Calculus Solutions Ex#1.6

 

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Q.01 Find the discontinuities if any
Solution:
$\quad f(x)=\sin \left(x^{2}-2\right)$ This a composition of two functions. Let $$g(x)=x^{2}-2$$ and $$h(x)=\sin x$$
so we can write $$f(x)=h(g(x))$$ $g(x)$ is a polynomial which is continuous everywhere, and $h(x)=\sin x$ is continuous on $(-\infty,+\infty)$. Therefore, their composition is continuous ie $f(x)$ is continuous everywhere. There is no point of discontinuity.
Q.02
Solution:
$f(x)=\cos \left(\frac{x}{x-\pi}\right)$ we can write $f(x)$, as a composition of two functions.
Let
$$g(x)=\frac{x}{x-\pi}$$ and $$h(x)=\cos x$$ so that $f(x)=h(g(x))=\cos \left(\frac{x}{x-2}\right)$.
$g(x)$ is discontinuous at those $x$ where the denominator becomes zero. Put denominators equal to zero. $\Rightarrow x-\pi=0 \Rightarrow x=\pi$ is point of discontinuity of $g(x) \Rightarrow x=\pi$ is a point of discontinuity of $f(x)$
Q.03
Solution:
$$\begin{aligned} & f(x)=|\cot x| \\ & f(x)=\left|\frac{\cos x}{\sin x}\right| \end{aligned}$$ $f(x)$ is continuous at $x$, where the denominator and numerator are both continuous and the denominator is not zero. Since $\cos x$ and $\sin x$ both are continuous everywhere. Therefore $f(x)$ will be discontinuous at all those points where the denominator $\sin x=0$. $$\begin{aligned} & \sin x=0 \\ & x=\sin ^{-1}(0) \\ & x=n \pi, n=0, \pm 1 \pm 2,3, \cdots \end{aligned}$$
Q.04
Solution:
$f(x)=\sec x$ $$f(x)=\frac{1}{\cos x}$$
where$$\sec x=\frac{1}{\cos x}$$ $f(x)$ will be discontinuous at all those points where the denominator cos $x=0$ $$\begin{aligned} & \cos x=0 \\ & x=\cos ^{-1}(0) \\ & x= \pm \frac{\pi}{2}, \pm 3 \frac{\pi}{2}, \pm 5\frac{\pi}{2}, \ldots. \\ & x=\frac{\pi}{2}+n\pi, \quad n=0, \pm 1 \pm 2, \pm 3, \ldots . \end{aligned}$$
Q.06
Solution:
$$f(x)=\frac{1}{1+\sin ^{2} x}$$
It is continuous everywhere because the denominator $1+\sin ^{2} x \neq 0$. for all $x$.
Q.07
Solution:
$$f(x)=\frac{1}{1-2 \sin x}$$
$f(x)$ will be discontinuous at all points where the denominator $1-2 \sin x=0$
$$1-2 \sin x=0$$
$$1=2 \sin x$$
$$\sin x=\frac{1}{2}$$
$$x=\sin ^{-1}\left(\frac{1}{2}\right)$$
$$x=\frac{\pi}{6}+2 n \pi, x=\frac{5 \pi}{6}+2 n \pi$$
$$n=0, \pm 1 \pm 2, \pm 3, \ldots$$
are the points where the function $f(x)$ is discontinuous.
Q.08
Solution:
$$f(x)=\sqrt{2+\tan ^{2} x}$$ $$f(x)=\sqrt{2+\frac {\sin ^{2} x}{\cos^{2}x}$$
$f(x)$ will be discontinuous at all those points where the denominator $\cos ^{2} x=0$
$$\begin{aligned} & \cos ^{2} x=0 \\ \\ & \cos x=0 \\ \\ & x=\cos ^{-1}(0) \\ \\ & x= \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \pm \frac{5 \pi}{2}, \cdots \\ \\ & \text {Thus the points where $f(x)$ is discontinuous are } \\ \\ & x=\frac{\pi}{2}+n \pi, \quad n=0, \pm 1 \pm 2, \pm 3, \cdots \end{aligned}$$