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1.

`f(x)={{:((1+2x)^(2)",","जब",x ne 2),(e^(2)",","जब",x = 0):}x = 0` पर सांतत्य की जाँच कीजिए ।

Answer» (i) `f(0) = e^(2)`
(ii) `f(0+0)=underset(h rarr 0)(lim)f(0+h)=underset(h rarr 0)(lim)[1 + 2(0+h)]^((1)/(0+h))`
`=underset(h rarr 0)(lim)[1+2h]^((1)/(h))`
`=underset(h rarr 0)(lim)[1+2h]^((1)/(2h)xx2)`
माना 2h = t, जब `h rarr 0`, तब `t rarr 0`
`therefore" "f(0+0)=underset(t rarr 0)(lim)[1+t]^((1)/(t)xx2)`
`rArr" "f(0+0)=e^(2)," "[because underset(t rarr 0)(lim)(1+x)^((1)/(x))=e]`
(iii) `f(0-0)=underset(h rarr 0)(lim)f(0-h)underset(h rarr 0)(lim)[1 + 2(0-h)]^((1)/(0-h))`
`=underset(h rarr 0)(lim)[1-2h]^((1)/(-h))`
`=underset(h rarr 0)(lim)[1-2h]^((1)/(-2h)xx2)`
माना `-2h = t` जब `h rarr 0`, तब `t rarr 0`
`therefore" "f(0-0)=underset(t rarr 0)(lim)[1+t]^((1)/(t)xx2)," "[because underset(x rarr 0)(lim)(1+x)^((1)/(x))=e]`
`rArr" "f(0-0)=e^(2)`
`because f(0 + 0)=f(0-0)=f(0)`
अत: दिया गया फलन x = 0 पर संतत है ।
Discussion
2.

`f(x)={{:((sin^(-1))/(x)",","जब",x ne 0),(1",","जब",x = 0):}x = 0` पर सांतत्य की जाँच कीजिए ।

Answer» (i) `f(0)=1`
(ii) `f(0-0)=underset(h rarr 0)(lim)f(0+h)=underset(h rarr 0)(lim)(sin^(-1)(0+h))/(0+h)`
`=underset(h rarr d)(lim)(sin^(-1)h)/(h)`
`=1," "[because underset(x rarr 0)(lim)(sin^(-1)x)/(x)=1]`
(ii) `f(0-0)=underset(h rarr 0)(lim)f(0-h)=underset(h rarr 0)(lim)(sin^(-1)(0-h))/(0-h)`
`=underset(h rarr 0)(lim)(sin^(-1)(-h))/(-h)`
`=underset(h rarr 0)(lim)(sin^(-1)h)/(-h)," "[because sin^(-1)(-x)=-sin^(-1)x]`
`=underset(h rarr 0)(lim)(sin^(-1)h)/(h)," "[because underset(x rarr 0)(lim)(sin^(-1)x)/(x)=1]`
=1
`therefore" "f(0-0)=f(0+0)=f(0)`
इसलिए फलन f(x) बिन्दु x = 0 पर संतत है ।
Discussion
3.

यदि `f(x)={{:((|x|-1)/(x-1)",",x ne 1),(-1",",x = 1):}`, तब दर्शाइये कि f बिन्दु x = 1 पर असंतत है ।

Answer» (i) `f(1) =-1`
(ii) R.H.L. `=underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1+h)`
`=underset(h rarr 0)(lim)(|1+h|-1)/((1+h)-1)`
`=underset(h rarr 0)(lim)(0+h)/(h)`
`=underset(h rarr 0)(lim)(h)/(h)`
= 1.
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(1-h)`
`=underset(h rarr 0)(lim)(|1-h|-1)/((1-h)-1)`
`=underset(h rarr 0)(lim)(|-h|)/(-h)`
`=underset(h rarr 0)(lim)(h)/(-h)," "[because |-x|=x]`
= -1.
`therefore" "underset(x rarr 1^(+))(lim)f(x) ne underset(x rarr 1^(-))(lim)f(x)`
अत: फलन f(x) बिन्दु x = 0 पर असंतत है ।
Discussion
4.

(a) दर्शाइये कि फलन `f(x) = sin(x^(2))` द्वारा परिभाषित फलन एक संतत फलन है । (b) दर्शाइये कि फलन `f(x) = |1 - x + |x||` द्वारा परिभाषित फलन f जहाँ x एक वास्तविक संख्या है, एक संतत फलन है । (c) दर्शाइये कि फलन `sin|x|` एक संतत फलन है ।

Answer» (a) माना `g(x) = x^(2)` और `h(x)=sin x` अब `(hog)(x)=h(g(x))=h(x^(2))=sin(x^(2))=f(x)`
`rArr" "hog = f`
चूँकि g(x) और h(x) दोनों संतत फलन हैं और f फलनों h और g का संयोजन है । अत: f एक संतत फलन है क्योकि दो संतत फलनों का संयोजन एक संतत फलन होता है ।
(b) माना `g(x)=1-x+|x|` और `h(x)=|x|`
चूँकि `g(x)=1-x+|x|` दो संतत फलनों क्रमश: बहुपद फलन और मापांक फलन का योगफल है इसलिए g(x) एक संतत फलन है । साथ ही h(x) एक संतत फलन है ।
अब `(hog)(x)=h(g(x))=h(1-x+|x|)`
`= |1-x+|x||`
`rArr" "(hog)(x)=f(x)`
`rArr" "hog=f`
अत: f दो संतत फलनों का संयुक्त फलन है इसलिए f भी एक संतत फलन है ।
(c) माना `f(x)=|x|` और `g(x)=sin x` तब f और g संतत फलन हैं ।
`therefore" "(gof)(x)=g(f(x))=g(|x|)=sin|x|`
चूँकि f और g संतत हैं इसलिए gof भी संतत है । अत: `sin|x|` एक संतत फलन है ।
Discussion
5.

यदि एक फलन f निम्नानुसार परिभाषित है - `f(x)={{:((|x-4|)/(x-4)",",x ne 4),(" "0",",x = 4):}` दर्शाइये कि f बिन्दु x = 4 के अतिरिक्त प्रत्येक बिन्दु पर संतत है ।

Answer» `f(x)={{:((|x-4|)/(x-4)",",x ne 4),(" "0",",x = 4):}`
`{{:((-(x-4))/(x-4)=-1",",x lt 4),((x-4)/(x-4)=1",",x gt 4),(" "0",",x = 4):}`
`[because |x-4|={{:(-(x-4)","x lt 4),(" "x-4","x ge 4):}}]`
स्थिति I. `x lt 4` के लिए `f(x) = -1`, जो कि अचर फलन है । अत: यह संतत फलन है ।
स्थिति II. `x gt 4` के लिए `f(x)=1`, जो कि अचर फलन है । अत: यह संतत फलन है ।
स्थिति III. x = 4 के लिए `f(4) = 0`
अब `underset(x rarr 4^(+))(lim)f(x)=underset(h rarr 0)(lim)f(4+h)=1`
`underset(x rarr 4^(-))(lim)f(x)=underset(h rarr 0)(lim)f(4-h)=-1`
`therefore" "underset(x rarr 4^(+))(lim)f(x) ne underset(x rarr 4^(-))(lim)f(x)`
`rArr f(x)` बिन्दु x = 4 पर संतत नहीं है ।
अत: f(x) बिन्दु x = 4 के अतिरिक्त सभी बिन्दुओ x = 4 पर संतत है ।
Discussion
6.

