Fx- -2sin x if x- -2 a sin x-b if -2-x-2 cos x if x-2- and fx is continuous everywhere then a-b -

The correct option is B ( 1,1) continuity at π2 f( π #160; 2 + ) =f( π #160; 2 ) #160; ⇒ asin π2+b= 2sin π2 ⇒ a b= 2 cont ...

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Derivative of Standard Functions

fx= -2sin x ...

Question

$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x & i f & x \leq - \frac{π}{2} a sin x + b & i f & - \frac{π}{2} < x < \frac{π}{2} cos x & i f & x \geq \frac{π}{2} \end{matrix}$ and $f (x)$ is continuous everywhere then $(a, b)$ =

(1, 1)

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(- 1, 1)

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(1, - 1)

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(- 1, - 1)

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x & i f & x \leq - \frac{π}{2} A sin x + B & i f & - \frac{π}{2} < x < \frac{π}{2} cos x & i f & x \geq \frac{π}{2} \end{matrix}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x i f x \leq - \frac{π}{2} A sin x + B i f - \frac{π}{2} < x < \frac{π}{2}; cos x i f x \geq \frac{π}{2} \end{matrix}

A

B

f (x)

f (x) = ⎧ ⎨ ⎩ \begin{matrix} - 2 sin x & if & x \leq - π / 2 A sin x + B & if & - π / 2 < x < π / 2 cos x, & if & x \geq π / 2 \end{matrix}

A

B

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, x \leq - \frac{π}{2} A sin x + B, - \frac{π}{2} < x < \frac{π}{2} cos x, x \geq \frac{π}{2} \end{matrix}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, - π \leq x \leq - \frac{π}{2} a sin x + b, - \frac{π}{2} < x < \frac{π}{2} cos x, \frac{π}{2} \leq x \leq π \end{matrix}

f (x)

[- π, π]

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