IIT JEE Advanced 2006 Paper 1

We have the cubic polynomial function
f(x) = ax3+bx2 + cx + d
f(1) = −1 = a + b + c + d
f(2) = 18 = 8a + 4b + 2c + d
19 = 7a + 3b + c (1)
Therefore,
f ′(x) = 3ax2 + 2bx + c
f ′(-1) = 0
3a – 2b + c = 0
3a + c = 0
c = −3a (2)
and f ′′(x) = 6ax + 2b
f ′′(0) = 0
b = 0 (3)
Substituting Eqs. (2) and (3) in Eq. (1), we get
19 = 7a – 3a
⇒ a =

19

4

Therefore, c = −

57

4

and d = −1 – a – b – c
= - 1 -

19

4

- 0 -

57

4

= -

(4+19+57)

4

= -

80

4

Therefore, f (x) =

19

4

x3−

57x

4

−

80

4

=

1

4

(19x3−57x−80)
and f' (x) = 0 ⇒

1

4

(57x2−57) = 0
(x2 – 1) = 0 ⇒ x = 1, −1
f' (x) =

57

4

(x - 1) (x + 1)
Therefore, f" (x) = (

57

2

)2x
⇒ f ′′(1) > 0
The point of minima is x = 1. Therefore,
f (1) =

1

4

(19 - 57 - 80) =

59

2

=

−59

2

Point (1,

−59

2

) distance from point (−1, 2) is
√(1+1)2(

−59

2

−2)2 = √4+

(63)2

4

=

1

2

√632+42
Hence, f(x) is increasing, that is, x ≥ 1.