Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is Also find the maximum volume in terms of volume of the sphere.

OR

Find the intervals in which is strictly increasing or strictly decreasing.

Given: A sphere of fixed radius r

To prove: the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is

To find: maximum volume in terms of volume of the sphere

Let R and h be the radius and height of cone respectively

Formula used:

The volume of the cone is given by,

B is center of the sphere

Using Pythagoras theorem in right triangle Δ BCD,

AB is also the radius of the sphere

∴ AC = AB + BC

Differentiating both sides with respect to R:

To find the maximum volume, put

Squaring both sides:

If

Differentiating both sides again:

On putting

Therefore, the volume is maximum when

Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is

As the radius of sphere = r

Therefore, the Volume of cone times of volume of the sphere

OR

Given: f(x) = sin 3x – cos 3x, 0 < x < π

To find: intervals in which function f(x) is strictly increasing or decreasing

f(x) = sin 3x – cos 3x

Differentiating with respect to x:

{∵ sin A cos B + cos A sin B = sin (A + B)}

For interval where f(x) is increasing: f’(x) > 0

We know,

sin x > 0

0 < sin x < π and 2π < sin x < 3π

Therefore,

Therefore, f(x) is strictly increasing in

For interval where f(x) is decreasing: f’(x) < 0

We know,

sin x < 0

π < sin x < 2π and 3π < sin x < 4π

Therefore,

Therefore, f(x) is strictly decreasing in