JEE Mains · Maths · STD 12 - 5. continuity and differentiation
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 :
- A \(62\)
- B \(73\)
- C \(54\)
- D \(51\)
Answer & Solution
Correct Answer
(D) \(51\)
Step-by-step Solution
Detailed explanation
\( \mathrm{f}(\mathrm{x})=\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+41 \) \( \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cx} \) \( \Rightarrow \mathrm{f}^{\prime}(1)=3 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=2 \ldots \ldots (1)\)…
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