Let f x be a cubic polynomial with f 1 = - 10 ,   f - 1 = 6 , and has a local minima at x = 1 , and f '…

Let fx be a cubic polynomial with f1=-10, f-1=6, and has a local minima at x=1, and f'x has a local minima at x=-1. Then f3 is equal to                .

Solution

Let, fx=ax3+bx2+cx+d

Diff w. r. t.  'x'

f'x=3ax2+2bx+c

Again diff. w. r. t.  'x'

f"x=6ax+2b

f"-1=0

-6a+2b=0

b=3a

f'1=0

3a+6a+c=0

c=-9a

f1=-10 

-5a+d=-10      i

f(-1)=6 

11a+d=6       ii

From equations i-ii

We get a=1, d=-5, b=3, c=-9

Then, fx=x3+3x2-9x-5

Hence,  f3=27+27-27-5

f3=22

Asked in: JEE Main 2021 (31 Aug Shift 2)

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