if f(y) = log y, then f(y) + f(1/y) is equal to: - logarithm and function (entrance exam)
if f(y) = log y, then f(y) + f(1/y) is equal to:
0
`f(x)=log(logx),g (x)=(1)/(logx)`
`f(x)=logx,g(x)=(1)/(logx)`
`f(x)=(1)/(logx),(x)=log(logx)`
`f(x)=(1)/(xlogx),g(x)=(1)/(logx)`
answer: a
solution: we have , `i=int{log(logx)+(1)/((logx)^(2))}dx`   
 `rarr i= inte^(t)(logt+(1)/(t^(2)))dt`,where t = log x   
 `rarri=inte^(t)(logt+(1)/(t))dt+inte^(t)(-(1)/(t)+(1)/(t^(2)))dt`   
 `rarri=e^(t)logt +e^(t)(-(1)/(t))+c`   
 `rarri=x(log(logx)-(1)/(logx))+c`   
 `:. f(x) =logx and g(x) = (1)/(logx)` Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. Let $y=f(x)$ be a function. Suppose that $a$ and $\mathbf{L}$ are numbers such that:


 


* whenever $\boldsymbol{x}$ is close to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ is close to $\mathbf{L} ;$
* as $\boldsymbol{x}$ gets closer and closer to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ gets closer and closer to $\mathbf{L}$; and
* suppose that $f(x)$ can be made as close as we want to $\mathbf{L}$ by making $x$ close to $a$ but not equal to $a$.


Then we say that the limit of $f(x)$ as $x$ approaches a is $L$ and we write $\lim \_{x \rightarrow a} f(x)=\mathbf{L}$ doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. 10.1 sequences polynomials: easy to evaluate, integrate, diﬀerentiate general functions: -a sequence is a list of #s -sequences are written as: -many series correspond to functions formally, the sequence converges to l if for every positive number , there corresponds an integer n such that for all slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved log 80 can be written as; log (8 ✕ 10) log (4 ✕ 2 ✕ 10) here we can use the formula log (a ✕ b) = log a + log b log 4 + log 2 + log 10  (1) log10 10 = 1 log 4 = 0.6020 log 2 = ½ log 4 = ½ ✕ 0.6020 = 0.3010 substituting the values in equation (1) we get the answer 1.9030 the sum of a pmf over its entire domain must equal 1.
so cς(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = 1.
 
ς(i=1 to ∞)[1/n^i] = 1/(n-1). so
 
ς(y=1 to ∞)[(1/3^y)] = 1/(3-1) = 1/2 and
ς(x=1 to ∞)[(1/4^x)] = 1/(4-1) =1/3.
 
so, σ(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = (1/2)*(1/3) = 1/6.
 
to make the sum = 1, then c=6. given = let, logx = z⇒ so,where c is the integrating constant.