The cordinates of the turning point of the graph of y=f(x) occurs at (1.1) and (5/3, 23/27)
2
(x−2)+1
Now, differentiating w.r.t. x, we get
f
′
(x)=2(x−1)(x−2)+(x−1)
2
=(x−1)[2(x−2)+x−1]
=(x−1)[3x−5]
Hence,
f
′
(x)=0 implies x=1 and x=
3
5
Corresponding values of y are y=1 and y=
27
23
respectively.
∴ Co-ordinates of turning point of the graph of f(x) occurs at (1,1) and (
3
5
,
27
23
)
Now,
f(x)=(x−1)
2
(x−2)+1
=(x
2
−2x+1)(x−2)+1
=x
3
−2x
2
−2x
2
+4x+x−2+1
=x
3
−4x
2
+5x−1
The value of p for which the equation f(x)=p has 3 distinct solutions lies in interval (
27
23
,1)
Area enclosed by f(x),x=0,y=1 and x=1 is
A=∫
0
1
(1−f(x))dx
⇒∫
0
1
(4x
2
−x
3
−5x+2) dx
⇒A=
12
7\
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