Answer
Verified
35.1k+ views
Hint: In order to prove that the two functions are not equal we need to find a value of x for which the two functions give different values. Hence first construct these functions and hence find a value of x for which fog and gof give different values. Hence prove that the functions are not equal.
Complete step-by-step solution -
Two functions f and g are not equal if Domain of f and Domain of g are different or Range of f and Range of g are different or there exists a value of x for which f(x) is not equal to g(x).
Consider $f\left( x \right)=\cos x$ and $g\left( x \right)=3{{x}^{2}}$
Now we have $fog\left( x \right)=f\left( 3{{x}^{2}} \right)=\cos \left( 3{{x}^{2}} \right)$
We have Domain of fog = R
Range of fog =[-1,1]
$gof\left( x \right)=g\left( \cos x \right)=3{{\cos }^{2}}x$
We have Domain of gof = R
Range of gof = [0,3]
Since Range of gof is not equal to range of fog, the two functions are not equal. Hence we have
$fog\ne gof$.
Note:[1] Finding the range
$\begin{align}
& -1\le \cos x\le 1 \\
& \Rightarrow 0\le {{\cos }^{2}}x\le 1 \\
& \Rightarrow 0\le 3{{\cos }^{2}}x\le 3 \\
\end{align}$
Hence Range of gof(x) is [0,3]
[2] This example serves as a proof that composition of two functions is not commutative, i.e. in general fog(x) is not the same as gof(x).
Graph of fog(x)
Graph of gof(x)
As is evident from the graphs of the two functions, we have the two functions are not equal.
Complete step-by-step solution -
Two functions f and g are not equal if Domain of f and Domain of g are different or Range of f and Range of g are different or there exists a value of x for which f(x) is not equal to g(x).
Consider $f\left( x \right)=\cos x$ and $g\left( x \right)=3{{x}^{2}}$
Now we have $fog\left( x \right)=f\left( 3{{x}^{2}} \right)=\cos \left( 3{{x}^{2}} \right)$
We have Domain of fog = R
Range of fog =[-1,1]
$gof\left( x \right)=g\left( \cos x \right)=3{{\cos }^{2}}x$
We have Domain of gof = R
Range of gof = [0,3]
Since Range of gof is not equal to range of fog, the two functions are not equal. Hence we have
$fog\ne gof$.
Note:[1] Finding the range
$\begin{align}
& -1\le \cos x\le 1 \\
& \Rightarrow 0\le {{\cos }^{2}}x\le 1 \\
& \Rightarrow 0\le 3{{\cos }^{2}}x\le 3 \\
\end{align}$
Hence Range of gof(x) is [0,3]
[2] This example serves as a proof that composition of two functions is not commutative, i.e. in general fog(x) is not the same as gof(x).
Graph of fog(x)
Graph of gof(x)
As is evident from the graphs of the two functions, we have the two functions are not equal.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
Other Pages
The correct IUPAC name of the given compound is A isopropylbenzene class 11 chemistry JEE_Main
1mol of ferric oxalate is oxidized by x mol of MnO4in class 11 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main
What is the volume of water that must be added to a class 11 chemistry JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main