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Find the value of a for which 4cosec 2 ( π (a + x)) +a 2 - 4a = 0 has a real solution.


Find the value of a for which

4cosec2  ( π (a + x)) +a2 - 4a = 0 has a real solution.

Grade:12

3 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 years ago

Dear Nirabhra,

Solution:- 4cosec2(n(a+x)) + a2 - 4a =0

[4cosec2(n(a+x)) -4] + (a2 - 4a + 4) =0

4[cosec2(n(a+x)) -1] + (a-2)2 =0

[2cot(n(a+x))]2 + (a-2)2 =0 which means a =2 and cot(n(a+x)) =0

Ans: a= 2

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Priyansh Bajaj

kirankumar kishor
31 Points
10 years ago

 

Dear Nirabhra,

Solution:- 4cosec2(n(a+x)) + a2 - 4a =0

[4cosec2(n(a+x)) -4] + (a2 - 4a + 4) =0

4[cosec2(n(a+x)) -1] + (a-2)2 =0

[2cot(n(a+x))]2 + (a-2)2 =0 which means a =2 and cot(n(a+x)) =0

AA
12 Points
2 years ago
 - 4a = 02 (π (a+x) ) + a24 cosec 

4 cosec 2 (π (a+x) ) =- a2 + 4a 

L .H .S is always greater than 4 so for real solution

 -a2 + 4a ≥4

or (a-2)2≤0

so only a=2 is possible

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