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:

Correct answer: C. 51 aligned & f(x)=a x^3+b x^2+c x+41 & f^ (x)=3 a x^2+2 b x+c x & f^ (1)=3 a+2 b+c=2 (1) aligned aligned & f^ (n)=6 ax+2 ~b & f^ (1)=6 a+2 ~b=4 & 3 a+b=2 .(2) aligned aligned & (1)-(2) & b+c=0...(3) aligned aligned & f(1)=40 & a+b+c+41=40 aligned use (3) a+41=40 by (2) aligned & -3+b=2 b=5 & c=-5 & a^2+b^2+c^2=1+25+25=51 aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differentiation

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Differentiation

Let f(x)=a x^3+b x^2+c x+41 be such that f(1)=40, f^ (1)=2 and f^...

Let

f(x)=a x^3+b x^2+c x+41

be such that

f(1)=40, f^{\prime}(1)=2

and

f^{\prime}(1)=4

. Then

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

is equal to:

Options

Correct answer

C. 51

Step-by-step solution

\begin{aligned} & f(x)=a x^3+b x^2+c x+41 \\ & f^{\prime}(x)=3 a x^2+2 b x+c x \\ & \Rightarrow f^{\prime}(1)=3 a+2 b+c=2 \ldots(1)\end{aligned}

\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{n})=6 \mathrm{ax}+2 \mathrm{~b} \\ & \Rightarrow \mathrm{f}^{\prime \prime}(1)=6 \mathrm{a}+2 \mathrm{~b}=4 \\ & 3 \mathrm{a}+\mathrm{b}=2 \ldots \ldots \ldots .(2)\end{aligned}

\begin{aligned} & (1)-(2) \\ & b+c=0...(3)\end{aligned}

\begin{aligned} & \mathrm{f}(1)=40 \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}+41=40 \end{aligned}

use (3)

a+41=40

by (2)

\begin{aligned} & -3+b=2 \Rightarrow b=5 & c=-5 \\ & a^2+b^2+c^2=1+25+25=51 \end{aligned}

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