IIT JEE Advanced 2006 Paper 1

We have the cubic polynomial function f(x) = $ax^3 + bx^2$ + 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 = $\frac{19}{4}$ Therefore, c = $-\frac{57}{4}$ and d = −1 – a – b – c = - 1 - $\frac{19}{4}$ - 0 - $\frac{57}{4}$ = - $\frac{4+19+57}{4}$ = - $\frac{80}{4}$ Therefore, f (x) = $\frac{19}{4}x^3 - \frac{57x}{4} - \frac{80}{4}$ = $\frac{1}{4}(19x^3-57x-80)$ and f' (x) = 0 ⇒ $\frac{1}{4}(57x^2-57)$ = 0 ($x^2$ – 1) = 0 ⇒ x = 1, −1 f' (x) = $\frac{57}{4}$ (x - 1) (x + 1) Therefore, f" (x) = $\left(\frac{57}{2}\right)2x$ ⇒ f ′′(1) > 0 The point of minima is x = 1. Therefore, f (1) = $\frac{1}{4}$ (19 - 57 - 80) = $\frac{59}{2}$ = $\frac{-59}{2}$ Point $\left(1,\frac{-59}{2}\right)$ distance from point (−1, 2) is $\sqrt{(1+1)^2\left(\frac{-59}{2}-2\right)^2}$ = $\sqrt{4+\frac{(63)^2}{4}}$ = $\frac{1}{2}\sqrt{63^2+4^2}$ Hence, f(x) is increasing, that is, x ≥ 1.