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Question Number 133994 by benjo_mathlover last updated on 26/Feb/21

Given a function f satisfies

f (x) = {\begin{cases} x + a \sqrt{2} \sin x; 0 ⩽ x < \frac{π}{4} \\ 2 x \cot x + b; \frac{π}{4} ⩽ x ⩽ \frac{π}{2} \\ a \cos 2 x - b \sin x; \frac{π}{2} < x ⩽ π \end{cases}

continuous in [0, π], then find

the value of a and b .

Answered by EDWIN88 last updated on 26/Feb/21

(1) f (x) must be continuous at x = \frac{π}{4}

\lim_{x \to π / 4^{-}} f (x) = \underset{x \to π / 4^{+}}{\lim f} (x)

\Rightarrow \frac{π}{4} + a \sqrt{2} . \frac{\sqrt{2}}{2} = 2 . \frac{π}{4} . 1 + b

\Rightarrow a - b = \frac{π}{4} \dots (i)

(2) f (x) also must be continuous at x = \frac{π}{2}

\underset{x \to π / 2^{-}}{\lim f} (x) = \underset{x \to π / 2^{+}}{\lim f} (x)

\Rightarrow 0 + b = a (- 1) - b; a + 2 b = 0 \dots (i i)

(i) - (i i) : - 3 b = \frac{π}{4}; \underset{―}{b = - \frac{π}{12} and a = \frac{π}{6}}

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