Problem 161 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The largest interval lying in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for which the function \(f(x)=4^{-x^{2}}+\cos ^{-1}\left(\frac{x}{2}-1\right)+\log (\cos x)\) is defined, is (A) \([0, \pi]\) (B) \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) (C) \(\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)\) (D) \(\left[0, \frac{\pi}{2}\right)\)

Step-by-Step Solution

Answer

(D) \([0, \frac{\pi}{2})\)

Step 1: Analyze Domain of Each Component

The function has three different components: \(4^{-x^2}\), \(\cos^{-1}\left(\frac{x}{2} - 1\right)\), and \(\log (\cos x)\). We need to find the domain for each component separately and then find the common domain.

Step 2: Domain for Exponential Function

The exponential function \(4^{-x^2}\) is defined for all real \(x\). Thus, it does not restrict the domain of the function \(f(x)\).

Step 3: Domain for Inverse Cosine Function

The function \(\cos^{-1}\left(\frac{x}{2} - 1\right)\) requires that \(-1 \leq \frac{x}{2} - 1 \leq 1\). Solving these inequalities, we find \(-1 \leq x \leq 3\).

Step 4: Domain for Logarithmic Function

The logarithmic function \(\log(\cos x)\) is defined only when \(\cos x > 0\). In the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), this means \(-\frac{\pi}{2} < x < \frac{\pi}{2}\).

Step 5: Intersection of Domains

Combine the domains found: from Step 3, \([-1, 3]\), and from Step 4, \((-\frac{\pi}{2}, \frac{\pi}{2})\). The overlap and restriction to \((-\frac{\pi}{2}, \frac{\pi}{2})\) give \([-1, \frac{\pi}{2})\).

Step 6: Conclusion

The largest interval from the given options that fits within and aligns with \([-1, \frac{\pi}{2})\) in \((-\frac{\pi}{2}, \frac{\pi}{2})\) is \([0, \frac{\pi}{2})\) which corresponds to option (D).

Key Concepts

Exponential Function

The exponential function in the given exercise is represented by the expression \(4^{-x^2}\). Exponential functions are mathematical functions of the form \(a^{bx}\), where \(a\) is a positive constant, and \(b\) is any real number. These functions are defined for all real numbers \(x\). This makes the domain of an exponential function very broad. In our specific case of \(4^{-x^2}\), the function is defined for every real value of \(x\), meaning this component itself doesn't limit or restrict the domain of the overall function \(f(x)\). Exponential functions form the backbone of many mathematical models, particularly in growth and decay scenarios. However, for the function \(f(x)\), the exponential part is quite accommodating, allowing room for real number inputs without constraints.

Inverse Cosine Function

The inverse cosine function, denoted as \(\cos^{-1}(x)\), is an important trigonometric function which retrieves the angle whose cosine is a given number. Because cosine values range from -1 to 1, \(\cos^{-1}(x)\) is only defined when its argument lies within that range. For the exercise, \(\cos^{-1}\left(\frac{x}{2} - 1\right)\) ensures that \(-1 \leq \frac{x}{2} - 1 \leq 1\). This results in the solution of the inequality \[ -1 \leq \frac{x}{2} - 1 \leq 1, \] by substituting and manipulating the expression:

• First, \(\frac{x}{2} - 1 \geq -1\).
• Leading to \(\frac{x}{2} \geq 0\) and thus, \(x \geq 0\).
• Second, \(\frac{x}{2} - 1 \leq 1\).
• Give us \(\frac{x}{2} \leq 2\) and thus, \(x \leq 4\).

These constraints combine to form \(x\) values from 0 to 4, narrowing the possible inputs for the inverse cosine part of \(f(x)\). This defines a partial domain, which we will further narrow down when considering the logarithmic function.

Logarithmic Function

The logarithmic component of the exercise, \(\log(\cos x)\), adds another layer of complexity in determining the domain of \(f(x)\). The key restriction here is the condition \(\cos x > 0\), because the logarithm function is only defined for positive arguments. Within the range \((-\frac{\pi}{2}, \frac{\pi}{2})\), the condition holds, as \(\cos x\) remains positive. However, when intersecting with the previous domain constraint of \([-1, 3]\), we focus on the common segment, which is \([0, \frac{\pi}{2})\). When engaging with logarithmic functions, remember that they are the inverse operations of exponentiation, akin to how division is the inverse of multiplication. This means for a positive real number \(a\), if \(a = e^b\), then \(b = \log(a)\). These inverse characteristics lead to logarithmic scales in many practical and theoretical applications. However, due to the strict requirement for positivity in the argument, this component can dramatically limit the domain of a composite function such as \(f(x)\).