Q41 / 90 · JEE Main 2017 (09 Apr Online) · Mathematics

Let f be a polynomial function such that f3x=f′x. f′′x, for all x∈R. Then :

  • A
    f2+f′2=28
  • Correct answer
    B
    f′′2-f′2=0
  • C
    f2-f′2+f′′2= 10
  • D
    f′′2-f2=4

Solution

Degree of fx will be 3

f(x)=ax3+bx2+cx+d

f3x=27ax3+9bx2+3cx+d

f'(x)=3ax2+2bx+c

f′′x=6ax+2b

f3x=f′x f′′x

Comparing the coefficient, we get

27a=18a2 ⇒a= 3 2

Also b=0, c=0, d=0

fx=32x3, f( 2 )=12

⇒ f′x=92x2, f′′x=9x

Hence, f'(2)=18, f''(2)=18

Hence, f''2−f'(2)=0

Hence, the correct answer is option: 'B'.

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