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Find the value of a for which 4cosec2 ( π (a + x)) +a2 - 4a = 0 has a real solution.
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= 2Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Nirabhra!!!
Regards,Askiitians ExpertsPriyansh Bajaj
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
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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