11i think [2√2,∞) [just a wild tukka]
find the range of y
y=1/|sinx| +1/|cosx|
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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, ∞)Upvote·0· Reply ·2017-08-16 21:47:17
6 Answers
11THE 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(Ï€/4) = 2√2
So 2√2 ≤y<∞
11by 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
11/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!!!
1We 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, ∞)