दर्शाइये कि फलन `f(x) = {{:(x^(3)+3",",x ne 0),(" "1",",x = 0):} x = 0` पर संतत नहीं है ।

Answer» (i) `f(0)=1`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(0+h)^(3)+3," "[because x = 0 + h ne 0]`
`=underset(h rarr 0)(lim)h^(3)+3`
`=0 + 3 = 3`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(0-h)^(3)+3" "[because x = 0 - h ne 0]`
`=underset(h rarr 0)(lim)(-h)^(3)+3`
`= 0 + 3 = 3`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x) ne f(0)`
अत: f(x) बिन्दु x = 0 पर संतत नहीं है ।
Discussion
7.

दर्शाइये कि फलन f(x) जहाँ `f(x)={{:((sin x)/(x)+cos x",",x ne 0),(" "2",",x = 0):} x = 0` पर संतत है ।

Answer» (i) `f(0)=2`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)[(sin (0+h))/(0+h)+cos(0+h)]`
`=underset(h rarr 0)(lim)[(sin h)/(h)+cos h]`
`=underset(h rarr 0)(lim)(sin h)/(h)+underset(h rarr 0)(lim)cos h`
`= 1 + 1 = 2`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)[(sin (0-h))/((0-h))+cos (0-h)]`
`=underset(h rarr 0)(lim)[(sin(-h))/(-h)+cos(-h)]`
`=underset(h rarr 0)(lim)[(-sin h)/(-h)+cos h]," "[because sin (-theta)=-sin theta,cos(-theta)=cos theta]`
`=underset(h rarr 0)(lim)[(sin h)/(h)+cos h]`
`=underset(h rarr 0)(lim)(sin h)/(h) + underset(h rarr 0)(lim)cos h`
`= 1 + 1 = 2`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)=2`
अत: फलन f(x) बिन्दु x = 0 पर संतत है ।
Discussion
8.

दर्शाइये कि फलन `f(x) = |x|` एक संतत फलन है ।

Answer» मापांक फलन कि परिभाषा से,
`f(x)=|x|={{:(x",",x ge 0),(-x",",x le 0):}`
स्पष्टत: dom (f) = R.
माना a एक स्वेच्छ वास्तविक संख्या है ।
स्थिति I. `a lt 0` के लिए, `f(a) =-a`
अब, `underset(x rarr a^(+))(lim)f(x)=underset(h rarr 0)(lim)f(a+h)`
`= underset(h rarr 0)(lim)-(a+h)`,
[`because a + h lt 0` जबकि `a lt 0` और h बहुत ही सूक्ष्म धनात्मक वास्तविक संख्या है]
`= -a + underset(h rarr 0)(lim)(-h)`
`=-a + 0 =-a`
और `underset(x rarr a^(-))(lim)f(x)=underset(h rarr 0)(lim)f(a-h)`
`= underset(h rarr 0)(lim)-(a-h)`,
[`because a - h lt 0` जबकि `a lt 0` और h बहुत ही सूक्ष्म धनात्मक वास्तविक संख्या है]
`=-a + underset(h rarr 0)(lim)h`
`=-a + 0 =-a`
`therefore" "underset(x rarr a^(+))(lim)f(x)=underset(x rarr a^(-))(lim)f(x)=f(a)`
अत: f(x) बिन्दु प्रत्येक `a lt 0` के लिए संतत है ।
स्थिति II. a = 0 के लिए `f(a) = f(0)=0`
अब `underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`= underset(h rarr 0)(lim)f(h)`
`= underset(h rarr 0)(lim)h = 0`
और `underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`= underset(h rarr 0)(lim)f(-h)`
`= underset(h rarr 0)(lim)-(-h)`,
[`because -h lt 0`, इसलिए `f(-h)=-(-h)`]
`=underset(h rarr 0)(lim)h = 0`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0^(-))(lim)f(x)=f(0)`
अत: f(x) बिन्दु x = 0 के लिए संतत है ।
स्थिति III. `a gt 0` के लिए, `f(a)=a`
अब `underset(x rarr a^(+))(lim)f(x)=underset(h rarr 0)(lim)f(a+h)`
`= underset(h rarr 0)(lim)(a+h)`,
[`because a + h gt 0`, इसलिए `f(x) =x`]
`= a + 0`
= a
और `underset(x rarr a^(-))(lim)=underset(h rarr 0)(lim)f(a-h)`
`= underset(h rarr 0)(lim)(a-h)`
[`because a-h lt 0` जबकि `a gt 0` और h बहुत ही सूक्ष्म धनात्मक वास्तविक संख्या है]
`= a-0=a`
`therefore" "underset(x rarr a^(+))(lim)f(x)=underset(h rarr a^(-))(lim)f(x)=f(a)`
अत: f(x) प्रत्येक `a gt 0` के लिए संतत है ।
अत: `f(x)=|x|` संतत है ।
Discussion
9.

दर्शाइये कि फलन f(x), जहाँ `f(x)={{:((sin x)/(x)+cos x",",x gt 0),(2",",x = 0),((4(1-sqrt(1-x)))/(x)",",x lt 0):}` बिन्दु x = 0 पर संतत है ।

Answer» (i) `f(0)=2`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(sin(0+h))/((0+h))+cos(0+h)," "[because x = 0 + h gt 0]`
`=underset(h rarr 0)(lim)[(sin h)/(h)+cos h]`
`=underset(h rarr 0)(lim)(sin h)/(h)+underset(h rarr 0)(lim)cos h`
`=1 + 1 = 2`
(iii) L.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)[(sin (0-h))/((0-h))+cos(0-h)]`
`=underset(h rarr 0)(lim)[(sin(-h))/(-h)+cos h]`
`=underset(h rarr 0)(lim)[(-sin h)/(-h)+cos h]`
`=underset(h rarr 0)(lim)[(sin h)/(h)]+underset(h rarr 0)(lim)cos h`
`=1 + 1 =2`
`therefore" "underset(x rarr 0)(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)=2`
अत: फलन f(x) बिन्दु x = 0 पर संतत है ।
Discussion
10.

निम्नलिखित फलनों की सांतत्यता की जाँच x = 0 पर कीजिए - (a) `f(x) = {{:((|sin x|)/(x)",",x ne 0),(1",",x = 0):}` (b) `f(x) = {{:(x "sin"(1)/(x)",",x ne 0),(0",",x = 0):}`

Answer» (a) यहाँ `f(x)=(|sin x|)/(x),`
(i) `f(0)=1`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(|sin (0 + h)|)/((0+h))," "[because x = 0 + h ne 0]`
`=underset(h rarr 0)(lim)(|sin h|)/(h)`
`=underset(h rarr 0)(lim)(sin h)/(h)`
`= 1" "[because underset(theta rarr 0)(lim)(sin theta)/(theta)=1]`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(|sin (0-h)|)/((0-h))`
`=underset(h rarr 0)(lim)(|sin (-h)|)/(-h)`
`=underset(h rarr 0)(lim)(|-sin h|)/(-h)," "[because sin (-theta)=sin theta]`
`=underset(h rarr 0)(lim)(sin h)/(-h)`
`=-1" "[because underset(theta rarr 0)(lim)(sin theta)d/(theta)=1]`
`therefore" "underset(x rarr 0^(-))(lim)f(x) ne underset(x rarr 0^(+))(lim)f(x)=f(0)`
अत: `f(x), x = 0` पर असंतत है ।
(b) यहाँ `f(x)=x "sin"(1)/(x)`
(i) `f(0)=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(0+h)"sin"(1)/((0+h))`
`=underset(h rarr 0)(lim)h sin((1)/(h))`
`= 0 xx` परिमिति राशि, [`because "sin"(1)/(h)` का मान -1 और +1 के मध्य स्थित होता है]
= 0
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(0-h)sin((1)/(0-h))`
`=underset(h rarr 0)(lim)(-h)sin(-(1)/(h))`
`=underset(h rarr 0)(lim)(-h)(-sin((1)/(h)))" "[because sin(-theta)=-sin theta]`
`=underset(h rarr 0)(lim)hsin((1)/(h))`
`=0 xx` अपरिमित राशि,
[`because "sin"(1)/(h)` का मान -1 और +1 के मध्य स्थित होता है]
= 0
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)`
अत: `f(x), x = 0` पर संतत है ।
Discussion
11.

