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Question

Mathematics Question on Application of derivatives

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 252 \sqrt{5}

B

f(x) is increasing for x[1,25]x \in [1, 2 \sqrt{ 5} ]

C

f(x) has local minima at x = 1

D

the value of f(0) = 15

Answer

f(x) has local minima at x = 1

Explanation

Solution

Let f(x)=ax3+bx2+cx+df\left(x\right) =ax^{3} +bx^{2}+cx +d
Then, f(2)=188a+4b+2c+d=18 f\left(2\right)=18 \Rightarrow 8a + 4b +2c +d =18 ....(1)
f(1)=1a+b+c+d=1f\left(1\right) = -1 \Rightarrow a+b+c+d =-1 ....(2)
f(x)f\left(x\right) has local max. at x=1x=-1
3a2b+c=0f(x)x=0b=0\Rightarrow 3a-2b+c=0 f'\left(x\right) x=0 \Rightarrow b=0 ......(4)
Solving (1), (2), (3) and (4), we get
f(x)=14(19x357x+34)f(0)=172f\left(x\right) = \frac{1}{4} \left(19x^{3} -57 x +34\right)\Rightarrow f\left(0\right) = \frac{17}{2}
Also f(x)=574(x21)0,x>1f'\left(x\right) = \frac{57}{4}\left(x^{2}-1\right)0, \forall x>1
Also f(x)=0x=1,1f'\left(x\right) =0 \Rightarrow x=1, -1
f"(1)<0,f"(1)>0x=1f" \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 1325\sqrt{13} \neq 2 \sqrt{5}