Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.1
Question: (1) Prove that the function `f(x)=5x-3` is continuous at `x=0`, at `x=-3` and at `x=5`.
Solution:
Here `f(x)=5x-3`
At `x=0`
`:. lim_(x->0)f(x)=lim_(x->0)(5x-3)`
`=5xx0-3=-3`
`f(0)=5xx0-3=-3`
`:.lim_(x->0)(x)=f(0)`
Thus `f(x)` is continuous at `x=0`
At `x=-3`
`:.lim_(x->-3) f(x)`
`lim_(x->-3)(5x-3)`
=`5xx(-3)-3`
=`-15-3=-18`
`f(-3)=5xx-3-3=-18`
`:. lim_(x->-3)f(x)=f(-3)`
Thus `f(x)` is continuous at `x=-3`
`At x=5`
`:. lim_(x->5)f(x)=lim_(x->5)(5x-3)`
`=5xx5-3=25-3=22`
`f(5)= 5xx5-3=25-3=22`
`:.lim_(x->5)f(x)=f(5)`
Thus `f(x)` is continuous at `x=5`
Question: (2) Examine the continuity of the functio `f(x)=2x^2-1` at `x-3`.
Solution:
Here `f(x)=2x-1`
`:.lim_(x->3)f(x)=lim_(x->3)xx2x^2-1`
`=2xx(3)^2-1=18-1=17`
`f(3)=2xx(3)^2-1=18-1=17`
`:. lim_(x->3)f(x)=f(3)`
Thus `f(x)` is continuous at `x=3`
Question: (3) Examine the following functions for continuity
(a)`f(x)=x-5`
Solution:
`f(x)=1/(x-5)` is a quotient function of two polynomial functions, so `f(x)` is continuous at. all values of `x` provided `x!=5`
(b)`f(x)=1/(x-5),x!=5`
Solution:
`f(x)=1/(x-5)` is a quotient function of two polynomial functions, so`f(x)` is continuous at all values of x provided`x!=5.`
(c) `f(x)= (x^2-25)/(x+5), x!=5`
Solution:
`f(x)= (x^2-25)/(x+5)`
`=((x+5)(x-5))/(x+5)=x-5`
`:.f(x)=x-5` is a polynomial function, so `f(x)` is continuous at all values of `x`
(d)`f(x)=|x-5|`
Solution:
`f(x)=|x-5|={(x-5, text(for), x>=5), (5-x, text(for), x<5):}`
For `x->5^+ lim_(x->5^+)f(x)`
`lim_(x->5^+)(x-5)=5-5=0`
For `x->5^, lim_(x->5^-)f(x)`
`=lim_(x->5^-)(5-x)=5-5=0`
`f(5)=5-5=0`
∵ L.H.L.= R.H.L.=`f(5)=>` The function is continuous at `x=5`
Question: (4) Prove that the function `f(x)=x^n` is continuous of `x=n`, where n is a positive integer.
Solution:
Here`f(x)=x^n`
`:.lim_(x->n) f(x)= lim_(x->n)x^n=n^n`
`f(n)=n^n`
`[∵ f(x)=x^n]`
`:.lim_(x->n)f(x)=f(n)`
Thus `f(x)` is continuous at `x=n`, where n is a positive integer
Question:5 .
Is the function f defined by
`f(x)={(x, if x>=1), (5, if x>1):}` continuous at `x=0?` at `x=1?` at `x=2?`
Solution:
Here `f(x)={(x, if x>=1), (5, if x>1):}`
At `x=0`
L.H.L.`=lim_(x->0^-)f(x)`
`=lim_(x->0^-)(x)`
Put `x=0-h` as `x->0^-,h->0`
`lim_(h->0^-)(0-h)=0-0=0`
R.H.L.=`lim_(x->0^+)f(x)`
=`lim_(x->0^+)(x)`
Put `x=0+h` as `x->0^+,h->0`
`lim_(h->0)(0+h)=0+0=0`
`[∵ f(x)=x]`
`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^-)(x)`
`x=1-hasx->1^-,h->0`
`lim_(h->0)(1-h)=1-0=1`
R.H.L.=`lim_(x->1^+)f(x)=5`
`:.L.H.L !=R.H.L.`
Thus `f(x)` is discontinuous at `x=1`
At `x=2`
`f(2)=5`
`:.L.H.L.=R.H.L.=f(2)`
Thus all points of discontinuity of f, where f is defined by
Question:6 .