निम्नलिखित फलनों के सांतत्यता की जाँच x = 0 पर कीजिए - (a) `f(x)={{:(e^(1//x)",",x ne 0),(0",",x = 0):}` `f(x)={{:(e^(1//x)/(1+e^(1//x))",",x ne 0),(0",",x = 0):}` (c) `f(x) = {{:(e^(1//x-1)/(e^(1//x)+1)",",x ne 0),(0",",x = 0):}`

Answer» (a) यहाँ `f(x)=e^(1//x)`
(i) `f(0)=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)e^(1//(0+h))," "[because x = 0 + h ne 0]`
`=underset(h rarr 0)(lim)e^(1//h)`
`=e^(1//10)=e^(oo)=e^(oo)`
(iii) L.H.L. `underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)e^(1//(0-h))," "[because x = 0 - h ne 0]`
`=underset(h rarr 0)(lim)e^(-1//h)`
`=e^(-oo)=0," "[because e^(-oo)=0]`
`therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x) = f(0)`
अत: `f(x), x = 0` पर संतत नहीं है ।
(b) यहाँ `f(x) = (e^(1//x))/(1+e^(1//x))`
(i) `f(0)=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(e^(1//(0+h)))/(1+e^(1//(0+h)))" "[because x = 0 + h ne 0]`
`=underset(h rarr 0)(lim)(e^(1//h))/(1+e^(1//h))`
`=underset(h rarr 0)(lim)(e^(1//h))/(e^(1//h)[e^(-1//h)+1])`
`=underset(h rarr 0)(lim)(1)/(e^(-1//h)+1)`
`= (1)/(e^(-oo)+1)=(1)/(0+1)=1`
(iii) L.H.L. `underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`underset(h rarr 0)(lim)(e^(1//(0-h)))/(1+e^(1//(0-h)))," "[because x = 0 - h ne 0]`
`=underset(h rarr 0)(lim)(e^(-1//h))/(1+e^(-1//h))`
`=(e^(-oo))/(1+e^(-oo))=(0)/(1+0)=0," "[because e^(-oo)=0]`
`therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x) = f(0)`
अत: f(x) बिन्दु x = 0 पर संतत नहीं है ।
(c) यहाँ `f(x) = (e^((1)/(x))-1)/(e^((1)/(x))+1)`
(i) `f(0)=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0^(+))(lim)(e^(1//0+h)-1)/(e^(1//0+h)+1)," "[because x = 0 + h ne 0]`
`=underset(h rarr 0)(lim)(e^(1//h)-1)/(e^(1//h)+1)`
`=underset(h rarr 0)(lim)(e^(1//h)[1-e^(-1//h)])/(e^(1//h)[1 + e^(-1//h)])`
`=underset(h rarr 0)(lim)(1-e^(-1//h))/(1+e^(-1//h))`
`=(1-e^(-oo))/(1+e^(-oo))=(1-0)/(1+0)=1`
(iii) L.H.L. `= underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`underset(h rarr 0)(lim)(e^(1//(0-h))-1)/(e^(1//0-h)+1)," "[because x = 0 - h ne 0]`
`=underset(h rarr 0)(lim)(e^(-1//h)-1)/(e^(-1//h)+1)`
`=(e^(-oo)-1)/(e^(-oo)+1)=(0-1)/(0+1)=-1`
`therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x) ne f(0)`
अत: `f(x), x = 0` पर संतत नहीं है ।
Discussion
12.

फलन f(x) के सांतत्यता की जाँच `x = (1)/(2)` पर कीजिए जहाँ f(x) निम्नानुसार परिभाषित है - `f(x) = {{:((1)/(2)+x",",x le x lt (1)/(2)),(1",",x=(1)/(2)),((3)/(2)+x",",(1)/(2) lt x le 1):}`

Answer» (i) `f((1)/(2))=1`
(ii) R.H.L. `=underset(x rarr (1^(+))/(2))(lim)f(x)=underset(h rarr 0)(lim)f((1)/(2)+h)`
`=underset(h rarr 0)(lim)(3)/(2)+((1)/(2)+h)," "[because x = (1)/(2)+h gt (1)/(2)]`
`=underset(h rarr 0)(lim)(2+h)`
`=2 + 0 = 2`
(iii) L.H.L. `=underset(x rarr (1^(-))/(2))(lim)f(x)=underset(h rarr 0)(lim)f((1)/(2)-h)`,
`=underset(h rarr 0)(lim)(1)/(2)+((1)/(2)-h)" "[because x = (1)/(2)-h lt (1)/(2)]`
`=underset(h rarr 0)(lim)(1-h)`
`=1 - 0= 1`
`therefore" "underset(x rarr (1^(+))/(2))(lim)f(x)ne underset(x rarr (1^(+))/(2))(lim)f(x)=f((1)/(2))`
अत: फलन f(x) बिन्दु `x = (1)/(2)` पर संतत है ।
Discussion
13.

फलन f(t) के सांतत्यता की जाँच `t = (pi)/(2)` पर कीजिए - `f(t) = {{:((cos t)/((pi)/(2)-t)",",t ne (pi)/(2)),(1",",t = (pi)/(2)):}`

Answer» (i) `f((pi)/(2))=1`
(ii) R.H.L. `=underset(t rarr (pi^(+))/(2))(lim)f(t)=underset(h rarr 0)(lim)f((pi)/(2)+h)`
`=underset(h rarr 0)(lim)(cos((pi)/(2)+h))/((pi)/(2)-((pi)/(2)+h))`
`=underset(h rarr 0)(lim)(-sin h)/(-h)`
`=underset(h rarr 0)(lim)(sin h)/(h)=1`
(iii) L.H.L. `=underset(t rarr (pi^(-))/(2))(lim)f(t)=underset(h rarr 0)(lim)f((pi)/(2)-h)`
`=underset(h rarr 0)(lim)(cos((pi)/(2)-h))/((pi)/(2)-((pi)/(2)-h))`
`= underset(h rarr 0)(lim)(sin h)/(h)`
= 1.
`therefore" "underset(t rarr (pi^(+))/(2))(lim)f(t)=underset(t rarr (pi^(-))/(2))(lim)f(t)=1`
अत: `f(t), t=(pi)/(2)` पर संतत है ।
Discussion
14.

