range

find the range of y

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

  • Ankit Kumar We can easily see that function is infinite : lim[x→0] 1/|sinx| + 1/|cosx| = 1/0 + 1 = ∞ lim[x→π/2] 1/|sinx| + 1/|cosx| = 1 + 1/0 = ∞.. We just need to find minimum value of function: y = 1/|sinx| + 1/|cosx| y = |cscx| + |secx| y' = −cotx |cscx| + tanx |secx| = 0 tanx |secx| = cotx |cscx| tanx |secx| / (cotx |cscx|) = 1 tan²x |tanx| = 1 |tan³x| = 1 |tanx| = 1 tanx = +- 1 When tanx = 1 or −1, then sinx = ±1/√2, cosx = ±1/√2 Minimum value: y = 1/|sinx| + 1/|cosx| = 1/(1/√2) + 1/(1/√2) = √2 + √2 = 2√2 Range: [2√2, ∞)

6 Answers

11
Anirudh Narayanan ·

i think [2√2,∞) [just a wild tukka]

11
Mani Pal Singh ·

it is [2√2,∞)

1
Tushar Gautam ·

We just need to find minimum value of function:

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

y = |cscx| + |secx|

y' = −cotx |cscx| + tanx |secx| = 0

tanx |secx| = cotx |cscx|

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

tan²x |tanx| = 1

|tan³x| = 1

|tanx| = 1

tanx = ± 1

When tanx = 1 or −1, then sinx = ±1/√2, cosx = ±1/√2

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

Range: [2√2, ∞)

11
Anirudh Narayanan ·

THE FUNCTION WILL BE THE SAME AS

1/sinx +1/cosx for x ε [0,π/2]

FOR SIMPLICITY'S SAKE WE'LL

TAKE y = 1/sinx +1/ cosx {x ε [0,π/2]}

= secx +cosecx

y' = secx tanx - cosecx cotx

0 = secx tanx - cosecx cotx

cosx/sin2x = sinx/cos2x

tan3x = 1 => x = π/4

y(π/4) = 2√2

So 2√2 ≤y<∞

11
Mani Pal Singh ·

by symmetry we can see that function will have the largest value at n∩+∩/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=2√2
so min 1/mod(sinx) + 1/mod(cosx)=2√2
and max=∞ (any one of them=0)
hence the range, [2√2,∞)
cheers!!!

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