106This is easy.
Let f(x) = (1+1/x)x (I am assuming a different function whose domain is x=R)
lnf(x) = x + ln(1+1/x)
=> f'(x) = f(x)[1 - 1/(x+1)(x)]
= f(x)(x2+x-1)/x(x+1)
f'(x) > 0 for x≥2
So the function is increasing in the interval [2,∞)
So max value will be when n-->∞ = e (from limits chptr)
min value will be when n=2 = 2.25
So for all values of x (inclusive of integers) 2< f(x)<3
21from bino theorem
(1+1/n)n=1+ n.1n + n(n-1)n22! + n(n-1)(n-2)n33! + ............. n(n-1)(n-2)..........[n-(n-1)]nnn!
=1+1+12!(1-1n) + 13!(1-1n)(1-2n) + 14!(1-1n)(1-2n)(1-3n)...............1n!(1-1n)(1-2n)(1-3n)......... (1-n-1n)
<1+1+12!+ 13!...............1n!
<1+1+12+122+ 123...........................12n-1
=3-12n-1
check here n ≥1 for this 2<f(n)<3
1another bino-inequality :
if a,b,c,d are consequtive coefficientss of expansion (1+x)n; x>0;
show that :
\left(\frac{b}{b+c} \right)^{2}\geq \left(\frac{ac}{(a+b)(c+d)} \right)
11Let a = n C r-1 , b = n C r , c = n C r+1
and d = n C r+2
Therefore, a+b = n+1 C r , b+c = n+1 C r+1 , c+d = n+1 C r+2
a+ba = n+1 C rn C r-1 = n+1r ........................(i)
and b+cb = n+1 C r+1 n C r = n+1r+1..............................(ii)
and c+dc = n+1r+2 ...............(iii)
Frm (i), (ii), and (iii); we get, aa+b , bb+c, cc+d are in A.P
Now Apply, A.M≥ G.M, we get
bb+c ≥ √ac(a+b) (c+d)
Therefore, { (bb+c) 2 - ac(a+b) (c+d) } ≥ 0
Therefore, thus proved.
24in post#2
f(x) = (1+1/x)x
lnf(x) = x + ln(1+1/x)
How ??
shouldnt it be lnf(x) = x.ln(1+1/x)
1not a doubt :)
f(x,y)=x2+3y2+2xy-6x-2y
find fmin
note:calculus in two variable may be used if explanation is given for making partial derivates 0 ,
11Ans) f(x,y) = x2+3y2+2xy-6x-2y
f x = 2x +2y - 6
f y = 6y + 2x - 2
Now Put f x = 0 ........ that implies x+y = 3 .................(i)
Now Put f y = 0 ......... that implies x+3y = 1 ....................(ii)
On solving (i) and (ii), we get
x = 4 and y=-1 (Stationary points)
Now, r = fxx = 2 ;
S= f xy = 2 ;
and similarily t = fyy = 6
Since rt - s 2 = 12 - 4 = 8 > 0 and r > 0 , then the function f(x,y) has a minimum value at point (4, -1)
Therefore, f min = 16 + 3 -8 - 24 +2 = - 11
Please correct me if wrong [ /hide]