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suppose a is a positive real number such that a to the power of 5 minus a cubed plus a equals 2 then a) a to the power of 6  <2 b)2< a to the power of 6 <3 c)3< a to the power of 6 <4 d)4<=  a to the power of 6
Asked by Girijeshpandey.rjil | 28 Mar, 2019, 06:20: PM
answered-by-expert Expert Answer
f(a) = a5 -a3 +a -2
 
if a =1, f(1) = -1 ;  if a=2, f(2) = 24  ;  hence root of f(a) is between 1 and 2  , i.e.,  1< a <2
 
let us check middle value for root. let a = 3/2,  f(3/2) = (243/32) -(27/8) +(3/2) -2 = 119/32   or   f(a) >3
 
let us check middle value between 1 and 1.5, let a = 1.25 = 5/4,   f(5/4) = (3125/1024) -(125/64) +(5/4) -2  ≈ 0.4
 
hence the root is near to 1.25. Let us consider  a = 1+Δa  , where Δa ≈ 0.25
 
then a6 = (1+Δa)6 ≈ 1+ 6C1 Δa  + 6C2 (Δa)2  ..................................(1)
 
[ powers of Δa3 and higher orders  are neglected in eqn.(1) ]
 
substituting Δa in eqn.(1), we get,  a6 ≈ (1+0.25)6 ≈ 1 + 6×0.25 + 15×(1/16) ≈ 3.4
 
hence  3 < a6 < 4
Answered by Thiyagarajan K | 29 Mar, 2019, 07:56: AM

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