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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Differentiation

NDA 2016 · Mathematics · Differentiation

Let f: R arrow R be a function such that f(x)=x^3+x^2 f^ (1)+x f f^...

Let f:R→R\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R} be a function such that f(x)=x3+x2f′(1)+xff′′(2)+f′′′(3)f(x)=x^3+x^2 f^{\prime}(1)+x f f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) for x∈R\mathrm{x} \in \mathrm{R} What is \mathrmf′(1)\mathrmf^{\prime}(1) equal to?

Options

  1. A–6
  2. B–5
  3. C1
  4. D0

Correct answer

B. –5

Step-by-step solution

f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3)...(1)f′(x)=3x2+2xf′(1)+f′′(2)...(2)f′′(x)=6x+2f′(1)...(3)f′′′(x)=6...(4)f′(1)=3+2f′(1)+f′′(2)...(5)f′(2)=12+2f′(1)...(6)\begin{aligned} & f(x)=x^3+x^2 f^{\prime}(1)+\mathrm{x} f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)...(1) \\ & f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)...(2) \\ & f^{\prime \prime}(x)=6 \mathrm{x}+2 f^{\prime}(1)...(3) \\ & f^{\prime \prime \prime}(\mathrm{x})=6...(4) \\ & f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)...(5) \\ & f^{\prime}(2)=12+2 f^{\prime}(1)...(6) \end{aligned} Using (6) in (5), we get f′(1)=3+2f′(1)+12+2f′′(1)−3f′(1)=15f′(1)=−5\begin{aligned} & f^{\prime}(1)=3+2 f^{\prime}(1)+12+2 f^{\prime \prime}(1)-3 f^{\prime}(1)=15 \\ & f^{\prime}(1)=-5 \end{aligned} Using this value in eqn (6) we get f′(2)=12+2×(−5)f′′(2)=2\begin{aligned} & f^{\prime}(2)=12+2 \times(-5) \\ & f^{\prime \prime}(2)=2 \end{aligned} Using x=3x=3 in eqn (4), f′′′(3)=6f^{\prime \prime \prime}(3)=6 Putting value of f′(1)+f′′(2)andf′′′(3)f^{\prime}(1)+f^{\prime \prime}(2) and f^{\prime \prime \prime}(3) in eqn (1) We get f(x)=x3+x2(−5)+x(2)+6=x3−5x2+2x+6\begin{aligned} f(x) & =x^3+x^2(-5)+x(2)+6 \\ & =x^3-5 x^2+2 x+6 \end{aligned} Putting x=1x=1f(1)=(1)3−5(1)2+2(1)+6f(1)=4\begin{aligned} & f(1)=(1)^3-5(1)^2+2(1)+6 \\ & f(1)=4 \end{aligned}$$
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