If the function f(x) = ${\left[\tan \left(\frac{\mathrm{\pi }}{4} + \mathrm{x}\right)\right]}^{\frac{1}{\mathrm{x}}}$ for $\mathrm{x} \ne 0$ is  = K for x = 0 continuous at x = 0, then K = ?

• e

• e^{- 1}

• e²

• e^{- 2}

If x = f(t) and y = g(t) are differentiable functions of t, then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ is

• $\frac{\mathrm{f}\text{'}\left(\mathrm{t}\right) . \mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{g}\text{'}\left(\mathrm{t}\right) . \mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)}{{\left[\mathrm{f}\text{'}\left(\mathrm{t}\right)\right]}^3}$

• $\frac{\mathrm{f}\text{'}\left(\mathrm{t}\right) . \mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{g}\text{'}\left(\mathrm{t}\right) . \mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)}{{\left[\mathrm{f}\text{'}\left(\mathrm{t}\right)\right]}^2}$

• $\frac{\mathrm{g}\text{'}\left(\mathrm{t}\right) . \mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{f}\text{'}\left(\mathrm{t}\right) . \mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right)}{{\left[\mathrm{f}\text{'}\left(\mathrm{t}\right)\right]}^3}$

• $\frac{\mathrm{g}\text{'}\left(\mathrm{t}\right) . \mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right) + \mathrm{f}\text{'}\left(\mathrm{t}\right) . \mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right)}{{\left[\mathrm{f}\text{'}\left(\mathrm{t}\right)\right]}^3}$

If f(x) = $$\begin{gathered} \left\{= \log {\left({\sec }^2\left(\mathrm{x}\right)\right)}^{{\cot }^2\left(\mathrm{x}\right)} \mathrm{for} \mathrm{x} \ne 0 \\ = \mathrm{K} \mathrm{for} \mathrm{x} = 0\right\} \end{gathered}$$ is continuous at x = 0, then K is

• e^{- 1}

• 1

• e

• 0

If $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) \left\{= \mathrm{x} \mathrm{for} \mathrm{x} \le 0 \\ = 0 \mathrm{for} \mathrm{x} > 0\right\} \end{gathered}$$, then f(x) at x = 0 is

• continuous but not differentiable

• not continuous but differentiable

• continuous and differentiable

• not continuous and not differentiable

Let f(x) = $$\begin{gathered} \left\{- 2\sin \left(\mathrm{x}\right) \mathrm{if} \mathrm{x} \le - \frac{\mathrm{\pi }}{2} \\ \mathrm{Asin}\left(\mathrm{x}\right) + \mathrm{B} \mathrm{if} - \frac{\mathrm{\pi }}{2} < \mathrm{x} < \frac{\mathrm{\pi }}{2} \\ \cos \left(\mathrm{x}\right) \mathrm{if} \mathrm{x} \ge \frac{\mathrm{\pi }}{2}\right\} \end{gathered}$$

For what values of A and B, the function f(x) is continuous throughout the real line ?

• A = - 1, B = 1

• A = - 1, B = - 1

• A = 1, B = - 1

• A = 1, B = 1

Let f(x) = $$\begin{gathered} \left\{\mathrm{\alpha }\left(\mathrm{x}\right)\sin \left(\frac{\mathrm{\pi x}}{2}\right) \mathrm{for} \mathrm{x} \ne 2 \\ 1 \mathrm{for} \mathrm{x} = 0\right\} \end{gathered}$$

where $\mathrm{\alpha }\left(\mathrm{x}\right)$ is such that $\underset{\mathrm{x}\to 0}{\lim }\left|\mathrm{\alpha }\left(\mathrm{x}\right)\right| = \infty$.

Then the function f(x) is contonuous at x = 0 if $\mathrm{\alpha }\left(\mathrm{x}\right)$ is chosen as

• $\frac{2}{\mathrm{\pi x}}$

• $\frac{1}{{\mathrm{x}}^2}$

• $\frac{2}{{\mathrm{\pi x}}^2}$

   

• $\frac{1}{\mathrm{x}}$

Let the equation ofa curve is given in implicit form as y = $\tan \left(\mathrm{x} + \mathrm{y}\right)$. Then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ in terms of y is

• $\frac{2\left(1 + {\mathrm{y}}^2\right)}{{\mathrm{y}}^6}$

• $\frac{- 2\left(1 + {\mathrm{y}}^2\right)}{{\mathrm{y}}^6}$

• $\frac{- 2\left(1 + {\mathrm{y}}^2\right)}{{\mathrm{y}}^5}$

• $\frac{2{\left(1 + {\mathrm{y}}^2\right)}^2}{{\mathrm{y}}^5}$

The function f(x) = ${\mathrm{xtan}}^{-1}\left(\frac{1}{\mathrm{x}}\right) \mathrm{for} \mathrm{x} \ne 0$, f(0) = 0 is

• differentiable at x = 0

• neither continuous at x = 0 nor differentiable at x = 0

• not continuous at x = 0

• continuous at x = 0 but not differentiable at x = 0

If function f(x) = $$\begin{gathered} \left\{\mathrm{xsin}\left(\frac{1}{\mathrm{x}}\right); \mathrm{x} \ne 0 \\ \mathrm{a}; \mathrm{x} = 0\right\} \end{gathered}$$ is continuous at x = 0, then the value of a is

• 0

• 1

• - 1

• None of these

230.

At which point the function f(x) = $\frac{{\mathrm{x}}^2}{\left[\mathrm{x}\right]}$, where [.] is greatest integer function, is discontinuous ?

• Only positve integers

• All postive and negative integers and (0,1)

• all rational numbers

• None of these