R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈{1, 2, 3, 4}.
∴ R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
∴R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
The correct answer is B.
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 A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
Determine order and degree(if defined) of differential equation \begin{align}\frac{d^2y}{dx^2}=\cos3x + sin3x\end{align}
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.
y = x2 + 2x + C : y' - 2x - 2 = 0
Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0