`f(x)={:{(2x+3, if x<=2), (2x-3, ifx>2):}`
Solution:
Here `f(x)={:{(2x+3, if x<=2), (2x-3, ifx>2):}`
L.H.L=`lim_(x->2)f(x)`
=`lim_(x->2)(2x+3)`
Put `x=2-h\ as\ x->2,h->0`
`:.lim_(h->0)[2(2-h)+3]`
=`lim_(h->0)(7-2h)=7-2xx0=7`
R.H.L=`lim_(x->2^+)f(x)`
`lim_(x->2^+)(2x-3)`
Put `x=2-h` as `x->2^-`, `h->0`
Put x`-2+h\ as\ x->2^+,h->0`
`:. lim_(h->0)[2(2+h)-3]`
`=lim_(h->0)(1+2h)`
`=1+2xx0=1`
`:.`L.H.L `!=` R.H.L
Question: 7 .
`f(x)={:{(|x|+3, if x<= -3), (-2x, if -3<\x\<\3), (6x+2, if x>=3):}`
Solution:
Here `f(x)={:{(|x|+3, if x<= -3), (-2x, if -3<\x\<\3), (6x+2, if x>=3):}`
At `\ x =-3`
L.H.L.`lim_(x->-3^-)f(x)`
`=lim_(x->-3^-)\(|x|+3)`
Put `x=(-3-h)as x->-3-, h->0`
`:.lim_(h->0)[|-3-h|+3]`
=`lim_(h->0)(6+h)=6+0=6`
R.H.L=`lim_(x->3^+)f(x)`
=`lim_(x->3^+)(-2x)`
put `x=-3+hasx->3^+,h->0`
`lim_(h->0)-2(-3+h)`
`lim_(h->0)(6-2h)`
=`-2xx0=6`
`f(-3)=|-3|+3=3+3`
`=6[:.f(x)=|x|+3]`
`:.` L.H.L = R.H.L.= `f(-3)`
Thus `f(x)` is continuoud at `x=-3`
At `x=3`
L.H.L.=`lim_(x->-3^-)f(x)`
`=lim_(x->-3^-)(-2x)`
Put`x=3-h as x->3^-,h->0`
`:. =lim_(h->0)-2(3-h)`
=`lim_(h->0)(-6+2h)`
R.H.L=`=lim_(x->-3^+)f(x)`
=`lim_(x->-3^+)(6x+2)`
Put `x=3+h as x->3^+,h->0`
`:.lim_(h->0)[6(3+h)+2]`
=`lim_(h->0)(18+6h+2)`
=`lim_(h->0)(20+6h)`
`=20+6xx0+20`
Thus `f(x)` is discontinuous at `x=3`
Question:8 .
`f(x)={:{(|x|/x, if x!=0), (0, if x=0):}`
Solution:
Here `f(x)={:{(|x|/x, if x!=0), (0, if x=0):}`
L.H.L=`lim_(x->0^-)f(x)`
`=lim_(x->0^-)|x|/x`
Put `x=0-h as x->0^-,h->0`
`:. lim_(h->0)(|0-h|)/(0-h)`
`lim_(h->0)h/-h =-1`
R.H.L.=`lim_(x->0^+)f(x)`
=`lim_(x->0^+)(|x|)/x`
Put `x= 0+h as x->0^+,h->0`
`:. lim_(h->0)(|0+h|)/(0+h)`
=`lim_(h->0)h/h=1`
`:.` L.H.L `!=` R.H.L.
