Limits Continuity And Differentiability Question 108
If $ y={{(1+1/x)}^{x}} $ then $ \frac{2\sqrt{y(2)+1/8}}{(log(3/2)-1/3)} $ is equal to-
Options:
A) 3
B) 4
C) 1
D) 2
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y={{( 1+\frac{1}{x} )}^{x}} $
Taking logarithm of both sides, we get
$ \log y=x[ \log ( 1+\frac{1}{x} ) ] $
$ \Rightarrow \frac{1}{y}y_1(x)=\frac{x^{2}}{x+1}( -\frac{1}{x^{2}} )+\log \left( 1+\frac{1}{x} \right) $
$ =-\frac{1}{x+1}+\log ( 1+\frac{1}{x} )…(1) $
Since, $ y(2)={{(1+1/2)}^{2}}=25/4 $
So, $ y_1(2)=(9/4)( -\frac{1}{3}+\log \frac{3}{2} ) $
Again differentiate eq. (1) w.r.t (x), we get
$ \frac{y(x)y_2(x)-{{[y_1(x)]}^{2}}}{{{(y(x))}^{2}}}=\frac{1}{{{(1+x)}^{2}}}-\frac{1}{x(x+1)} $
By putting $ x=2 $ , we get
$ \frac{y(2)y_2(2)-{{(y_1(2))}^{2}}}{{{(y(2))}^{2}}}=\frac{-1}{18} $
Now, put the value of y(2) and $ y_1(2) $
$ \Rightarrow y_2(2)=( \frac{9}{4} ){{( -\frac{1}{3}+\log \frac{3}{2} )}^{2}}-\frac{1}{8} $
$ {{( y_2(2)+\frac{1}{8} )}^{4}}=9{{( \log \frac{3}{2}-\frac{1}{3} )}^{2}} $
$ \Rightarrow $ Required expression = 3
BETA
