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}$