If f(x)=1−sin3⁡x3cos2⁡x, if x π2, so that f(x) is continuous at x=π2, then

At x=π2, L.H.L=limx→π2 1-sin3x3cos2x(00) =limx→π2 -3sin2xcosx-6cosxsinx( ∵ By L–Hospital rule) =12 R.H.L=limx→π2 b(1-sinx)(π-2x)2(00) =limx→π2 -bcosx-4(π-2x)(00)( ∵ By L–Hospital rule) =limx→π2 bsinx8( ∵ By L–Hospital rule) =b8 since f is continuous at x=π2 then L.H.L=R.H.L=f(π2) ⇒12==b8=a ⇒a=12,b=4

If f(x)=1−sin3⁡x3cos2⁡x, if x π2, so that f(x) is continuous at x=π2, then

At x=π2, L.H.L=limx→π2 1-sin3x3cos2x(00) =limx→π2 -3sin2xcosx-6cosxsinx( ∵ By L–Hospital rule) =12 R.H.L=limx→π2 b(1-sinx)(π-2x)2(00) =limx→π2 -bcosx-4(π-2x)(00)( ∵ By L–Hospital rule) =limx→π2 bsinx8( ∵ By L–Hospital rule) =b8 since f is continuous at x=π2 then L.H.L=R.H.L=f(π2) ⇒12==b8=a ⇒a=12,b=4

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $f (x) = \{\begin{cases} \frac{1 - \sin^{3} x}{3 \cos^{2} x}, if x < \frac{π}{2} \\ a, if x = \frac{π}{2} \\ \frac{b (1 - \sin x)}{(π - 2 x)^{2}}, if x > \frac{π}{2} \end{cases},$
so that $f (x)$ is continuous at $x = \frac{π}{2},$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$a = \frac{1}{2}$

$b = 5$

$a = 1$

$b = - 4$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$A t x = \frac{π}{2}, L . H . L = \lim_{x \to {\frac{π}{2}}^{}} \frac{1 - \sin^{3} x}{3 \cos^{2} x} (\frac{0}{0}) = \lim_{x \to {\frac{π}{2}}^{}} \frac{- 3 \sin^{2} x \cos x}{- 6 \cos x \sin x} (∵ B y L - H o s p i t a l r u l e) = \frac{1}{2} R . H . L = \lim_{x \to {\frac{π}{2}}^{}} \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}} (\frac{0}{0}) = \lim_{x \to {\frac{π}{2}}^{}} \frac{- b \cos x}{- 4 (π - 2 x)} (\frac{0}{0}) (∵ B y L - H o s p i t a l r u l e) = \lim_{x \to {\frac{π}{2}}^{}} \frac{b \sin x}{8} (∵ B y L - H o s p i t a l r u l e) = \frac{b}{8} \sin c e f i s c o n t i n u o u s a t x = \frac{π}{2} t h e n L . H . L = R . H . L = f (\frac{π}{2}) \Rightarrow \frac{1}{2} = = \frac{b}{8} = a \Rightarrow a = \frac{1}{2}, b = 4$