Mathematics Tricks & Tips of Continuity of functions For NDA
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Tricks & Tips of Continuity of functions For NDA

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Continuity of functions defined differently in different intervals

See examples
Q 1866691575

For what value of `lambda` is the function defined by

`f(x) = {tt( (lambda(x^2 -2x) , text(if) x le 0 ), (4x+1, text(if) x > 0) )` continuous at `x = 0?` What about continuity at `x = 1?`


Class 12 Exercise 5.1 Q.No. 18
Solution:

At `x = 0 , L.H.L.= Lim_(x->0^-) lambda (x^2 -2x) =0`,

`R.H.L. = Lim_(x->0^+) (4x+1) =1, f(0) = 0`

`f (0) = L.H.L. ne R.H.L. => f` is not continuous at `x = 0`, whatever value of `lambda in R` may be

At `x =1, Lim_(x->1^-) f(x) = Lim_(x->1^+) f(x) = Lim_(x->1) (4x+1) =f(1) = 5`

`f` is continuous at `x = 1` for all values of `lambda`

but `=>f` is not continuous at `x = 0` for any value of `lambda`
Q 2884801757

Let `f(x) = { tt ( ( 3x -4 , 0 le x le 2 ), (2x +p , 2 < x le 9) )`

If f is continuous at x = 2, then what is the value
of p?

(A)

0

(B)

2

(C)

-2

(D)

-1

Solution:

`:. f(x) = { tt ( (3x-4 , 0 le x le 2), (2x + p , 2 < x le 9) )`

and f(x) is continuous at x = 2

`:. lim_(x-> 2^-) f(x) = f(2) = 2`

`=> lim_(x->2^+) (2x +p) = f(2)`

`=> 2(2) + p = 2`

`=> p = -2`
Correct Answer is `=>` (C) -2
Q 1877012886

Find all points of discontinuity of `f`, where.

`f(x) = {tt ( ((sin x)/x , text(if) x < 0), (x+1, text(if) x ge 0) )`
Class 12 Exercise 5.1 Q.No. 23
(A)

only one point

(B)

only two point

(C)

one o two point

(D)

None of these

Solution:

At `x = 0 , L.H.L. = lim_(x->0^-) f(x) = (sin (-h) )/ (-h) =1`,

`:. f(0)= 1, R.H.L. = lim_(x->0^+) f(x) =lim_(h->0) 0+h + 1 = 1`

`:. f` is continuous at `x = 0` when `x < 0, sin x` and `x` both are continuous

`:. (sin x)/x ` is also continuous


when `x> 0, f(x) =x+1` is a polynomial

`:. f` is continuous.

So `=> f` is not discontinuous at any point
Correct Answer is `=>` (D) None of these
Q 1817112989

Determine if `f` defined by

`f(x) = {tt ( (x^2 sin (1/x), text(if) x ne 0), (0, if x= 0) )` is a continuous function?
Class 12 Exercise 5.1 Q.No. 24
Solution:

`L.H.L. =lim_(x->0^-) (x^2 sin (1/x)) = lim_(h->0) (-h)^2 sin (1/(-h))`

`= - lim_(h->0) h^2 (sin (1/h))`

`sin (1/h) ` lies between `-1` and `1`, a finite quantity

`:. h^2 sin (1/h) ->0 ` as `h->0`

`:. L.H.L. =0` , similarly `lim_(x->0^+ ) (x^2 sin (1/x)) =0`

Also `f(0) = 0` (given)

`:. L.H.L = R.H.L. = f(0)`

`:. f` is continuous for all `x in R`.
Q 2803091848

The function `f(x) = { tt ( (x^2/a, 0 le x < 1 ), ( a, 1 le x < sqrt 2), ( (2b^2 - 4b)/x^2 , sqrt2 le x < oo) )` is

continuous for `0 le x < oo`, then the most suitable
values of a and b are

(A)

`a= 1, b = - 1`

(B)

