$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \frac{\mathrm{x}}{1 + \left|\mathrm{x}\right|}, \mathrm{for} \mathrm{x} \in \mathrm{R}, \mathrm{then} \mathrm{f}\text{'}\left(0\right) \mathrm{is} \mathrm{equal} \mathrm{to} :$

• 0

• 1

• 2

• 3

If u(x, y) = ylog(x) + xlog(y), then

u_xu_y - u_xlog(x) - u_ylog(y) + log(x)log(y) is equal to

• 0

• - 1

• 1

• 2

$\mathrm{If} \mathrm{y} = {\sin }^{-1} \mathrm{x}, \mathrm{then} \left(1 - {\mathrm{x}}^2\right)\frac{{\mathrm{dy}}^2}{{\mathrm{dx}}^2} \mathrm{is} \mathrm{equal} \mathrm{to}$

• $- \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}$

• 0

• $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}$

• $\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2$

$$\begin{gathered} \mathrm{If} \mathrm{f} : \mathrm{R} \to \mathrm{R} \mathrm{is} \mathrm{defined} \mathrm{by} \\ \mathrm{f}\left(\mathrm{x}\right) = \left\{\frac{\mathrm{x} - 2}{{\mathrm{x}}^2 - 3\mathrm{x} + 2 } \mathrm{if} \mathrm{x} \in \mathrm{R} - \left\{1, 2\right\} \\ 2 \mathrm{if} \mathrm{x} = 1 \\ 1 \mathrm{if} \mathrm{x} = 2\right\} \mathrm{then} \\ \underset{\mathrm{x}\to 2}{\lim } \frac{\mathrm{f}\left(\mathrm{x}\right) - \mathrm{f}\left(2\right)}{\mathrm{x} - 2} = \end{gathered}$$

• 0

• - 1

• 1

• - $\frac{1}{2}$

$$\begin{gathered} \mathrm{If} \mathrm{f} :\hspace{0.17em}\mathrm{R} \to \mathrm{R} \mathrm{is} \mathrm{defined} \mathrm{by} \\ \mathrm{f}\left(\mathrm{x}\right) = \left\{\frac{\mathrm{x} - 2}{{\mathrm{x}}^2 + 3\mathrm{x} + 2 } \mathrm{if} \mathrm{x} \in \mathrm{R} - \left\{- 1, - 2\right\} \\ - 1 \mathrm{if} \mathrm{x} = - 2 \\ 0 \mathrm{if} \mathrm{x} = - 1\right\} \\ \mathrm{Then} \mathrm{f} \mathrm{is} \mathrm{contineous} \mathrm{on} \mathrm{the} \mathrm{set} \end{gathered}$$

• R

• $\mathrm{R} - \left\{- 2\right\}$

• $\mathrm{R} - \left\{- 1\right\}$

• $\mathrm{R} - \left\{- 1, 2\right\}$

If f : R $\to$ R is an even function which is twice differentiable on R and f''($\mathrm{\pi }$) = 1, then f''(- $\mathrm{\pi }$) is equal to

• - 1

• 0

• 1

• 2

Observe the following statements

I. f(x) = ${\mathrm{ax}}^{41} +\hspace{0.17em}{\mathrm{bx}}^{- 40} \Rightarrow \frac{\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)} = 1640{\mathrm{x}}^{- 2}$

II. $\frac{\mathrm{d}}{\mathrm{dx}}{\tan }^{-1}\left(\frac{2\mathrm{x}}{1 - {\mathrm{x}}^2}\right) = \frac{1}{1 + {\mathrm{x}}^2}$

which of the following is correct ?

• I is true, but II is false

• Both I and II are true

• Neither I nor II is true

• I is false, but II is true

If f(x) = $10\cos \left(\mathrm{x}\right) + \left(13 + 2\mathrm{x}\right)\sin \left(\mathrm{x}\right), \mathrm{then} \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right) + \mathrm{f}\left(\mathrm{x}\right)$ is equal to

• cos(x)

• 4cos(x)

• sin(x)

• 4sin(x)

If $\mathrm{x}\sqrt{1 + \mathrm{y}} + \mathrm{y}\sqrt{1 +\hspace{0.17em}\mathrm{x}}= 0, \mathrm{then} \frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• $\frac{1}{{\left(1 +\hspace{0.17em}\mathrm{x}\right)}^2}$

• $- \frac{1}{{\left(1 +\hspace{0.17em}\mathrm{x}\right)}^2}$

• $\frac{1}{1 +\hspace{0.17em}{\mathrm{x}}^2}$

• $\frac{1}{1 -\hspace{0.17em}{\mathrm{x}}^2}$

410.

If $\mathrm{u} = {\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) + {\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$, then the value of $\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}} + \mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}$ is

• 0

• 1

• 2

• None of these