KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q9-Q12)
Exercise 9.1
(Q1-Q4) (Q5-Q8) (Q9-Q12) (Q13-Q16) (Q17-Q20) (Q21-Q24) (Q25-Q28) (Q29-Q33)
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]
f (0)=0f (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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