Home
Class 12
MATHS
The function f(x) = (1-sin x + cos x)/(1...

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

A

`1/2`

B

`1/2`

C

`-1`

D

1

Text Solution

AI Generated Solution

The correct Answer is:
To find the value of \( f(\pi) \) such that the function \( f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \) is continuous at \( x = \pi \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( \pi \). ### Step-by-Step Solution: 1. **Identify the Function**: The function is given as: \[ f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \] 2. **Evaluate the Function at \( x = \pi \)**: First, we need to check if \( f(\pi) \) is defined: \[ f(\pi) = \frac{1 - \sin(\pi) + \cos(\pi)}{1 + \sin(\pi) + \cos(\pi)} = \frac{1 - 0 - 1}{1 + 0 - 1} = \frac{0}{0} \] Since this results in an indeterminate form \( \frac{0}{0} \), we need to find the limit as \( x \) approaches \( \pi \). 3. **Finding the Limit**: We will compute the limit: \[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \] As \( x \) approaches \( \pi \), both the numerator and denominator approach \( 0 \). Therefore, we can apply L'Hôpital's Rule. 4. **Differentiate the Numerator and Denominator**: Differentiate the numerator: \[ \frac{d}{dx}(1 - \sin x + \cos x) = -\cos x - \sin x \] Differentiate the denominator: \[ \frac{d}{dx}(1 + \sin x + \cos x) = \cos x - \sin x \] 5. **Apply L'Hôpital's Rule**: Now we apply L'Hôpital's Rule: \[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} \frac{-\cos x - \sin x}{\cos x - \sin x} \] 6. **Evaluate the Limit at \( x = \pi \)**: Substitute \( x = \pi \): \[ \cos(\pi) = -1, \quad \sin(\pi) = 0 \] Therefore, \[ \lim_{x \to \pi} f(x) = \frac{-(-1) - 0}{-1 - 0} = \frac{1}{-1} = -1 \] 7. **Conclusion**: To make \( f(x) \) continuous at \( x = \pi \), we set: \[ f(\pi) = -1 \] ### Final Answer: The value of \( f(\pi) \) so that \( f(x) \) is continuous at \( x = \pi \) is \( -1 \).
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CONTINUITY

    TARGET PUBLICATION|Exercise Evaluation Test|10 Videos
  • CONTINUITY

    TARGET PUBLICATION|Exercise Critical Thinking|80 Videos
  • BINOMIAL DISTRIBUTION

    TARGET PUBLICATION|Exercise EVALUTION TEST|12 Videos
  • DEFINITE INTEGRALS

    TARGET PUBLICATION|Exercise EVALUATIO TEST|30 Videos

Similar Questions

Explore conceptually related problems

f(x)= (1-sinx+cosx)/(1+sinx+cosx) discontinuous at x=pi . Find f(pi) so that f(x) is continuous x=pi

If f(x)=(sqrt(2)cos x-1)/(cot x-1),x!=(pi)/(4). Find the value of f((pi)/(4)) so that f(x) becomes continuous at x=pi/4

Knowledge Check

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

    A
    `-1//2`
    B
    `1//2`
    C
    `-1`
    D
    1
  • The function f(x)=(1-sinx+cosx)/(1+sinx+cosx) is not defined at x=pi . The value of f(pi) so that f(x) is continuous at x=pi is

    A
    `-(1)/(2)`
    B
    `(1)/(2)`
    C
    `-1`
    D
    1
  • The function f(x)=(1-sinx+cosx)/(1+sinx+cosx) is nto defined at x=pi . The value of f(pi) so that f(x) is continuous at x=pi is:

    A
    `-(1)/(2)`
    B
    `(1)/(2)`
    C
    `-1`
    D
    1
  • Similar Questions

    Explore conceptually related problems

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

    The function f(x)=(sin 2x)^(tan^(2)2x) is not defined at x=(pi)/(4) . The value of f(pi//4) , so that f is continuous at x=pi//4 , is

    The function f(x)=(sinx)^(tan^(2)x) is not defined at x=(pi)/(2) . The value of f((pi)/(2)) such that f is continuous at x=(pi)/(2) is

    Let f(x)=(sin (pi cos^2 x))/x^2 , x ne 0 . The value of f(0) so that f is a continuous function is

    Let the function f(x) = tan^(1-) (sin x + cos x) be defined on [ 0, 2 pi] Then f(x) is