Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.1-Q22-34
Question:22 .Discuss the continuity of the cosine, cosecant, secont and cotangent functions.
Solution:
Here `f(x)=cos x`
At `x=a`
`=lim_(x->a)f(x)`
`=lim_(x->a)cos x=cos a`
`=f(a)=cos a`
`[∵f(x)=cos x]`
`:. lim_(x->a)f(x) =f(a)`
thus `f(x)` is continuous at `x=a,` But a is an arbitrary point so `f(x)` is continuous at all points.
So `f(x)` is continuous at all points.
Here `f(x) = cosec x`
`f(x)` is not defined at `x=n pi, n in Z`
Here `f(x)=sec x`
`f(x)` is not defined at `x=(2n+1) pi/2, n in Z`
Thus `f(x)` is continuous at all points except `x=(2n+1) pi/2,n in Z`
Here `f(x) =cot x`
`f(x)` is not defined at `x=n pi, n in Z`
Thus `f(x)` is continuous at all points except `x=n pi, n in Z`,
Question :23 . Find all points of discontinuity of where
`f(x)={(sinx/x, if x<0), (x+1, if x>=0):}`
Solution:
`f(x)={(sinx/x, if x<0), (x+1, if x>=0):}`
L.H.L=`lim_(x->0^-)f(x)`
`=lim_(x->0^-)sin x/x`
Put `x=0-h` as `x->0^-,h->0`
`=lim_(h->0)(sin(0-h))/(0-h)`
`= lim_(h->0)(-sinh)/-h`
`=lim_(h->0)(sin h)/h=1`
R.H.L.`=lim_(x->0^+)f(x)`
`=lim_(x->0^+)(x+1)`
Put `x=0+h` as `x->0^+,h->0`
`:. lim_(h->0)(0+h+1)`
`=0+1=1`
`=f(0)=0+1=1`
`[∵f(x)=x+1]`
`:.`L.H.L. =R.H.L =`f(0)`
Thus `f(x)` is continuous at `x=0` There is no point of discontinuity for this function `f(x)`
Question: 24 .Determine if `f` defined by
`f(x)={(x^2 sin\ 1/x, ifx=!=0), (0, ifx=0):}`Is a continuous function?
Solution:
Here `f(x)={(x^2 sin\ 1/x, if\ x=!=0), (0, ifx=0):}`
L.H.L.`lim_(x->0^-)f(x)`
`=lim_(x->0^-)x^2 sin\ 1/x`
Put `x=0-h` as `x->0^-,h->0`
`:.lim_(h->0)(0-h)^2 sin\ 1/(o-h)`
`lim_(h->0)-h^2 sin\ 1/h`
`=lim_(h->0)h.sin\ (1/h)/(1/h)`
`=-0xx1=0`
R.H.L. `lim_(x->0^+)f(x)`
` =lim_(x->0^+)x^2 sin\ 1/x`
Put`x=0+h` as `x->0^+, h->0`
`:. lim_(h->0)(0+h)^2 sin\ 1/(0+h)`
`=lim_ (h->0)h^2 sin\ 1/h `
`=lim_ (h->0)h. (sin1//h)/(1//h)`
=`0xx1=0`
`f(0)=0`
`:.`L.H.L. = R.H.L.=`f(0)`
Thus`f(x)` is continuous at `x=0`
Question:25 .Examine the continuity of `f` Where `f` is defined by
`f(x)= {(sinx-cosx, if x!=0), (-1, ifx=0):}`
Solution:
Here `f(x)= {(sinx-cosx, if x!=0), (-1, ifx=0):}`
L.H.L.` =lim_(x->0^-)f(x)`
` =lim_(x->0^-)(sin x-cos x)`
Put `x=0-h` as `->0^-, h->0`
`:.=lim_(h->0)[sin(0-h)-cos (0-h)]`
`=lim_(h->0)(-sin h-cos h)`
`=lim_(h->0)(-sin h - cos h)`
`=0-1=-1`
R.H.L`=lim_(x->0^+)f(x)`
=`lim_(x->0^+)(sinx-cosx)`
Put `x=0+h` as `x->0^+,h->0`
`:. lim_(h->0)[sin(0+h)-cos(0+h)]`
=`lim_(h->0)(sin h-cos h)`
`=0-1=-1`
`f(0)=-1`
`:.`L.H.L.=R.H.L.=`f(0)`
Thus `f(x)` is continuous at `x=0`
Find the value of `k` so that the function `f` is contieous at the indicated point in Exercise 26 to 29.
