The minimum value of the function f (x) = $2\left|\mathrm{x} - 1\right| + \left|\mathrm{x} - 2\right|$ is

• 0

• 1

• 2

• 3

Maximum value of the function f(x) = $\frac{\mathrm{x}}{8} + \frac{2}{\mathrm{x}}$ on the interval [1, 6] is

• 1

• $\frac{9}{8}$

• $\frac{13}{12}$

• $\frac{17}{8}$

For  $- \frac{\mathrm{\pi }}{2} < \mathrm{x} < \frac{3\mathrm{\pi }}{2}$, the avlue of $\frac{\mathrm{d}}{\mathrm{dx}}\left({\tan }^{-1}\left(\frac{\cos \left(\mathrm{x}\right)}{1 + \sin \left(\mathrm{x}\right)}\right)\right)$ is equal to

• $\frac{1}{2}$

• - $\frac{1}{2}$

• 1

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

If the area of a rectangle is 64 sq unit, find the minimum value possible for its perimeter

Let f(x) = x³e^{- 3x}, x > 0. Then, the maximum value of f(x) is

• e^{- 3}

• 3e^{- 3}

• 27e^{- 9}

• $\infty$

The point in the interval [0, 2$\mathrm{\pi }$], where f(x) = e^x sin(x) has maximum slope, is

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

• $\mathrm{\pi }$

• $\frac{3\mathrm{\pi }}{2}$

The minimum value of f(x) = ${\mathrm{e}}^{\left({\mathrm{x}}^4 - {\mathrm{x}}^3 + {\mathrm{x}}^2\right)}$

• e

• - e

• 1

• - 1

If the line ax + by + c = 0 is a tangent to the curve xy = 4, then 

• a < 0, b > 0

• $\mathrm{a} \le 0, \mathrm{b} > 0$

• a < 0, b < 0

• $\mathrm{a} \le 0, \mathrm{b} < 0$

If the normal to the curve y = f(x) at the point (3, 4) make an angle 3$\mathrm{\pi }$/4 with the positive x-axis, then f'(3) is

• 1

• - 1

• - $\frac{3}{4}$

• $\frac{3}{4}$

The equation of normal of x² + y² - 2x + 4y - 5 = 0 at (2, 1) is

• y = 3x - 5

• 2y = 3x - 4

• y = 3x + 4

• y = x + 1