If the function g(x) = is differntiable, then the value of k+m is

• 2

• 16/5

• 10/3

• 10/3

If a ε R and the equation - 3(x-[x]2 + 2(x-[x] +a² = 0(where,[x] denotes the greatest integer ≤ x) has no  integral solution, then all possible value of  lie in the interval

• (-1,0) ∪ (0,1)

• (1,2)

• (-2,-1)

• (-∞,-2) ∪ (2, ∞)

Consider the following relations:
R = {(x, y)| x, y are real numbers and x = wy for some rational number w}; S = {(m/p, p/q)| m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then

• R is an equivalence relation but S is not an equivalence relation

• neither R nor S is an equivalence relation

• S is an equivalence relation but R is not an equivalence relation

• S is an equivalence relation but R is not an equivalence relation

4.

Let p(x) be a function defined on R such that  = 1, p'(x)  p'(1-x),for all x∈[0,1] p(0) = 1 and p(1) = 41. Then  equals

• √41

• 21

• 41

• 41

The function 

• neither injective nor surjective.

• invertible

• injective but not surjective.

• injective but not surjective.

Let W denote the words in the English dictionary. Define the relation R by :
R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is

• not reflexive, symmetric and transitive

• reflexive, symmetric and not transitive

• reflexive, symmetric and transitive

• reflexive, symmetric and transitive

The graph of the function y = f(x) is symmetrical about the line x = 2, then

• f(x + 2)= f(x – 2)

• f(2 + x) = f(2 – x)

• f(x) = f(-x)

• f(x) = f(-x)

The domain of the function 

• [2, 3]

• [2, 3)

• [1, 2]

• [1, 2]

$\left\{\mathrm{x} \in \mathrm{R}: \left|\cos \left(\mathrm{x}\right)\right| \ge \sin \left(\mathrm{x}\right)\right\} \cap \left[0, \frac{3\mathrm{\pi }}{2}\right]$ is equal to

• $\left[0, \frac{\mathrm{\pi }}{4}\right] \cup \left[\frac{3\mathrm{\pi }}{4}, \frac{3\mathrm{\pi }}{2}\right]$

• $\left[0, \frac{\mathrm{\pi }}{4}\right] \cup \left[\frac{\mathrm{\pi }}{2}, \frac{3\mathrm{\pi }}{2}\right]$

• $\left[0, \frac{\mathrm{\pi }}{4}\right] \cup \left[\frac{5\mathrm{\pi }}{4}, \frac{3\mathrm{\pi }}{2}\right]$

• $\left[0, \frac{3\mathrm{\pi }}{2}\right]$

If ${\log }_{0.2}\left(\mathrm{x} - 1\right) > {\log }_{0.04}\left(\mathrm{x} + 5\right)$, then

• - 1 < x < 4

• 2 < x < 3

• 1 < x < 4

• 1 < x < 3