R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴R is reflexive.
Now,
Let (P, Q) ∈ R.
⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.
⇒ (Q, P) ∈ R
∴R is symmetric.
Now,
Let (P, Q), (Q, S) ∈ R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ R
∴R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Let f: X → Y be an invertible function. Show that the inverse of f –1 is f, i.e., (f–1)–1 = f.
y = x2 + 2x + C : y' - 2x - 2 = 0
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}
Find the direction cosines of a line which makes equal angles with the coordinate axes.