Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen that (1, 1), (2, 2), (3, 3) ∉R.
∴ R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R
However,
(1, 1) ∉ R
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
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 : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto.
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Determine order and degree(if defined) of differential equation \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0\end{align}
Represent graphically a displacement of 40 km, 30° east of north.
If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.
Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).
Answer the following as true or false.
\begin{align}(i) \overrightarrow{a}\; and\; \overrightarrow{-a}\; are\; collinear.\end{align}
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}
y = Ax : xy' = y (x ≠ 0)
\begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align}
y = x2 + 2x + C : y' - 2x - 2 = 0
y = cosx + C : y' + sinx = 0