71This is the one thing that has haunted me always. Now that you have posted it, I'll surely learn something from this post!
Let us break the rules ..
Let the co domain of sin-1x be (pi2 ,3pi2)
tan-1x be (pi2 ,3pi2)
cos-1x be (pi, 2pi)
cot-1x be (pi,2pi)
1)Then prove that f(x) = 3sin-1x - 2cos-1x is an odd function.
2) Find the minimum value of (sin-1x)3 + (cos-1x)3
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10 Answers
71
30According to mathematician Shubhodip's new rules [3],
sin-1x + cos-1x = 5 pi2
for question 1,
f(x)= 3\sin^{-1}x - 2 \cos^{-1} x
= 3\sin^{-1}x - 2\left(\frac{5\pi}{2} - \sin^{-1}x\right)
= 5\left( \sin^{-1}x - \pi \right)
It is easy to note sin-1x is pi at x=0, taking (0,pi) as origin, function is clearly odd.
30shubhodip I'll try the next one too.. don't post the solution any time soon.
302. (\sin^{-1} x)^3 + (\cos^{-1})^3 = \left(\sin^{-1} x\right) ^3 + \left(\frac{5\pi }{2} - \sin^{-1} x\right)^3
= \left( \sin^{-1}x\right)^3 + \left( \frac{5\pi }{2}\right)^3 - \left( \sin^{-1}x\right)^3 - \frac{15\pi }{2}\sin^{-1}x\left(\frac{5\pi }{2} - \sin^{-1}x \right)
= \frac{125\pi ^3}{8} - \frac{75\pi^2 }{4}\sin^{-1}x + \frac{15\pi }{2}\sin^{-1}x^2
Since \frac{\pi }{2}\leq \sin^{-1}x \leq \frac{3\pi }{2} ,
= \frac{125\pi ^3}{8} - \frac{75\pi^3 }{8} + \frac{15\pi^3 }{8} = \boxed{\frac{65\pi ^3}{8}}
@shubhodip - can you post your solution now? [7]
49first of all sin-1x and cos-1x are positive numbers as described by the sum!
thus,
(sin-1x)3+(cos-1x)3 ≥ 2√(sin-1x)3(cos-1x)3 (simple AM GM)
min value when sin-1x=cos-1x
and min value = 2(sin-1x)3 for that x!
sin and cos become equal at every nπ+π4
so, in given ranges they become equal only at x=5Ï€/4
so, min value=2.125Ï€3/64=12532Ï€3
koi calculation mistakes ho to shama kare!
49yaar answer mil gaya subhodip se! :D
JEE k baad pehli baar confidence re building! :)
JEE me maine 80 marks ka silly mistake kar aaya...and Seoni said "No mistake is a silly mistake!!" :'( :'( :((
30My approach to the 2nd one was wrong I guess. Will keep this in mind! [1]
49perfecting Ashish's proof to first sum...
after f(x)=5[sin-1x - π]
f(-x)=5[sin-1(-x) - π]
let sinθ=x
π2≤θ≤3π2
sin(m-θ)=-x
-π2≥-θ≥-3π2
m-π2≥m-θ≥m-3π2
for m-θ to lie within principle value, m must be equal to 2π
thus, sin-1(-x)=2Ï€-sin-1x
so, f(-x) = 5[2π - sin-1x - π] = 5[π - sin-1x] = -5[sin-1x - π]
or, f(-x) = -f(x)
thus, f(x) is an odd function! [1]