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  • If <img alt="\mathrm{f(x)=\left\{\begin{array}{c}x+2, x \in(1,2] \\ p x^2-q, x \in[0,1)\end{array}\right. and f(1)=3}" src="/latex-image/?%5Cmathrm%7Bf%28x%29%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc

If \mathrm{f(x)=\left\{\begin{array}{c}x+2, x \in(1,2] \\ p x^2-q, x \in[0,1)\end{array}\right. and f(1)=3}, then the possible value of the pair (p, q) for which f(x) cannot be continuous at x=1 is

 

 

Option: 1

 (3,0)
 


Option: 2

 (1,-2)
 


Option: 3

 (5,2)
 


Option: 4

(1,1)


Answers (1)

best_answer

 f(x) is continuous at x=1, if
\mathrm{\begin{aligned} & \lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} f(1-h)=f(1)=3 \\ & \lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(1+h+2)=3 \\ & \lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left(p(1-h)^2-q\right)=p-q \\ & \therefore f(x) \text { is not continuous at } x=1 \text { if } p-q \neq 3 \end{aligned} }
 

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vishal kumar

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