KC sinha solutions | myhelper: KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q9-Q12)

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KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q9-Q12)

Exercise 9.1

Question 9

सिद्ध करें कि

$f(x)=x^{n}, x=0$ पर संतत है, जहाँ n एक धन पूर्णांक है।

[Prove that the function

$f(x)=x^{n}, x=0$ is continuous at x=0 , where n is a positive integer.]
Sol :
x=0 ,

f(0)=

$=0^{n}$ =0

f(x), x=0 पर परिभाषित है ।

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{n}$

$=0^{n}$
=0

$\lim _{x \rightarrow 0} f(x)=f(0)$

f(x) ,x=0 पर संतता(continuous) है

Question 10

क्या

$f(x)=x^{2}-\sin x+5$ से परिभाषित फलन

$x=\pi$ पर सतत है ?

Is the function defined by

$f(x)=x^{2}-\sin x+5$ continuous at

$x=\pi ?$
Sol:
At x = 𝜋

L.H.S

$\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi}\left(x^{2}-\sin x+5\right)$

$=\lim _{h \rightarrow 0}\left[(\pi-h)^{2}-\sin(\pi-h)+5\right]$

$=\lim _{h \rightarrow 0}\left[x^{2}-2 \pi h+h^{2}-\sin h+5\right]$

$= \pi^{2}-2 \pi(0)+0^{2}-\sin \theta+5$

$=\pi^{2}+5$

R.H.L

$\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi}\left(x^{2}-\sin x+5\right)$

$=\lim _{h \rightarrow 0}\left[(\pi+h)^{2}-\sin (\pi+h)+5\right]$

$=\lim _{h \rightarrow 0}\left[(\pi+h)^{2}-\sin (\pi+h)+5\right]$

$=\lim _{h \rightarrow 0}\left(\pi^{2}+2 \pi h+h^{2}+\operatorname{sin} h+5\right)$

$= \pi^{2}+2 \pi(0)+0^{2}+\sin 0+5$

$=\pi^{2}+5$

$f(\pi)=\pi^{2}-\sin \pi+5$

$=\pi^{2}+5$

∴

$\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)=f(x)$

f(x) , x=π पर संतता(continuous) है

Question 11

(i) f(x) का x=0 पर सांतत्य की जाँच करें, जहाँ

[Test the continuity of f(x) at x=0 where]

$\begin{matrix}f(x)=\frac{x e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right.\\=0,x=0\end{matrix}$
Sol :
At x=0

L.H.L

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{x e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{(0-h) \cdot e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}\right)$

$=\lim _{h \rightarrow 0} \frac{-h \cdot e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}$

$=-0 \cdot \frac{e^{-\infty}}{1+e^{-\infty}}=0$

R.H.S

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{x \cdot e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{(0+h) \cdot e^{\frac{1}{0+h}}}{1+e^{\frac{1}{0+h}}}\right)$

$={\lim}_{h \rightarrow 0} \frac{h \cdot e^{\frac{1}{h}}}{1+e^{\frac{1}{h}}}$

$=0 \cdot \frac{e^{\infty}}{1+e^{\infty}}=0$

f(0)=0

∵

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}}f(x)=f(0)$

f(x),x=0 पर संतता(continuous) है

(ii) माना कि [Let ]

$\begin{aligned} f(x) &=\frac{e^{\frac{1}{x}}-1}{1+e^{\frac{1}{x}}}\text{when }x\neq0 \left\{\begin{matrix}x<0\\x>0\end{matrix}\right. \\ &=0 ,\text{when }x=0 \end{aligned}$

दिखाएँ कि

$f(x), x=0$ पर असंतत है।

[Show that f(x) is discontinuous at x=0]
Sol :
At x=0 ,

L.H.L

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}}-1}{1+e^{\frac{1}{x}}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{e^{\frac{1}{0-h}-1}}{1+e^{\frac{1}{0-h}}}\right)$

$=\lim _{h \rightarrow 0} \frac{e^{\frac{-1}{h}}-1}{1+e^{-\frac{1}{h}}}$

$=\frac{e^{-\infty}-1}{1+e^{-\infty}}$

$=\frac{\frac{1}{e^{\infty}}-1}{1+\frac{1}{e^{\infty}}}$

$=\frac{\frac{1}{\infty}-1}{1+\frac{1}{\infty}}$

$=\frac{0-1}{1+0}=\frac{-1}{1}$

=-1

R.H.L

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}-1}}{1+e^{\frac{1}{x}}}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}-1}}{e^{\frac{1}{x}+1}}\right)$

