Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.1-Q13-21
Question:(13) Is the function defined by
`f(x)={(x+5, if x<=1), (x-5, if x>1):}` a continuous function?
Solution:
Here `f(x)={(x+5, if x<=1), (x-5, if x>1):}`
L.H.L. =`lim_(x->1^-)f(x)`
`=lim_(x->1^-)(x+5)`
Put `x=1-h` as `x->1^-, h->0`
`:.lim_(h->0)[1-h+5]`
`=lim_(h->0)(6-h)=6-0=6`
R.H.L.=`lim_(x->1^+)f(x)`
`=lim_(x->1^+)(x-5)`
Put `x=1+h` as `x->1^+, h->0`
`:.lim_(h->0)[1+h-5]`
`=lim_(h->0)(h-4)`
`=0-4=-4`
`:.` L.H.L.`!=`R.H.L.
Thus `f(x)` is discontinuous at `x=1`.
Discuss the continuity of the function f, where `f` is defined by
Question: 14.
`f(x)={(3, if 0<=x\<=1), (4, if 1>x\<3), (5, if 3<=x\<=10):}`
Solution:
Here `f(x)={(3, if 0<=x\<=1), (4, if 1>x\<3), (5, if 3<=x\<=10):}`
At `x=1`
L.H.L. `lim_(x->1^-)f(x)`
`=lim_(x->1^+)(3)=3`
R.H.L.= `lim_(x->1^+)f(x)`
`=lim_(x->1^+)(4)=4`
`:.` L.H.L.`!=` R.H.L.
Thus `f(x)` is discontinuous at `x=1`.
L.H.L=`lim_(x->3^-)f(x)`
`lim_(x->3^-)(4)=4`
R.H.L=`lim_(x->3^+)f(x)`
`lim_(x->3^+)(5)=5`
`:.` L.H.L. `!=` R.H.L.
Thus `f(x)` is discontinuous at `x=3`
Question:15 .
`f(x)={(2x, if x<0), (0, if 0<=x\<=1), (4x, if x>1):}`
Solution:
Here `f(x)={(2x, if x<0), (0, if 0<=x\<=1), (4x, if x>1):}`
At `x=0`
L.H.L. `lim_(x->0^-)f(x)`
`=lim_(x->0^-)(2x)`
Put `x=0-h` as `x->0^-,h->0`
`:. lim_(h->0)[2(0-h)]`
`= lim_(h->0)(-2h)`
`=-2xx0=0`
R.H.L=`lim_(x->0^+)f(x)`
`=lim_(x->0^+)(0)=0`
`f(0)=0`
`:.`L.H.L.= R.H.L.=`f(0)`
Thus `f(x)` is continuous at `x=0`
At `x=1`
L.H.L. `lim_(x->1^-)f(x)`
`lim_(x->1^-)(0)=0`
R.H.L.` =lim_(x->1^+)f(x)`
`lim_(x->1^+)(4x)`
Put `x=1+h` as `x-> 1^+, h->0`
`:.lim_(h->0)[4(1+h)]`
`=lim_(h->0)(4+4h)`
`4+4xx0=4`
`:.` L.H.L.`!=` R.H.L.
Thus `f(x)` is discontinuous at `x=1`
Question:16 .
`f(x)={(-2, if x\<=\-1), (2x, if -1<\x\<=1), (2, if x\>1):}`
Solution:
Here`f(x)={(-2, if x\<=\-1), (2x, if -1<\x\<=1), (2, if x\>1):}`
At `x=-1`
L.H.L. `lim_(x->-1^-)f(x)`
`=lim_(x->-1^-)(-2)=-2`
R.H.L. `=lim_(x->-1^+)f(x)`
`=lim_(x->-1^+)(2x)`
Put `x=-1+h` as `x-> -1^+, h->0`
`:. lim_(h->0)2(-1+h)`
`=lim_(h->0)(-2+2h)`
`=-2+0=-2`
`=f(-1)=-2`
L.H.L. =R.H.L.`=f(-1)`
Thus `f(x)` is continuous at `x=-1`
At `x=1`
L.H.L.` =lim_(x->1^-)f(x)`
`=lim_(x->1^-)(2x)`
Put `x=1-h` as `x->1^-, h->0`
`:.lim_(h->0)[2(1-h)]`
`=lim_(h->0)(2-2h)`
`=2-2xx0=2`
R.H.L.`=lim_(x->1^+)f(x)`
`=lim_(x->1^+)(2)=2`
`=f(1)=2xx1=2`
`:.` L.H.L.=R.H.L. =`f(1)`
Thus `f(x)` is continuous at `x=1`
Question:17 .Findthe relationship between a and b so that the function f defined by
`f(x)={(ax+1, if x\<=3), (bx+3, if x>3):}` is continuous of `x=3`
Solution:
Here `f(x)={(ax+1, if x\<=3), (bx+3, if x>3):}`
L.H.L. `lim_(x->3^-)f(x)`
`=lim_(x->3^-)f(ax+1)`
Put `x=3-h` as `x-> 3^-, h->0`
`:.lim_(h->0)[a(3-h)+1]`
`=lim_(h->0)[(3a-ah+1)]=3a+1`
R.H.L.`=lim_(x->3^+)f(x)`
`=lim_(x->3^+)(bx+3)`
Put `x=3+h` as `x->3^+,h->0`
`:.lim_(h->0)[b(3+h)+3]`
`=lim_(h->0)(3b+bh+3)` `=3b+3`
`=f(3)=3a+1`
since `f(x)` is continuous at `x=3`
`:.` L.H.L =R.H.L =`f(3)`
`3a+1= 3b+3`
`=>3a=3b+2`
`=>a=b+2/3`
Question:18 .For what value of `lambda` is the function defined by
`f(x){(lambda(x^2-2x), if x<=0), (4x+1, if x>0):}`
continuous at `x=0`? What about continuity at `x=1`?
