Exercise 9.1
Question 21
निम्नलिखित फलनों के असांत्यता के बिन्दुओं को ज्ञात करें ।
[Find the points of discontinuity of the following functions]KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q21-Q24)
(i) f(x)=\left\{\begin{array}{l}x+1,(\text { if }) x \geq 1 \\ \left.x^{2}+1, \text {(if }\right) x<1\end{array}\right.
Sol :
At x=1
L.H.L
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}+1\right)
=1^{2}+1
=2
R.H.L
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x+1)
=1+1
=2
f(1)=1+1=2
∴\lim _{x \rightarrow 1^{-}} f(x)=\lim_{x+1^+}f(x)=f(1)
f(x) <to be added>
(ii) f(x)= \begin{array}{ll}2 x+3, & \text {(if) } x \leq 2 \\ 2 x-3, & \text {(if) } x>2\end{array}
Sol :
(iii) Find the points of discontinuity of the following functions:
f(x)=\left\{\begin{array}{ll}x, 0 \leq x<\frac{1}{2} \\ 1, x=\frac{1}{2} \\ 1-x, \frac{1}{2}<x \leq 1\end{array}\right.
Sol :
At x=\frac{1}{2}
L.H.L
\lim _{x \rightarrow \frac{1}{2}^-} f(x)=\lim _{x \rightarrow \frac{1}{2}}(x)=\frac{1}{2}
R.H.L
\lim_{x \rightarrow \frac{1}{2}+} f(x)=\lim _{x \rightarrow \frac{\pi}{2}}(1-x)
=1-\frac{1}{2}=\frac{1}{2}
f\left(\frac{1}{2}\right)=1
∴\lim _{x \rightarrow \frac{1}{2}-}f(x)=\lim _{x \rightarrow \frac{1}{2}+}f( x)\neq f\left(\frac{1}{2}\right)
f(x) , x=\frac{1}{2} पर असंतता(discontinuous) है
(iv) f(x)= \begin{array}{ll}x^{3}-3, & \text { (if) } x \leq 2 \\ x^{2}+1, & \text { (if) } x>2\end{array}
Sol :
(v) f(x)= \begin{array}{ll}x^{3}-3, & \text {(if) } x \leq 2 \\ x^{2}+1, & \text {(if) } x>2\end{cases}
Sol :
(vi) f(x)=\left\{\begin{array}{l}\frac{\sin x}{x},\text{if }x<0 \\ x+1,\text{if }x\geq 0\end{array}\right.
Sol :
At x=0,
L.H.L
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin x}{x}
=1
R.H.L
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(x+1)
=0+1
=1
f(0)=0+1
=1
∴\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)
f(x), <to be added>
(vii) f(x)=\left\{\begin{array}{ll}2x+3, -3 \leq x<-2 \\ x+1,-2\leq x<0 \\ x+2, 0\leq x \leq 1\end{array}\right.
Sol :
At x=-2
At x=0
Question 22
k का मान तिनकालें ताकि निम्नलिखित फलन दिए हुए बिन्दु पर संतत हों ।
[Find the values of k such that the following functions are continuous at the indicated point](i) f(x)=\left\{\begin{array}{l}k x+1,\text{if }x\leq 5 \\ 3 x-5,\text{if }x>5\end{array}\right.
Sol :
At x=5
L.H.L
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 3}(k x+1)
=k(5)+1
=5k+1
R.H.L
\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(3 x-5)
=3(5)-5
=10
∵f(x),x=5 पर संतता है
5k+1=10
5k=9
k=\frac{9}{5}
(ii) f(x)=\left\{\begin{array}{ll}k x^{2}, \text {(if ) } x \leq 2 \\ 3, \text {(if) } x>2\end{array}\right.
Sol :
At x=2 ,
L.H.L
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(k x^{2}\right)
=k(2)^{2}
=4k
R.H.L
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(3)=3
∵f(x),x=2 पर संतता है ।
\lim _{x \rightarrow 2^{-}} f(x)=\lim_{x\rightarrow 2^{+}} f (x)
4k=3
k=\frac{3}{4}
(iii) f(x)=\left\{\begin{array}{ll}k x+1, \text {(if ) } x \leq \pi \\\cos x, \text {(if) } x>\pi\end{array}\right.
