Which is greater in the following pairs?
(i) e^\pi or \pi^e (that's old)
(ii) 2^{\sqrt{2}} or e
(iii) \ln 8 or 2.
In each case your answer must be accompanied by a proof.
3x1/x=f(x)
f'(x)is a increasing function so as Î >e
therefore Î 1/Ï€>e1/e
so Î e>eÎ
62rocky are you sure f(x) is an increasing function?
@Kaymant sir,
Is a proof like √2 ln 2 <1 allowed for the 2nd one? or do you have another proof in mind? (Ditto for 3 in terms of logic)
1The second one -
Let f { x } = 2 x ln x - x 2 + 1
f ' { x } = 2 + 2 ln x - 2 x
You can always say from the graphs of y = x and y = ln x that ,
ln x + 1 ≤ x for all x .
So f ' { x } ≤ 0 for all x ---- > implies that f { x } is a decreasing function for all x .
Hence , f { √2 } ≤ f { 1 }
or , 2 √2 ln √2 - 2 + 1 ≤ 0 - 1 + 1 = 0
or , ln 2 √2 ≤ 1
or , 2 √2 ≤ e
1For the first one ,
Let f { x } = x 1 / x
To maximise or minimise f { x } , lets put f ' { x } = 0 ,
{ 1 / x } x 1 / x - 1 - { x 1 / x ln x } / x 2 = 0
or , x 1 / x - 2 = x1 / x - 2 ln x
or , ln x = 1
or , x = e ;
Check out f ' ' { x } at x = e , it comes out to be - ve .
So at x = e , f { x } is maximised .
Hence , f { e } > f { Î }
or , e 1 / e > Î 1 / Î
or , e Î > Î e
3412nd one, in a different way:
Using the well known (or easily proved :D) inequality for x>0
e^x \ge 1+x.
Hence for x≥1, we have e^{x-1} \ge x \Rightarrow x-1 \ge \ln x \Rightarrow x \ge 1 + \ln x
Now integrate between the limits 1 and √2 to obtain
\left|\frac{x^2}{2} \right|_1^{\sqrt 2} > |x \ln x|_1^{\sqrt 2}
or
\frac{1}{2} > \sqrt 2 \ln \sqrt 2 \Rightarrow 1 > 2 \sqrt 2 \ln \sqrt 2
It now follows that e^ 1 > e^{2 \sqrt 2 \ln \sqrt 2} = 2^{\sqrt 2}
3413rd one:
\frac{1}{x} > 1- \ln x for x>1 (easily proved by verifying its increasing for x>1)
Integrating between limits 1 and 2, we obtain:
\ln 2 > 2- 2\ln 2
or \ln 8>2
111 query, prophet sir.
Ur soln for the 3rd one is gr8 (as usual)
But how do u hit at that inequality? I mean how do we get to that in the exam hall? Any hints?
1Q1)
let f(x) = x1/x = e^{\frac{1}{x}lnx}
f'(x) = e^{\frac{1}{x}lnx}\left(\frac{1}{x^{2}}-\frac{lnx}{x^{2}} \right)
making f'(x) = 0 → 1-lnx = 0 → x = e
for x>e, lnx > lne → lnx > 1
→ lnx - 1 >0
so f'(x) < 0
hence f(x) is decreasing function
since \pi > e
f(\pi) < f(e)
\pi ^{1/\pi } < e^{1/e} \Rightarrow \pi ^{e}<e^{\pi }
341I was attempting to get at it the same way as in #2. The inequality x≥1+ln x as such was not yielding the inequality as it is required here, the direction was wrong. Then i realized if I write 1/x for x, the direction reverses.
Such things could strike, but it depends on the time available and willingness to experiment.
