At x=0, f(x)={cases} x/|x|+2x², & x ≠ 0 \\ k, & x = 0 {cases} is

Correct Answer is : Discontinuous for all real values of k

Discontinuous for all real values of k

Correct Answer is : Discontinuous for all real values of k

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TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 63 of 160

Marks: +1, -0

At x=0, f(x)=

\begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}

Solution:

Given that,

f(x)=\begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0. \end{cases}

\text{LHL} = \lim\limits_{x \to 0^{-}} f(x) = \lim\limits_{h \to 0} f(0-h)

= \lim\limits_{h \to 0} \frac{-h}{\left| -h \right| + 2(-h)^{2}} = \frac{-h}{h+2h^{2}} = \frac{-h}{h(1+2h)}

= \lim\limits_{h \to 0} \frac{-1}{1+2h} = \frac{-1}{1+2 \times 0} = -1

\text{RHL} = \lim\limits_{x \to 0^{+}} f(x) = \lim\limits_{h \to 0} f(0+h)

= \lim\limits_{h \to 0} \frac{h}{\left| h \right| + 2h^{2}} = \frac{h}{h(1+2h)}

= \lim\limits_{h \to 0} \frac{1}{1+2h} = \frac{1}{1+2 \times 0} = 1

Also,

f(0)=k

\because

LHL

\neq

RHL, therefore given function is discontinuous for all values of

k

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