$\mathrm{The} \mathrm{preposition} \mathrm{p}\Rightarrow \left(~\left(\mathrm{p} \wedge ~ \mathrm{q}\right)\right) \mathrm{is} \mathrm{equivalent} \mathrm{to}$

• q

• $~ \mathrm{p} \vee \mathrm{q}$

• $\mathrm{p} \vee ~ \mathrm{q}$

• $~ \mathrm{p} \vee ~ \mathrm{q}$

Ler R₁ and R₂ be two relations defined as follows :

R₁ = {(a, b) $\in$ R₂ : a₂ + b₂ $\in$ Q} and R₂ = {(a, b) $\in$R2: a₂ + b₂ $\in$Q}, where Q is the set of all rational numbers, then

• R₁ is transitive but R₂ is not transitive.

• R₂ is transitive but R₁ is not transitive.

• Neither R₁ nor R₂ is transitive

• R₁ and R₂ are both transitive.

Suppose f(x) is a polynomial of degree four, having critical points at – 1, 0, 1. If T = {x $\in$ R |f(x) = f(0)}, then the sum of squares of all the elements of T is :

• 4

• 6

• 2

• 8

$$\begin{gathered} \mathrm{The} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = \left\{\frac{\mathrm{\pi }}{4} + {\tan }^{-1}\left(\mathrm{x}\right), \left|\mathrm{x}\right| \le 1 \\ \frac{1}{2}\left(\left|\mathrm{x}\right| - 1\right), \left|\mathrm{x}\right| > 1 \right\} \mathrm{is} : \end{gathered}$$

• $\mathrm{continuous} \mathrm{on} \mathrm{R} – \{ – 1\} \mathrm{and} \mathrm{differentiable} \mathrm{on} \mathrm{R} – \left\{ – 1, 1\right\}$

• $\mathrm{both} \mathrm{continuous} \mathrm{and} \mathrm{differentiable} \mathrm{on} \mathrm{R} – \left\{– 1\right\}$

• $\mathrm{both} \mathrm{continuous} \mathrm{and} \mathrm{differentiable} \mathrm{on} \mathrm{R} – \left\{1\right\}$

• $\mathrm{continuous} \mathrm{on} \mathrm{R} – 1 \mathrm{and} \mathrm{differentiable} \mathrm{on} \mathrm{R} – \left\{– 1, 1\right\}$

$$\begin{gathered} \mathrm{If} 3^{2\sin \left(2\mathrm{\alpha }\right) - 1}, 14 \mathrm{and} 3^{4 - 2\sin \left(2\mathrm{\alpha }\right)} \mathrm{are} \mathrm{the} \mathrm{first} \mathrm{three} \mathrm{terms} \mathrm{of} \mathrm{an} \mathrm{A}.\mathrm{P}. \\ \mathrm{for} \mathrm{some} \mathrm{\alpha }, \mathrm{then} \mathrm{the} \mathrm{sixth} \mathrm{term} \mathrm{of} \mathrm{this} \mathrm{A}.\mathrm{P}. \mathrm{is} \end{gathered}$$

• 81

• 65

• 66

• 78

616.

If the minimum and the maximum values of the function

$\mathrm{f} :\hspace{0.17em}\left[\frac{\mathrm{\pi }}{4}, \frac{\mathrm{\pi }}{2}\right] \to \mathrm{R}, \mathrm{defined} \mathrm{by} \mathrm{f}\left(\mathrm{\theta }\right) = \left|\begin{array}{ccc} - {\sin }^2\left(\mathrm{\theta }\right)& - 1 - {\sin }^2\left(\mathrm{\theta }\right)& 1\\ {\cos }^2\left(\mathrm{\theta }\right)& - 1 - {\cos }^2\left(\mathrm{\theta }\right)& 1\\ 12& 10& - 2\end{array}\right|$

are m and M respectively, then the ordered pair (m, M) = : 

• ( - 4, 4)

• $\left(0, 2\sqrt{2}\right)$

• ( - 4, 0)

• (0, 4)

If x = 1 is a critical point of the function f(x) = (3x² + ax –2 – a) e^x, then

• x = 1 and x = $- \frac{2}{3}$ are local minima of f

• x = 1 is a local maxima and x =  $- \frac{2}{3}$is a local minima of f.

• $\mathrm{x} = 1 \mathrm{is} \mathrm{a} \mathrm{local} \mathrm{minima} \mathrm{and} \mathrm{x} = - \frac{2}{3} \mathrm{is} \mathrm{a} \mathrm{local} \mathrm{maxima} \mathrm{of} \mathrm{f}.$

• $\mathrm{x} = 1 \mathrm{and} \mathrm{x} = - \frac{2}{3}\mathrm{are} \mathrm{local} \mathrm{maxima} \mathrm{of} \mathrm{f}.$

$$\begin{gathered} \mathrm{Let} \mathrm{A} = \left\{\mathrm{a}, \mathrm{b}, \mathrm{c}\right\} \mathrm{and} \mathrm{B} = \left\{1, 2, 3, 4\right\}.\mathrm{Then} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{elements} \mathrm{in} \mathrm{the} \mathrm{set} \\ \mathrm{C}=\left\{\mathrm{f} : \mathrm{A} \to \mathrm{B}|2 \in \mathrm{f}(\mathrm{A}) \mathrm{and} \mathrm{f} \mathrm{is} \mathrm{not} \mathrm{one}–\mathrm{one}\right\}\mathrm{is}... \end{gathered}$$

The position of moving car at time tis given by f(t) = at² + bt + c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t₁, t₂] is attained at the point

• $\frac{{\mathrm{t}}_1 + {\mathrm{t}}_2}{2}$

• $\frac{{\mathrm{t}}_2 - {\mathrm{t}}_1}{2}$

• $2\mathrm{a}\left({\mathrm{t}}_1 +\hspace{0.17em} {\mathrm{t}}_2\right) \hspace{0.17em}+ \mathrm{b}$

• $\mathrm{a}\left({\mathrm{t}}_2 - {\mathrm{t}}_1\right) + \mathrm{b}$

$$\begin{gathered} \mathrm{For} \mathrm{a} \mathrm{suitably} \mathrm{chosen} \mathrm{real} \mathrm{constant} \mathrm{a}, \mathrm{let} \mathrm{a} \mathrm{function}, \mathrm{f} :\mathbf{ }\mathbf{R}\mathbf{ }\mathbf{-}\mathbf{ }\left\{\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{a}\right\}\mathbf{ }\mathbf{\to }\mathbf{ }\mathbf{R}\mathbf{ }\mathrm{be} \mathrm{defined} \mathrm{by} \\ \mathrm{f}\left(\mathrm{x}\right) = \frac{\mathrm{a} - \mathrm{x}}{\mathrm{a} + \mathrm{x}}. \mathrm{Further} \mathrm{suppose} \mathrm{that} \mathrm{for} \mathrm{any} \mathrm{real} \mathrm{number} \mathrm{x} \ne 0 \mathrm{and} \\ \mathrm{f}\left(\mathrm{x}\right) \ne - \mathrm{a}, \left(\mathrm{fof}\right)\left(\mathrm{x}\right) = \mathrm{x}. \mathrm{Then} \mathrm{f}\left( - \frac{1}{2}\right) = ? \end{gathered}$$

• 3

•  - 3

• $\frac{1}{3}$

• $- \frac{1}{3}$