`a= -1 , b = 1+ sqrt 2`

(C)

`a=-1 , b =1`

(D)

None of these

Solution:

For continuity at x 1,
RHL = LHL = V

`=> 1/a = a => a =1 , -1`

For continuity at `x = sqrt 2`

RHL = LHL = V => `a = b^2 - 2b`

When, `a = 1, b^2 - 2b - 1 = 0`

`:. b = 1 pm sqrt 2`

when `a= -1 , b^2 -2b + 1 = 0`

`:. b =1`
Correct Answer is `=>` (C) `a=-1 , b =1`

Continuity of composite function

.If `f` is continuous at `x = a` & `g` is continuous at `x =a` then the composite `g(f(x))` is continuous at `x = a`

e.g. . `f (x)= (x sin x)/(x^2+2)` & `g (x) = | x |` are continuous at `x = 0` hence the composite

`(gof) (x) = |(x sin x)/(x^2+2)|` will also be continuous at `x = 0`.
Q 2613145940

If `f(x) = sgn(x)` and `g(x) = x(1 - x^2)`, then the number of points of discontinuity of function `f(g(x))` is

(A)

exact two

(B)

exact three

(C)

finite and more than `3`

(D)

infinitely many

Solution:

`f (g(x) ) = { tt ( (1,x < -1 ) , ( 0, x = -1 ), ( -1, -1 < x < 0 ), ( 0, x= 0) , (1, 0 < x < 1), (0, x =1), ( -1, x> 1) )`

`:.` points of discontinuity are `x =- 1, 0, 1`
Correct Answer is `=>` (B) exact three
Q 1887434387

Show that the function defined by `f(x) = | cos x |` is a continuous function.
Class 12 Exercise 5.1 Q.No. 32
Solution:

Let `g(x) = |x| ` and `h(x) = cos x,

f(x) = goh(x)= g (h(x)) = g (cos x) = | cos x |`

Now `g(x) = |x| ` and `h(x) = cos x` both are continuous for all values of `x in R`.

`:. (goh) (x)` is also continuous.

Hence, `f(x) = goh (x) = | cos x |` is continuous for all values of `x in R`.
Q 1847434383

Show that the function defined by `f (x) =cos (x^2)` is a continuous function.
Class 12 Exercise 5.1 Q.No. 31
Solution:

Now, `f(x) = cos x^2`,

let `g (x) = cos x` and `h(x) = x^2`

`:. goh (x) = g (h(x)) = cos x^2`

Now `g(x) = cosx` and `h(x) = x^2` both are continuous `AA x in R`.

`:.f (x) = goh (x) = cos x^2` is also continuous at all `x in R`.

Continuity of function involving |x|

See examples
Q 1846780673

Find all points of discontinuity of `f,` where `f` is defined by
`f(x) = {tt ( ( |x|/x, text(if) x ne 0 ), (0, text(if) x =0) )`
Class 12 Exercise 5.1 Q.No. 8
Solution:

`f(x) = {tt ( ( |x|/x, text(if) x ne 0 ), (0, text(if) x =0) )`


At `x=0`

`L.H.L. =lim_(x->0^-) (|x|/x) = -1, f(0) =0`

`R.H.L. =lim_(x->0^+) (|x|/x) =1`

`:. L.H.L ne R.H.L. ne f(0)`

`:. f` is discontinuous at `x = 0`

`:.` The point of discontinuity is `x = 0`
Q 2874312256

Consider the following statement

I. `f(x) = | x |` is continuous ` AA x in R`

II. `f(x) = x^3 + x^2 - 1` is not continuous `AA x in R`

Which of the above statement(s) is I ate correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

For `f(x) = | x | , f(x) = { tt ( (-x , x < 0 ), ( x , x ge 0) )`

For continuous

`lim_(x -> 0^-) f(x) = lim_(x->0^+) f(x) = f(0)`

`lim_(x -> 0^-) (-x) = lim_(x-> 0^+) f(x) = f(0) = 0`

`0 =0 = 0`

Hence, `f(x) = |x |` is continuous for all
x for `f(x) = x^3 + x^2- 1`

Clearly, `x^3 + x^2 - 1` is defined for all
value of x.

