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 | is a continuous function.
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Answer
It is given function is $\text{f(x)} = |\cos\text{x}|$
The given function f is defined for real number and f can be written as the composition of two functions, as
f = goh, where, $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \cos\text{x}$
First we have to prove that $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \cos\text{x}$ 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
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
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) = cosx
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
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\cos\text{x}$
$ = ^{\ \ \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}$
$= \cos \text{k}\cos0 - \sin\text{k}\sin0$
$= \cos \text{k} \times 1 - \sin \times\ 0$
$=\cos \text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =\text{h(k)}$
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,$\text{ f(x)} = \text{(gof)(x)} = \text{g(h(x))} = \text{g}(\cos \text{x)}= |\cos\text{x}|$ is a continuous function.
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