Find C of Lagrange's mean value theorem for the function f(x) = 3x² + 5x + 7 in the interval [1, 3].

• $\frac{7}{3}$

• 2

• $\frac{3}{2}$

• $\frac{4}{3}$

For f(x) = (x - 1)^2/3, the mean value theorem is applicable to f(x) in the interval

• [2, 4]

• [0, 2]

• [- 2, 2]

• any finite interval

The differential coefficient of f(log(x)) with respect to x, where f(x) = log(x) is

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

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

• ${\left(\mathrm{xlog}\left(\mathrm{x}\right)\right)}^{- 1}$

• None of these

If f(x) = $$\begin{gathered} \left[\left[\mathrm{x}\right] + \left[- \mathrm{x}\right], \mathrm{x} \ne 2 \\ \mathrm{\lambda }, \mathrm{x} = 2\right] \end{gathered}$$, then f is continuous at x = 2, provided $\mathrm{\lambda }$ is equal to

• 1

• 0

• - 1

• 2

985.

The derivative of ${\sec }^{-1}\left(\frac{1}{2{\mathrm{x}}^2 - 1}\right)$ with respect to $\sqrt{1 - {\mathrm{x}}^2}$ at x = 1/2 will be

• 1/4

• ${\sec }^{-1}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)$

• 4

• 0

If $\mathrm{x} = {\mathrm{acos}}^3\left(\mathrm{\theta }\right), \mathrm{y} = {\mathrm{asin}}^3\left(\mathrm{\theta }\right), \mathrm{then} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}$ is equal to

• ${\tan }^2\left(\mathrm{\theta }\right)$

• $\sec \left(\mathrm{\theta }\right)$

• ${\sec }^2\left(\mathrm{\theta }\right)$

• $\left|\sec \left(\mathrm{\theta }\right)\right|$

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

• $\frac{1}{1 + {\mathrm{x}}^2}$

• $\frac{- 1}{1 + {\mathrm{x}}^2}$

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

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

The value of $\sqrt{12}$. upto three places of decimals using the method of Newton-Raphson, will be

• 3.463

• 3.462

• 3.467

• None of these

If $\mathrm{f}\left(\mathrm{x}\right) = \log \left(\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right) \mathrm{and} \mathrm{g}\left(\mathrm{x}\right) = \frac{3\mathrm{x} +\hspace{0.17em}{\mathrm{x}}^3}{1 + 3{\mathrm{x}}^2}$, then fog(x) is equal to

• - f(x)

• 3f(x)

• [f(x)]³

• None of these

Let f(x) be differentiable on the interval (0, $\infty$) such that f(1) =1 and $\underset{\mathrm{t}\to \infty }{\lim }\frac{{\mathrm{t}}^2\mathrm{f}\left(\mathrm{x}\right) - {\mathrm{x}}^2\mathrm{f}\left(\mathrm{t}\right)}{\mathrm{t} - \mathrm{x}} = 1$ for each x > 0. Then, f(x) is equal to

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

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

• $- \frac{1}{\mathrm{x}}$

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