Solution:
Here `f(x){(lambda(x^2-2x), if x<=0), (4x+1, if x>0):}`
L.H.L. `lim_(x->0^-)f(x)`
`=lim_(x->0^-)lambda(x^2-2x)`
Put `x=0-h` as `x-> 0^-, h->0`
`:.lim_(h->0)lambda[(0-h)^2-2(0-h)]`
`=lim_(h->0)[lambda(h^2+2h)]=0`
R.H.L.`=lim_(x->0^+)f(x)`
`lim_(x->0^+)(4x+1)`
Put `x=0+h` as `x->0^+,h->0`
`:.=lim_(h->0)[4(0+h)+1]`
`=lim_(h->0)[4h+1]=0+1=1`
`:.` L.H,L `!=` R.H.L
Thus `f(x)` is not continuous at `x=0` for any value of `lambda`
At `x=1`
L.H.L. `lim_(x->1^-)f(x)`
`=lim_(x->1^-)(4x+1)`
Put `x=1-h` as `x-> 1^-, h->0`
`:.lim_(h->0)[4(1-h)+1]`
`=lim_(h->0)[5-4h]=5-0=5`
R.H.L `lim_(x->1^+)f(x)`
`=lim_(x->1^+)(4x+1)`
Put `x=1+h` as `x-> 1^+, h->0`
`:.lim_(h->0)[4(1+h)+1]`
`=lim_(h->0)(5+4h)=5+0=5`
`=f(1)=4xx1+1=5`
Thus `f(x)` is continuous at `x-1` for all values of `lambda.`
Question:19 .Show that the function defined by `g(x)=x-[x]` is discontinuous at all integral points. Here `[x]`denotes the greatest integer less than or equal to `x`
Solution:
Here `g(x)=x-[x]`
Let a be an integer then `[a-h=a-1,[a+h]=a` and `[a]=a`
At `x=a`
L.H.L.`lim_(x->a^-)g(x)`
`=lim_(x->a^-)(x-[x])`
Put `x=a-h` as `x->a^-,h->0`
`:.lim_(h->0)(a-h-[a-h])`
`=lim_(h->0)[a-h-(a-1)]`
`=a-a+1=1`
R.H.L.`=lim_(x->a^+)g(x)`
`lim_(x->a^+)(x-[x])`
Put `x=a+h` as `x->a^+,h->0`
`:.lim_(h->0)(a+h-[a+h])`
`lim_(h->0)(a+h-a)`
`=lim_(h->0)h=0`
`:.`L.H.L`!=`R.H.L.
Thus `g(x)` is discontinuous at all integral points.
Question:20 .Is the function defined by `f(x)=x^2-sinx+5` continuous at `x=pi?`
Solution:
Here `f(x)= x^2- sinx+5`
`=lim_(x->pi)f(x)`
`=lim_(x->pi)(x^2-sinx+5)`
`=sin^2pi+pi^2-2pi`
`=0+pi^2-2pi`
`=pi^2-2pi`
`=f(pi)=sin^2pi+pi^2-2pi`
`=0+pi^2-2pi`
`=pi^2-2pi`
`:. lim_(x->pi)f(x)=f(pi)`
Thus `f(x)` is continuous at `x=pi`
Question: 21 .Discuss the continuity of the following functions:
(a)`f(x)=sinx+cos x`
Solution:
(a)Here `f(x)=sinx+cosx`
`=sqrt2(1/(sqrt2)sinx+1/sqrt2 cosx)`
`=sqrt2(sin x cos\ pi/4 + cos x sin\ pi/4)`
`=sqrt2 sin (x+\pi/4)`
At `x=a`
`=lim_(x->a)f(x)`
`=lim_(x->a)sqrt2 sin(x+pi/4)`
`=sqrt2 sin(a+pi/4)`
`=f(a)=sqrt2 sin(a+pi/4)`
`[∵ f(x)=sqrt2 sin(x+pi/4)] `
`:.lim_(x->a)f(x)=f(a)`
Thus `f(x)` is continuous at all points.
(b)`f(x)=sin x-cos x`
Solution:
Here `f(x)=sin x-cos x`
`=sqrt2(1/sqrt2 sin x-1/sqrt2 cos x)`
`=sqrt2(sin x cos\pi/4-cos x sin\pi/4)`
`=sqrt2 sin(x-x/4)`
At ` x=a`
`lim_(x->a)f(x)`
`=lim_(x->a)sqrt2 sin(x-pi/4)`
`=sqrt 2 sin(a-pi/4)`
`=f(a)=sqrt2 sin (a-pi/4)`
`[∵ f(x)=sqrt2 sin(x-pi/4)]`
`:. lim_(x->a)f(x)=f(a)`
Thus `f(x)` is continuous at all points
(c) `f(x)=sin xcos x`
Solution:
Here `f(x)=sin xcos x`
`=1/2 (2sinxcosx)=1/2 sin2x`
At `x=a`
`=lim_(x->a)f(x)`
`=lim_(x->a)sin2x=1/2 sin2x`
`=1/2sin 2a`
`=f(a)=1/2\sin 2a`
`[∵ f(x)=1/2\sin 2x]`
`:. lim(x->a)f(x)=f(a)`
Thus `f(x)` is continuous at all points.
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