Question: 26 .
`f(x)= {((kcosx)/(pi-2x), at x!=pi/2), (3, atx=pi/2):}`
Here `f(x)= {((kcosx)/(pi-2x), at x!=pi/2), (3, atx=pi/2):}`
L.H.L =`lim_(x-> pi^-/2)f(x)`
=`lim_(x-> pi^-/2)(k cos x)/(pi-2x)`
Put` =pi/2 -h` as ` x->pi/2, h->0`
`:. lim_(h->0)(kcos(pi/2 -h))/(pi-2(pi/2 -h))`
=`lim_(h->0)(k sin h)/(2h)`
`=lim_(h->0)k/2xx(sin h)/h`
`=k/2xx1=k/2`
R.H.L.`=lim_(x-> pi^+/2)f(x)`
=`lim_(x-> pi^+/2)(k cos x)/(pi-2x)`
Put `x=pi/2+h` as `x-> pi^+/2, h->0`
`:. lim_(h->0)(kcos(pi/2 +h))/(pi-2(pi/2 +h))`
=`lim_(h->0)(-k sin h)/(-2h)`
`=lim_(h->0)k/2xx(sin h)/h`
`=k/2xx1=k/2`
`=f(pi/2)=3`
Since `f(x)` is continuous at
`=x=pi/2`
`:.` L.H.L = R.H.L.
=`f(pi/2)=k/2=3=>k=6`
Question:27 .
`f(x)= {(kx^2, if x<=2), (3, ifx>2):}at x=2`
Solution:
`f(x)= {(kx^2, if x<=2), (3, ifx>2):}`
L.H.L = `lim_(x->2^-)f(x)`
`=lim(x->2^-)kx^2`
Put `x=2-h` as `x->2,h->0`
`:.lim_(h->0)(2-h)^2`
` lim_(h->0)k(4+h^2-4h)=4k`
R.H.L =`lim_(x->2^+)f(x)`
`lim(x->2^+)(3)=3`
`f(2) =kxx(2)^2=4k`
Since `f(x)` is continuous at `x=2=2`
`:.` L.H.L.=R.H.L.
`=> 4k=3=>k=3/4`
Question:28 .
`f(x)= {(kx+1, if x<=pi), (cos x, ifx>pi):}at x=pi`
Solution:
Here `f(x)= {(kx+1, if x<=pi), (cos x, ifx>pi):}at x=pi`
L.H.L = `lim_(x->pi^-)f(x)`
`=lim(x->pi^-)(kx+1)`
Put `x=pi-h` as `x->pi,h->0`
`:.lim_(h->0)k(pi-h)+1`
` lim_(h->0)kpi-kh+1=kpi+1`
R.H.L =`lim_(x->pi^+)f(x)`
`lim_(x->pi+)cos x`
put `x=pi+h` as `x->^+,h->0`
`:. lim_(h->0)cos (pi+h)`
`=lim_(h->0)-cos h=-1`
`=f(pi)=kpi+1`
since `f(x)` is continuous at`x=pi``:.`L.H.L.=R.H.L.
`=f(x)=>kpi+1=-1=>k=-2/11`
Question: 29 .
`f(x)= {(kx+1, if x<=5), (3x-5, ifx>5):}at x=5`
Solution:
Here `f(x)= {(kx+1, if x<=5), (3x-5, ifx>5):}`
L.H.L = `lim_(x->5^-)f(x)`
`=lim_(x->5^-)(kx+1)`
Put `x=5-h` as `x->5,h->0`
`:.lim_(h->0)[k(5-h)+1]`
` =lim_(h->0)[5k-kh+1]=5k+1`
R.H.L =`lim_(x->5^+)f(x)`
`=lim_(x->5+)(3x-5)`
put `x=5+h` as `5->^+,h->0`
`:. lim_(h->0)[3(5-h)-5]`
` =lim_(h->0)(10+3h)=10`
`=f(5)=5k+1`
`[∵ f(x) =kx+1]`
Since `f(x)`is continuous at `x=5`
`:.` L.H.L.=R.H.L.
