Lf fx- 1-2sin x-4x x-4- a-x-4 - is continuous at x-4 then a-

The correct option is D 14 Since it is continuous at x=π4 Therefore, f( π #160;/ 4 #160;) =lim _ x→π #160;/ 4 #160; 1 √( 2 )sin #160 ...

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Points of Discontinuity

lf fx=[ 1-√...

Question

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

4

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\frac{1}{4}

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Solution

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Similar questions

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) = \frac{1 - tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{4}]

f (x)

[0, \frac{π}{2}]

f (\frac{π}{4})

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) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \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 =

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}

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