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JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Differentiation

Let

f (x) = ax^{3} + bx^{2} + cx + 41

be such that

f (1) = 40, f^{'} (1) = 2

and

f^{''} (1) = 4

.
Then

a^{2} + b^{2} + c^{2}

is equal to :

(A)62
(B)73
(C)54
(D)51✓ Correct

Answer: (D)

Solution

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

,

f^{''} (1) = 6 a + 2 b = 4 \Rightarrow 3 a + b = 2

, and

f (1) = a + b + c + 41 = 40 \Rightarrow a + b + c = - 1

. Solving:

a = - 1, b = 5, c = - 5

. Hence

a^{2} + b^{2} + c^{2} = 1 + 25 + 25 = 51

— option (4).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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