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If f(x)={{:((1-sqrt2sinx)/(pi-4x)",",ifx...

If `f(x)={{:((1-sqrt2sinx)/(pi-4x)",",ifxne(pi)/(4)),(a",",if x=(pi)/(4)):}` in continuous at `(pi)/(4)`, then a is equal to :

A

2

B

2

C

3

D

`1/4`

Text Solution

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The correct Answer is:
To determine the value of \( a \) such that the function \[ f(x) = \begin{cases} \frac{1 - \sqrt{2} \sin x}{\pi - 4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases} \] is continuous at \( x = \frac{\pi}{4} \), we need to ensure that \[ \lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right) = a. \] ### Step 1: Calculate the limit as \( x \) approaches \( \frac{\pi}{4} \) We start by calculating the limit: \[ \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{1 - \sqrt{2} \sin x}{\pi - 4x}. \] ### Step 2: Substitute \( x = \frac{\pi}{4} \) Substituting \( x = \frac{\pi}{4} \) directly into the limit gives: \[ 1 - \sqrt{2} \sin\left(\frac{\pi}{4}\right) = 1 - \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 - 1 = 0, \] and \[ \pi - 4\left(\frac{\pi}{4}\right) = \pi - \pi = 0. \] This results in the indeterminate form \( \frac{0}{0} \). ### Step 3: Apply L'Hôpital's Rule Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator. The derivative of the numerator \( 1 - \sqrt{2} \sin x \) is: \[ -\sqrt{2} \cos x, \] and the derivative of the denominator \( \pi - 4x \) is: \[ -4. \] Thus, we can rewrite the limit as: \[ \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x}{4}. \] ### Step 4: Evaluate the limit Now we substitute \( x = \frac{\pi}{4} \): \[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. \] So we have: \[ \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{4} = \frac{1}{4}. \] ### Step 5: Set the limit equal to \( a \) Since the function is continuous at \( x = \frac{\pi}{4} \), we set: \[ a = \lim_{x \to \frac{\pi}{4}} f(x) = \frac{1}{4}. \] ### Conclusion Thus, the value of \( a \) is \[ \boxed{\frac{1}{4}}. \]

To determine the value of \( a \) such that the function \[ f(x) = \begin{cases} \frac{1 - \sqrt{2} \sin x}{\pi - 4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases} ...
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