Hence,.f(x) is continuous, `AA x in R`
Correct Answer is `=>` (A) Only I
Q 2410634519

Which one of the following is correct in respect of

the function `f (x) = x^2/|x|` for `x ne 0` and `f(0) = 0`?
NDA Paper 1 2012
(A)

`f(x)` is discontinuous everywhere

(B)

`f(x)` is continuous everywhere

(C)

`f(x)` is continuous at `x = 0` only

(D)

`f(x)` is discontinuous at `x = 0` only

Solution:

`f(x) = x^2/|x| , x ne 0`

`f(0) = 0`

Now, redefined the function `f(x)`.

or `f(x) = { tt ( (x^2/x = x, text (if) x > 0 ), (x^2/(-x) = -x , text (if) x < 0 ) )`

`f(0) = 0`

Clearly, it is a modulus function and modulus function is
continuous everywhere.
Correct Answer is `=>` (B) `f(x)` is continuous everywhere

Continuity of function involving |f(x)|

Discuss the continuity of `|cosx|`

Since `cos x` is continuous in `R.`

on taking modulus it will be continuous in `R^+` because modulus returns only positive value.

Continuity of function involving [x] GIF and {x} Fractional Part Function

See examples
Q 1425491361

The function, `f (x) = [| x | ]- | [ x ] |`, where ` [x]` denotes greatest integer function

(A)

is continuous for all positive integers

(B)

is discontinuous for all non-positive integers

(C)

has finite number of elements in its range

(D)

is such that its graph does not lie above the `x`-axis

Solution:

`f(x)= [ |x|]-| [x] |`

`(A)` Let `x= k` ; `k in I^+`

`f(k) =0`

`f(k^-) = lim_(h->0) ((k-1)-(k-1))=0`

`f(k^+) = lim_(h->0) (k-k) =0`

`=> f(x) `is continuous for all positive integers.

`(B)` Let `x= t` ; `t in I^-`

`f(t) =0`
`f(t^+) =lim_(h->0) (-t-1+t)=-1`
`f(t^-) = lim_(h->0) (-t-1+t)=-1`

`=> f(x)` is discontinuous for all negative integers and discontinuous also at `x = 0`.
Also it's range includes only finite number of elements which are either ` 0` or `- 1`.
Correct Answer is `=>` (A)
Q 1316501479

If `f(x)=[x^2]-[x]^2` where `[.]` is the largest integer function then

(A)

`f(x)` is discontinuous for all integral value of ` x`

(B)

`f(x)` is discontinuous only at `x=0,1`

(C)

`f(x)` is continuous only at `x=1`

(D)

None of these

Solution:

`f(x)=[x^2]-[x]^2`

`f(1^+)=lim_{h to 0}([(1+h)^2]-[1+h]^2)`

`Rightarrow 1-1=0`

`f(1^-)=lim _{h to 0}([(1-h)^2]-[1-h]^2)`

`Rightarrow 0-0=0`

`therefore f(x)` is continuous at `x=1`

Hence `3` is the correct answer.
Correct Answer is `=>` (C) `f(x)` is continuous only at `x=1`
Q 2410256110

At how many points is the function `{f(x)} = [x]`
discontinuous?
NDA Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`3`

(D)

infinite

Solution:

By the graph of the greatest integer function, we know
that `f(x) = [x]` is discontinuous at infinite points. (i.e., all integer
points)
Correct Answer is `=>` (D) infinite
Q 2304823758

If `f(x) = [x sin (px)]`, then `f(x)` is not
BITSAT Mock
(A)

continuous at `0`

(B)

continuous in `(0, 1)`

(C)

derivable at `1`

(D)

none of these

Solution:

We note that for all `x in [- 1, 1]`,

`1 > x sin (pi x) ge 0`

`=> f(x) = [x sin (pi x)]`

`= 0`, for all `x in [- 1, 1]`

So `f(x)` is a constant function for

`- 1 le x le 1`. So `f` is continuous in `(- 1, 1)`.

