Question

The function g is defined by
g(x) = sin x, for 0≤x<1/2π
px-1, for 1/2π≤x<2π
q , for x≥2π
where p and q are constants. Find the values of p and q if g is continuous at x=1/2π and x=2π

Answer 1

if g is continuous at x=1/2π then
\(\Large \lim \limits_{x\rightarrow \pi/2^-}g(x) =\lim \limits_{x\rightarrow \pi/2^+}g(x)\)

Answer 2

what will u choose for \(\pi/2^-\)
?
sin x or px-1 ?

Answer 3

px-1 maybe

Answer 4

let me tell you what \(\pi/2^-\)
it is a number which is very near to pi/2 but LESS then pi/2
so, you will choose the function which has \(x<\pi/2\)

Answer 5

now once again,
what will u choose for π/2−
?
sin x or px-1 ?

Answer 6

sin x :D

Answer 7

correct! :)
so you will choose px-1 for x->pi/2+
now can u find those 2 limits ? in my first comment ?

Answer 8

can you write the all solution please ..

Answer 9

\(\Large \lim \limits_{x\rightarrow \pi/2^-}g(x) =\lim \limits_{x\rightarrow \pi/2^+}g(x) \\ \Large \lim \limits_{x\rightarrow \pi/2^-}\sin x =\lim \limits_{x\rightarrow \pi/2^+}px-1\)
you know how to solve limits questions, right ?

Answer 10

wait i try it first

Answer 11

sure, go ahead :)
take your time....

Answer 12

i got the wrong answer .. the correct answer is p=4/π but i got 2.055/π .. what should i do ..

Answer 13

did u put x = pi/2 in sin x and px-1 ?

Answer 14

what is sin pi/2 = .... ?

Answer 15

0.027

Answer 16

no!
you used your calculator in degree mode! you need to use it in radians mode :)
because pi/2 is in radians

Answer 17

sin pi/2 = 1
actually.

Answer 18

so, now can you get p from
1 = p (pi/2) -1
?

Answer 19

wait

Answer 20

ok i got the correct answer for p ,, i use wrong mode in calculator .. can u give me the other equations for find q ?

Answer 21

sure,

Answer 22

\(\Large \lim \limits_{x\rightarrow 2\pi^-}g(x) =\lim \limits_{x\rightarrow 2\pi^+}g(x) \\ \Large \lim \limits_{x\rightarrow 2\pi^-}px-1 =\lim \limits_{x\rightarrow 2\pi^+}q\)

Answer 23

ok i try it now ..

Answer 24

when you are done, tell me what u got for q

Answer 25

i got q= 7

Answer 26

good work! i got 7 too :)

Answer 27

Thank a lot sir .. :D

Answer 28

welcome ^_^
and since you are new here
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Answer 29

Thank For helping me .. if i got any problem's and i 'm sure i will discuss in OPEN STUDY .. Such A Good website for student like me ^_^

Answer 30

i am glad to hear that! :)
its good for everyone ;)