Value of A and B, so that the function $$f(x)$$ defined by $$f(x)=x+A2sin$$⁡x if $$0$$≤x

Question

Value of A and B, so that the function $$f(x)$$ defined by $$f(x)=x+A2sin$$⁡x  if $$0$$≤x<π$$4$$,$$2x+$$$$cot$$⁡$$x+B$$  if π$$4$$≤x<π$$2$$, becomes continuous are Acos⁡$$2x$$−Bsin⁡x if π$$2$$≤x≤π

Infinity Learn · Accepted Answer

Here we will check continuity at $$x=$$π$$/4$$ and $$x=$$π$$/2At$$          $$x=$$π$$/4$$.               limx→π−$$4fx=limx$$→π−$$4x+A2sinx=$$π$$4+Aand$$     limx→π$$+4$$  $$fx=limx$$→π$$+42xcotx+B=$$π$$2+BFor$$ confinuity at $$x=$$π$$/4$$                π$$4+A=$$π$$2+B$$⇒A−$$B=$$π$$4Again$$, limx→π−$$2$$  $$fx=limx$$→π−$$22xcotx+B=Band$$,     limx→π$$+2$$  $$fx=limx$$→π$$+22Acos2x$$−$$Bsinx=$$−A−B,so f is continuous at this point if              −A−$$B=B$$⇒$$A=$$−$$2BSolving$$ (i) and (ii), $$A=$$π$$/6$$ and $$B=$$−π$$/12Hence$$ $$(1)is$$ the correct answer.

$Value of A and B, so that the function f (x) defined by$ $f (x) = \{\begin{cases} x + A \sqrt{2} \sin x if 0 \leq x < \frac{π}{4}, \\ 2 x + \cot x + B if \frac{π}{4} \leq x < \frac{π}{2}, becomes continuous are \\ A \cos 2 x - B \sin x if \frac{π}{2} \leq x \leq π \end{cases}$

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Value of A and B, so that the function f(x) defined by f(x)=x+A2sin⁡x if 0≤x<π4,2x+cot⁡x+B if π4≤x<π2, becomes continuous are Acos⁡2x−Bsin⁡x if π2≤x≤π

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$Value of A and B, so that the function f (x) defined by$ $f (x) = \{\begin{cases} x + A \sqrt{2} \sin x if 0 \leq x < \frac{π}{4}, \\ 2 x + \cot x + B if \frac{π}{4} \leq x < \frac{π}{2}, becomes continuous are \\ A \cos 2 x - B \sin x if \frac{π}{2} \leq x \leq π \end{cases}$