Suppose f(x) is differentiable x = 1 and 

• 3

• 4

• 5

• 5

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

• –f(x)

• f(x)

• f(a) + f(a – x)

• f(a) + f(a – x)

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

• a function

• reflexive

• not symmetric

• not symmetric

The range of the function ^7-xP_x-3 is

• {1, 2, 3}

• {1, 2, 3, 4, 5}

• {1, 2, 3, 4}

• {1, 2, 3, 4}

Let f and g be differentiable functions satisfying g′(a) = 2, g(a) = b and fog = I (identity function). Then f ′(b) is equal to

• 1/2

• 2

• 2/3

• 2/3

Let S = {t ∈ R: f(x) = |x-π|.(e^|x| - 1) sin |x| is not differentiable at t}. Then the set S is equal to

• {0,π}

• $\varphi (an empty set)$

• {0}

• {π}

Let S = { x ∈ R : x ≥ 0 and $2|\sqrt{x}-3| + \sqrt{x}(\sqrt{x}-6) + 6 = 0\}$ Then S:

• Contains exactly four elements

• Is an empty set

• Contains exactly one element

• Contains exactly two elements

On the set R of real numbers we define xPy if and only if $\mathrm{xy} \ge 0$. Then, the relation P is

• reflexive but not symmetric

• symmetric but not reflexive

• transitive but not reflexive

• reflexive and symmetric but not transitive

On R, the relation p be defined by 'x$\mathrm{\rho }$y holds if and only if x- y is zero or irrational'. Then,

• $\mathrm{\rho }$ s reflexive and transitive but not symmetric

• $\mathrm{\rho }$ s reflexive and symmetric but not transitive 

• $\mathrm{\rho }$ s symmetric and transitive but not reflexive

• $\mathrm{\rho }$ is equivalence relation

70.

Mean of n observations x₁, x₂, ..., x_n, is $\overline{\mathrm{x}}$. If an observation ${\mathrm{x}}_{\mathrm{q}}$, is replaced by x_q^', then the new mean is

• $\overline{\mathrm{x}} - {\mathrm{x}}_{\mathrm{q}} + {\mathrm{x}}_{\mathrm{q}}\text{'}$

• $\frac{\left(\mathrm{n} - 1\right)\overline{\mathrm{x}} + {\mathrm{x}}_{\mathrm{q}}\text{'}}{\mathrm{n}}$

• $\frac{\left(\mathrm{n} - 1\right)\overline{\mathrm{x}} - {\mathrm{x}}_{\mathrm{q}}\text{'}}{\mathrm{n}}$

• $\frac{\mathrm{n}\overline{\mathrm{x}} - {\mathrm{x}}_{\mathrm{q}} + {\mathrm{x}}_{\mathrm{q}}\text{'}}{\mathrm{n}}$