Limit Continuity And Differentiability Ques 139
- 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 continuous. 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: 139.(d)
Solution:
- We have,
$$ \begin{alignedat} & f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\ \Rightarrow & f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime}(2) \ \Rightarrow & f^{\prime \prime}(x)=6 x+2 f^{\prime}(x) \ \Rightarrow & f^{\prime \prime \prime}(x)=6 \\ \Rightarrow & f^{\prime \prime \prime}(3)=6 \end{aligned} $$
Putting $x=1$ in Eq. (i), we get
$$ f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime}(2) $$
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
$$ \begin{alignedat} & f^{\prime}(1)=3+2 f^{\prime}(1)+\left(12+2 f^{\prime}(1)\right) \\ & \Rightarrow \quad 3 f^{\prime}(1)=-15 \\ & \Rightarrow \quad f^{\prime}(1)=-5 \\ & \Rightarrow \quad f^{\prime \prime}(2)=12+2(-5)=2 \quad \text { [using Eq. (v)] } \\ & \therefore \quad f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\ & \Rightarrow \quad f(x)=x^{3}-5 x^{2}+2 x+6 \\ & \Rightarrow \quad f(2)=2^{3}-5(2)^{2}+2(2)+6=8-20+4+6=-2 \end{aligned} $$
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