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

• 3

•  - 3

• $\frac{1}{3}$

• $- \frac{1}{3}$

Let f : R $\to$ R be a function defined by f(x) = max {x, x²}. Let S denote the set of all points in R, where f is not differentiable. Then:

• $\mathrm{\varphi }(\mathrm{an} \mathrm{empty} \mathrm{set})$

• $\left\{1\right\}$

• $\left\{0\right\}$

• $\left\{0, 1\right\}$

For all twice differentiable functions f : R $\to$ R, with f(0) = f(1) = f'(0) = 0

• f''(x) = 0, for some x $\in$ (0, 1)

• $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right) = 0, \mathrm{at} \mathrm{every} \mathrm{point} \mathrm{x} \in (0, 1)$

• f''(0) = 0

• $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right) \ne 0, \mathrm{at} \mathrm{every} \mathrm{point} \mathrm{x} \in (0, 1)$

The area (in sq. units) of the region enclosed by the curves y = x² – 1 and y = 1 – x² is equal to :

• $\frac{8}{3}$

• $\frac{7}{2}$

• $\frac{4}{3}$

• $\frac{16}{3}$

$$\begin{gathered} \mathrm{The} \mathrm{probabilities} \mathrm{of} \mathrm{three} \mathrm{events} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{are} \mathrm{given} \mathrm{P}\left(\mathrm{A}\right) = 0.6, \mathrm{P}\left(\mathrm{B}\right) = 0.4 \mathrm{and} \\ \mathrm{P}\left(\mathrm{C}\right) = 0.5. \mathrm{If} \mathrm{P}(\mathrm{A} \cup \mathrm{B}) = 0.8, \mathrm{P}(\mathrm{A} \cap \mathrm{C}) = 0.3, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) = 0.2, \mathrm{P}(\mathrm{B} \cap \mathrm{C})= \mathrm{\beta } \\ \mathrm{and} \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) = \mathrm{\alpha }, \mathrm{where} 0.85 \le \mathrm{x} \le 0.95, \mathrm{then} \mathrm{\beta } \mathrm{lies} \mathrm{in} \mathrm{the} \mathrm{interval} : \end{gathered}$$

• $\left[0.36, 0.40\right]$

• $\left[0.35, 0.36\right]$

• $\left[0.25, 0.35\right]$

• $\left[0.20, 0.25\right]$

$\mathrm{The} \mathrm{integral} {\int }_1^2{\mathrm{e}}^{\mathrm{x}} . {\mathrm{x}}^2\left(2 + {\log }_{\mathrm{e}}\left(\mathrm{x}\right)\right)d\mathrm{x} = ?$

• e(2e - 1)

• e(4e + 1)

• 4e² - 1

• e(4e - 1)

The common difference of the A.P. b₁,b₂,....,bm is 2 more than common difference of A.P. a₁,a₂,...,a_n. If a₄₀ = –159, a₁₀₀ = – 399 and b₁₀₀ = a₇₀, then b₁is equal to : 

• 127

• 81

• - 127

• - 81

$$\begin{gathered} \mathrm{If} \mathrm{y} = \left(\frac{2}{\mathrm{\pi }}\mathrm{x} - 1\right)\csc \left(\mathrm{x}\right) \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{differential} \mathrm{equation}, \\ \frac{\mathrm{dy}}{\mathrm{dx}} + \mathrm{p}\left(\mathrm{x}\right)\mathrm{y} = \frac{2}{\mathrm{\pi }}\csc \left(\mathrm{x}\right), 0 < \mathrm{x} < \frac{\mathrm{\pi }}{2} \mathrm{then} \mathrm{the} \mathrm{function} \mathrm{p}\left(\mathrm{x}\right) \mathrm{is} \mathrm{equal} \mathrm{to}: \end{gathered}$$

• $\tan \left(\mathrm{x}\right)$

• $\csc \left(\mathrm{x}\right)$

• $\cot \left(\mathrm{x}\right)$

• $\sec \left(\mathrm{x}\right)$

19.

Let z = x + iy be a non-zero complex number such that z² = i |z|², where i = $\sqrt{ - 1}$, then z lies on the:

• real axis

• line y = x

• line y = - x

• imaginary axis

A plane P meets the coordinate axes at A, B and C respectively. The centroid of $∆$ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is:

• $\frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} - 1}{1} = \frac{\mathrm{z} - 2}{1}$

• $\frac{\mathrm{x} - 1}{1} = \frac{\mathrm{y} - 1}{2} = \frac{\mathrm{z} - 2}{2}$

• $\frac{\mathrm{x} - 1}{1} = \frac{\mathrm{y} - 1}{1} = \frac{\mathrm{z} - 2}{2}$

• $\frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} - 1}{2} = \frac{\mathrm{z} - 2}{1}$