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6 votes
6 votes

The equation

    $\frac{1}{3}+\frac{1}{2}s^{2}+\frac{1}{6}s^{3}=s$

has

  1. exactly three solution in $[0.1]$
  2. exactly one solution in $[0,1]$
  3. exactly two solution in $[0,1]$
  4. no solution in $[0,1]$

2 Answers

Best answer
10 votes
10 votes

$S^{3}+3S^{2}-6S+2=0$
$(S-1)(S^2+4S-2)=0$

$S_{1}=1 \hspace{1 cm} S_{2},S_{3}=\dfrac{-4\pm \sqrt{16+8}}{2}$

$\hspace{3.6 cm}=-2\pm \sqrt{6}$

$\hspace{3cm}S_{2}=-2 - \sqrt{6}\\ \hspace{3.5 cm}=-4.4$ $\hspace{2 cm}S_{3}=-2+\sqrt{6}\\\hspace{2.5 cm} =.44$

$S_{1},S_{3}\in [0,1]$ Hence,exactly two solutions.

c is ans.

edited by
2 votes
2 votes
It can also be solved using Intermediate value theorem

s^ 3 + 3* s^ 2 -6*s + 2 =0

s=0:   +ve

s=1     0 +ve

s=0.9    -ve value

Thus from 0 to 0.9 , there is transition from +ve to -ve , here one root exist

From 0.9 to 1 , again transition from -ve to +ve , here root exist which is 1.

(C) two solution
Answer:

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