381.

f(x) and g(x) are differentiable in the interval [0, 1] such that f(0) = 2, g(0) = 0, f(1) = 6, g(1) = 2, then Rolle's theorem is applicable for which of the following in [0, 1] ?

• f(x) - g(x)

• f(x) - 2g(x)

• f(x) + 3g(x)

• None of the above

The positive root of equation x² - 2x - 5 = 0 lies in the interval

• {0, 1}

• (1, 2)

• (2, 3)

• (3, 4)

One root of the equation x² - 4x+ 1 = 0 is between 1 and 2. The value ofthis root using Newton-Raphson method will be

• 1.775

• 1.850

• 1.875

• 1.950

The point/points of discontinuity of the function f(x) = $$\begin{gathered} \left\{\left|\mathrm{x}\right| + 3, \mathrm{if} \mathrm{x} \le - 3 \\ - 2\mathrm{x}, \mathrm{if} - 3 < \mathrm{x} < 3 \\ 6\mathrm{x} + 2, \mathrm{if} \mathrm{x} \ge 3\right\} \end{gathered}$$ is/are

• 3, - 3

• 3

• - 3

• None of these

The value of $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin \left({\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)\right)\right)$ at x = 0  is

• 0

• $- \sqrt{2}$

• $\frac{- 1}{2\sqrt{2}}$

• $- \frac{1}{\sqrt{2}}$

If f(x) = $\frac{{\mathrm{x}}^2 - 10\mathrm{x} + 25}{{\mathrm{x}}^2 - 7\mathrm{x} + 10}$ and f is continuous at x = 5, then f(5) is equal to

• 0

• 5

• 10

• 25

If h(x) = ${\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}}}$, then at x = 1, $\frac{\mathrm{h}\text{'}\left(\mathrm{x}\right)}{\mathrm{h}\left(\mathrm{x}\right)}$ is equal to

• h(x)

• $\frac{1}{\mathrm{h}\left(\mathrm{x}\right)}$

• $1 + \log \left(\mathrm{h}\left(\mathrm{x}\right)\right)$

• $- \log \left(\mathrm{h}\left(\mathrm{x}\right)\right)$

$\frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^{-1}\left(3\mathrm{x} - 4{\mathrm{x}}^3\right)\right)$ is equal to

• $\frac{3}{\sqrt{4 - {\mathrm{x}}^2}}$

• $\frac{3}{\sqrt{1 - {\mathrm{x}}^2}}$

• $\frac{1}{\sqrt{4 - {\mathrm{x}}^2}}$

• $- \frac{1}{\sqrt{4 - {\mathrm{x}}^2}}$

If f(x) = $\frac{{\mathrm{x}}^2}{\mathrm{x} + \mathrm{a}}$, then f''(a) is equal to

• 4a

• $\frac{1}{8\mathrm{a}}$

• $\frac{1}{4\mathrm{a}}$

• 8a

If u = ${\mathrm{e}}^{{\mathrm{x}}^2 - {\mathrm{y}}^2}$, then

• xu_x = yu_y

• yu_x = xy_y

• yu_x + xu_y = 0

• x²u_y + y²u_x = 0