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The solution set of (5+4 )(2 +1)=0 in the interval [0,2 ] , is :

Options

  1. A3 , 2 3
  2. B3 ,
  3. C2 3 , 4 3
  4. D2 3 , 5 3

Correct answer

C. 2 3 , 4 3

Step-by-step solution

We have, (5+4 )(2 +1)=0 ...(i) = 1- ^2 2 1+ ^2 2 = 1-t^2 1+t^2 [ . put . 2 =t ] Then, Eq. (i) becomes [5+4 ( 1-t^2 1+t^2 ) ] [2 ( 1-t^2 1+t^2 )+1 ]=0 [5+5 t^2+4-4 t^2 ] [2-2 t^2+1+t^2 ]=0 (t^2+9 ) (3-t^2 )=0 t= 3 2 = 3 or 2 =- 3 2 = 3 or 2 = 2 3 = 2 3 or 4 3

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