If f(x)={1−2sinxπ−4xif x≠π4aif x=π4 is continuous at π4 then a =

f is continuous at π4∴f(π4)=Ltx→π4f(x)a=Ltx→π4 −2sinxπ−4xa=Ltx→π4 −2cosx−4=−2×12−4a=14

If f(x)={1−2sinxπ−4xif x≠π4aif x=π4 is continuous at π4 then a =

f is continuous at π4∴f(π4)=Ltx→π4f(x)a=Ltx→π4 −2sinxπ−4xa=Ltx→π4 −2cosx−4=−2×12−4a=14

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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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$\frac{1}{4}$

answer is D.

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Detailed Solution

f is continuous at $\frac{π}{4}$

$\begin{array}{l} ∴ f (\frac{π}{4}) = \underset{x \to \frac{π}{4}}{L t} f (x) \\ a = \underset{x \to \frac{π}{4}}{L t} \frac{- \sqrt{2} \sin x}{π - 4 x} \\ a = \underset{x \to \frac{π}{4}}{L t} \frac{- \sqrt{2} \cos x}{- 4} = \frac{- \sqrt{2} \times \frac{1}{\sqrt{2}}}{- 4} \\ a = \frac{1}{4} \end{array}$