Manipal Entrance Test 2017 Math Paper

The equation of given curve is y=x4−2x3+x2+3
On differentiating w.r.t. x, we get

dy

dx

=4x3−6x2+2x
Again, differentiating, we get

d2y

dx2

=12x2−12x+2
Put,

dy

dx

=0 for maxima or minima.
⇒2x(2x2−3x+1)=0
⇒x=0,1,

1

2

∴(

d2y

dx2

)x=0=2
∴ Function is minimum at x=0
and (

d2y

dx2

)x=1 =12−12+2=2
∴ Function is minimum at x=1
Also, (

d2y

dx2

)(x=

1

2

)=−1<0
∴ Function is maximum at x=

1

2

∴ Required area =

1

∫

0

(x4−2x3+x2+3)dx
=[

x5

5

−

2x4

4

+

x3

3

+3x]01
=

1

5

−

1

2

+

1

3

+3 =

91

30

sq unit