Limit Continuity and Differentiability 7 Question 8

8. Let $f(x)=\begin{gathered}\max {|x|, x^{2} }, \quad|x| \leq 2 \ 8-2|x|, \quad 2<|x| \leq 4\end{gathered}$

Let $S$ be the set of points in the interval $(-4,4)$ at which $f$ is not differentiable. Then, $S$

(2019 Main, 10 Jan I)

(a) equals ${-2,-1,0,1,2}$

(b) equals ${-2,2}$

(c) is an empty set

(d) equals ${-2,-1,1,2}$

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Answer:

Correct Answer: 8. (d)

Solution:

  1. We have,

$$ \begin{aligned} & f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\ \Rightarrow & f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\ \Rightarrow & f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\ \Rightarrow & f^{\prime \prime \prime}(x)=6 \\ \Rightarrow & f^{\prime \prime \prime}(3)=6 \end{aligned} $$

Putting $x=1$ in Eq. (i), we get

$$ f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2) $$

and putting $x=2$ in Eq. (ii), we get

$$ f^{\prime \prime}(2)=12+2 f^{\prime}(1) $$

From Eqs. (iv) and (v), we get

$$ \begin{aligned} & f^{\prime}(1)=3+2 f^{\prime}(1)+\left(12+2 f^{\prime}(1)\right) \\ & \Rightarrow \quad 3 f^{\prime}(1)=-15 \\ & \Rightarrow \quad f^{\prime}(1)=-5 \\ & \Rightarrow \quad f^{\prime \prime}(2)=12+2(-5)=2 \quad \text { [using Eq. (v)] } \\ & \therefore \quad f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\ & \Rightarrow \quad f(x)=x^{3}-5 x^{2}+2 x+6 \\ & \Rightarrow \quad f(2)=2^{3}-5(2)^{2}+2(2)+6=8-20+4+6=-2 \end{aligned} $$



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