Q. 225.0( 1 Vote )

# Show that height

Let VAB be the cone of height ‘h’ and semi-vertical angle α

Let x be the radius of the base of the cylinder A’B’CD which is inscribed in the right circular cone and H be the height

Then, Height of the cylinder, H = VO – VO’

H = (h – x cot α) …(i)

Now, Volume of a cylinder, V = πx2H

= πx2(h – x cot α)

= πx2h – πx3 cot α

Differentiating with respect to x, we get

…(ii)

For maxima or minima V,

2πxh - 3πx2 cot α = 0

2πxh = 3πx2 cot α

2h = 3x cot α

Now, we again differentiate the eq. (ii), we get

When , we have

V is maximum when

Putting the value of x in eq. (i), we get

Height of the cylinder = of height of cone

Now, the maximum volume of the cylinder is:

V = πx2H

OR

Given:

x [0, 2π]

Step 1: First we find the f’(x)

Differentiate with respect to x, we get

[ cos2x + sin2x = 1]

Step 2: Putting f’(x) = 0

cos x (4 – cos x) = 0

cos x = 0 or 4 – cosx = 0

x = cos-1(0) or cos x = 4

or But -1 ≤ cos x ≤ 1

So, cos x = 4 is not possible

We know that,

Putting n=0

Putting n=1

Putting n=2

since, x (0, 2π)

So, value of x are

Step 3: plotting of value of x

Thus, we divide the interval (0, 2π) into three disjoint intervals

Step 4: now, we find the interval at which the given function is strictly increasing or strictly decreasing:

and

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