Sol :
At x=π
L.H.L
\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi}(k x+1)=
=kπ+1
R.H.S
\lim_{x\rightarrow\pi^+}f(x)=\lim _{x \rightarrow \pi}(\cos x)
=cosπ
=-1
∵f(x),x=π
\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)
kπ+1=-1
kπ=-2
k=-\frac{2}{\pi}
(iv) f(x)=\left\{\begin{array}{ll}\frac{1-\cos 4 x}{8 x^{2}}, & x \neq 0\left\{\begin{array}{ll}x< 0\\x>0\end{array}\right. \\ k ;x=0\end{array}\right.
Sol :
At x=0
L.H.L
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos {4} x}{8 x^{2}}\right)
=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{{8 x^{2}}}
=\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)^{2}
=(1)^{2}=1
R.H.L
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos 4 x}{8 x^{2}}\right)
=1
f(0)=k
∵f(x),x=0 पर संतता है
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)
1=1=k
∴k=1
(v) f(x)=\left\{\begin{array}{cl}\frac{2^{x+2}-16}{4^{x}-16} & x \neq 2\left\{\begin{array}{l}x<2 \\ x>2\end{array}\right. \\ k, x=2\end{array}\right.
Sol :
At x=2 ,
L.H.L
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}
=\lim_{x{\rightarrow 2}} \frac{2^{x} \cdot 2^{2}-16}{4^{x}-16}
=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}\right)^{2}-4^{2}}
=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}
R.H.L
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=\frac{1}{2}
f(2)=k
∵ f(x), x=2 पर संतता है ।
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)
\frac{1}{2}=\frac{1}{2}=k
∴k=\frac{1}{2}
(यदि) If f(x)=\left\{\begin{array}{ll}1, \text {(if ) } x<5 \\ a x+b, \text { (if ) } 3<x<5 \\ 7, \text { (if ) } x \geq 5\end{array}\right.
Sol :
At x=5
L.H.L
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 3}(k x+1)
=k(5)+1
=5k+1
R.H.L
\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(3 x-5)
=3(5)-5
=10
∵f(x),x=5 पर संतता है
5k+1=10
5k=9
k=\frac{9}{5}
(ii) f(x)=\left\{\begin{array}{ll}k x^{2}, \text {(if ) } x \leq 2 \\ 3, \text {(if) } x>2\end{array}\right.
Sol :
At x=2 ,
L.H.L
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(k x^{2}\right)
=k(2)^{2}
=4k
R.H.L
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(3)=3
∵f(x),x=2 पर संतता है ।
\lim _{x \rightarrow 2^{-}} f(x)=\lim_{x\rightarrow 2^{+}} f (x)
4k=3
k=\frac{3}{4}
(iii) f(x)=\left\{\begin{array}{ll}k x+1, \text {(if ) } x \leq \pi \\\cos x, \text {(if) } x>\pi\end{array}\right.
Sol :
At x=π
L.H.L
\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi}(k x+1)=
=kπ+1
R.H.S
\lim_{x\rightarrow\pi^+}f(x)=\lim _{x \rightarrow \pi}(\cos x)
=cosπ
=-1
∵f(x),x=π
\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)
kπ+1=-1
kπ=-2
k=-\frac{2}{\pi}
(iv) f(x)=\left\{\begin{array}{ll}\frac{1-\cos 4 x}{8 x^{2}}, & x \neq 0\left\{\begin{array}{ll}x< 0\\x>0\end{array}\right. \\ k ;x=0\end{array}\right.
