Exercise 9.1
Question 1
निम्नलिखित फलनों के सांतत्य (संतता) की जाँच दुए हुए बिन्दुओं पर कीजिए ।[Examine the continuity of the following functions at indicated points]
[Note : f(x), x=a
f(a)𝜖R
\lim _{x \rightarrow a} f(x)=f(a)]
(i) f(x)=2x+3 , (at) x=1 पर
Sol :
x=1 पर ,
f(1)=2(1)+3
=5𝜖R
∴f(x) , x=1 पर परिभाषित है ।
\lim _{x \rightarrow 1} f(x) \equiv \lim _{x \rightarrow 1}(2 x+3)
=2(1)+3
=5
\lim _{x \rightarrow 1} f(x)=f(1)
∴ f(x) , x=1 पर संतता(continuous) है
(ii) f(x)=2x2-1 , (at) x=3 पर
Sol :
x=3 पर ,
f(3)=2(3)2-1
=17𝜖R
∴ f(x) , x=3 पर परिभाषित है ।
\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(2 x^{2}-1\right)
=2(3)2-1=17
\lim _{x \rightarrow 3} f(x)=f(3)
∴ x=3 , f(x) संतता है ।
(iii) f(x)=5x-3 , (at) x=0 , -3 , 5 पर
Sol :
x=0 पर ,
f(0)=5(0)-3
=-3
f(x),x=0 पर परिभाषित है ।
\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0}(5 x-3)
=5(0)-3
=-3
\lim _{x \rightarrow 0} f(x)=f(0)
∴ f(x),x=0 पर संतता है ।
(b) x=-3 पर ,
f(-3)=5(-3)-3
=-18∈R
∴f(x),x=-3 पर संतता है ।
=\lim _{x \rightarrow-3} f(x)=\lim_{x\rightarrow-3}(5x-3)
=5(-3)-3
=-18
\lim _{x \rightarrow-3} f(x)=f(-3)
∴ f(x),x=-3 पर संतता है ।
(iv) f(x)=x^{3}+1. (at) x=1 पर
Sl :
(v) f(x)=x^{2}, (at) x=0 पर
Sol :
Question 2
साबित करें कि फलन f(x), जहाँ
[Prove that the function f(x) where]f(x)=4x+3 when x
=3x+7 , when x=4
x=4 पर संतता है । (is continuous at x=4)
Sol :
f(x)=\left\{\begin{array}{l}4 x+3, \text {when } x \neq 4\left\{\begin{array}{l}x<4 \\ x>4\end{array}\right. \\ 3 x+7, \text{when }x=4\end{array}\right.
x=4 ,
L.H.L
\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{-}}(4 x+3)
=\lim _{h \rightarrow 0}[4(4-h)+3]
=\lim _{h \rightarrow 0}[16-4 h+3]
=\lim _{h \rightarrow 0}(19-4h)
=19-4(0)=19
R.H.L
\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{+}}(4 x+3)
=\lim _{h \rightarrow 0}[4(4+h)+3]
=\lim_{h \rightarrow 0}(16+4 h+3]
=\lim_{h \rightarrow 0}(19+4h)
=19+4(0)
=19
f(4)=3(4)+7
=19
∴\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(4)
f(x), x=4 पर संतता(continuous) है
ALTERNATE METHOD
f(x)=\left\{\begin{array}{l}4 x+3, \text {when } x \neq 4\left\{\begin{array}{l}x<4 \\ x>4\end{array}\right. \\ 3 x+7, \text{when }x=4\end{array}\right.
L.H.L
\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4}(4 x+3)
=4(4)+3=19
R.H.L
\lim_{x\rightarrow4^{-}}f(x)=\lim _{x \rightarrow 4^+}(x)=f( 4)
f(x),x=4 पर संतता(continuous) है
Question 3
दिखाएँ कि फलन
[Show that the function]f(x)=\left\{\begin{array}{r}x^{3}+3, \text { if } x \neq 0 \\ 1,\text { if } x=0\end{array}\right.
x=4 पर असंतत है [is discontinuous at x=4]
Sol :
At x=0
L.H.L
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(x^{3}+3\right)
=\lim _{h \rightarrow 0}\left[(0-h)^{3}+3\right]
=\lim _{h \rightarrow 0}\left[-h^{3}+3\right]
=-0^{3}+3
=3
R.H.L
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(x^{3}+3\right)
=\lim _{h \rightarrow 0}\left[(0+h)^{3}+3\right]
f(0)=1
∴\lim _{x \rightarrow 0^{-}} f(2)=\lim _{x \rightarrow 0^{+}} f(x)≠f(0)
<to be added>
Question 4
=\frac{1}{2}-x ,when 0<x<\frac{1}{2}
=\frac{1}{2} when x=\frac{1}{2}
तो x=0 तथा x=\frac{1}{2} पर f(x) के सातत्य की जाँच करें ।
[then test the continuity of f(x) at x=0 and x=\frac{1}{2}. ]
Sol :
At 2=0
R.H.L
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1}{2}-x\right)
=\frac{1}{2}-0=\frac{1}{2}
f(0)=0
\therefore \lim _{x \rightarrow 0^{+}} f(x) \neq f(0)
Hence , f(x) is discontinuous at x=0
At x=\frac{1}{2}
L.H.L
\lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}-x\right)
=\frac{1}{2}-\frac{1}{2}=0
f\left(\frac{1}{2}\right)=\frac{1}{2}
∴\lim _{x \rightarrow \frac{1}{2}^{-}}f(x)\neq f\left(\frac{1}{2}\right)
<to be added>
At 2=0
R.H.L
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1}{2}-x\right)
=\frac{1}{2}-0=\frac{1}{2}
f(0)=0
\therefore \lim _{x \rightarrow 0^{+}} f(x) \neq f(0)
Hence , f(x) is discontinuous at x=0
ALTERNATE METHOD
At x=\frac{1}{2}
L.H.L
\lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}-x\right)
=\frac{1}{2}-\frac{1}{2}=0
f\left(\frac{1}{2}\right)=\frac{1}{2}
∴\lim _{x \rightarrow \frac{1}{2}^{-}}f(x)\neq f\left(\frac{1}{2}\right)
<to be added>
Gaurav singh
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