fx=-2 sin x, for -π≤x≤-π2 a sin x+b, for -π2 -π2, fx=a sin x+blimx→-π2+ a sin x+b=2a sin -π2+b=2-a+b=2 _________2Adding equation 1 and 22b=2b=1a+b=0a=-b=-1

Qno 1

Qno 1 Exervise for Session 2 continuity -

f (x) = \{\begin{matrix} - 2 \sin x, f o r - π \leq x \leq - \frac{π}{2} \\ a \sin x + b, f o r - \frac{π}{2} < x < \frac{π}{2} \\ \cos x, f o r \frac{π}{2} \leq x \leq π \end{matrix} F o r f (x) t o b e c o n t i u o u s a t x = k \lim_{x \to k^{+}} f (x) = \lim_{x \to k^{-}} f (x) = f (k) F o r f (x) t o b e c o n t i n u o u s a t x = \frac{π}{2} \lim_{x \to {\frac{π}{2}}^{+}} f (x) = \lim_{x \to {\frac{π}{2}}^{-}} f (x) = f (\frac{π}{2}) f (\frac{π}{2}) = \cos \frac{π}{2} = 0 \lim_{x \to {\frac{π}{2}}^{-}} f (x) = 0 F o r x < \frac{π}{2}, f (x) = a \sin x + b \lim_{x \to {\frac{π}{2}}^{-}} (a \sin x + b) = 0 (a \sin \frac{π}{2} + b) = 0 a + b = 0_________(1) F o r f (x) t o b e c o n t i n u o u s a t x = - \frac{π}{2} \lim_{x \to - {\frac{π}{2}}^{+}} f (x) = \lim_{x \to - {\frac{π}{2}}^{-}} f (x) = f (- \frac{π}{2}) f (- \frac{π}{2}) = - 2 \sin (- \frac{π}{2}) = 2 \sin (\frac{π}{2}) = 2 \lim_{x \to - {\frac{π}{2}}^{+}} f (x) = 2 F o r x > - \frac{π}{2}, f (x) = a \sin x + b \lim_{x \to - {\frac{π}{2}}^{+}} (a \sin x + b) = 2 (a \sin (- \frac{π}{2}) + b) = 2 - a + b = 2_________(2) A d d i n g e q u a t i o n 1 a n d 2 2 b = 2 b = 1 a + b = 0 a = - b = - 1

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Qno 1 Qno 1 Exervise for Session 2 continuity -

Qno 1

Qno 1 Exervise for Session 2 continuity -