Let f ( x ) = a x 3 + b x 2 + c x + 41 be such that f ( 1 ) = 40 , f ' ( 1 ) = 2 and f ' ' ( 1 ) = 4 . Then a 2 + b 2 + c 2 is equal to [2024]

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Solution

Q.
Let $f (x) = a x^{3} + b x^{2} + c x + 41$ be such that $f (1) = 40,$ $f' (1) = 2$ and $f'' (1) = 4 .$ Then $a^{2} + b^{2} + c^{2}$ is equal to [2024]

1 73

2 62

3 54

4 51

Ans.
(4)

$We have, f (x) = a x^{3} + b x^{2} + c x + 41$

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

$f' (1) = 2 \Rightarrow 3 a + 2 b + c = 2$ ...(ii)

$Also, f'' (x) = 6 a x + 2 b$

$f'' (1) = 4 \Rightarrow 6 a + 2 b = 4$ ...(iii)

$On solving (i), (ii), and (iii), we get a = - 1, b = 5, c = - 5$

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

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