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

Angle between y² = x and x² = y at the origin is

• 2 tan^-1(3/4)

• tan^-1(4/3)

• π/2

• π/4

The distance covered by a particle in t seconds is given by x = 3 + 8t - 4t². After 1 s its velocity will be

• 0 unit/s

• 3 unit/s

• 4 unit/s

• 7 unit/s

479.

If the tangent to the parabola y = x(2 - x) at the point (1, 1) intersects the parabola at P. Find the coordinate of P.

The maximum value of xy subject to x + y = 8

• 8

• 16

• 20

• 24