Question

f(x) is cubic polynomial with f(2) = 18 and f(1) = -1. Also f(x) has local maxima at x = -1 and f '(x) has local minima at x = 0, then

• A. the distance between (-1, 2) and (a f(a)), where x = a is the point of local minima is $2 \sqrt{5}$
• B. f(x) is increasing for $x \in [1, 2 \sqrt{ 5} ]$
• C. f(x) has local minima at x = 1
• D. the value of f(0) = 15

Answer 1

Answer: (B), (C)

Let $f\left(x\right) =ax^{3} +bx^{2}+cx +d$
Then, $ f\left(2\right)=18 \Rightarrow 8a + 4b +2c +d =18$ ....(1)
$ f\left(1\right) = -1 \Rightarrow a+b+c+d =-1$ ....(2)
$ f\left(x\right) $ has local max. at $x=-1 $
$\Rightarrow 3a-2b+c=0 f'\left(x\right) x=0 \Rightarrow b=0 $ ......(4)
Solving (1), (2), (3) and (4), we get
$f\left(x\right) = \frac{1}{4} \left(19x^{3} -57 x +34\right)\Rightarrow f\left(0\right) = \frac{17}{2} $
Also $f'\left(x\right) = \frac{57}{4}\left(x^{2}-1\right)0, \forall x>1 $
Also $f'\left(x\right) =0 \Rightarrow x=1, -1$
$ f" \left(-1\right)<0 , f"\left(1\right) >0 \Rightarrow x = -1 $ is a point of local max. and x = 1 is a point of local min. Distance between (- 1, 2) and (1, f (1)), i.e. (1, -1) is $\sqrt{13} \neq 2 \sqrt{5}$