Thus `f(x)`is discontinuous at `x=0`
Question:9 .
`f(x)={(x/|x|, if x<0), (-1, ifx>=0):}`
Solution:
Here `f(x)={(x/|x|, if x<0), (-1, ifx>=0):}`
L.H.L.`=lim_(x->0^-)f(x)`
`=lim_(x->0^-)x/|x|`
Put `x=0-h` as `x->0^-,h->0`
`:.=lim_(h->0)(0-h)/(|0-h|)`
`=lim_(h->0)(-h)/h=-1`
R.H.L.`=lim_(x->0^+)f(x)=-1`
`f(0)=-1`
`:.` L.H.L=R.H.L= `f(0)`
Thus `f(x)` is continuous at `x=0` ,There is no point of disconinuity for this function `f(x)`
Question:(10)
`f(x)={(x+1, if x>=1), (x^2+1, if x<1):}`
Solution:
Here `f(x)={(x+1, if x>=1), (x^2+1, if x<1):}`
L.H.L. =`=lim_(x->1^-)f(x)`
`=lim_(x->1^-)(x^2+1)`
Put` x=1+h` as `x->1^+,h->0`
`lim_(h->0)[(1-h)^2+1]`
`=lim_(h->0)[1+h^2-2h+1]`
`=lim_(h->0)[2+h^2-2h]`
`=2+0-0=2`
R.H.L.`=lim_(x->1^+)f(x)`
`lim_(x->1^+)(x+1)`
Put `x=1+h` as `x->1^+, h->0`
`:.lim_(h->0)(1+h+1)`
`lim_(h->0)(2+h)`
`=2+0=2`
`f(1)=1+1=2`
`[∵ f(x) =x+1]`
`:.` L.H.L.=R.H.L.=`f(1)`
Thus `f(x)` is continuous at `x=1`. There is no point of discontinuity for this function `f(x)` .
Question: 11 .
`f(x)={(x^3-3, if x>=2), (x^2+1, if x<2):}`
Solution:
Here `f(x)={(x^3-3, if x>=2), (x^2+1, if x<2):}`
L.H.L. =`lim_(x->2^-)f(x)`
`=lim_(x->2^-)f(x^3-3)`
Put `x=2-h` as `x->2^-, h->0`
`:.lim_(h->0)[(2-h)^3-3]`
`=lim_(h->0)[(8-h^3-12h+6h^2-3]`
`lim_(h->0)(5-h^3-12h+6h^2)=5`
R.H.L.`lim_(x->2^+)f(x)`
`=lim_(x->2^+)(x^2+1)`
Put `x=2+h` as `x->2^+,h->0`
`:. lim_(h->0)[(2+h)^2+1]`
`=lim_(h->0)(4+h^2+4h+1]`
` lim_(h->0)(5+h^2+4h]=5`
`f(2)=(2)^3-3`
`=8-3=5`
`[∵ f(x)=x^3-3]`
`:.` L.H.L.=R.H.L. `=f(2)`
Thus `f(x)` is continuous at `x=2` There is no point of discontinuity for this function `f(x)`.
Question: 12 .
`f(x)={:{(x^10-1, if x<=1), (x^2, if x>1):}`
Solution:
Here`f(x)={(x^10-1, if x<=1), (x^2, if x>1):}`
L.H.L`=lim_(x->1^-)f(x)`
`=lim_(x->2^-)(x^10-1)`
Put `x=1-h` as `x->1^-,h->0`
`:. lim_(h->0)[(1-h)^10-1]`
`=(1-0)^10-1`
`=1-1=0`
R.H.L.=` lim_(x->1^+)f(x)`
`= lim_(x->1^+)(x^2)`
Put `x=1+h` as `x->1^+,h->0`
`lim_(h->0)(1+h)^2`
`lim_(h->0)1+h^2+2h=1`
`:.L.H.L. `!=` R.H.L.
Thus `f(x)` is not continuous at `x=1`
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