Let y = $\left(\frac{3^{\mathrm{x}} - 1}{3^{\mathrm{x}} + 1}\right)\sin \left(\mathrm{x}\right) + {\log }_{\mathrm{e}}\left(1 + \mathrm{x}\right), \mathrm{x} > - 1$. Then, at x = 0, $\frac{\mathrm{dy}}{\mathrm{dx}}$ equals

• 1

• 0

• - 1

• - 2

672.

If f is a real-valued differentiable function such that f(x)f' (x) < 0 for all real x, then

• f(x) must be an increasing function

• f(x) must be a decreasing function

• $\left|\mathrm{f}\left(\mathrm{x}\right)\right|$ must be an increasing function

• $\left|\mathrm{f}\left(\mathrm{x}\right)\right|$ must be a decreasing function

Rolle's theorem is applicable in the interval [- 2, 2] for the function

• f(x) = x³

• f(x) = 4x⁴

• f(x) = 2x³ + 3

• f(x) = $\mathrm{\pi }\left|\mathrm{x}\right|$

The solution of $25\frac{d^2\mathrm{y}}{d{\mathrm{x}}^2} - 10\frac{d\mathrm{y}}{d\mathrm{x}} + \mathrm{y} = 0$, y(0) = 1, y(1) = $2{\mathrm{e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ is

• $\mathrm{y} = {\mathrm{e}}^{5\mathrm{x}} + {\mathrm{e}}^{- 5\mathrm{x}}$

• $\mathrm{y} = \left(1 + \mathrm{x}\right){\mathrm{e}}^{5\mathrm{x}}$

• $\mathrm{y} = \left(1 + x\right){\mathrm{e}}^{\raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

• $\mathrm{y} = \left(1 + \mathrm{x}\right){\mathrm{e}}^{\raisebox{1ex}{$- \mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

If f(x) and g(x) are twice differentiable functions on (0, 3) satisfying f"(x) = g''(c), f'(1) = 4g'(D) = 6, f(2) = 3, g(2) = 9, then f(1) - g(1) is

• 4

• - 4

• 0

• - 2

For function $\mathrm{f}\left(\mathrm{x}\right) = {\mathrm{e}}^{\cos \left(\mathrm{x}\right)}$, Rolle's theorem is

• applicable, when $\frac{\mathrm{\pi }}{2} \le \mathrm{x} \le \frac{3\mathrm{\pi }}{2}$

• applicable, when $0 \le \mathrm{x} \le \frac{\mathrm{\pi }}{2}$

• applicable, when $0 \le \mathrm{x} \le \mathrm{\pi }$

• applicable, when $\frac{\mathrm{\pi }}{4} \le \mathrm{x} \le \frac{\mathrm{\pi }}{2}$

f(x) = $$\begin{gathered} \left\{0, \mathrm{x} = 0 \\ \mathrm{x} - 3, \mathrm{x} > 0\right\} \end{gathered}$$ the function f(x) is

• increasing when x $\ge 0$

• strictly increasing when x > 0

• strictly increasing at x = 0

• not continuous at x = 0 and so it is not increasing when x > 0

The function f(ax) = ax + b is strictly increasing for all real x, if

• a > 0

• a < 0

• a = 0

• a $\le$ 0

If y = 2x³ - 2x² + 3x - 5, then for x = 2 and $∆$x = 0.1, value of $∆$y is

• 2.002

• 1.9

• 0

• 0.9

If the function

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{\frac{{\mathrm{x}}^2 - \left(\mathrm{A} + 2\right)\mathrm{x} + \mathrm{A}}{\mathrm{x} - 2}, \mathrm{for} \mathrm{x} \ne 2 \\ 2, \mathrm{for} \mathrm{x} = 2\right\} \end{gathered}$$

is continuous at x = 2, then

• A = 0

• A = 1

• A = - 1

• A = 2