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 $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
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 $x=-1$
$\Rightarrow 3a-2b+c=0f^{\prime }\left(x\right)x=0\Rightarrow b=0$ ......(4)
Solving (1), (2), (3) and (4), we get
$f\left(x\right)=\frac{1}{4}\left(19x^3-57x+34\right)\Rightarrow f\left(0\right)=\frac{17}{2}$
Also $f^{\prime }\left(x\right)=\frac{57}{4}\left(x^2-1\right)0,\forall x>1$
Also $f^{\prime }\left(x\right)=0\Rightarrow x=1,-1$
$f"\left(-1\right)<0,f"\left(1\right)>0\Rightarrow x=-1$ 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 $\sqrt{13}\ne 2\sqrt{5}$