JEE Main 10 Jan 2019 Shift 1 Solved Paper

f (x) = x2+x2f′(1) + xf'' (2) + f''' (3)
⇒ f' (x) = 3x2 + 2xf' (1) + f'' (x) ... (1)
⇒ f'' (x) = 6x + 2f' (x) ... (2)
⇒ f''' (x) = 6 ... (3)
put x = 1 in equation (1) :
ƒ'(1) = 3 + 2ƒ'(1) + ƒ''(2) .....(4)
put x = 2 in equation (2) :
ƒ''(2) = 12 + 2ƒ'(1) .....(5)
from equation (4) & (5) :
–3 – ƒ'(1) = 12 + 2ƒ'(1)
⇒ 3ƒ'(1) = –15

⇒ ƒ'(1) = –5 Þ ƒ''(2) = 2 ....(2)
put x = 3 in equation (3) :
ƒ'''(3) = 6
∴ ƒ(x) = x3–5x2 + 2x + 6
ƒ(2) = 8 – 20 + 4 + 6 = –2