Also, `f(1) = [1 sin pi] = 0`

`Lt_(x->1^-) f(x) = Lt_(x->1^-) [x sin pi x] = 0`

`Lt_(x->1^+) f(x) = Lt_(x->1^+) [x sin pi x] = -1`

{In the immediate right neighbourhood
of `1, - 1 < x sin pi x < 0}`

`=> f` is not continuous at `x = 1`.

Consequently `f` is not derivable at `1`.
Correct Answer is `=>` (C) derivable at `1`

.continuity of signum functions

.`sgn (x) = (|x|)/x` if `x ne 0`

`=0` if `x=0`

i.e., `sgn(x) = +1 ` if `x > 0`

`=-1` if `x < 0`

`=0 ` if `x=0`

e.g., Discuss the continuity of `f(x) = sgn (x^3-x)`

Here, `x^3-x=0=> x=0 , -1 , 1`

Here `f(x)` is discontinuous at `x = 0, - 1, 1`

.Continuity of functions in which `f (x)` is defined differently for rational and irrational values of `x`

E.g. Discuss the continuity of the following function `f(x) = { tt (( 1, if , x, is, rational),(0, if , x , is , irrational))`

For any `x = a`,


L.H.L. `= lim_(x->a^-) f(x) = lim_(x-> a^(-)) f(a-h) =0` or `1` [as `lim_(h->0) (a - h)` can be rational or irrational}

R.H.L `=lim_(x-> a^(+)) f(x) = lim_(h->0) f(a+h) =0` or `1`

Hence, `f (x)` oscillates between `0` and `1` as for all values of `a` .

`:.` L.H.L. and R.H.L. do not exist.

`=> f (x)` is discontinuous at a point `x = a` for all values of `a`
Q 1847434383

Show that the function defined by `f (x) =cos (x^2)` is a continuous function.
Class 12 Exercise 5.1 Q.No. 31
Solution:

Now, `f(x) = cos x^2`,

let `g (x) = cos x` and `h(x) = x^2`

`:. goh (x) = g (h(x)) = cos x^2`

Now `g(x) = cosx` and `h(x) = x^2` both are continuous `AA x in R`.

`:.f (x) = goh (x) = cos x^2` is also continuous at all `x in R`.

.

Q 2410634519

Which one of the following is correct in respect of

the function `f (x) = x^2/|x|` for `x ne 0` and `f(0) = 0`?
NDA Paper 1 2012
(A)

`f(x)` is discontinuous everywhere

(B)

`f(x)` is continuous everywhere

(C)

`f(x)` is continuous at `x = 0` only

(D)

`f(x)` is discontinuous at `x = 0` only

Solution:

`f(x) = x^2/|x| , x ne 0`

`f(0) = 0`

Now, redefined the function `f(x)`.

or `f(x) = { tt ( (x^2/x = x, text (if) x > 0 ), (x^2/(-x) = -x , text (if) x < 0 ) )`

`f(0) = 0`

Clearly, it is a modulus function and modulus function is
continuous everywhere.
Correct Answer is `=>` (B) `f(x)` is continuous everywhere

Continuity of function `xsin (1/x)` and `x cos (1/x)` at `x=0`

Let's check for `f(x) = xsin (1/x)`

`lim_(x->0)f(x) = lim_(x->0) xsin (1/x)`

LHL : `lim_(x->0^-)f(x) = lim_(h->0) (0-h)sin (1/(0-h)) = lim_(h->0) (0-h)sin (1/(0-h)) = lim_(h->0) (-h xx ("value between -1 to +1 ")) = 0`

