21.

All the points on the X-axis have

• x = 0

• y = 0

• x = 0, y = 0

• y = 0, z = 0

If xy = e^{y - x}, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ =

• $\frac{\mathrm{y}\left(\mathrm{x} - 1\right)}{\mathrm{x}\left(\mathrm{y} + 1\right)}$

• $\frac{\mathrm{y}\left(\mathrm{x} + 1\right)}{\mathrm{x}\left(\mathrm{y} - 1\right)}$

• $\frac{\mathrm{x} + 1}{\mathrm{y} - 1}$

• $\frac{\mathrm{x} - 1}{\mathrm{y} + 1}$

If y = ${\cos }^{-1}\left(\frac{\sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right)}{\sqrt{2}}\right), - \frac{\mathrm{\pi }}{4} < \mathrm{x} < \frac{\mathrm{\pi }}{4}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ =

• - 1

• 1

• 0

• None of these

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

If $\left[\begin{array}{ccc}0& 2\mathrm{\beta }& \mathrm{\gamma }\\ \mathrm{\alpha }& \mathrm{\beta }& - \mathrm{\gamma }\\ \mathrm{\alpha }& - \mathrm{\beta }& \mathrm{\gamma }\end{array}\right]$ is orthagonal, then the values of $\mathrm{\alpha }, \mathrm{\beta } \mathrm{and} \mathrm{\gamma }$ will be

• $\mathrm{\alpha } = \pm \frac{1}{\sqrt{5}}, \mathrm{\beta } = \pm \frac{1}{\sqrt{3}}, \mathrm{\gamma } = \pm \frac{1}{\sqrt{2}}$

• $\mathrm{\alpha } = \pm \frac{1}{\sqrt{6}}, \mathrm{\beta } = \pm \frac{1}{\sqrt{7}}, \mathrm{\gamma } = \pm \frac{1}{\sqrt{3}}$

• $\mathrm{\alpha } = \pm \frac{1}{\sqrt{3}}, \mathrm{\beta } = \pm \frac{1}{\sqrt{3}}, \mathrm{\gamma } = \pm \frac{1}{\sqrt{3}}$

• $\mathrm{\alpha } = \pm \frac{1}{\sqrt{2}}, \mathrm{\beta } = \pm \frac{1}{\sqrt{6}}, \mathrm{\gamma } = \pm \frac{1}{\sqrt{3}}$

The value of $\sin \left(2{\tan }^{-1}\left(\frac{1}{3}\right)\right) + \cos \left({\tan }^{-1}\left(2\sqrt{2}\right)\right)$ will be

• $\frac{13}{12}$

• $\frac{14}{15}$

• $- \frac{14}{15}$

• $\frac{12}{13}$

The area of the triangle formed by the points (a, b + c), (b, c + a) and (c, a + b) will be

• abc

• a² + b² + c²

• ab + bc + ca

• 0

Let f : R $\to$ R be the function defined by f(x) = $\sqrt{\mathrm{x} - 3} \forall , \mathrm{x} \in \mathrm{R}$. Then f^-1(x) = ?

• x + 3

• x² + 3

• $\frac{\mathrm{x} + 3}{2}$

• $\frac{{\mathrm{x}}^2 + 3}{2}$

${\tan }^{-1}\left(\frac{\mathrm{x} - 2}{\mathrm{x} - 3}\right) + {\tan }^{-1}\left(\frac{\mathrm{x} + 2}{\mathrm{x} + 3}\right) = \frac{\mathrm{\pi }}{4}$, then x = ?

• $\pm \frac{1}{\sqrt{2}}$

• $\pm \sqrt{\frac{7}{2}}$

• $\pm \sqrt{\frac{2}{7}}$

• None of these

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