Consider f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f –1 and show that (f –1)–1 = f.
Function f: {1, 2, 3} → {a, b, c} is given by,
f(1) = a, f(2) = b, and f(3) = c
If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:
gof = Ix and fog = Iy where X = {1, 2, 3} and Y= {a, b, c}.
Thus, the inverse of f exists and f - 1 = g.
∴f - 1: {a, b, c} → {1, 2, 3} is given by,
f - 1(a) = 1, f - 1(b) = 2, f-1(c) = 3
Let us now find the inverse of f - 1 i.e., find the inverse of g.
If we define h: {1, 2, 3} → {a, b, c} as
h(1) = a, h(2) = b, h(3) = c, then we have:
∴, where X = {1, 2, 3} and Y = {a, b, c}.
Thus, the inverse of g exists and g - 1 = h ⇒ (f - 1) - 1 = h.
It can be noted that h = f.
Hence, (f - 1) - 1 = f.
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\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}
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