UPSEE 2017 Solved Paper

Given,
y=

x

∫

0

(t2−3t+2)dt ......(i)
On differentiating w.r.t. x′, we get

dy

dx

=x2−3x+2 ........(ii)
Again, on differentiating w.r.t. 'x', we get

d2y

dx2

=2x−3 .......(iii)
We know that, at point of inflection

d2y

dx2

=0
∴ From Eq. (iii), we get
2x−3=0⇒x=

3

2

Now, we have to check behaviour of

d2y

dx2

at point x=

3

2

x=

3

2

.
Clearly, at x=

3

2

sign at

d2y

dx2

changes
∴(

3

2

,

3

4

) is point of inflection.