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  • <img alt="\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} \frac{x^4-5 x^2+4}{|(x-1)(x-2)|}, & x \neq 1,2 \\ 6, & x=1 \\ 12, & x=2 \end{array}\right.}" src="https://entrancecorner.on

\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} \frac{x^4-5 x^2+4}{|(x-1)(x-2)|}, & x \neq 1,2 \\ 6, & x=1 \\ 12, & x=2 \end{array}\right.}       Then f(x)  is continuous on the set 

Option: 1

R


Option: 2

R-[1]


Option: 3

R-[2]


Option: 4

R-[1,2]


Answers (1)

best_answer

For any  \mathrm{x \neq 1,2}  we find that f(x) is the quotient of two polynomials and a polynomial is everywhere continuous. Therefore f(x) is continuous for all  \mathrm{x \neq 1,2} . Continuity at x=1 : We have,

\mathrm{ \lim _{x \rightarrow 1^{-}} f(x) =\lim _{h \rightarrow 0} f(1-h) \\ }

\mathrm{ =\lim _{h \rightarrow 0} \frac{(1-h-2)(1-h+2)(1-h+1)(1-h-1)}{|(1-h-1)(1-h-2)|} \\ }

\mathrm{ =\lim _{h \rightarrow 0} \frac{(3-h)(2-h)(-1-h)(-h)}{|(-h)(-1-h)|} . \\ }

\mathrm{ =\lim _{h \rightarrow 0} \frac{(3-h)(2-h) h(h+1)}{h(h+1)}=6 . }

\mathrm{ \underset{x \rightarrow 1^{+}}{\text {and, } \lim _{h \rightarrow 0} f(x)} =\lim _{h \rightarrow 0} f(1+h) \\ }

\mathrm{ =\lim _{h \rightarrow 0} \frac{(1+h-2)(1+h+2)(1+h+1)(1+h-1)}{\left|(1+h-1)\left(1+h^{\prime}-2\right)\right|} \\ }

\mathrm{ =\lim _{h \rightarrow 0} \frac{(h-1)(3+h)(2+h)(h)}{|h(h-1)|} }

\mathrm{ =-\lim _{h \rightarrow 0} \frac{(h-1)(3+h)(2+h) h}{h(1-h)}=-6 \\ }
\mathrm{ \therefore \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x) }

So f(x) is not continuous at x=1. Similarly, f(x) is not continuous at x=2.

 

Posted by

Sanket Gandhi

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