`f(x) = |x|` द्वारा परिभाषित महत्तम पूर्णांक फलन के असांतत्य के समस्त बिन्दुओ को ज्ञात कीजिए, जहाँ ।x। उस महत्तम पूर्णांक को प्रकट करता है, जो x से कम या उसके बराबर है ।

Answer» फलन `f(x) = [x]` सभी वास्तविक संख्याओं के लिए परिभाषित है ।
माना a कोई वास्तविक संख्या है ।
स्थिति I. यदि a पूर्णांक नहीं है, तो `f(x) = [a]`
अब `underset(x rarr a)(lim)f(x)=underset(x rarr a)(lim)[x]=[a]=f(a)`
अत: f(x) सभी वास्तविक संख्याओं के लिए संतत है, जो कि पूर्णांक नहीं है ।
स्थिति II. यदि a पूर्णांक हो, तो `f(a) = [a]`
`underset(x rarr a^(+))(lim)f(x)=underset(h rarr 0)(lim)f(a+h)=underset(h rarr 0)(lim)[a+h]=a`
और `underset(x rarr a^(-))(lim)f(x)=underset(h rarr 0)(lim)f(a-h)=underset(h rarr 0)(lim)[a-h]=a-1`
`therefore" "underset(x rarr a^(+))(lim)f(x) ne underset(h rarr a^(-))(lim)f(x)`
अत: f(x) सभी पूर्णांक संख्याओं के लिए संतत नहीं है ।
Discussion
15.

दर्शाइये कि फलन `f(x)={{:(5x-4",","जब",0 lt x le 1),(4x^(2)-3x",","जब",1 lt x lt 2):}x = 1` पर संतत है ।

Answer» (i) `f(1)=5 xx 1-4=1`
(ii) R.H.L. `=underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1+h)`
`=underset(h rarr 0)(lim)4(1+h)^(2)-3(1+h)," "[because x = 1 + h gt 1]`
`=4(1+0)^(2)-3(1+0)`
`= 4 - 3 = 1`
(iii) L.H.L. `=underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1-h)`
`=underset(h rarr 0)(lim)5(1-h)-4," "[because x = 1 - h lt 1]`
`=5(1-0)-4`
`=5-4=1`
`therefore" "underset(x rarr 1^(+))(lim)f(x)=underset(x rarr 1^(-))(lim)f(x)=f(1)`
अत: फलन f(x) बिन्दु x = 1 पर संतत है ।
Discussion
16.

k का मान ज्ञात कीजिए यदि फलन f बिन्दु `x = pi` पर संतत हो, जहाँ `f(x)={{:(kx+1",","यदि",x le pi),(cos x",","यदि",x gt pi):}`

Answer» (i) `f(pi)=k pi + 1`
(ii) R.H.L. `underset(x rarr pi^(+))(lim)f(x)=underset(h rarr 0)(lim)f(pi + h)`
`=underset(h rarr 0)(lim)cos(pi+h)," "[because x = pi + h gt pi]`
`= cos pi`
= - 1
(iii) L.H.L. `underset(x rarr pi^(-))(lim)f(x)=underset(h rarr 0)(lim)f(pi - h)`
`= underset(h rarr 0)(lim)k(pi - h)+1`
`underset(h rarr 0)(lim)(k pi - kh + 1)`
`= k pi + 1`
चूँकि फलन f(x) बिन्दु `x = pi` पर संतत है ।
`therefore" ""R.H.L. = L.H.L."=f(pi)`
`rArr" "k pi + 1 = -1`
`rArr" "k pi = -2`
`rArr" "k = -(2)/(pi)`
Discussion
17.

k का मान ज्ञात कीजिए यदि फलन `f(x)={{:(kx^(2)",","यदि",x le 2),(3",","यदि",x gt 2):}` संतत है ।

Answer» `x lt 2` के लिए `f(x)=kx^(2)` एक बहुपद फलन है इसलिए f(x) संतत फलन है ।
`x gt 2` के लिए `f(x) = 3` एक अचर फलन इसलिए f(x) संतत फलन है ।
अब हम फलन f(x) के सांतत्य की जाँच बिन्दु x = 2 पर करेंगे ।
`f(2) = k(2)^(2)=4k`
अब `underset(x rarr 2^(+))(lim)f(x) = underset(h rarr 0)(lim)f(2+h)`
`=underset(h rarr 0)(lim(3),)," "[because x = 2 + h gt 2]`
= 3.
और `underset(x rarr 2^(-))(lim)f(x) = underset(h rarr 0)(lim)f(2-h)`
`=underset(h rarr 0)(lim) k(2-h)^(2)," "[because x = 2 - h lt 2]`
= 4k
चूँकि फलन `f(x) AA x in R` संतत फलन है । अतएव f(x) बिन्दु x = 2 पर संतत होगा ।
`therefore" "underset(x rarr 2^(+))(lim)f(x)=underset(x rarr 2^(-))(lim)f(x)=f(2)`
`rArr" "3 = 4k`
`rArr" "k = (3)/(4)`
Discussion
18.

`f(x) = {{:((|x|)/(x)",",x ne 0),(0",",x = 0):}`

Answer» Correct Answer - असांतत्य बिन्दु x = 0 पर
संकेत : `f(x) = {{:(1,"यदि",x gt 0),(0,"यदि",x = 0),(-1,"यदि",x lt 0):}`
चूँकि `|x| = x, x gt 0, |x| = - x, x lt 0`
Discussion
19.

यदि प्रदत फलन बिन्दु x = 0 पर संतत हो, तो a, b व c के मान ज्ञात कीजिए - `f(x)={{:((sin(a+1)x+sin x)/(x)",","यदि",x lt 0),(" "c",","यदि",x = 0),((sqrt(x+bx^(2))-sqrt(x))/(bx^(3//2))",","यदि",x gt 0):}`

Answer» (i) `f(0)=c`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(sqrt(h+bh^(2))-sqrt(h))/(bh^(3//2))`
`=underset(h rarr 0)(lim)(sqrt(h(1+bh))-sqrt(h))/(bh^(3//2))`
`=underset(h rarr 0)(lim)(sqrt(h)[sqrt((1+bh))-1])/(b sqrt(h)h)`
`=underset(h rarr 0)(lim)(sqrt(1+bh)-1)/(bh)xx(sqrt(1+bh)+1)/(sqrt(1+bh)+1)`, [परिमेयीकरण द्वारा]
`=underset(h rarr 0)(lim)((1+bh)-1)/(bh(sqrt(1+bh)+1))`
`=underset(h rarr 0)(lim)(bh)/(bh(sqrt(1+bh)+1))`
`=underset(h rarr 0)(lim)(1)/(sqrt(1+bh)+1)`
`rArr" ""R.H.L."=(1)/(1+1)=(1)/(2)`
(iii) L.H.L. `=underset(h rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(sin[(a+1)(0-h)]+sin(0-h))/((0-h))`
`=underset(h rarr 0)(lim)(sin[-(a+1)h]+sin(-h))/(-h)`
`=underset(h rarr 0)(lim)(-sin(a+1)h-sin h)/(-h)" "[because sin(-theta)=-sin theta]`
`=underset(h rarr 0)(lim)(sin(a+1)h + sin h)/(h)`
`=underset(h rarr 0)(lim)(sin(a+1)h)/(h)+underset(h rarr 0)(lim)(sin h)/(h)`
`=(a+1)underset(h rarr 0)(lim)(sin(a+1)h)/((a+1)h)+1`
`rArr" ""L.H.L."=(a+1)xx 1 + 1 = a + 2`
चूँकि फलन f(x) बिंदु x = 0 पर संतत है ।
`therefore" ""R.H.L.=L.H.L." = f(0)`
`rArr" "(1)/(2)=a+2=c`
`rArr" "a+2=(1)/(2)"और" c = (1)/(2)`
`rArr" "a=(1)/(2)-"2 और c"=(1)/(2)`
`rArr" "a=(-3)/(2)"और c"=(1)/(2)`
अत: `a=-(3)/(2),c=(1)/(2)` और b कोई भी वास्तविक मान ग्रहण कर सकता है ।
Discussion
20.

