If f x - - 1 - -2sin x- - 4x- if x -4- a- if x - -4 - is continuous at -4 then a -

The correct option is D 14 lim _ x →π 4 1 √( 2 )sin x π 4 x using L'Hospital rule lim _ x →π 4 √( 2 )cos x 4 = √( 2 )cos #1 ...

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Differentiation of Inverse Trigonometric Functions

If f x = [ ...

Question

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

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\frac{1}{4}

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

x = \frac{π}{4},

a

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}

x = \frac{π}{4}

b

f (x) = \frac{1 - t a n x}{1 - \sqrt{2} s i n x}

x \neq \frac{π}{4}

x = \frac{π}{4}

f (\frac{π}{4})

f

x = \frac{π}{4}

f

f (x) = \frac{1 - \sqrt{2} sin x}{π - 4 x}

x \neq \frac{π}{4}

= \frac{1}{4}

x \neq \frac{π}{4}

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

x = \frac{π}{4}

a =

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