Home
Class 12
MATHS
The function f(x) = sin x - x cos pi is...

The function f(x) = sin x - x cos `pi` is

A

maximum or minimum for all integral multiple of `pi`

B

maximum if x is an odd positive or even negative integral multiple of `pi`

C

minimum of x is an even positive or odd negative integral multiple of `pi`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the maxima and minima of the function \( f(x) = \sin x - x \cos x \), we will follow these steps: ### Step 1: Differentiate the function We first need to find the first derivative of the function \( f(x) \). \[ f'(x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}(x \cos x) \] Using the product rule for the second term: \[ \frac{d}{dx}(x \cos x) = \cos x - x \sin x \] Thus, the derivative becomes: \[ f'(x) = \cos x - (\cos x - x \sin x) = x \sin x \] ### Step 2: Set the derivative to zero To find the critical points, we set the first derivative equal to zero: \[ f'(x) = x \sin x = 0 \] This equation holds true when either \( x = 0 \) or \( \sin x = 0 \). ### Step 3: Solve for critical points 1. From \( x = 0 \), we have one critical point. 2. From \( \sin x = 0 \), we get \( x = n\pi \), where \( n \) is any integer. Thus, the critical points are: \[ x = 0, \, n\pi \quad (n \in \mathbb{Z}) \] ### Step 4: Determine the nature of critical points To determine whether these critical points are maxima or minima, we can use the second derivative test. We need to find the second derivative \( f''(x) \). ### Step 5: Differentiate again We differentiate \( f'(x) = x \sin x \): \[ f''(x) = \frac{d}{dx}(x \sin x) = \sin x + x \cos x \] ### Step 6: Evaluate the second derivative at critical points 1. For \( x = 0 \): \[ f''(0) = \sin(0) + 0 \cdot \cos(0) = 0 \] Since \( f''(0) = 0 \), we cannot conclude anything about the nature of this critical point from the second derivative test. 2. For \( x = n\pi \): \[ f''(n\pi) = \sin(n\pi) + n\pi \cos(n\pi) = 0 + n\pi (-1)^n = n\pi (-1)^n \] - If \( n \) is even, \( f''(n\pi) > 0 \) (minimum). - If \( n \) is odd, \( f''(n\pi) < 0 \) (maximum). ### Conclusion - At \( x = 0 \), we need further analysis (potential inflection point). - At \( x = n\pi \): - Minima at even \( n \) (e.g., \( n = 0, 2, 4, \ldots \)) - Maxima at odd \( n \) (e.g., \( n = 1, 3, 5, \ldots \))
Promotional Banner

Topper's Solved these Questions

  • MAXIMA AND MINIMA

    MOTION|Exercise EXERCISE - 3|32 Videos
  • MAXIMA AND MINIMA

    MOTION|Exercise EXERCISE -4 (LEVEL - I)|10 Videos
  • MAXIMA AND MINIMA

    MOTION|Exercise EXERCISE -2 (LEVEL- I)|26 Videos
  • MATRICES

    MOTION|Exercise Exercise - 4 (Level-II)|28 Videos
  • METHOD OF DIFFERENTIATION

    MOTION|Exercise EXERCISE - 4 LEVEL -II|5 Videos

Similar Questions

Explore conceptually related problems

The function f(x) = sin x + cos x will be

In the interval ((pi)/(4), (11 pi)/(12)) , then function f (x) = sin 3x - cos 3x, 0 lt x lt pi is:

Verify Rolle's Theorem for the function : f(x) = sin x + cos x in the interval [0,2pi]

If [sin x]+[2cos x]=-3 then the range of the function f(x)=sin x+cos x is

Let K be the set of all values of x, where the function f(x) = sin |x| - |x| + 2(x-pi) cos |x| is not differentiable. Then, the set K is equal to

The function f(x) = (1-sin x + cos x)/(1+sin x + cosx) is not defined at x = pi . The value of f(pi) , so that f(x) is continuous at x = pi is

Let the function f(x) = sin x + cos x , be defined in [0, 2pi] , then f(x)

The function f(x)=cos x, 0 le x le pi is

Verify Rolle's theorem for each of the following functions : f(x) = sin x + cos x "in " [0, (pi)/(2)]