यदि `f(x)={{:((1-cos 4 x)/(x^(2))",","जब",x lt 0),(" "a",","जब",x = 0),(((sqrt(x)))/((16+sqrt(x))-4))",","जब",x gt 0):}` और f बिन्दु x = 0 पर संतत है, तो k का मान ज्ञात कीजिए ।

Answer» (i) `f(0) = a`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(sqrt((0+h)))/((sqrt(16+sqrt((0+h))))-4)," "[because x = 0 + h gt 0]`
`=underset(h rarr 0)(lim)(sqrt(h))/(sqrt(16+sqrt(h)))-4`
`=underset(h rarr 0)(lim)(sqrt(h))/((sqrt(16+sqrt(h)))-4)xx(sqrt(16+sqrt(h))+4)/((sqrt(16+sqrt(h)))+4)`
`=underset(h rarr 0)(lim)(sqrt(h)[sqrt(16+sqrt(h))+4])/(sqrt(h))`
`=underset(h rarr 0)(lim)(sqrt(h)[sqrt(16+sqrt(h))+h])/(sqrt(h))`
`=underset(h rarr 0)(lim)[sqrt(16+sqrt(h))+4]`
`=sqrt(16+0)+4`
`=4 + 4 = 8`
(iii) L.H.L. `underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(1-cos 4(0-h))/((0-h)^(2))," "[because x = 0 - h lt 0]`
`=underset(h rarr 0)(lim)(1-cos 4h)/(h^(2))`
`=underset(h rarr 0)(lim)(1-(1-2 sin^(2)2h))/(h^(2))," "[because sin 2 theta = 1 - 2 sin^(2)theta]`
`=underset(h rarr 0)(lim)(2 sin^(2)2h)/(h^(2))`
`=2 underset(h rarr 0)(lim)((sin 2h)/(2h))^(2)xx 4`
`= 2 xx 1 xx 4 = 8`
चूँकि दिया गया फलन x = 0 पर संतत है ।
`therefore" ""R.H.L.=L.H.L."=f(0)`
`rArr" "a = 8`
Discussion
21.

अंतराल `[-1,2]` में फलन `f(x) = |x| + |x - 1|` के सांतत्यता की विवेचना कीजिए ।

Answer» चूँकि `x in [-1,2]`, तब
`f(x)=-x-(x-1)=1-2x` जबकि `-1 le x lt 0`
`f(x) = x -(x-1)=1` जबकि `0 lt x le 1`
`f(x) = x + x - 1 = 2x -1` जबकि `1 lt x le 2`
`f(x)={{:(1-2x",",-1 le x lt 0),(" "1",",0 le x le 1),(2x-1",",1 lt x le 2):}`
`-1 lt x lt 0` के लिए `f(x)=1-2x` एक बहुपद फलन है इसलिए यह एक संतत फलन है ।
`0 lt x lt 1` के लिए `f(x)=1` एक अचर फलन है इसलिए यह एक संतत फलन है ।
`1 lt x lt 2` के लिए `f(x)=2x-1` एक बहुपद फलन है इसलिए यह एक संतत फलन है ।
अब हम f(x) के सांतत्यता कि जाँच `x =-1,0,1` और 2 पर जाँच करेंगे ।
`x =-1` पर : `f(-1)=1-2(-1)=3`
अब `underset(x rarr (-1)^(+))(lim)f(x)=underset(h rarr 0)(lim)f(-1+h)`
`=underset(h rarr 0)(lim)1-2(-1+h)," "[because -1+h lt "0 और f(x)"= 1-2x,-1 le x lt 0]`
`= underset(h rarr 0)(lim)(3-2h)=3`
`therefore" "underset(x rarr (-1)^(+))(lim)f(x)=f(-1)`
`rArr f(x)` बिन्दु x = -1 पर संतत है ।
x = 0 पर : `f(0) =1`
`underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)f(h)`
`=underset(h rarr 0)(lim)1," "[because h gt "0 और f(x)"= 1, 0 le x le "1 के लिए"]`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=f(0)`
`rArr f(x)` बिन्दु x = 0 पर संतत है ।
x = 1 पर : `f(1) = 1`
`underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1+h)`
`=underset(h rarr 0)(lim)2(1+h)-1," "[because 1 + h gt "0 और f(x)"=2x -1, 1 lt x le "2 के लिए"]`
`= underset(h rarr 0)(lim)(1+2h)=1`
`therefore" "underset(x rarr 1^(+))(lim)f(x)=f(1)`
`rArr f(x)` बिन्दु x = 1 पर संतत है ।
x = 2 पर :`f(2) = 2(2)-1=3`
`underset(x rarr 2^(-))(lim)f(x)=underset(h rarr 0)(lim)f(2-h)`
`=underset(h rarr 0)(lim)2(2-h)-1," "[because 2 - h lt "2 और f(x)"=2x -1, 1 lt x le "2 के लिए"]`
`=underset(h rarr 0)(lim)(3-2h)=3`
`therefore" "underset(x rarr 2^(-))(lim)f(x) = f(2)`
`rArr f(x)` बिन्दु x = 2 पर संतत है ।
अत: f(x) अंतराल (-1, 2) के प्रत्येक बिन्दु तथा x = -1 और 2 पर संतत है ।
अतएव f(x) अंतराल [-1, 2] में संतत है ।
Discussion
22.

निम्नलिखित फलन के सांतत्य पर विचार कीजिए - `f(x) = {{:(x",","यदि",x ge 0),(x^(2)",","यदि",x lt 0):}`

Answer» स्पष्टत: f(x) सभी वास्तविक संख्याओं के लिए परिभाषित है ।
माना a कोई वास्तविक संख्या है ।
स्थिति I. जब `a gt 0`
चूँकि `f(x) = x`, यदि `x ge 0`
तब `f(a) = a`
अब `underset(x rarr a)(lim)f(x) = underset(x rarr a)(lim) x = a = f(a)`
अत: शून्य से बड़ी सभी वास्तविक संख्याओं के लिए संतत है ।
स्थिति II. जब `a lt 0`
चूँकि `f(x) = x^(2)`, यदि `x lt 0`
तब `f(a) = a^(2)`
अब `underset(x rarr a)(lim)f(x) = underset(x rarr a)(lim)x^(2)=a^(2)=f(a)`
अत: शून्य से छोटी वास्तविक संख्याओं के लिए संतत है ।
स्थिति III. जब a = 0
चूँकि `f(x) = x`, यदि `x = 0`
तब `f(a) = a` अर्थात `f(0) = 0`
अब `underset(x rarr 0^(-))(lim)f(x) = underset(h rarr 0)(lim) f(0-h)`
`= underset(h rarr 0)(lim)(0-h)^(2)," "[because x = 0 - h lt 0]`
= 0
और `underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0 + h)`
`=underset(h rarr 0)(lim)(0+h)," "[because x = 0 + h gt 0]`
= 0
`therefore" "underset(x rarr 0^(-))(lim)f(x)=underset(x rarr 0^(+))(lim)f(x)=f(0)`
अत: f(x) बिन्दु x = 0 पर संतत है । अतएव f(x) सभी वास्तविक संख्याओं के लिए संतत है ।
Discussion
23.

