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CBSE Class 11-science Answered

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Asked by | 28 Sep, 2012, 07:11: AM
answered-by-expert Expert Answer

4 cosec2 A + 9 sin2 A = 4/ sin2 A + 9 sin2 A = (4 + 9 sin4 A)/ sin2 A

Let F(A) = (4 + 9 sin4 A)/ sin2 A

(A) = [(4 × 9 sin3 A. cos A) (sin2 A) - (4 + 9 sin4 A) × 2 cos A. sin A] / (sin2 A)2

 

This on further simplification gives

(A) =  2cos A. (9 sin4 A - 4)/ sin3 A

 

To find the maximum or the minimum value, it is required to consider, 2cos A. (9 sin4 A - 4)/ sin3 A = 0

This gives cos A = 0 or sin2 A = 2/3.

 

The maximum value of cosec2 A is infinite. In that case sin2 A = 0.

So, maximum value of F(A) is infinite.

 

When cos A = 0 i.e. sin2 A = 1, then F(A) = 4 × 1 + 9 × 1 = 13

When sin2 A = 2/3, i.e. cosec2 A = 3/2, then F(A) = 4 × 3/2 + 9 × 2/3 = 12.

Among 12 and 13, the minimum value is 12.

So, the minimum value of F(A) is 12.

Answered by | 28 Sep, 2012, 11:13: AM
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