NDA Math Functions

f(x)=‌‌

27

4

(x

2

3

−x)
=‌‌

27

4

((x2)

1

3

−x)
For f(x)=1
‌‌

27

4

((x2)

1

3

−x)=1
(x

1

3

)2−x=‌‌

4

27

Put x

1

3

=y
x=y3
y2−y3=

4

27

⇒y3−y2+‌‌

4

27

=0
⇒27y3−27y2+4=0
This is a cubic equation
If we put y=−

1

3

then (3y+1)=0 is a factor of cubic equation
    27y3+9y2
    −‌‌‌‌‌−
   –––––––––––––––
      −36y2+4
      −36y2−12y
      +‌‌‌‌‌‌‌+
   ––––––––––––––––
       12y+4
       12y+4
       −‌‌‌‌−
    ––––––––––––––––
        ×
    –––––––––––––––––
(3y+1)(9y2−12y+4)=0
(3y+1)(3y−2)2=0
Hence y=−

1

3

,

2

3

Thus f(x)=1 has two solution