Question

If the function f, g, h are defined from the set of real numbers R to R such that f(x) = \[{{x}^{2}}\]-1 , g (x) = \[\sqrt{x{}^{2}+1}\], h(x) = 0, if \[x\ge 0\] = x if ${{x}^{2}}$ $x\le 0$, then (hofog) (x) is
(1) 1
(2) 0
(3) -1
(4) $x^{2}$

Answer 1

Hint: This question is based on functions. To solve the question, we must know how to multiply the three functions. Substitute the value of g (x) then we substitute the value of f (x) by taking the value of x equal to the value of g (x). similarly we put the value of h (x) to solve our q

Complete step by step solution: 
We have given the three functions f, g, h which are defined from the set of real numbers R to R.
We have to find the value of (hofog)(x).
We know (hofog)(x) = h (f (g (x) ))) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know value of g (x) = \[\sqrt{x{}^{2}+1}\] ( given in the question )
Put the value of g(x) in equation (1), we get
h (f (g (x) ))) = h (f (\[\sqrt{x{}^{2}+1}\])) . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now we know the value of f(x) = \[{{x}^{2}}\] -1 ( given in the question )
Put the value of f (x) in equation (2), we get
h (f (\[\sqrt{x{}^{2}+1}\])) = h ( ${{\left( \sqrt{{{x}^{2}}+1} \right)}^{2}}$- 1 )
Solving the above equation, we get
h ( ${{\left( \sqrt{{{x}^{2}}+1} \right)}^{2}}$- 1 ) = h$\left( {{x}^{2}}+1-1 \right)$
which is equal to $h\left( {{x}^{2}} \right)$. . . . . . . . . . . . . .. . . . .. . . .. . (3)
Now we have given the value of h (x) in the question
Put the value of h(x) in equation (3), we get
 $h\left( {{x}^{2}} \right)$ = ${{x}^{2}}$
Hence the value of (hofog)(x) = ${{x}^{2}}$
Option (4) is correct.
Option (4) is correct.

Note: In these types of questions, students must take care in substituting the values. Substitute the variable x that’s there within the outside function with the within function by taking the individual functions as a reference.