R = {(a, b): a ≤ b2}
It can be observed that
\begin{align} \left(\frac{1}{2},\frac{1}{2}\right) ∉ R , since \frac{1}{2}>\left(\frac{1}{2}\right)^2 = \frac{1}{4}\end{align}
∴R is not reflexive.
Now, (1, 4) ∈ R as 1 < 42
But, 4 is not less than 12.
∴(4, 1) ∉ R
∴R is not symmetric.
Now,
(3, 2), (2, 1.5) ∈ R
(as 3 < 22 = 4 and 2 < (1.5)2 = 2.25)
But, 3 > (1.5)2 = 2.25
∴(3, 1.5) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, 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.
If f(x) = , show that fof(x) = x, for all x ≠ 2/3. What is the inverse of f ?
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.
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Let f, g and h be functions from R to R. Show that
(f + g)oh = foh + goh
(f . g)oh = (foh) . (goh)
Sir please help me?