Sol :
At x=0
L.H.L
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos {4} x}{8 x^{2}}\right)
=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{{8 x^{2}}}
=\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)^{2}
=(1)^{2}=1
R.H.L
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos 4 x}{8 x^{2}}\right)
=1
f(0)=k
∵f(x),x=0 पर संतता है
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)
1=1=k
∴k=1
(v) f(x)=\left\{\begin{array}{cl}\frac{2^{x+2}-16}{4^{x}-16} & x \neq 2\left\{\begin{array}{l}x<2 \\ x>2\end{array}\right. \\ k, x=2\end{array}\right.
Sol :
At x=2 ,
L.H.L
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}
=\lim_{x{\rightarrow 2}} \frac{2^{x} \cdot 2^{2}-16}{4^{x}-16}
=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}\right)^{2}-4^{2}}
=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}
R.H.L
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=\frac{1}{2}
f(2)=k
∵ f(x), x=2 पर संतता है ।
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)
\frac{1}{2}=\frac{1}{2}=k
∴k=\frac{1}{2}
Question 23
a और b का मान ज्ञात करें जिसके लिए f एक संतत फलन है।
find the value of a and be for which f(x) si continuous function
Sol :
At x=3
L.H.L
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}(1)=1
R.H.L
\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}(a x+b)
=a(3)+b
=3a+b
∵f(x),x=3 पर संतता है ।
\lim _{x \rightarrow 3}-f(x)=\lim _{x \rightarrow 3^{+}} f(x)
1=3a+b
3a+b=1..(i)
f(x)=\left\{\begin{array}{ll}1, \text {(if ) } {x \leq 3} \\ a x+b, \text { (if }) 3<x<5 \\ 7, \text {(if ) } x \geq 5\end{array}\right.
At x=5,
L.H.L
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(a x+b)
=5a+b
R.H.L
\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(7)=7
∵f(x),x=5 पर संतता है
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)
5a+b=7..(ii)
(i) तथा (ii) को जोड़ने पर
\begin{array}3 a+b=1\\5 a+b=7 \\\hline -2a=-6 \end{array}
a=3
a का मान (i) मे रखने पर,
3a+b=1
3(3)+b=1
b=1-9
b=-8
∴a=3 , b=-8
(यदि) If \begin{aligned} f(x) &=a x^{2}-b, 0 \leq x<1 \\ &=2, x=1 \\ &=x+1, 1<x \leq 2 \end{aligned}
x=1 पर संतत है तो a और b में सम्बन्ध ज्ञात करें ।
find the value of a and be for which f(x) si continuous function
Sol :
At x=3
L.H.L
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}(1)=1
R.H.L
\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}(a x+b)
=a(3)+b
=3a+b
∵f(x),x=3 पर संतता है ।
\lim _{x \rightarrow 3}-f(x)=\lim _{x \rightarrow 3^{+}} f(x)
1=3a+b
3a+b=1..(i)
f(x)=\left\{\begin{array}{ll}1, \text {(if ) } {x \leq 3} \\ a x+b, \text { (if }) 3<x<5 \\ 7, \text {(if ) } x \geq 5\end{array}\right.
At x=5,
L.H.L
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(a x+b)
=5a+b
R.H.L
\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(7)=7
∵f(x),x=5 पर संतता है
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)
5a+b=7..(ii)
(i) तथा (ii) को जोड़ने पर
\begin{array}3 a+b=1\\5 a+b=7 \\\hline -2a=-6 \end{array}
a=3
a का मान (i) मे रखने पर,
3a+b=1
3(3)+b=1
b=1-9
b=-8
∴a=3 , b=-8
Question 24
x=1 पर संतत है तो a और b में सम्बन्ध ज्ञात करें ।
[is continuous at x=1. find the relation between a and b]
Sol :
At x=1
L.H.L
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(a x^{2}-b\right)
=a(1)^{2}-b
=a-b
R.H.L
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x+1)
=1+1
=2
∵f(x),x=1 पर संतत है।
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)
a-b=2 or a=2+b
Sol :
At x=1
L.H.L
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(a x^{2}-b\right)
=a(1)^{2}-b
=a-b
R.H.L
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x+1)
=1+1
=2
∵f(x),x=1 पर संतत है।
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)
a-b=2 or a=2+b
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