1) Let f(x) = sin^-1{(|x|-5)/x} + (log5(5-x))^-1 then find the domain of f.
2) If 2f(x) + 3f(1/x) = x^2 – 1
a) f‘(2) = ?
b) limit f(x)/x = ?
(x tends to infinity )
c) The number of solutions of the equation f(x) = 5^x + 5^-x is ?
d) the number of solutions of the equation f(x) – [x] is ? ([] denotes the greatest integer function)
please tell me the method of solving such questions
i can tell u the answer if u are online because we need to interact reply if u are online
The general method to solve the second question can be to first find f(x). In this particular case,
2f(x) + 3f(1/x) = x2-1 .........*
Putting 1/x in place of x,
2f(1/x) + 3f(x) = 1/x2-1 .........#
The equations * and # are equations in two variables f(x) and f(1/x) and hence one can solve for f(x).
In the first question, log is in the denominator. Hence,
\(log 5(5-x)\neq 0\).
For log to be defined 5(5-x) > 0.
For sin inverse to be defined, \(-1\le\frac {|x|-5}{x}\le 1\).
2f(x) + 3f(1/x) = x2-1 .........*
Putting 1/x in place of x,
2f(1/x) + 3f(x) = 1/x2-1 .........#
The equations * and # are equations in two variables f(x) and f(1/x) and hence one can solve for f(x).
In the first question, log is in the denominator. Hence,
\(log 5(5-x)\neq 0\).
For log to be defined 5(5-x) > 0.
For sin inverse to be defined, \(-1\le\frac {|x|-5}{x}\le 1\).
1) The domain is (5,-infinity) excluding x = -5,0.
2) Solving the way the educator did u get f(x)=[3/x2-2x2-1]/5 -----------(1)
a) U can differentiate (1) to get f‘(2).
b) The limit = (- infinity).
c) Now f(x) has a maxima (+ infinity) at x=0 and it's roots are +-1. This function is continous at all points except x = 0. y = 5x + 5-x is continous at all points and has a minima at x = 0 and rises continously on both sides of the y axis toward x = +-(infinity). Therefore the graph of 5x + 5-x cuts f(x) at exactly 2 points and hence f(x) = 5x + 5-x has 2 roots.
d) Do u want the number of solutions of f(X)-[x]=0? Please elaborate.