JEE Main 2019 — Differentiation Question with Solution

f(x) = x³ + x²f '(1) + xf ''(2) + f '''(3)

$$\Rightarrow$$  f '(x) = 3x² + 2xf '(1) + f ''(x)     . . . . . (1)

$$\Rightarrow$$  f ''(x) = 6x + 2f '(1)     . . . . . . (2)

$$\Rightarrow$$  f '''(x) = 6      . . . . . .(3)

put x = 1 in equation (1) :

f '(1) = 3 + 2f '(1) + f ''(2)     . . . . .(4)

put x = 2 in equation (2) :

f ''(2) = 12 + 2f '(1)     . . . . .(5)

from equation (4) & (5) :

$$-$$3 $$-$$ f '(1) = 12 + 2f'(1)

$$\Rightarrow$$  3f '(1) = $$-$$ 15

$$\Rightarrow$$  f '(1) = $$-$$ 5 $$\Rightarrow$$  f ''(2) = 2      . . . . .(2)

put x = 3 in equation (3) :

f ''' (3) = 6

$$\therefore$$  f(x) = x³ $$-$$ 5x² + 2x + 6

f(2) = 8 $$-$$ 20 + 4 + 6 = $$-$$ 2