# range

find the range of y

y=1/|sinx| +1/|cosx|

• Ankit Kumar We can easily see that function is infinite : lim[xâ†’0] 1/|sinx| + 1/|cosx| = 1/0 + 1 = âˆž lim[xâ†’Ï€/2] 1/|sinx| + 1/|cosx| = 1 + 1/0 = âˆž.. We just need to find minimum value of function: y = 1/|sinx| + 1/|cosx| y = |cscx| + |secx| y' = âˆ’cotx |cscx| + tanx |secx| = 0 tanx |secx| = cotx |cscx| tanx |secx| / (cotx |cscx|) = 1 tanÂ²x |tanx| = 1 |tanÂ³x| = 1 |tanx| = 1 tanx = +- 1 When tanx = 1 or âˆ’1, then sinx = Â±1/âˆš2, cosx = Â±1/âˆš2 Minimum value: y = 1/|sinx| + 1/|cosx| = 1/(1/âˆš2) + 1/(1/âˆš2) = âˆš2 + âˆš2 = 2âˆš2 Range: [2âˆš2, âˆž)

11
Anirudh Narayanan ·

i think [2√2,âˆž) [just a wild tukka]

11
Mani Pal Singh ·

it is [2√2,âˆž)

11
Anirudh Narayanan ·

THE FUNCTION WILL BE THE SAME AS

1/sinx +1/cosx for x Îµ [0,Ï€/2]

FOR SIMPLICITY'S SAKE WE'LL

TAKE y = 1/sinx +1/ cosx {x Îµ [0,Ï€/2]}

= secx +cosecx

y' = secx tanx - cosecx cotx

0 = secx tanx - cosecx cotx

cosx/sin2x = sinx/cos2x

tan3x = 1 => x = Ï€/4

y(Ï€/4) = 2√2

So 2√2 â‰¤y<âˆž

11
Mani Pal Singh ·

by symmetry we can see that function will have the largest value at nâˆ©+âˆ©/4
so it will have max value of 1/1/√2+1/1/√2=2√2

1
gordo ·

1/mod(sinx) , 1/mod(cosx) >0
=> AM>=GM
1/mod(sinx) + 1/mod(cosx)>2{2/mod(sin2x)}1/2
we know max of the RHS=2âˆš2
so min 1/mod(sinx) + 1/mod(cosx)=2âˆš2
and max=âˆž (any one of them=0)
hence the range, [2âˆš2,âˆž)
cheers!!!

1
Tushar Gautam ·

We just need to find minimum value of function:

y = 1/|sinx| + 1/|cosx|

y = |cscx| + |secx|

y' = âˆ’cotx |cscx| + tanx |secx| = 0

tanx |secx| = cotx |cscx|

tanx |secx| / (cotx |cscx|) = 1

tanÂ²x |tanx| = 1

|tanÂ³x| = 1

|tanx| = 1

tanx = Â± 1

When tanx = 1 or âˆ’1, then sinx = Â±1/âˆš2, cosx = Â±1/âˆš2

Minimum value: y = 1/|sinx| + 1/|cosx| = 1/(1/âˆš2) + 1/(1/âˆš2) = âˆš2 + âˆš2 = 2âˆš2

Range: [2âˆš2, âˆž)