Limit Continuity And Differentiability Ques 63
- Let $f: R \rightarrow R$ be a function such that $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in R$.
Then, $f(2)$ equals
(2019 Main, 10 Jan I)
(a) $30$
(b) $-4$
(c) $-2$
(d) $8$
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Answer:
Correct Answer: 63.(c)
Solution: (c) We have,
$ \begin{aligned} & f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime} (3)\\ \Rightarrow \quad & f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \quad ……..(i) \\ \Rightarrow \quad & f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \quad ……..(ii) \\ \Rightarrow \quad & f^{\prime \prime}(x)=6 \quad ……..(iii) \\ \Rightarrow \quad & f^{\prime \prime \prime}(3)=3 \end{aligned} $
Putting $x=1$ in Eq. (i), we get
$ f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2) \quad ……..(iv) $
and putting $x=2$ in Eq. (ii), we get
$ f^{\prime}(2)=12+2 f^{\prime}(1) \quad ……..(v) $
From Eqs. (iv) and (v), we get
$ \begin{array}{rlrl} & f^{\prime}(1) =3+2 f^{\prime}(1)+\left(12+2 f^{\prime}(1)\right) \\ \Rightarrow & 3 f^{\prime}(1) =-15 \\ \Rightarrow & f^{\prime}(1) =-5 \\ \Rightarrow & f^{\prime \prime}(2) =12+2(-5)=2 \\ \therefore& (x) =x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime}(3) \\ \Rightarrow & f(x) =x^3-5 x^2+2 x+6 \\ \Rightarrow & f(2) =2^3-5(2)^2+2(2)+6=8-20+4+6=-2 \end{array} $
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