Problem 161
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
Verified Answer
The largest interval is (D) \([0, \frac{\pi}{2})\).
1Step 1: Analyze the domain for the term \(4^{-x^2}\)
The expression \(4^{-x^2}\) is defined for all real numbers since any real number can be squared and then used as an exponent of 4. Therefore, there are no restrictions from this term.
2Step 2: Define the domain for the term \(\cos^{-1}\left(\frac{x}{2} - 1\right)\)
The expression \(\cos^{-1}(u)\) is defined for \(-1 \leq u \leq 1\). Thus, for \(\cos^{-1}\left(\frac{x}{2} - 1\right)\) to be defined, we require:\[-1 \leq \frac{x}{2} - 1 \leq 1\]Solving the inequalities:1. \(-1 \leq \frac{x}{2} - 1\) becomes \(0 \leq \frac{x}{2}\) or \(x \geq 0\).2. \(\frac{x}{2} - 1 \leq 1\) simplifies to \(x \leq 4\).Thus, \(x\) must satisfy \(0 \leq x \leq 4\).
3Step 3: Investigate the domain for the term \(\log(\cos x)\)
The expression \(\log(\cos x)\) is defined when \(\cos x > 0\). In the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), \(\cos x\) is positive for \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\). This does not add extra restrictions, but confirms \(x\) must stay within \((-\frac{\pi}{2}, \frac{\pi}{2})\).
4Step 4: Intersect all conditions with the original interval
From Step 2, we have the condition \(0 \leq x \leq 4\). Considering the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\) from Step 3, the intersection gives \(x \in [0, \frac{\pi}{2})\). This ensures all terms in \(f(x)\) are defined within the original interval.
5Step 5: Determine the largest interval answer option
Comparing all given options, the interval \([0, \frac{\pi}{2})\) matches (D), which does not exceed restrictions found in previous steps.
Key Concepts
Trigonometric FunctionsLogarithmic FunctionInverse Trigonometric Functions
Trigonometric Functions
Trigonometric functions, commonly referred to as trig functions, play a crucial role in various mathematical fields. They are functions of an angle and are typically related to the ratios of sides within a right triangle. The primary trigonometric functions are sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)). Understanding these functions is essential for solving problems in trigonometry, calculus, and even engineering.
- The sine function (\(\sin\)) is defined as the ratio of the length of the opposite side to the hypotenuse in a right triangle.
- The cosine function (\(\cos\)) is the ratio of the adjacent side to the hypotenuse.
- The tangent function (\(\tan\)) is the ratio of the opposite side to the adjacent side.
Logarithmic Function
The logarithmic function, often denoted as \(\log\), is another fundamental concept in mathematics. It is the inverse operation to exponentiation, meaning that \(\log_b(a) = c\) if and only if \(b^c = a\), where \(b\) is the base of the logarithm.
In this particular case, the expression \(\log(\cos x)\) is part of the function we're analyzing. The \(\log\) function is defined only for positive real numbers. Therefore, to determine the domain of \(\log(\cos x)\), we need to ensure that \(\cos x > 0\).
We found that within the given interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), \(\cos x\) remains positive, satisfying this condition. Thus, no additional restrictions are placed by the logarithmic component, apart from what is already determined by the cosine function.
In this particular case, the expression \(\log(\cos x)\) is part of the function we're analyzing. The \(\log\) function is defined only for positive real numbers. Therefore, to determine the domain of \(\log(\cos x)\), we need to ensure that \(\cos x > 0\).
We found that within the given interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), \(\cos x\) remains positive, satisfying this condition. Thus, no additional restrictions are placed by the logarithmic component, apart from what is already determined by the cosine function.
Inverse Trigonometric Functions
Inverse trigonometric functions are used to find the angles that correspond to specific trigonometric function values. The inverse function allows us to retrieve an angle from given sine, cosine, or tangent values. Here, we specifically examine the inverse cosine function, denoted as \(\cos^{-1}(x)\).
The inverse cosine function \(\cos^{-1}\) has a domain of \([-1, 1]\) for the input value and ranges over angles between \([0, \pi]\). Hence, for \(\cos^{-1}\left(\frac{x}{2} - 1\right)\) to be defined, the expression inside the inverse function must lie within \([-1, 1]\).
The inverse cosine function \(\cos^{-1}\) has a domain of \([-1, 1]\) for the input value and ranges over angles between \([0, \pi]\). Hence, for \(\cos^{-1}\left(\frac{x}{2} - 1\right)\) to be defined, the expression inside the inverse function must lie within \([-1, 1]\).
- We start by ensuring \(\frac{x}{2} - 1 = u\) satisfies \(-1 \leq u \leq 1\).
- This translates to the condition \(0 \leq x \leq 4\).
Other exercises in this chapter
Problem 159
Let \(f:(-1,1) \rightarrow B\), be a function defined by \(f(x)=\tan ^{-1} \frac{2 x}{1-x^{2}}\), then \(f\) is both one-one and onto when \(B\) is the interval
View solution Problem 160
A real valued function \(f(x)\) satisfies the functional equation \(f(x-y)=f(x) f(y)-f(a-x) f(a+y)\) where \(a\) is a given constant and \(f(0)=1, f(2 a-x)\) is
View solution Problem 163
For real \(x\), let \(f(x)=x^{3}+5 x+1\), then \(\quad\) (A) \(f\) is one-one but not onto \(R\) (B) \(f\) is onto \(R\) but not one-one (C) \(f\) is one-one an
View solution Problem 164
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is (A) \((0, \infty)\) (B) \((-\infty, 0)\) (C) \((-\infty, \infty)-\\{0\\}\) (D) \((-\infty, \infty)
View solution