$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\tan }^{-1}\left(\frac{\sqrt{1 + {\mathrm{x}}^2} + \sqrt{1 - {\mathrm{x}}^2}}{\sqrt{1 + {\mathrm{x}}^2} - \sqrt{1 - {\mathrm{x}}^2}}\right)\right]$ is equal to

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

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

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

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

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

• $\mathrm{y}\left[\frac{1}{2\mathrm{x}} + \frac{4}{2\mathrm{x} + 3} - \frac{1}{2\left(\mathrm{x} + 1\right)}\right]$

• $\mathrm{y}\left[\frac{1}{3\mathrm{x}} + \frac{4}{2\mathrm{x} + 3} - \frac{1}{2\left(\mathrm{x} + 1\right)}\right]$

• $\mathrm{y}\left[\frac{1}{3\mathrm{x}} + \frac{4}{2\mathrm{x} + 3} - \frac{1}{\mathrm{x} + 1}\right]$

• None of the above

53.

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

• x²

• y²

• xy²

• xy

The function y = 2x³ - 9x² + 12x - 6 is monotonic decreasing when

• 1 < x < 2

• x > 2

• x < 1

• None of these

Rolle's theorem is not applicable for the function f(x) = $\left|\mathrm{x}\right|$, where x $\in$ [- 1, 1] because

• the function f(x) is not continuous in the interval [- 1, 1]

• the function f(x) is not differentiable in the interval (- 1, 1)

• $\mathrm{f}\left(- 1\right) \ne \mathrm{f}\left(1\right)$

• $\mathrm{f}\left(- 1\right) = \mathrm{f}\left(1\right) \ne 0$

If the function f(x) = x³ - 6ax² + 5x satisfies the conditions of Lagrange's Mean Value theorem for the interval [1, 2] and the tangent to the curve y = f(x) at x = 7/4 is parallel to the chord that join the points ofintersection of the curve with the ordinates x = 1 and x = 2 . Then, the value of a is

• $\frac{35}{16}$

• $\frac{35}{48}$

• $\frac{7}{16}$

• $\frac{5}{16}$

Maximum slope of the curve y = - x³ + 3x² + 9x - 27 is

• 0

• 12

• 16

• 32

If 2a + 3b + 6c = 0, then atleast one root of the equation ax² + bx + c = 0, lies in the interval

• (0, 1)

• (1, 2)

• (2, 3)

• None of these

If the lines $\frac{\mathrm{x} - 1}{- 3} = \frac{\mathrm{y} - 2}{2\mathrm{k}} = \frac{\mathrm{z} - 3}{2}$ and $\frac{\mathrm{x} - 1}{3\mathrm{k}} = \frac{\mathrm{y} - 5}{1} = \frac{\mathrm{z} - 6}{- 5}$ are at right angles, then the value of k will be

• $- \frac{10}{7}$

• $- \frac{7}{10}$

• - 10

• 7

The coordinates of the point where the line $\frac{\mathrm{x} - 6}{- 1} = \frac{\mathrm{y} + 1}{0} = \frac{\mathrm{z} + 3}{4}$ meets the plane x + y - z = 3, are

• (2, 1, 0)

• (7, - 1, 7)

• (1, 2, - 6)

• (5, - 1, 1)