फलन `f(x) = |x - 1| + |x - 2|` के सांतत्य की विवेचना x = 1 और x = 2 पर कीजिए ।

Answer» दिया गया है -
`f(x) = |x - 1| + |x - 2|`
`rArr" "f(x)={{:(-(x-1)-(x-2)",","यदि",x lt 1),((x-1)-(x-2)",","यदि",1 le x lt 2),((x-1)+(x-2)",","यदि",x ge 2):}`
`rArr" "f(x)={{:(-2x + 3",","यदि",x lt 1),(1",","यदि",1 le x lt 2),(2x-3",","यदि",x ge 2):}`
x = 1 पर सांतत्यता -
(i) `f(1)=1`
(ii) R.H.L. `=underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1+h)`
`=underset(h rarr 0)(lim)1," "[because x = 1 + h gt 1]`
= 1
(iii) L.H.L. `=underset(x rarr 1^(-))(lim)f(x)=underset(h rarr 0)(lim)f(1-h)`
`=underset(h rarr 0)(lim)-2(1-h)+3," "[because x = 1 - h lt 1]`
`=underset(h rarr 0)(lim)1 + 2h`
= 1
`therefore" ""R.H.L.=L.H.L."=f(1)`
अत: f(x) बिन्दु x = 1 पर संतत है ।
x = 2 पर सांतत्यता -
(i) `f(2)=2 xx 2 - 3 = 1`
(ii) R.H.L. `=underset(x rarr 2^(+))(lim)f(x)=underset(h rarr 0)(lim)f(2+h)`
`=underset(h rarr 0)(lim)2(2+h)-3," "[because x = 2 + h gt 0]`
`=underset(h rarr 0)(lim)1+2h`
`=1 + 0 = 1`
(iii) L.H.L. `= underset(x rarr 2^(-))(lim) f(x) = underset(h rarr 0)(lim)f(2-h)`
`=underset(h rarr 0)(lim)1," "[because x = 2 - h lt 2]`
= 1
`therefore" ""R.H.L.=L.H.L."=f(2)`
अत: f(x) बिन्दु x = 2 पर संतत है ।
Discussion
24.

k का मान ज्ञात कीजिए यदि फलन x = 0 पर संतत है, जहाँ `f(x)={{:((1-cos 4x)/(8x^(2))",","यदि",x ne 0),(" "k",","यदि",x = 0):}`

Answer» (i) `f(0)=k`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(1-cos 4(0+h))/(8(0+h)^(2))" "[because x = 0 + h ne 0]`
`=underset(h rarr 0)(lim)(1-cos 4h)/(8h^(2))`
`=underset(h rarr 0)(lim)(2sin^(2)2h)/(8h^(2))`
`=underset(h rarr 0)(lim)((sin 2h)/(2h))^(2)`
`=1^(2)=1," "[because underset(theta rarr 0)(lim)(sin theta)/(theta)=1]`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(1-cos 4(0-h))/(8(0-h)^(2))`
`=underset(h rarr 0)(lim)(1-cos 4h)/(8h^(2))," "[because cos (-theta)=cos theta]`
= 1
चूँकि फलन f(x) बिन्दु x = 0 पर संतत है ।
`therefore" ""R.H.L.=L.H.L."=f(0)`
`rArr" "k = 1`
Discussion
25.

a तथा b के मानो को ज्ञात कीजिए जबकि `f(x)={{:(" "5",","यदि",x le 2),(ax+b",","यदि",2 lt x lt 10),(" "21",","यदि",x ge 10):}` द्वारा परिभाषित फलन एक संतत फलन है ।

Answer» स्थिति I. `x lt 2` के लिए `f(x) = 5` एक अचर फलन है इसलिए यह एक संतत फलन है ।
स्थिति II. `2 lt x lt 10` के लिए `f(x) = ax + b` एक बहुपद फलन है इसलिए यह एक संतत फलन है ।
स्थिति III. `x gt 10` के लिए `f(x) = 21` एक अचर फलन है इसलिए यह एक संतत फलन है ।
स्पष्टत: फलन f(x) बिन्दुओ x = 2 और x = 10 के अतिरिक्त सभी `x in R` पर संतत है ।
x = 2 पर : `f(2) = 5`
अब `underset(x rarr 2^(+))(lim)f(x)=underset(h rarr 0)(lim)f(2+h)`
`=underset(h rarr 0)(lim)a(2+h)+b," "[because f(x) = ax + "b यदि 2" lt x lt 10]`
= 2a + b
और `underset(x rarr 2^(-))(lim)f(x)=underset(h rarr 0)(lim)f(2-h)`
`=underset(h rarr 0)(lim)5," "[because f(x) = "5 यदि x" le 2]`
= 5
चूँकि f(x) बिन्दु x = 2 पर संतत है । तब
`underset(x rarr 2^(+))(lim)f(x)=underset(x rarr 2^(-))(lim)f(x)=f(2)`
`rArr" "2a + b = 5" "...(1)`
x = 10 पर : `f(10) = 21`,
अब, `underset(x rarr 10^(+))(lim)f(x)=underset(h rarr 0)(lim)f(10 + h)`
`=underset(h rarr 0)(lim) 21," "[because f(x) = "21 यदि x" x ge 10]`
= 21
और `underset(x rarr 10^(-))(lim)f(x)=underset(h rarr 0)(lim)f(10-h)`
`=underset(h rarr 0)(lim)a(10-h)+b," "[because f(x) = ax + "b यदि 2" lt x lt 10]`
= 10a + b
चूँकि f(x) बिन्दु x = 10 पर संतत है । तब
`underset(x rarr 10^(+))(lim)f(x)=underset(x rarr 10^(-))(lim)f(x)=f(10)`
`rArr" "10a + b = 21" "...(2)`
समी. (1) और (2) से,
`a = 2, b = 1`
Discussion
26.

यदि फलन f(x) जहाँ `f(x)={{:((log(1+ax)-log(1-bx))/(x)",",x ne 0),(" "k",",x = 0):}x = 0` पर संतत है, तब k का मान ज्ञात कीजिए ।

Answer» (i) `f(0) = k`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x) = underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(log(1+a(0+h))-log(1-b(0+h)))/((0+h))`
`underset(h rarr 0)(lim)(log(1+ah)-log(1-bh))/(h)`
`=underset(h rarr 0)(lim)(log(1+ah))/(h)-underset(h rarr 0)(lim)(log(1-bh))/(h)`
`=a xx underset(h rarr 0)(lim)(log (1+ah))/(ah)-(-b)underset(h rarr 0)(lim)(log (1-bh))/((-b)h)`
`= a xx 1 - (-b) xx 1," "[because underset(x rarr 0)(lim)(log (1+x))/(x)=1]`
`rArr" ""R.H.L." = a + b`
(iii) L.H.L. `underset(x rarr 0^(-))(lim)f(x) = underset(h rarr 0)(lim)f(0-h)`
`underset(h rarr 0)(lim)(log(1+a(0-h))-log(1-b(0-h)))/((0-h))`
`=underset(h rarr 0)(lim)(log(1-ah)-log(1+bh))/(-h)`
`=underset(h rarr 0)(lim)(log(1-ah))/(-h)+underset(h rarr 0)(lim)(log(1+bh))/(h)`
`= a xx underset(h rarr 0)(lim)(log(1-ah))/((-a)h)+b xx underset(h rarr 0)(lim)(log(1+x))/(bh)`
`rArr" ""L.H.L." = a xx 1 + b xx 1 = a + b, " "[because underset(x rarr 0)(lim)(log(1+x))/(x)=1]`
चूँकि फलन f(x) बिन्दु x = 0 पर संतत है ।
`therefore" ""R.H.L.=L.H.L." = f(0)`
`rArr" "a + b =k`.
Discussion
27.

