Question

if f(x) = 1-sin^3x/3cos^2x if x
a , if x = pi/2 is continous at x = pi/2

b(1-sinx)/(pi-2x)^2


FIND VALUES OF a , b

Answer 1

Step-by-step explanation:

Given if f(x) = 1-sin^3x/3cos^2x if a ,x =  pi/2 is continuous at x = pi/2

b(1-sinx)/(pi-2x)^2   FIND VALUES OF a , b

• Now let us write the function.
•                         So f(x) = {1 – sin^3 x / 3 cos^2 x when x < π/2
•                    When x = π/2 , it will be a
•                    Now b(1 – sin x) / (π – 2x)^2 when x > π/2
•              Now left hand side will be
•                              So lim x ->π/2-  1 – sin^3 x / 3 cos^2 x  
•  Now we have a^3 – b^3 = (a – b) (a^2 + ab + b^2)
•                       = lim x-> π/2 -  (1 – sinx) (1^2 + 1.sinx + sin^2 x) / 3 (1 – sin^2 x)
•                      = lim x-> π/2 -  (1 – sinx) (1^2 + 1.sinx + sin^2 x) / 3 (1 – sin x) (1 + sin x)
•                     =  now putting value of limit we get
•                              = 1 + 1 + 1 / 3(1 + 1)
•                             = ½  
•         Now f (π/2) = a
•                Or a = 1/2  
• Now taking the right hand side we get
•            =  Lim x ->π/2 +  b(1 – sin x) / (π – 2x)^2
•       Now converting sinx to cos x  
•            = lim x ->π/2 +         b (1 – cos (π/2 – x) ) / (π – 2x)^2
•           = lim x -> π/2 +        b . 2 sin^2 (π / 4 – x / 2) / [ 4(π/4 – x/2)]^2
•          = lim x -> π/2 +          2b . sin^2  (π / 4 – x / 2) / 16 (π / 4 – x / 2)^2
•                       Suppose lim x->0, sin x / x = 1, so x and x are same.
•        Similarly we get  
•                                               = 2b / 16
•                                               = b/8
•                      Therefore right hand side is b/8
• Now l. hs = ½ , f(π/2) = a and r. h.s = b/8
•                   So a = ½ and  
•                Also b/8 = ½
•                     Or b = 4

Reference link will be

https://brainly.in/question/16645685