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  •  If <img alt="\begin{aligned} \mathrm{f(x)}&=\mathrm{p x^2-q, x \in[0,1)}\\ &\mathrm{x+1, x \in(1,2]} \end{aligned}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cbegin%

 If \begin{aligned} \mathrm{f(x)}&=\mathrm{p x^2-q, x \in[0,1)}\\ &\mathrm{x+1, x \in(1,2]} \end{aligned}
and \mathrm{f(1)=2}  then the value of the pair \mathrm{(p, q)}  for which \mathrm{f(x)} cannot be continuous at \mathrm{ x=1} is

Option: 1

(2,0)


Option: 2

(1,-1)


Option: 3

(4,2)


Option: 4

(1,1)


Answers (1)

best_answer

\begin{aligned} &\mathrm{ f(x) is~ continuous~ at~ x=1 ~if}\\ &\mathrm{\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} f(1-h)=f(1)=2 .}\\ &\mathrm{ \lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\{1+h+1\}=2 }\\ &\mathrm{ \lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left\{p(1-h)^2-q\right\}=p-q}\\ & \therefore \quad \text{ f(x) is not continuous at x=1 if p-q} \neq 2. \end{aligned}

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Ritika Kankaria

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