RHL : `lim_(x->0^+)f(x) = lim_(h->0) (0+h)sin (1/(0+h)) = lim_(h->0) (0+h)sin (1/(0+h)) = lim_(h->0) (h xx ("value between -1 to +1 ")) = 0`

`f(0) = lim_(x->0) xsin (1/x) = 0 xx ("value between -1 to +1 ")`

Since LHL = RHL = f(0)

So function is continuous at `x = 0`

similarly for `x cos (1/x),` its also continuous at `x = 0`

Continuity of periodic functions

E.g. Find number of point of discontinuity in interval `[-3.2pi , 12 pi] ` where `f(x) = cosecx`

Now `f(x) = cosecx` will be discontinuous where `sinx = 0` so `x in n pi` where `n in I`

Since period of `sinx` is `2pi ` and `sinx` is zero two time in a period

Points of discontinuity = 3 points for `[-3.2 pi , 0) + (6xx2) ` point for ` [0, 12pi) = 15` points

Theorems on continuity

(1 ) If `f(x) ` is continuous and g(x) is continuous then,

`f(x) + g(x) =>` Continuous

`f(x) - g(x) =>` Continuous

`f(x) * g(x) =>` Continuous

`(f(x) )/(g(x) ) =>` Continuous (`because g(x) != 0`)

Eg :

`f(x) = sin x ` , Continuous in R

`g(x) = cosx `, Continuous in R

`f(x) + g(x) = sin x + cos x ` Continuous in R

`f(x) -g(x) = sin x - cos x` Continuous in R

`f(x) * g(x) = sin x cosx = (sin2x)/2 ` Continuous in R

`(f(x))/(g(x) ) = (sin x )/(cosx)` is Continuous `R -{((2n+1) pi)/2}`



(2) If `f(x) + g(x) ` is Continuous at `x = a` then f(x) & g(x) is not necessarily Continuous at x = a

Eg :

`f (x) = { tt ((1/x, x!= 0),(0, x = 0))`

`g(x) = { tt((-1/x , x=0 ),( 0, x = 0))`

Here f(x) & g(x ) is not Continuous at x =0

But `f(x) + g(x) = { tt(( 0, x!= 0),(0, x =0))`

`f(x) + g(x) ` will be Continuous

(3) Similarly If `f(x) - g(x) ` is Continuous at

`x =a ` , then `f(x) ` & g(x) is not necessarily Continuous at x = a

(4 ) If f (x ) is Continuous at x = a and g(x) is discontinuous at x =a then

a) `f(x) + g(x) =>` discontinuous at x = a

b) `f (x ) - g(x) =>` discontinuous at x = a

c) `f(x )* g(x) =>` not necessarily discontinuous at x =a

d) `(f(x))/(g(x)) =>` not necessarily discontinuous at x = a

Eg :

`f(x) = x` Continuous at x =0

`g(x) = 1/x` discontinuous at x =0

a) `f (x) + g(x) = x + 1/x = (x^2 +1 )/x` will discontinuous at x = 0

b) Similarly `f (x) - g(x)` will be discontinuous at x = 0

c) Now `f (x) * g(x) = x * 1/x = 1`

Here `f(x) * g(x) =>` is continuous at `x = 0` since `g(x)` is discontinuous at x =0

d) Let `f(x) = ax` Continuous at x =0

`g (x) = 1/xb` discontinuous at x =0

`(f(x))/(g(x)) = a/b` continuous at x =0

5)

a) If `f(x) pm g(x) ` is discontinuous at x = a,

then one [f(x) or g (x) ] or both [ f(x) and g(x) ] must be discontinuous

b) If `f(x) //g(x) => ` discontinuous

`f(x)* g(x) =>` discontinuous

then f(x) & g(x) is not necessarily discontinuous at x =a

Eg :

`f(x) = x` continuous at x =0

`g(x) = x^2` continuous at x = 0

`(f(x))/(g(x) ) = 1/x` discontinuous at x = 0 similarly proof other one.

