Home
Class 12
MATHS
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

AI Generated Solution

The correct Answer is:
To find 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 \( \frac{\pi}{4} \) First, we need to evaluate the limit: \[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \sqrt{2} \sin x}{\pi - 4x}. \] ### Step 2: Substitute \( x = \frac{\pi}{4} \) Substituting \( x = \frac{\pi}{4} \): - The numerator becomes \( 1 - \sqrt{2} \sin\left(\frac{\pi}{4}\right) = 1 - \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 - 1 = 0 \). - The denominator becomes \( \pi - 4\left(\frac{\pi}{4}\right) = \pi - \pi = 0 \). Thus, we have an indeterminate form \( \frac{0}{0} \). ### Step 3: Apply L'Hôpital's Rule Since we have \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that: \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \] if the limit on the right exists. #### Differentiate the numerator and denominator: - The derivative of the numerator \( 1 - \sqrt{2} \sin x \) is \( -\sqrt{2} \cos x \). - The derivative of the denominator \( \pi - 4x \) is \( -4 \). ### Step 4: Rewrite the limit Now we rewrite the limit using L'Hôpital's Rule: \[ \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} = \frac{\sqrt{2} \cos x}{4}. \] ### Step 5: Substitute \( x = \frac{\pi}{4} \) again Now substituting \( x = \frac{\pi}{4} \): \[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. \] Thus, we have: \[ \lim_{x \to \frac{\pi}{4}} f(x) = \frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{4} = \frac{1}{4}. \] ### Step 6: Set the limit equal to \( a \) Since \( f(x) \) 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 find 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} ...
Promotional Banner

Similar Questions

Explore conceptually related problems

If f (x) = {{:((1- sqrt2sin x )/(pi - 4 x )",",, x ne (pi)/(4)),(a" "",",,x = (pi)/(4)):} is continuous at x = (pi)/(4) , then a =

If f(x)={{:((1-sinx)/(pi-2x),,","xne(pi)/(2)),(lambda,,","x=(pi)/(2)):},"be continuous at "x=(pi)/(2), then value of lambda is

If f(x) ={((3sinx-sqrt(3)cosx)/(6x-pi),",",x != (pi)/6),(a, ",",x = pi/6):} is continuous at x = (pi)/6 , then a =

If f(x)={(x+a sqrt(2) sinx"," ,0 lt x lt (pi)/(4)),(2x cotx+b",",(pi)/(4) le x le (pi)/(2)),(a cos 2x-b sinx",", (pi)/(2) lt x le pi):} is continuous at x=(pi)/(4) , then a - b is equal to

Find the value of k, if the function f given by : {:(f(x)=(1-tanx)/(1-sqrt2sinx)",", "for" x nepi/4),(=k/2",","for"x=pi/4):} is continous at x =pi/4*

Let f(x)=(1-tan x)/(4x-pi),x!=(pi)/(4),x in[0,(pi)/(2)], If f(x) is continuous in [0,(pi)/(4)], then find the value of f((pi)/(4))

Let f (x)= {{:((1- tan x)/(4x-pi), x ne (pi)/(4)),( lamda, x =(pi)/(4)):}, x in [0, (pi)/(2)), If f (x) is continuous in [0, (pi)/(2)) then lamda is equal to: