Solve the Following Question.(4 Marks)

Question 14 Marks

Find $f(a)$, if $f$ is continuous at $x=a$ where, $f(x)=\frac{1+\cos (\pi x)}{\pi(1-x)^2}$, for $x \neq 1$ and at $a =1$

Answer

$f(x)$ is continuous at $x=1$
$\therefore f (1)=\lim _{x \rightarrow 1} f (x)$
$\therefore f (1)=\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\pi(1-x)^2}$
Put $1-x=h$
$\therefore x=1-h$
As $x \rightarrow 1, h \rightarrow 0$
$\therefore f(1)=\lim _{h \rightarrow 0} \frac{1+\cos [\pi(1-h)]}{\pi h^2}$
$=\lim _{h \rightarrow 0} \frac{1+\cos (\pi-\pi h)}{\pi h^2}$
$=\lim _{h \rightarrow 0} \frac{1-\cos \pi h}{\pi h^2}$
$=\lim _{h \rightarrow 0} \frac{1-\cos \pi h}{\pi h^2} \times \frac{1+\cos \pi h}{1+\cos \pi h}$
$=\lim _{h \rightarrow 0} \frac{1-\cos ^2 \pi h}{\pi h^2(1+\cos \pi h)}$
$=\frac{1}{\pi} \lim _{h \rightarrow 0} \frac{\sin { }^2 \pi h}{h^2(1+\cos \pi h)}$
$=\frac{1}{\pi} \lim _{h \rightarrow 0}\left(\frac{\sin \pi h}{h}\right)^2 \times \frac{1}{1+\cos \pi h}$
$=\frac{1}{\pi} \lim _{h \rightarrow 0}\left(\frac{\sin \pi h}{\pi h}\right)^2 \times \pi^2 \times \frac{1}{\lim _{h \rightarrow 0}(1+\cos \pi h)}$
$\begin{array}{l}=\frac{1}{\pi} \times(1)^2 \times \pi^2 \times \frac{1}{1+1} \ldots\left[\begin{array}{l}\text { As } h \rightarrow 0, \pi h \rightarrow 0 \\ \text { and } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\end{array}\right] \\ =\frac{\pi}{2}\end{array}$

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Question 24 Marks

Find $k$ if the following function is continuous at the point indicated against them:
$
\left.\begin{array}{rlrl}
f(x) & =\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}, & & \text { for } x \neq 2 \\
& =k_1 & & \text { for } x=2
\end{array}\right\} \text { at } x=2
$

Answer

$f(x)$ is continuous at $x=2$
$
\begin{aligned}
& \therefore \mathrm{f}(2)=\lim _{x \rightarrow 2} \mathrm{f}(x) \\
& \therefore \mathrm{k}=\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}
\end{aligned}
$
Put $x-2=h$
$
\therefore \mathrm{x}=2+\mathrm{h}
$
As $\mathrm{x} \rightarrow 2, \mathrm{~h} \rightarrow 0$
$
\begin{aligned}
& \therefore k=\lim _{h \rightarrow 0}\left[\frac{5(2+h)-8}{8-3(2+h)}\right]^{\frac{3}{2 h}} \\
& =\lim _{h \rightarrow 0}\left(\frac{10+5 h-8}{8-6-3 h}\right)^{\frac{3}{2 h}} \\
& =\lim _{h \rightarrow 0}\left(\frac{2+5 h}{2-3 h}\right)^{\frac{3}{2 h}} \\
& =\lim _{h \rightarrow 0}\left[\frac{2\left(1+\frac{5 h}{2}\right)}{2\left(1-\frac{3 h}{2}\right)}\right]^{\frac{3}{2 h}}
\end{aligned}
$
$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\left(1+\frac{5 h}{2}\right)^{\frac{3}{2 h}}}{\left(1-\frac{3 h}{2}\right)^{\frac{3}{2 h}}} \\ & =\frac{\lim _{h \rightarrow 0}\left[\left(1+\frac{5 h}{2}\right)^{\frac{2}{3 h}}\right]^{\frac{5}{2} \times \frac{3}{2}}}{\lim _{h \rightarrow 0}\left[\left(1-\frac{3 h}{2}\right)^{\frac{-2}{3 h}}\right]^{\frac{-3}{2} \times \frac{3}{2}}} \\ & =\frac{e^{\frac{15}{4}}}{e^{\frac{-9}{4}}} \cdots\left[\because h \rightarrow 0, \frac{5 h}{2} \rightarrow 0, \frac{-3 h}{2} \rightarrow 0\right] \\ & =e^{\frac{24}{4}} \text { and } \lim _{x \rightarrow 0}(1+x)^{\frac{1}{2}}=e \\ & =e^6\end{aligned}$

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Question 34 Marks

Discuss the continuity of the following function at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous:
$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$

Answer

$\begin{aligned}
& f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12} \\
& f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{(x-4)(x+3)}
\end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is not defined at for $\mathrm{x}=4$ and $\mathrm{x}=-3$
$\therefore$ The domain of function $f=R-\{-3,4\}$ for $x \neq-3,4$
$\begin{aligned}
& f(x)=\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)} \\
& \therefore f(x)=x-2, x \neq-3,4 \\
& \therefore f(-3)=-5 \text { and } f(4)=2
\end{aligned}$
$f(x)$ is discontinuous $x=4$ and $x=-3$
This discontinuity is removable.
$\therefore \mathrm{f}(\mathrm{x})$ can be redefined as
$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$
$=-5$, for $x=-3$

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Question 44 Marks

Discuss the continuity of the following function at the point(s) or on the interval indicated against them:
$
\begin{aligned}
& =\frac{x^2-3 x-10}{x-5}, & & \text { for } 3 \leq x \leq 6, x \neq 5 \\
f(x) & =10, & & \text { for } x=5 \\
& =\frac{x^2-3 x-10}{x-5}, & & \text { for } 6\end{aligned}
$

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Question 54 Marks

For what values of $a$ and $b$ is the function
$ f(x)=\frac{x^2-4}{x-2} \text {, for } x<2 \\
=a x^2-b x+3, \text { for } 2 \leq x<3 \\
=2 x-a+b, \text { for } x \geq 3$
continuous for every $x$ on $R$ ?

Answer

$f(x)$ is continuous for every $x$ on $R.....($given$)$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$ and $\mathrm{x}=3$.
As $f(x)$ is continuous at $x=2$,
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$
$\therefore \lim _{x \rightarrow 2^{-}} \frac{x^2-4}{x-2}=\lim _{x \rightarrow 2^{+}}\left(a x^2-b x+3\right)$
$\therefore \lim _{x \rightarrow 2^{-}} \frac{(x-2)(x+2)}{x-2}=\lim _{x \rightarrow 2^{+}}\left(a x^2-\mathrm{b} x+3\right)$
$\therefore \lim _{x \rightarrow 2^{-}}(x+2)=\lim _{x \rightarrow 2^{+}}\left(a x^2-\mathrm{b} x+3\right) \ldots\left[\begin{array}{l}
\because x \rightarrow 2, x \neq 2 \\
\therefore x-2 \neq 0
\end{array}\right] \\
\therefore 2+2=\mathrm{a}(2)^2-\mathrm{b}(2)+3 \ldots \text { (i) }$
$\therefore 4 \mathrm{a}-2 \mathrm{~b}+3=4$
$\therefore 4 \mathrm{a}-2 \mathrm{~b}=1 \ldots$
As $\mathrm{f}(x)$ is continuous at $x=3$, $\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$
$\therefore \lim _{x \rightarrow 3^{-}}\left(a x^2-b x+3\right)=\lim _{x \rightarrow 3^{+}}(2 x-a+b)$
$\therefore a(3)^2-b(3)+3=2(3)-a+b$
$ \therefore 9 a-3 b+3=6-a+b$
$ \therefore 10 a-4 b=3 \ldots .(\text { (ii) }$
Multiplying $(i)$ by $2$, we get
$8 a-4 b=2 \text {...(iii) }$
Subtracting $(ii)$ from $(iii),$ we get
$-2 \mathrm{a}=-1$
$ \therefore \mathrm{a}=\frac{1}{2}$
Substituting $a=\frac{1}{2}$ in $(i),$ we get
$4\left(\frac{1}{2}\right)-2 b=1$
$ \therefore 2-2 b=1$
$ \therefore 1=2 b$
$ \therefore b=\frac{1}{2}$
$ \therefore a=\frac{1}{2}$ and $b=\frac{1}{2}$

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Question 64 Marks

$\begin{aligned}
& \text {If } f(x)=\frac{24^x-8^x-3^x+1}{12^x-4^x-3^x+1} \text {, for } x \neq 0 \\
& =k \text {, for } x=0
\end{aligned}$
is continuous at $x=0$, then find $k$.

Answer

$f(x)$ is continuous at $x=0 ($given$)$
$\therefore \mathrm{f}(0)=\lim _{x \rightarrow 0} \mathrm{f}(x)$
$ \therefore \mathrm{k}=\lim _{x \rightarrow 0} \frac{24^x-8^x-3^x+1}{12^x-4^x-3^x+1}$
$ =\lim _{x \rightarrow 0} \frac{8^x \cdot 3^x-8^x-3^x+1}{4^x \cdot 3^x-4^x-3^x+1}$
$ =\lim _{x \rightarrow 0} \frac{8^x\left(3^x-1\right)-1\left(3^x-1\right)}{4^x\left(3^x-1\right)-1\left(3^x-1\right)}$
$ =\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)\left(8^x-1\right)}{\left(3^x-1\right)\left(4^x-1\right)}$
$ =\lim _{x \rightarrow 0} \frac{8^x-1}{4^x-1}$
$ \text { As } x \rightarrow 0,3^x \rightarrow 3^0, \\ \cdots\left[\begin{array}{l} 3^x \rightarrow 1,3^x \neq 1 \\ \therefore 3^x-1 \neq 0 \end{array}\right]$
$=\lim _{x \rightarrow 0}\left(\frac{\frac{8^x-1}{x}}{\frac{4^x-1}{x}}\right)$
$ \ldots[\because x \rightarrow 0, x \neq 0]$
$=\frac{\lim _{x \rightarrow 0} \frac{8^x-1}{x}}{\lim _{x \rightarrow 0} \frac{4^x-1}{x}}$
$ =\frac{\log 8}{\log 4} \cdots\left[\because \lim _{x \rightarrow 0}\left(\frac{\mathrm{a}^x-1}{x}\right)=\log \mathrm{a}\right]$
$ =\frac{\log (2)^3}{\log (2)^2}$
$ =\frac{3 \log 2}{2 \log 2}$
$ \mathrm{k}=\frac{3}{2}$

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Question 74 Marks

If $f(x)=\frac{\sqrt{2+\sin x}-\sqrt{3}}{\cos ^2 x}$, for $x \neq \frac{\pi}{2}$, is continuous at $x=\frac{\pi}{2}$ then find $f\left(\frac{\pi}{2}\right)$.

Answer

$f(x)$ is continuous at $x=\frac{\pi}{2}, ($given$)$
$\mathrm{f}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \mathrm{f}(x)$
$ =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2+\sin x}-\sqrt{3}}{\cos ^2 x}$
$ =\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{\sqrt{2+\sin x}-\sqrt{3}}{1-\sin ^2 x} \times \frac{\sqrt{2+\sin x}+\sqrt{3}}{\sqrt{2+\sin x}+\sqrt{3}}\right]$
$ =\lim _{x \rightarrow \frac{\pi}{2}} \frac{2+\sin x-3}{(1-\sin x)(1+\sin x)(\sqrt{2+\sin x}+\sqrt{3})}$
$ =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x-1}{-(\sin x-1)(1+\sin x)(\sqrt{2+\sin x}+\sqrt{3})}$
$ =\lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{-(1+\sin x)(\sqrt{2+\sin x}+\sqrt{3})}$
$ \ldots\left[
\sin x \neq \frac{\pi}{2}, \sin x \rightarrow 1, \therefore \sin x-1 \neq 0
\right]$
$=\frac{-1}{\lim _{x \rightarrow \frac{\pi}{2}}(1+\sin x)(\sqrt{2+\sin x}+\sqrt{3})}$
$ =\frac{-1}{\lim _{x \rightarrow \frac{\pi}{2}}(1+\sin x) \cdot \lim _{x \rightarrow \frac{\pi}{2}}(\sqrt{2+\sin x}+\sqrt{3})}$
$ =\frac{-1}{(1+1)(\sqrt{2+1}+\sqrt{3})}$
$=\frac{-1}{2 \times 2 \sqrt{3}}f \left(\frac{\pi}{2}\right)$
$=\frac{-1}{4 \sqrt{3}}$

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Question 84 Marks

Which of the following functions has a removable discontinuity?
$\begin{array}{rlr}
f(x)=\frac{x^3-8}{x^2-4}, & & \text { for } x >2 \\
=3, & \text { for } x=2 \\
=\frac{e^{3(x-2)^2}-1}{2(x-2)^2}, & \text { for } x<2
\end{array}
$

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Question 94 Marks

Discuss the continuity of the following functions at the points indicated against them. $\begin{aligned} f(x) & =\frac{\sqrt{3}-\tan x}{\pi-3 x}, x \neq \frac{\pi}{3} \\ & =\frac{3}{4}, \quad \text { for } x=\frac{\pi}{3}, \text { at } x=\frac{\pi}{3} .\end{aligned}$

Answer

$f \left(\frac{\pi}{3}\right)=\frac{3}{4} \quad \ldots$ (given)
$
\lim _{x \rightarrow \frac{\pi}{3}} f(x)=\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{3}-\tan x}{\pi-3 x}
$
Put $\frac{\pi}{3}-x= h$,
$
\therefore \quad x=\frac{\pi}{3}- h
$
As $x \rightarrow \frac{\pi}{3}, h \rightarrow 0$
$
\therefore \quad \lim _{x \rightarrow \frac{\pi}{3}} f (x)=\lim _{ h \rightarrow 0} \frac{\sqrt{3}-\tan \left(\frac{\pi}{3}- h \right)}{\pi-3\left(\frac{\pi}{3}- h \right)}
$
$
=\lim _{h \rightarrow 0} \frac{\sqrt{3}-\frac{\tan \frac{\pi}{3}-\tan h}{1+\tan \frac{\pi}{3} \tan h}}{3 h}
$
$=\lim _{h \rightarrow 0} \frac{\sqrt{3}-\frac{\sqrt{3}-\tan h}{1+\sqrt{3} \tan h}}{3 h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{3}(1+\sqrt{3} \tan h)-(\sqrt{3}-\tan h)}{3 h(1+\sqrt{3} \tan h)}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{3}+3 \tan h-\sqrt{3}+\tan h}{3 h(1+\sqrt{3} \tan h)}$
$=\lim _{h \rightarrow 0} \frac{4 \tan h}{3 h(1+\sqrt{3} \tan h)}$
$=\lim _{h \rightarrow 0} \frac{4}{3(1+\sqrt{3} \tan h)} \times \frac{\tan h}{h}$
$=\frac{4}{3}\left(\lim _{h \rightarrow 0} \frac{1}{1+\sqrt{3} \tan h}\right)\left(\lim _{h \rightarrow 0} \frac{\tan h}{h}\right)$
$=\frac{4}{3}\left[\frac{1}{1+\sqrt{3}(0)}\right]$
$=\frac{4}{3}$
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{3}} f (x) \neq f \left(\frac{\pi}{3}\right)$
$\therefore \quad f (x)$ is discontinuous at $x=\frac{\pi}{3}$.

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Question 104 Marks

Find all the points of discontinuities of f(x) = [x] on the interval (-3, 2).

Answer

f(x) = [x], x ∈ (-3, 2)
i.e., f(x) = -3, x ∈ (-3, -2)
= -2, x ∈ [-2, -1)
= -1, x ∈ [- 1, 0)
= 0, x ∈ [0, 1)
= 1, x ∈ [1, 2)

Similarly, f(x) is discontinuous at the points x = -1, x = 0, x = 1.
Thus all the integer values of x in the interval (-3, 2),
i.e., the points x = -2, x = -1, x = 0 and x = 1 are the required points of discontinuities.

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Maths Solve the Following Question.(4 Marks)

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