फलन `f(x) = |x|` के सांतत्य की विवेचना x = 0 पर कीजिए ।

Answer» यहाँ `f(x)=|x|, a = 0`.
(i) `f(0)=|0|=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)|0+h|`
`=underset(h rarr 0)(lim)|h|`
`=|0|=0`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)|0-h|`
`=underset(h rarr 0)(lim)|-h|`
`= |0|`
= 0
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)`
अत: `f(x) = |x|` बिन्दु x = 0 पर संतत है ।
Discussion
28.

x = 1 पर फलन `f(x) = 2x + 3` के सांतत्य की जाँच कीजिए ।

Answer» यहाँ `f(x)=2x + 3, x = 1` पर,
(i) `f(1) = 2 xx 1 + 3 = 5`
(ii) R.H.I. `=underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1+h)`
`=underset(h rarr 0)(lim)2(1+h)+3`
`=underset(x rarr 0)(lim)(2+2h)+3`
`=underset(h rarr 0)(lim)5 + 2h`
`= 5 + 2 xx 0 = 5`
(iii) L.H.L. `=underset(x rarr 1^(-))(lim)f(x)=underset(h rarr 0)(lim)f(1-h)`
`=underset(h rarr 0)(lim)2(1-h)+3`
`=underset(h rarr 0)(lim)(2-2h)+3`
`=underset(h rarr 0)(lim) 5 - 2h`
`= 5 - 2 xx 0 = 5`
`therefore` R.H.L. = L.H.L. `= f(1) = 5`
अत: f(x) बिन्दु x = 1 पर संतत है ।
Discussion
29.

a और b के मध्य संबंध स्थापित कीजिए जिनके लिए `f(x)={{:(ax + 1",","यदि",x le 3),(bx + 3",","यदि",x lt 3):}` द्वारा परिभाषित फलन x = 3 पर संतत है ।

Answer» (i) `f(3)=a xx 3 + 1 = 3a + 1`
(ii) R.H.L. `=underset(x rarr 3^(+))(lim)f(x)=underset(h rarr 0)(lim)f(3+h)`
`=underset(h rarr 0)(lim)b(3+h)+3," "[because x = 3 + h gt 3]`
`=underset(h rarr 0)(lim)3b + bh + 3`
`=3b + 0 + 3`
`rArr" ""R.H.L."=3b + 3`
(iii) L.H.L. `=underset(x rarr 3^(-))(lim)f(x)=underset(h rarr 0)(lim)f(3-h)`
`=underset(h rarr 0)(lim)a(3-h)+1," "[because x = 3 - h lt 3]`
`=underset(h rarr 0)(lim)3a - ah + 1`
`=3a - 0 + 1`
`rArr" "L.H.L."=3a + 1`
चूँकि दिया गया फलन x = 3 पर संतत है,
R.H.L. = L.H.L. = f(3)
`rArr" "3b + 3 = 3a + 1`
`rArr" "3a - 3b = 2`
तब जो कि a और b के मध्य अभीष्ट संबंध है ।
Discussion
30.

दर्शाइये कि `f(x) = x^(2)` सभी वास्तविक संख्याओं के लिए संतत है ।

Answer» माना a एक स्वेच्छ वास्तविक संख्या है ।
`f(a)=a^(2)`
अब `underset(x rarr a^(+))(lim)f(x)=underset(h rarr 0)(lim)f(a+h)`
`= underset(h rarr 0)(lim)(a+h)^(2)=a^(2)`
और `underset(x rarr a^(-))(lim)f(x)=underset(h rarr 0)(lim)f(a-h)`
`= underset(h rarr 0)(lim)(a-h)^(2)=a^(2)`
`therefore" "underset(x rarrd a^(+))(lim)f(x)=underset(x rarr a^(-))(lim)f(x)=f(a)`
अत: f(x) बिन्दु x = a पर संतत है ।
चूँकि a एक स्वेच्छ वास्तविक संख्या है इसलिए f(x) सभी वास्तविक संख्याओं के लिए संतत है ।
विकल्पत: चूँकि `f(x)=x^(2)` बहुपदीय फलन है इसलिए `f(x) = x^(2)` सभी वास्तविक संख्याओं के लिए संतत है ।
Discussion
31.

k का मान ज्ञात कीजिए यदि फलन f(x) बिन्दु `x = (pi)/(2)` पर संतत है, जहाँ `f(x)={{:((k cos x)/(pi - 2x)",","यदि",x ne (pi)/(2)),(3",","यदि",x = (pi)/(2)):}`

Answer» (i) `f((pi)/(2))=3`
(ii) R.H.L. `=underset(x rarr (pi^(+))/(2))(lim)f(x)=underset(h rarr 0)(lim)f((pi)/(2)+h)`
`= underset(h rarr 0)(lim)(k cos((pi)/(2)+h))/(pi-2((pi)/(2)+h))," "[because x = (pi)/(2)+h ne (pi)/(2)]`
`=underset(h rarr 0)(lim)(k(-sin h))/(pi-(pi + 2h))," "[because cos ((pi)/(2)+theta)=-sin theta]`
`=underset(h rarr 0)(lim)(k sin h)/(2h)`
`= (k)/(2)underset(h rarr 0)(lim)(sin h)/(h)`
`=(k)/(2) xx 1=(k)/(2)," "[because underset(theta rarr 0)(lim)(sin theta)/(theta)=1]`
(iii) L.H.L. `=underset(x rarr (pi^(-))/(2))(lim)f(x)=underset(h rarr 0)(lim)f((pi)/(2)-h)`
`=underset(h rarr 0)(lim)(k cos ((pi)/(2)-h))/(pi-2((pi)/(2)-h))`
`=underset(h rarr 0)(lim)(k sin h)/(pi-(pi+2h))`
`=underset(h rarr 0)(lim)(k sin h)/(2h)`
`=(k)/(2)underset(h rarr 0)(lim)(sin h)/(h)=(k)/(2)xx 1 = (k)/(2)`
चूँकि फलन f(x) बिन्दु `x = (pi)/(2)` पर संतत है ।
`therefore" ""R.H.L.=L.H.L."=f((pi)/(2))`
`rArr" "(k)/(2)=3`
`rArr" "k = 6`
Discussion
32.

