CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Examine that sin |x| is a continuous function.
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Answer
It is given function is $\text{f(x)} = \sin|\text{x}|$ The given function f is defined for real number and f can be written as the composition of two functions, as f = goh, where, $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \sin\text{x}$ First we have to prove that $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \sin\text{x}$ are continuous functions. g(x) = lxl can be written as $\text{g(x)}=\begin{cases}-\text{x},&\text{if}\ \text{x}<{0}\\\text{x},& \text{if}\ \text{x}\geq0\end{cases}$ Now, g is defined for all real number. Let k be a real number. Case I: If k < 0, Then g(k) = -k And $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(-\text{x}) = -\text{k}$ Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$ Therefore, g is continuous at all points x, i.e. x > 0 Case II: If k > 0, Then g(k) = k and $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{x}=\text{k}$ Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$ Therefore, g is continuous at all points x, i.e. x < 0 Case III: If k = 0, Then, g(k) = g(0) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}(-\text{x}) = 0$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}(\text{x}) = 0$ $\therefore^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text g({\text x}) =\text{g}( 0)$ Therefore, g is continuous at x = 0 From the above 3 cases, we get that g is continuous at all points. h(x) = sinx We know that h is defined for every real number. Let k be a real number. Now, put x = k + h If x → k, then h → 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\sin\text{x}$ $ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin(\text{k}+\text{h})$ $ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}[\sin\text{k}\cos\text{h} + \cos\text{k}\sin\text{h}]$ $ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin\text{k}\cos\text{h} +^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos\text{k}\sin\text{h}$ $= \sin\text{k}\cos0 + \cos\text{k}\sin0$ $=\sin \text{k}$ $\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =\text{g(k)}$ Thus, h(x) = cos x is continuous function. We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k), Then (fog) is continuous at k. Therefore, $\text{ f(x)} = \text{(gof)(x)} = \text{g(h(x))} = \text{g}(\sin \text{x)}= |\sin\text{x}|$is a continuous function.
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