`=f(5)=>5k+1` `=10=>k=9/5`
Question: 30 .
`f(x)={(5, if x<=2), (ax+b, if 2<\x<\10), (21, if x >=10):}` is a continuous function.
Solution:
Here`f(x)={(5, if x<=2), (ax+b, if 2<\x\<10), (21, if x >=10):}`
At `x=2`
L.H.L`lim_(x->2^-)f(x)`
`=lim_(x->2^-)(5)=5`
R.H.L.=
`lim(x->2^-)f(x)`
`lim(x->2^-)(ax+b)`
Put `x=2+h` as `x->2^+, h->0`
`:. lim_(h->0)[a(2+h)+b]`
`= lim_(h->0)(2a+ah+b)=2a+b`
`=f(2) =5`
since `f(x)` is continuous at`x=2`
`:.` L.H.L=R.H.L.
`f(2)=>2a+b=5`
At ` x=10`
L.H.L.` = lim_(x->10^-)f(x)`
`= lim_(x->10^-)(ax+b)`
put `x=10=h `as `x->10^-,h->0`
`:. lim_(h->0)[a(10-h)+b] `
`=10a+b`
R.H.L.=`lim_(x->10^+)f(x)`
`=lim_(x->10^+)(21)=21`
`=f(10)=21`
since `f(x)` is continuous at `x=10`
L.H.L.=R.H.L. `=f(10)=>10a+b=21`
Solving (i)and (ii)we have
`a=1, b=1`
Question:31 .Show that the function defined by `f(x)=cos(x^2)` is a continuous function.
Solution:
Let `h(x)=x^2` and `g(x)=cosx`
Now `h(x)` is a polynomial function, so it is continuous
`g(x)` is cosine function, so it is continuous function.
`:. (goh)(x)=g[h(x)]`
`=g(x^2)=cos x^2`
since `g(x)`and `h(x)` are both continuous functions, so composition of `(x)` and `h(x)` is alsoa continuous function.
Thus`f(x)= cos (x^2)` is a continuous function.
Question:32 . Show that the function defined by `f(x)=|cos x|` is a continuous function.
Solution:
Let ` g(x)=cos x` and`h(x) =|x|`
Now `g(x)` is a cosine function, so it is Continuous function.
`h(x)=|x|` is the absolute value function, so it is a continuous function.
`:. (hog)(x)`
` = h[g(x)]=h(cos x)=|cos x|`
Since `g(x)` and `h(x)` are bith continuous function, so competition of `g(x)` and `h(x)` is also a continuous function.
Thus`f(x)=|cos x|` is a continuous function.
Question:33 .Examine that sin`|x|` is a continuous function.
Solution:
Let ` g(x)=|x|` and `h(x) =sin x`
Now `g(x)=|x|` is the absolute value function, so it is continuous function.
`h(x)=sin x` is the sine function, so it is a continuous function.
`:. (hog)(x)`
`= h[g(x)]`
` =h(|x|)=sin|x|`
Since `g(x)` and `h(x)` are both continuous functions, composition of `g(x)`and `h(x)` is also a continuous function.
Question:34 .Find all the points of discontinuity of `f` defined by `f(x)=|x|-|x+1|`.
Solution:
Let `g(x)=|x|` and `h(x)=|x+1|`
Now `g(x)=|x|` is the absolute value function, so it is a continuous function.`h(x)=|x+1|` is the absolute value function, so it is a continuous function.Since `g(x)`and `h(x)` are both continuous functions, so difference of two continuous functions is a contineous function.Thus `f(x)=sin|x|-|x+1|` is a continuous function at all points.There is no point at which `f(x)` is discontinuous.
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