$=\lim _{x \rightarrow 0} \frac{e^{\frac{1}{x}}\left(1-e^{-\frac{1}{x}}\right)}{e^{\frac{1}{x}}\left(1+e^{\frac{-1}{x}}\right)}$

$=\lim _{h \rightarrow 0}\left(\frac{1-e^{-\frac{1}{0+h}}}{1+e^{-\frac{1}{0+h}}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{1-e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}\right)$

$=\frac{1-e^{-\infty}}{1+e^{-\infty}}$

$=\frac{1-0}{1+0}=\frac{1}{1}=1$

$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)$

f(x), x=0 पर असंतता(discontinuous) है

Question 12

निम्नीलिखत फलनों के निदिष्ट बिन्दुओं पर सांतत्य की जाँच करें ।

[Test the continuity of the following functions at indicated points]
(i)

$f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, x \neq 1\left\{\begin{array}{l}x<1 \\ x>1\\\end{array}\right.\\ 2, x=1\end{array}\right.$
Sol :
At x=1

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)$

$=\lim _{x \rightarrow 1}\left(\frac{x^{2}-1^{2}}{x-1}\right)$

$={\lim}_{x\rightarrow1} \frac{(x-1)(x+1)}{x-1}$

R.H.L

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)$

=1+1
=2

f(1)=2

∴

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$

f(x) ,x=1 संतता(continuous) है

(ii)

$f(x)=\left\{\begin{array}{cc}\frac{x^{2}-x-6}{x-3}, x \neq 3 \left\{\begin{array}{c}x<3 \\x>3\end{array}\right. \\ 5, x=3\end{array}\right.$
Sol :
x=3

L.H.L

$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(\frac{x^{2}-x-6}{x-3}\right)$

$=\lim _{x \rightarrow 3}\left(\frac{x^{2}-3 x+2 x-6}{x-3}\right)$

$=\lim _{x \rightarrow 3}\left(\frac{x(x-3)+2(x-3)}{x-3}\right)$

$=\lim _{x \rightarrow 3} \frac{(x-3)(x+2)}{x-3}$

=3+2
=5

R.H.L

$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}\left(\frac{x^{2}-x-6}{x-3}\right)$

=5

f(3)=5

∴

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3)$

f(x) ,x=3 संतता(continuous) है

(iii)

$f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}},x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right. \\ 1, x=0\end{array}\right.$
Sol :
L.H.L

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)$

$=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4}} \times \frac{1}{4}$

$=\frac{1}{4}\times {2}\lim_{x\rightarrow0} \frac{\sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4}}$

$=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$

$=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$

$=\frac{1}{2}(1)^{2}=\frac{1}{2}$

R.H.L

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)=\frac{1}{2}$

f(0)=1

$\therefore \lim _{x \rightarrow 0^{-}} f(x)\neq\lim _{x \rightarrow 0^{+}} f(x)\neq f(0)$

f(x) ,x=0 असंतता(discontinuous) है

(iv)

$f(x)=\left\{\begin{array}{cl}\frac{|x-a|}{x-a},x \neq a\left\{\begin{array}{l}x<a \\ x>a\end{array}\right. \\ 1, x=a\end{array}\right.$
Sol :
At x=a

L.H.L

$\lim _{x \rightarrow a^{-}} f(x)=\lim_{x\rightarrow a^{-}} \frac{|x-a|}{x-a}$

$=\lim _{x \rightarrow 0} \frac{-(x-a)}{x-a}$

=-1

R.H.L

$\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}} \frac{|x-a|}{x-a}$

$=\lim_{x\rightarrow a} \frac{x-a}{x-a}$

=1

∴

$\lim _{x \rightarrow a^{-}} f(x) \neq \lim _{x\rightarrow a^+} f(x)$

f(x) ,x=a असंतता (discontinuous) है

(v)

$f(x)=\left\{\begin{array}{cc}\frac{\sin 3 x}{x}, & x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right. \\ 1, & x=0\end{array}\right.$
Sol :
At x=0

L.H.L

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin3 x}{x}$

$=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \times 3$

=1×3
=3

R.H.L

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}$

$=\lim_{x\rightarrow {0}} \frac{\sin 3 x}{3 x} \times 3$

=1×3
=3

f(0)=1

∴

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$

f(x) ,x=0 असंतता (discontinuous) है

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