If fx 1- 2 sin x - -4x -if x- 4 b-if x- 4 is continuous at x-4- find the value of b-

We need to have lim_x→π/4 =f(x=π/4)=b Now, lim_x→π/4 f(x)=lim_x→π/41 √(2)sin xπ 4x =lim_x→π/4√(2)cos x 4x =√(2)cos (π/4) 4 =14 ...

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Real Valued Functions

If fx 1-√ 2...

Question

If $f (x) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - \sqrt{2} sin x}{π - 4 x}, i f x \neq \frac{π}{4} b, i f x = \frac{π}{4} \end{matrix}$ is continuous at $x = \frac{π}{4}$ ., find the value of $b$ .

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - \sqrt{2} sin x}{π - 4 x}, & i f x \neq \frac{π}{4} a, & i f x = \frac{π}{4} \end{matrix}

\frac{π}{4}

a =

p

q

f (x) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, i f x < \frac{π}{2} p, i f x = \frac{π}{2} i s c o n t i n o u s a t x = \frac{π}{2} \frac{q (1 - sin x)}{{(π - 2 x)}^{2}}, i f x > \frac{π}{2} \end{matrix}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, i f x < \frac{π}{2} a, i f x = \frac{π}{2} \frac{b (1 - sin x)}{{(π - 2 x)}^{2}}, i f x > \frac{π}{2} \end{matrix}

f (x)

x = \frac{π}{2}

a

b

k

x = \frac{π}{2}

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, if x \neq \frac{π}{2} 3, if x = \frac{π}{2} \end{matrix}

k

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, i f x \neq \frac{π}{2} 3, i f x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

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