KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q5-Q8)

Exercise 9.1











Question 5

साचित करें कि f(x)=|x|, x=0 पर संतत है।
[Prove that f(x)=|x| is continuous at x=0]
Sol :
$f(x)=\left\{\begin{array}{l}-x \text{ if x<0} \\ 0\text{ if x=0}\\x\text{ if x>0}\end{array}\right.$

At x=0

L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim_{x\rightarrow0}(-x)$
=-0
=0

R.H.L
$\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0}(x)$
=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


Question 6

(क्या)Is $f(x)=(1+x)^{\frac{1}{x}}$ , when x≠0
f(x)=e , when x=0
x=0 पर संतत हैं ? (continuous at x=0 ?)
Sol :
At x=0

L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim_{x\rightarrow0}(1+x)^{\frac{1}{x}}$

[if$\frac{1}{x} \rightarrow 0 \Rightarrow x \rightarrow \infty$]

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

R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}$
[if$\frac{1}{x} \rightarrow 0 \Rightarrow x \rightarrow \infty$]

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

f(0)=e

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

x=0 पर f(x) संतता(continuous) है

Question 7

क्या f(x) , x=0 पर संतत है? जहाँ
[If  f(x) continuous at x=0 ? where]

$f(x)=\frac{\cos a x-\cos b x}{x^{2}}, \text{where }x\neq0 \left\{\begin{matrix}x<0\\x>0\end{matrix}\right.$
$f(x)=\frac{b^2-a^2}{2}, \text{where }x=0$
Sol :
At x=0 ,

L.H.L

=$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$
[$\cos C-\cos D=2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$]

$=\lim _{x \rightarrow 0} \frac{2 \sin \frac{a x+b x}{2} \operatorname{sin} \frac{b x-a x}{2}}{x^{2}}$

$=2 \lim _{x \rightarrow 0} \frac{\sin \left(\frac{a+b}{2}\right) x}{x} \cdot \lim _{x \rightarrow 0} \frac{\sin \left(\frac{b-9}{2}\right) x}{x}$

$=2 \lim _{x \rightarrow 0} \frac{\operatorname{sin}\left(\frac{a+b}{2}\right) x}{\left(\frac{a+b}{2}\right)x} \times\left(\frac{a+b}{2}\right) \cdot \lim _{x \rightarrow 0} \frac{\sin \left(\frac{b-a}{2}\right)x}{\left(\frac{b-a}{2}\right) x}\times \left(\frac{b-a}{2}\right)$

$=2 \times\left(\frac{a+b}{x}\right) \times \frac{b-a}{2}$

$=\frac{b^{2}-a^{2}}{2}$

R.H.L
=$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\cos a x-\cos bx}{x^{2}}$

$=\frac{b^{2}-a^{2}}{2}$

$f(0)=\frac{b^{2}-a^{2}}{2}$

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

f(x),x=0  संतता(continuous) है

Question 8

(यदि) If  $f(x)=\left\{\begin{array}{ll}\frac{|x^3|}{x},\text{when }x\neq0;\left\{\begin{matrix}x<0\\x>0\end{matrix}\right. \\ 0,\text{when }x=0\end{array}\right.$
तो f(x) का x=0 पर सांतत्य की जाँच करें ।
[then test the continuity of f(x) at x=0]
Sol :
At x=0 ,

L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\left|x^{3}\right|}{x}$

$=\lim _{x \rightarrow 0} \frac{-x^{2}}{x}=0^2$
=0

R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{| x^{-3}|}{x}$

$=\lim _{x \rightarrow 0} \frac{x^{3}}{x}=0^{2}$

=0

f(0)=0

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

f(x) ,x=0  संतता(continuous) है


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