KC sinha solutions | myhelper: KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q17-Q20)

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KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q17-Q20)

Exercise 9.1

Question 17

(यदि) If

$\begin{aligned} f(x) &=\frac{1}{2}-x, 0 \leq x<\frac{1}{2} \\ &=\frac{1}{2}, x=\frac{1}{2} \\ &=\frac{3}{2}-x, \frac{1}{2}<x \leq 1 \end{aligned}$

तो f(x) के सांतत्य की जाँच करें ।

[examine the continuity of f(x)]
Sol :
At

$x=\frac{1}{2}$

L.H.L

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

$=\frac{1}{2}-\frac{1}{2}=0$

R.H.L

$\lim _{x \rightarrow \frac{1}{2}^+} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{3}{2}-x\right)$

$=\frac{3}{2}-\frac{1}{2}=\frac{2}{2}$

=1

∴

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

f(x),

$x=\frac{1}{2}$ पर असंतता है ।

Question 18

यदि (If )

$\begin{aligned} f(x) &=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1 \left\{\begin{array}{l}x<1 \\ x>1\end{array}\right.\\ &=2, \quad x=1 \end{aligned}$

तो f(x) के सांतत्य की जाँच करें ।

[examine the continuity of f(x)]
Sol :
At x=1 ,
L.H.L

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$

$=\lim _{x \rightarrow 1}\left(\frac{x^{2}-3 x-2 x+3}{x^{2}-1^2}\right)$

$=\lim _{x \rightarrow 1}\left[\frac{x(x-3)-1(x-3)}{(x-1)(x+1)}\right]$

$=\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{(x-1)(x+1)}$

$=\frac{1-3}{1+1}$

$=\frac{-2}{2}$

=-1

R.H.L

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$

=-1

f(1)=2

f(x),x=1 पर असंतता है ।

Question 19

If

$f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$

तो f(x) के सांतत्य की जाँच करें ।

examine the continuity of f(x)
Sol :
At x=1

L.H.L

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

$=1^{2}$

=1

R.H.L

$\lim_{x\rightarrow1^+}f(x)=\lim _{x \rightarrow 1}(2 x-1)$

=2(1)-1

=1

f(1)=2(1)-1
=2-1
=1

∴

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

f(x),x=1 पर संतता है

Question 19

If

$f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$
examine the continuity of f(x)
Sol :

At x=2,

L.H.L

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

=2(2)-1

=4-1

=3

R.H.L

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(x+3)$

=2+3

=5

∴

$\lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$

f(x), x=2 पर असंतता है ।

Question 20

(यदि) If

$f(x)=\frac{x^{2}-4}{x-2}, \quad 0<x<2$
=x+1 , 2≤x≤5

तो f(x) के सांतत्यता की जाँच करें
test the continuity of f(x)
Sol :
At x=2 ,

L.H.L

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$

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

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$

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

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

=2+2

=4

R.H.L

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}$

=4

R.H.L

$\lim _{x \rightarrow 2}+f(x)=\lim _{x \rightarrow 2}(x+1)$

=2+1

=3

$\therefore \lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$

f(x), x=2 पर असंतता(discontinuous) है

Question 17

(यदि) If

$\begin{aligned} f(x) &=\frac{1}{2}-x, 0 \leq x<\frac{1}{2} \\ &=\frac{1}{2}, x=\frac{1}{2} \\ &=\frac{3}{2}-x, \frac{1}{2}<x \leq 1 \end{aligned}$

तो f(x) के सांतत्य की जाँच करें ।

[examine the continuity of f(x)]
Sol :
At

$x=\frac{1}{2}$

L.H.L

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

$=\frac{1}{2}-\frac{1}{2}=0$

R.H.L

$\lim _{x \rightarrow \frac{1}{2}^+} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{3}{2}-x\right)$

$=\frac{3}{2}-\frac{1}{2}=\frac{2}{2}$

=1

∴

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

f(x),

$x=\frac{1}{2}$ पर असंतता है ।

Question 18

यदि (If )

$\begin{aligned} f(x) &=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1 \left\{\begin{array}{l}x<1 \\ x>1\end{array}\right.\\ &=2, \quad x=1 \end{aligned}$

तो f(x) के सांतत्य की जाँच करें ।

[examine the continuity of f(x)]
Sol :
At x=1 ,
L.H.L

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$

$=\lim _{x \rightarrow 1}\left(\frac{x^{2}-3 x-2 x+3}{x^{2}-1^2}\right)$

$=\lim _{x \rightarrow 1}\left[\frac{x(x-3)-1(x-3)}{(x-1)(x+1)}\right]$

$=\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{(x-1)(x+1)}$

$=\frac{1-3}{1+1}$

$=\frac{-2}{2}$

=-1

R.H.L

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$

=-1

f(1)=2

f(x),x=1 पर असंतता है ।

Question 19

If

$f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$

तो f(x) के सांतत्य की जाँच करें ।

examine the continuity of f(x)
Sol :
At x=1

L.H.L

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

$=1^{2}$

=1

R.H.L

$\lim_{x\rightarrow1^+}f(x)=\lim _{x \rightarrow 1}(2 x-1)$

=2(1)-1

=1

f(1)=2(1)-1
=2-1
=1

∴

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

f(x),x=1 पर संतता है

Question 19

If

$f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$
examine the continuity of f(x)
Sol :

At x=2,

L.H.L

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

=2(2)-1

=4-1

=3

R.H.L

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(x+3)$

=2+3

=5

∴

$\lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$

f(x), x=2 पर असंतता है ।

Question 20

(यदि) If

$f(x)=\frac{x^{2}-4}{x-2}, \quad 0<x<2$
=x+1 , 2≤x≤5

तो f(x) के सांतत्यता की जाँच करें
test the continuity of f(x)
Sol :
At x=2 ,

L.H.L

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$

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

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$

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

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

=2+2

=4

R.H.L

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}$

=4

R.H.L

$\lim _{x \rightarrow 2}+f(x)=\lim _{x \rightarrow 2}(x+1)$

=2+1

=3

$\therefore \lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$

f(x), x=2 पर असंतता(discontinuous) है

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