6) Similarly If f(x) & g(x) both are discontinuous at x =a

a) `f(x) pm g(x) ` discontinuous at x =a

b) `f(x) * g(x) ` and `(f(x) )/ (g(x))` is not necessarily discontinuous at x = a
Q 2884612557

Consider the following statements

I. `lim_(x->0) x^2/x` exists. II. `(x^2/x)` is not continuous at x = 0.

III. `lim_(x->0) ( |x| )/x` does not exist.

Which of the above statement(s) is/are correct?

(A)

I, II and Ill

(B)

I and II

(C)

II and Ill

(D)

I and Ill

Solution:

I. `lim_(x->0) x^2/x = lim_(x->0 ) (x) = 0`

II. It is true that `x^2/x` is not continuous at x=0.

III. `LHL = lim_(b->0) ( | 0-b | )/(0-b) = lim_(b->0) = -1`

`RHL = lim_(b->0) ( | 0+b | )/(0+b) = lim_(b->0) b/b = 1`

`:. LHL ne RHL ` so, it does not exist.
Correct Answer is `=>` (A) I, II and Ill
Q 2028645501

If `f * g` is continuous at `x = a`, then check whether `f` and `g` are separately continuous
at `x = a`.
NCERT Exemplar
(A) True
(B) False
Solution:

Let `f(x) = sin x` and `g (x) = cot x`

`:. f(x) * g(x) = sin x * (cos x)/(sin x) = cos x`

which is continuous at `x = 0` but `cot x` is not continuous at `x = 0`.
Correct Answer is `=>` (B)
Q 1887101987

Discuss the continuity of the following
functions:

(a) `f (x) =sin x +cos x`

(b) `f(x)=sin x-cos x`

(c) `f(x)=sin x* cos x`
Class 12 Exercise 5.1 Q.No. 21
Solution:

(a) `f(x) = sin x + cos x`

`= sqrt 2 ( 1/(sqrt2) sinx + 1/(sqrt2) cos x )`

`sqrt (2) (sin x cos pi/4 + cos x sin pi/4)`

`= sqrt(2) sin (x + pi/4)`

At `x =c`

`L.H.L. = lim_(x->c^-) sqrt(2) sin (x+ pi/4)`

`= sqrt(2) sin (c+ pi/4)`

`R.H.L. = lim_(x->c^+) sqrt(2) sin (x+ pi/4)`

`=sqrt(2) sin (c+ pi/4) =f(c) `

`:. f` is continuous for all `x in R`



(b) `f(x) = sin x - cos x`

`= sqrt(2) (1/(sqrt2) sin x - 1/(sqrt2) cos x)`

`= sqrt(2) (sin x cos pi/4- cos x sin pi/4)`

`= sqrt(2) sin (x- pi/4)`

At `x= c`,

`L.H.L. =lim_(x->c^+) sqrt(2) sin (x- pi/4)`

`= sqrt(2) sin (c- pi/4)`

`R.H.L.= lim_(x->c^+) sqrt(2) sin (x- pi/4)`

`:. f` is continuous for all `x in R`.


(c) `f(x) = sin x cos x =1/2 (2 sin x cos x)`

`=1/2 sin 2x`. Again `f` is continuous for all `x in R`.
Q 2018245100

If `f(x) = 2x` and `g(x) = x^2/2 + 1`, then which of the following can be a

discontinuous function?
NCERT Exemplar
(A)

`f(x) + g (x)`

(B)

`f(x) - g (x)`

(C)

`f(x) . g (x)`

(D)

` (g (x))/( f (x))`

Solution:

We know that, if `f` and `g` be continuous functions, then

(a) `f + g` is continuous (b) `f - g` is continuous.

(c) `fg` is continuous (d) `f/g` is continuous at these points where `g(x) != 0`.

Here, `(g(x))/(f(x)) = ( x^2/2 + 1)/(2x) = (x^2 + 2)/(4x) `

which is discontinuous at `x = 0`.
Correct Answer is `=>` (D) ` (g (x))/( f (x))`
 
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