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Class-12-science»Maths

Let f(x)={ (1-sin³x)/3cos²x ,x< pi/2

p , x= pi /2

[q(1-sin x)]/(pi-2x)² , x>pi/2 }

If f(x) is continuous at x=pi/2,(p,q) = ______

(ans is (1/2 , 4)

Plz explain it

$G i v e n, f (x) i s c o n t i n u o u s a t x = \frac{π}{2} . S o, f (\frac{π}{2} - 0) = f (\frac{π}{2}) = f (\frac{π}{2} + 0) ...... (1) N o w, f (\frac{π}{2}) = p a n d, f (\frac{π}{2} - 0) = \lim_{h \to \frac{π}{2}} (\frac{1 - \sin^{3} h}{{3 \cos}^{2} h}) [f o r m \frac{0}{0}] = \lim_{h \to \frac{π}{2}} (\frac{- 3 \sin^{2} h \cosh}{- 6 \cos h \sinh}) [U \sin g L' h o s p i t a l' s R u l e] = \lim_{h \to \frac{π}{2}} (\frac{\sin h}{2}) = \frac{1}{2} \sin (\frac{π}{2}) = \frac{1}{2} \Rightarrow f (\frac{π}{2} - 0) = \frac{1}{2} A n d, f (\frac{π}{2} + 0) = \lim_{h \to \frac{π}{2}} (\frac{q (1 - \sinh)}{{(π - 2 h)}^{2}}) [f o r m \frac{0}{0}] = \lim_{h \to \frac{π}{2}} (\frac{- q \cos h}{2 (π - 2 h) (- 2)}) [U \sin g L' h o s p i t a l' s R u l e] = \lim_{h \to \frac{π}{2}} (\frac{q \cos h}{4 (π - 2 h)}) [f o r m \frac{0}{0}] = \lim_{h \to \frac{π}{2}} (\frac{- q \sin h}{- 8}) [U \sin g L' h o s p i t a l' s R u l e] = \frac{q}{8} \sin (\frac{π}{2}) = \frac{q}{8} \Rightarrow f (\frac{π}{2} + 0) = \frac{q}{8} S i n c e, f (\frac{π}{2} - 0) = f (\frac{π}{2}) = f (\frac{π}{2} + 0) T h e r e f o r e, \frac{1}{2} = p = \frac{q}{8} \Rightarrow p = \frac{1}{2}, a n d q = 4$

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