CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Show that the function defined by $f(x) = \cos (x^2 )$ is a continuous function.
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Answer
It is given function is $\text{f(x)} = \cos(\text{x}^{2})$
This function f is defined for every real number and f can be written as the composition of two function as,
f = goh, where, $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$
First we have to prove that $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$ are continuous functions.
We know that g is defined for every real number.
Let k be a real number.
Then, g(k) = cos k
Now, put $x = k + h$
lf $x \rightarrow k$, then $h \rightarrow 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\cos\text{x}$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos (\text{k}+\text{h})$
$= ^{\text{lim}}_{\text{h}\rightarrow\text{0}}-[\cos\text{k}\cos\text{h} - \sin\text{k}.\sin\text{h}]$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos\text{k}\cos\text{h} -^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}} \sin\text{k}\sin\text{h}$
$= \cos\text{k}\cos0 - \sin\text{k}\sin0$
$= \cos\text{k} \times 1 - \sin \times\ 0$
$= \cos\text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
Thus, g(x) = cosx is continuous function.
Now, $h(x) = x^2$
So, h is defined for every real number.
Let c be a real number, then $h(c) = c^2$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{x}^{2}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} =\text{h(c)}$
Therefore, h is a continuous function.
We know that for real valued functions g and h, Such that (fog) is continuous at c.
Therefore, $f(x) = (goh)(x) = \cos(x^2)$ is a continuous function.
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