarithm of a negative number or zero. In our exercise, \(\log(\sin^{-1}[x])\) requires that the input to the \(\log\) is positive. Since \(\sin^{-1}[x]\) must be greater than zero, this defines part of the domain.

Domain and Range

The domain and range are key aspects of functions in mathematics. The domain refers to all possible input values (or \(x\)-values), while the range refers to all possible output values (or \(y\)-values) produced by a function.

Our given function \(f(x) = \sin^{-1}(\log[x]) + \log(\sin^{-1}[x])\) combines the restrictions of both the inverse trigonometric and logarithmic functions. To find the domain, we must identify all \(x\) values where the function is valid. We rely on the greatest integer function \([x]\), which rounds \(x\) down to the nearest whole number. In this exercise, the domain is restricted by the necessity that the inputs to both the inverse sine and logarithmic functions are appropriate.

For the function to be defined, \[x\] must be in \{1, 2\}. This relates back to the explained domains of each part of the function: \(\sin^{-1}(x)\) must have a value between -1 and 1 and \(\log(x)\) must have a positive input. Consequently, by evaluating these conditions together, the domain is \([1, 2)\).

Meanwhile, the range of the function has been determined by evaluating the expression at valid inputs, resulting in a consistent output of \(\log(\frac{\pi}{2})\), hence the range \{\log(\frac{\pi}{2})\}\.