Answer :


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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