Question

If a root of the equation n² sin²x - 2sinx - ( 2n + 1 ) = 0 lies in [ 0 , π / 2 ] then minimum positive integer value of n is

Answer 1

see the attachment,
   we see that a is root of any quadratic whose lies between 0 and π/2 
now, we have to find out conditions f(x) = ax² +bx + c , when α is a root lies between k₁ and k₂. e.g., k₁ ≤ α ≤k₂
(1) D> 0 for all real 
(2) af(k₁) ≤ 0
(3)af(k₂)≥ 0

now, here f(sinx) = n²sin²x - 2sinx -(2n+1) , and let α is a root  between o to π/2 .
  hence, D = (2)^2 +4n^2(2n+1) > 0
                 = 4 + 8n^3 + 4n^2 > 0
                 = 2n^3 + n^2 + 1 > 0 
                 = 2n^3 + 2n^2 -n^2 -n +n+1 > 0
                 =2n^2(n+1) -n(n+1)+1(n+1) > 0
                 =(n+1)(2n^2-n+1) > 0
                 = n > -1 [ ∵2n^2 -n +1 be always positive ]----------------(1)
again, 
  af(k₁) < 0 
so, n^2f(sin0) < 0 ⇒ n^2f(0) ≤ 0 
   n^2{0 -0 -(2n+1)} ≤ 0 
    n ≥ -1/2 ---------------------------------------(2)

similarly,
n^2f(sinπ/2) > 0 ⇒n^2f(1) ≥ 0 
  n^2 [n^2 -2 -2n -1 ] ≥ 0
  n^2 (n^2 -2n -3 ) ≥ 0
   (n - 3)(n + 1) ≥ 0 
  n ≥ 3 or, n ≤ -1 ---------------------------(3)

from equation (1) , (2) and (3) 
n ≥ 3 
so, minimum integer value of n = 3 
  
     
             

Answer 2

★ QUADRATIC RESOLUTION ★

Given the function is having a root in the desired range , by putting the actual root in the range and solving the inequality considering the minimum range , we can obtain the minimum positive integral value of " n " ,
considering that the range isn't exclusive , and hence includes the minimum case , so , 3 is included in the answer and it's only the minimum value

Calculated steps are referred to attached above

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