निम्नलिखित फलन के सांतत्य पर विचार कीजिए - `f(x) = {{:(x+2",",x le 1),(x-2",",x gt 1):}`

Answer» स्पष्टत: f(x) सभी वास्तविक संख्याओं के लिए परिभाषित है ।
माना a कोई वास्तविक संख्या है ।
स्थिति I. जब `a lt 1`
चूँकि `f(x) = x+2` यदि `x le 1`
तब `f(a) = a + 2`
अब `underset(x rarr a)(lim)f(x)=underset(x rarr a)(lim)x + 2 = a + 2 = f(a)`
अत: 1 से छोटी वास्तविक संख्याओं के लिए f(x) संतत है ।
स्थिति II. जब `a gt 1`
चूँकि `f(x)=x-2` यदि `x gt 1`
तब `f(a)=a - 2`
अब `underset(x rarr a)(lim)f(x)=underset(x rarr a)(lim)x - 2 = a - 2 = f(a)`
अत: 1 से बड़ी वास्तविक संख्याओं के लिए f(x) संतत है ।
स्थिति III. जब a = 1
चूँकि `f(x)=x+2` यदि `x = 1`
तब `f(a) = a+2` या `f(1) = 1 + 2 = 3`.
अब `underset(x rarr 1^(-))(lim)f(x)=underset(h rarr 0)(lim)f(1-h)`
`= underset(h rarr 0)(lim) (1 - h + 2), " "[because x = 1 - h lt 1]`
= 3
और `underset(x rarr 1^(+))(lim)f(x) = underset(h rarr 0)(lim) f(1 + h)`
`= underset(h rarr 0)(lim)(1+h-2)," "[because x = 1 + h gt 1]`
= -1
`therefore" "underset(x rarr 1^(-))(lim)f(x) ne underset(h rarr 1^(+))(lim)f(x)`
अत: f(x) बिन्दु `x = 1` पर संतत नहीं है अर्थात x = 1 फलन का असांतत्य बिन्दु है ।
Discussion
33.

`lambda` के किस मान के लिए फलन `f(x)={{:(lambda(x^(2)-2x)",","यदि",x le 0),(4x + 1",","यदि",x gt 0):}x = 0` पर संतत है ?

Answer» (i) `f(0)=lambda(0-0)=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)4(0+h)+1`
`=underset(h rarr 0)(lim)4h + 1`
`= 4 xx 0 + 1 = 1`
(iii) L.H.L. `underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)lambda^(2)[(0-h)^(2)-2(0-h)]`
`=underset(h rarr 0)(lim) lambda^(2)[h^(2)+2h]`
`=lambda^(2)(0+0)=0`
अत: R.H.L. `ne` L.H.L.
परन्तु दिया गया है कि f(x), x = 0 पर संतत है ।
`therefore" ""R.H.L.=L.H.L."=f(0)`
`rArr" ""R.H.L.=L.H.L."` जो कि विरोधाभास है ।
अत: x = 0 पर `lambda` के किसी भी मान के लिए फलन f संतत नहीं है ।
Discussion
34.

दर्शाइये कि फलन `f(x) = 2x - |x|` बिन्दु x = 0 पर संतत है ।

Answer» यहाँ `f(x) = 2x - |x|, a = 0`
(i) `f(0) = 2 xx 0-|0|=0-0=0`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)2(0+h)-|0+h|`
`=underset(h rarr 0)(lim)2h-|h|`
`=0 - 0 = 0`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)2(0-h)-|0-h|`
`=underset(h rarr 0)(lim)-2h-|h|," "[because |-x|=|x|]`
`=0 - 0 = 0`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)`
अत: f(x) बिन्दु x = 0 पर संतत है ।
Discussion
35.

फलन `f(x) = (1)/(x), x ne 0` द्वारा परिभाषित फलन f के सांतत्यता पर विचार कीजिए ।

Answer» चूँकि फलन `f(x) = (1)/(x)` सभी शून्येतर वास्तविक संख्याओं के लिए परिभाषित है । माना `a ne 0` एक वास्तविक संख्या है, तब
`f(a) = (1)/(a)`
अब `underset(x rarr a)(lim)f(x)=underset(x rarr a)(lim)(1)/(x)=(1)/(a)=f(a)`
अत: `f(x)=(1)/(x)` में संतत फलन है ।
Discussion
36.

निम्नलिखित फलनों की सांतत्यता की जाँच x = 0 पर कीजिए - (a) `f(x) = {{:((1-cos x)/(x^(2))",",x ne 0),((1)/(2)",",x = 0):}` (b) `f(x) = {{:((sin ax)/(sin bx)",",x ne 0),((a)/(b)",",x = 0):}`

Answer» (a) यहाँ `f(x)={{:((1-cos x)/(x^(2))",",x ne 0),(" "(1)/(2)",",x = 0):}`
(i) `f(0) = (1)/(2)`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(1-cos(0+h))/((0+h)^(2))`
`=underset(h rarr 0)(lim)(1-cos h)/(h^(2))`
`=underset(h rarr 0)(lim)(1-(1-2 "sin"^(2) (h)/(2)))/(h^(2))" "[because cos x = 1 -2 "sin"^(2)(x)/(2)]`
`=underset(h rarr 0)(lim)(2"sin"^(2)(h)/(2))/(h^(2))`
`=2 underset(h rarr 0)(lim)(("sin"(h)/(2))/((h)/(2)))^(2)xx (1)/(4)`
`=2 xx (1)^(2)xx(1)/(4)," "[because underset(theta rarr 0)(lim)(sin theta)/(theta)=1]`
`=(1)/(2)`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(1-cos(0-h))/((0-h)^(2))`
`=underset(h rarr 0)(lim)(1-cos h)/(h^(2))" "[because cos (-theta)=cos theta]`
`=underset(h rarr 0)(lim)(2"sin"^(2)(h)/(2))/(h^(2))`
`=2 underset(h rarr 0)(lim)(("sin"(h)/(2))/((h)/(2)))^(2)xx(1)/(4)`
`=(1)/(2)`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)`
अत: `f(x), x = 0` पर संतत है ।
(b) यहाँ `f(x)={{:((sin ax)/(sin bx)",",x ne 0),((a)/(b)",",x = 0):}`
(i) `f(0) = (a)/(b)`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(sin a(0+h))/(sin b(0+h))`
`=underset(h rarr 0)(lim)(sin ah)/(sin bh)`
`=underset(h rarr 0)(lim)[(sin ah)/(ah)xxahxx(bh)/(sinbh)xx(1)/(bh)]`
`=(a)/(b)[underset(h rarr 0)(lim)(sin ah)/(ah)xx underset(h rarr 0)(lim)(1)/(((sin bh)/(bh)))]`
`=(a)/(b)xx 1 xx 1," "[because h rarr 0 rArr ah rarr "0 और bh" rarr 0]`
`=(a)/(b)`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(sin a(0-h))/(sin b(0-h))`
`=underset(h rarr 0)(lim)(sin(-ah))/(sin(-bh))`
`= underset(h rarr 0)(lim)(sin ah)/(sin bh)`
`=(a)/(b)`
`therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)=(a)/(b)`
अत: `f(x), x = 0` पर संतत है ।
Discussion
37.

दर्शाइये कि फलन `f(x)={{:((|x|)/(x)",",x ne 0),(1",",x = 0):} x = 0` पर असंतत है ।

Answer» (i) `f(0) = 1`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(|0+h|)/((0+h))`
`=underset(h rarr 0)(lim)(|h|)/(h)`
`=underset(h rarr 0)(lim)(h)/(h)=1," "[because |x|=x]`
(iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)`
`=underset(h rarr 0)(lim)(|0-h|)/((0-h))`
`=underset(h rarr 0)(lim)(|-h|)/(-h)`
`=underset(h rarr 0)(lim)(h)/(-h)," "[because |-x|=x]`
= -1.
`therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x)`
अत: फलन f(x) बिन्दु x = 0 पर असंतत है ।
Discussion