Prove that the function f defined by f(x) = * 20 l |x| + 2 x 2 , & x 0 k, & x = 0 . remains discontinuous at x = 0 , regardless the choice of k

We have f(x) = * 20 l |x| + 2 x 2 , & x 0 k, & x = 0 . . At x = 0, = x |x| + 2 x 2 = h 0 |0 - h| + 2 (0 - h) 2 = h 0 h + 2 h 2 = h 0 h(1 + 2h) = - 1 = x |x| + 2 x 2 = h 0 |0 + h| + 2 (0 + h) 2 = h 0 h + 2 h 2 = h 0 h(1 + 2h) = 1 and f(0) = k Since, for any value of . Hence, f(x) is discontinuous at x = 0 regardless the choice of k .

Continuity and Differentiability — Class 12 Maths Solution

exemplar sa SA NCERT Exemp. Ex.5.3 ,Q.15,Page 108

Question

Prove that the function $f$ defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{x}{{|x| + 2{x^2}}},}&{{\rm{ if }}x \ne 0}\\{k,}&{{\rm{ if }}x = 0}\end{array}} \right.$ remains discontinuous at $x = 0$, regardless the choice of $k$ }

Step-by-step Solution

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{x}{{|x| + 2{x^2}}},}&{{\rm{ if }}x \ne 0}\\{k,}&{{\rm{ if }}x = 0}\end{array}} \right.$.

At $x = 0,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{|x| + 2{x^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{(0 - h)}}{{|0 - h| + 2{{(0 - h)}^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h + 2{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h(1 + 2h)}} = - 1$

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{|x| + 2{x^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{0 + h}}{{|0 + h| + 2{{(0 + h)}^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{h}{{h + 2{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{h(1 + 2h)}} = 1$
and $f(0) = k$

Since, ${\rm{LHL}} \ne {\rm{RHL}}$ for any value of ${\rm{k}}$.

Hence, $f(x)$ is discontinuous at $x = 0$ regardless the choice of $k$.

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