Problem 14
Question
The value of \(a\) in order that \(f(x)=\sin x-\cos x-a x+b\) decreases for all real values is given by (A) \(a \geq \sqrt{2}\) (B) \(a<\sqrt{2}\) (C) \(a \geq 1\) (D) \(a<1\)
Step-by-Step Solution
Verified Answer
(A) \(a \geq \sqrt{2}\)
1Step 1: Find the Derivative
Determine the derivative of the function \( f(x) = \sin x - \cos x - ax + b \). The derivative is \( f'(x) = \cos x + \sin x - a \).
2Step 2: Ensure Derivative is Negative
For the function to be decreasing for all real values, the first derivative, \( f'(x) = \cos x + \sin x - a \), must be negative for all \(x\): \( \cos x + \sin x - a < 0 \).
3Step 3: Maximize Expression
The maximum value of \( \cos x + \sin x \) is found by considering \( \cos x = \frac{1}{\sqrt{2}} \) and \( \sin x = \frac{1}{\sqrt{2}} \), which leads to \( \cos x + \sin x = \sqrt{2} \).
4Step 4: Determine Condition for \(a\)
Assuming the maximum value of \( \cos x + \sin x \) is \( \sqrt{2} \), then \( \sqrt{2} - a < 0 \), or \( a > \sqrt{2} \). However, the condition for it to always be negative is \( a \geq \sqrt{2} \).
5Step 5: Conclusion
The inequality \( a \geq \sqrt{2} \) ensures that \( \cos x + \sin x - a < 0 \) for all \(x\). This implies that the correct choice is (A) \( a \geq \sqrt{2} \).
Key Concepts
Derivative of a functionDecreasing functionTrigonometric functions
Derivative of a function
In calculus, the derivative of a function represents the rate at which a function is changing at any given point. It's like looking at how steep a curve is at a particular moment. For the function \( f(x) = \sin x - \cos x - ax + b \), we determine its derivative by applying basic differentiation rules. Differentiating sine \( \sin x \) gives us \( \cos x \), and differentiating cosine \( \cos x \) gives us \(-\sin x\). The derivative of a constant \( b \) is zero, and the derivative of \(-ax\) is \(-a\). Thus, the overall derivative of the function, denoted as \( f'(x) \), is \( \cos x + \sin x - a \).
If you think of \( f'(x) \) as describing the slope of the tangent line to the curve at any point \( x \), it helps explain how the function is behaving: whether it's going up, down, or staying flat.
If you think of \( f'(x) \) as describing the slope of the tangent line to the curve at any point \( x \), it helps explain how the function is behaving: whether it's going up, down, or staying flat.
Decreasing function
A decreasing function has a derivative that is always negative, signifying that the function's value diminishes as \( x \) increases. For the function \( f(x) = \sin x - \cos x - ax + b \) to be decreasing for all real \( x \), its derivative \( f'(x) = \cos x + \sin x - a \) must be less than zero at every point. Conceptually, this means that the combination of the cosine and sine values needs to be countered by \( a \), maintaining the derivative's negativity.
To achieve this, the largest value that \( \cos x + \sin x \) can take, which is \( \sqrt{2} \), determines \( a \). When combined, these trigonometric functions reach their peak at 45 degrees or \( \pi/4 \) radians, where both \( \cos x \) and \( \sin x \) are \( \frac{1}{\sqrt{2}} \), summing to \( \sqrt{2} \). Thus, to ensure that \( f'(x) = \cos x + \sin x - a \) is negative for all \( x \), \( a \) must be at least \( \sqrt{2} \).
To achieve this, the largest value that \( \cos x + \sin x \) can take, which is \( \sqrt{2} \), determines \( a \). When combined, these trigonometric functions reach their peak at 45 degrees or \( \pi/4 \) radians, where both \( \cos x \) and \( \sin x \) are \( \frac{1}{\sqrt{2}} \), summing to \( \sqrt{2} \). Thus, to ensure that \( f'(x) = \cos x + \sin x - a \) is negative for all \( x \), \( a \) must be at least \( \sqrt{2} \).
- The condition is \( a \geq \sqrt{2} \) to guarantee that the function remains consistently decreasing.
Trigonometric functions
Trigonometric functions, such as \( \sin x \) and \( \cos x \), are fundamental in calculus due to their oscillating nature and periodic behavior. They describe the relationships in triangles and model phenomena in waves and circles. These functions are periodic with a period of \( 2\pi \), meaning they repeat their values every \( 2\pi \) units.
In the exercise at hand, the properties of \( \sin x \) and \( \cos x \) are crucial because they combine to influence the derivative of our given function. Together, they fluctuate between values, making the sum \( \cos x + \sin x \) sometimes significant or minor. The maximum value they can achieve together is \( \sqrt{2} \). This maximum arises from both functions peaking simultaneously, a rare but calculable occurrence.
In the exercise at hand, the properties of \( \sin x \) and \( \cos x \) are crucial because they combine to influence the derivative of our given function. Together, they fluctuate between values, making the sum \( \cos x + \sin x \) sometimes significant or minor. The maximum value they can achieve together is \( \sqrt{2} \). This maximum arises from both functions peaking simultaneously, a rare but calculable occurrence.
- Understanding the fluctuations of these functions helps us decide the condition for \( a \) in ensuring a decreasing function.
Other exercises in this chapter
Problem 12
If the function \(f(x)=3 \cos |x|-6 a x+b\) increases for all \(x \in R\), then the range of values of \(a\) is given by (A) \(a>-\frac{1}{2}\) (B) \(a
View solution Problem 13
The equation \(x+e^{x}=0\) has (A) only one real root (B) only two real roots (C) no real root (D) None of these
View solution Problem 17
The two tangents to the curve \(a x^{2}+2 h x y+b y^{2}=1\), \(a>0\) at the points where it crosses \(x\)-axis, are (A) parallel (B) perpendicular (C) inclined
View solution Problem 18
The curve \(y-e^{x y}+x=0\) has a vertical tangent at the point (A) \((1,1)\) (B) at no point (C) \((0,1)\) (D) \((1,0)\)
View solution