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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Differentiation

Let

f : R \to R

be a function such that

f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R

. Then

f (2)

equals:

(A)-4
(B)30
(C)-2✓ Correct
(D)8

Answer: (C)

Solution

Since

f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3)

, the three unknown constants are

a = f^{'} (1)

,

b = f^{''} (2)

and

c = f^{'''} (3)

. Differentiating,

f^{'} (x) = 3 x^{2} + 2 a x + b

,

f^{''} (x) = 6 x + 2 a

,

f^{'''} (x) = 6

, so at once

c = f^{'''} (3) = 6

. Then

a = f^{'} (1) = 3 + 2 a + b

gives

a + b = - 3

, and

b = f^{''} (2) = 12 + 2 a

gives

2 a - b = - 12

; solving,

a = - 5

and

b = 2

. Thus

f (x) = x^{3} - 5 x^{2} + 2 x + 6

and

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

.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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