Mathematics Must Do Problems of Continuity of functions for NDA
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Must Do Problems of Continuity of functions for NDA

Must do problem for NDA

Q 2783291147

Let `f(x)` be defined as follows :

`f(x) = { tt (( 2x+1 , -3 < x < -2),(x-1 , -2 le x < 0),(x+2 , 0 le x < 1))`

Which one of the following statements is correct in respect of the above function?
NDA Paper 1 2017
(A)

It is discontinuous at `x = -2` but continuous at every other point.

(B)

It is. continuous only in the interval ` (-3, -2).`

(C)

It is discontinuous at `x = 0` but continuous at every other point.

(D)

It is discontinuous at every point.

Solution:

`f(x) = { tt (( 2x+1 , -3 < x < -2),(x-1 , -2 le x < 0),(x+2 , 0 le x < 1))`

By Graph It is discontinuous at `x = 0` but continuous at every other point.
Correct Answer is `=>` (C) It is discontinuous at `x = 0` but continuous at every other point.
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 2446356273

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

is continuous for `0 le x < oo`,then the most
suitable values of `a` and `b` are
UPSEE 2013
(A)

`a= 1, b = -1`

(B)

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

(C)

`a= -1 , b =1`

(D)

none of the above

Solution:

Since, `f` is continuous for `0 le x < oo , f` is

continuous at `x = 1`.

`:. a = pm 1`

Since, `f` is continuous at `x = sqrt (2)`

`:. (2b^2 -4b)/2 =a`

`=> b^2 -2b =a`

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

`=> b= 1 pm sqrt (2)`

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

`=> (b-1)^2 =0 => b =1`

Hence, `a = -1` and `b = 1` are most suitable
values.
Correct Answer is `=>` (C) `a= -1 , b =1`
Q 2436301272

Let `f (x+ y) = f(x) + f(y)` for all `x` and `y`. If the
function `f (x)` is Continuous at `x = 0`, then `f (x)` is
continuous
UPSEE 2014
(A)

only at `x = 0`

(B)

at `x in R - { 0}`

(C)

for all x

(D)

None of these

Solution:

Given, `f(x + y) = f(x) + f(y);` for all `x` and `y`.

Since, `f(x)` is continuous at `x = 0`, we have

`lim_(x -> 0) f(x) = f(0)`.

To show that `f(x)` is continuous at any point a,

we shall prove that

`lim_(x -> a) f(x) = f(a)`

`=> lim_(h -> 0) f(a +h) = f(a)`

Indeed, `lim_(h -> 0) f(a +h) = lim_(h -> 0) [f(a) + f (h)]`

`= f(a) + lim_(h -> 0) f(h) = f(a) + f(0)`

`= f(a + 0) = f(a)`
Correct Answer is `=>` (D) None of these
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 2685156967

Draw the graph and find the points of discontinuity for `f (x) = [2 cos x], x in [0, 2 pi]`,
(` [.]` represents the gratest integer function).

Solution:

`f (x) = [2cos x]`

Clearly from the graph given in figure

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

and when `2 cos x = ± 1`

or `x = 0` and when `2 cos x = ±1`

or `x = 0` and `cos x = ±1/2`

or `x = 0` and `x = pi/3, (2 pi)/3 ,(5 pi)/3`
Q 1618880700

Consider the following statements

I. `f(x) = [x]`, where `[.]` is the greatest integer
function, is discontinuous at `x = n`, where `n in Z`.

II. `f(x) = cotx` is discontinuous at `x = n pi`, where
`n in Z.`

Which of the above statements is/are correct?
NDA Paper 1 2015
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. We have, `f(x) = [x]`

Clearly, `lim_(x-> n^+) f(x) = lim_(h->0) f (n + h)`

`= lim_(h->0) [n + h] = n`

and `lim_(x-> n^-) f(x) = lim_(h->0) (n- h)`

`= lim_(h->0) [n- h] = n- 1`

`:. lim_(x-> n^+) f(x) != lim_(x-> n^-) f(x)` for all `n in Z`'

So, `f` is discontinuous at `x = n`

II. We have, `f(x) = cot x`

consider, `lim_ (x-> npi+) f(x) = lim_(h->0) f (npi + h) = lim_(h->0) cot (n pi + h)`

`= lim_(h->0) cot hpi` which does not exist

Thus, `f` is discontinuous at `x = n pi, n in Z`.
Correct Answer is `=>` (C) Both I and II
Q 2643045843

`f(x) = { tt ( (x+2 , x < 0 ), ( -x^2 -2 , 0 le x < 1 ), ( x, x ge 1) )`

then the number of points of discontinuity of ` | f(x) |` is

(A)

`1`

(B)

`2`

(C)

`3`

(D)

None of these

Solution:

`f(x) = { tt ( (x+2 , x < 0 ), ( -x^2 -2 , 0 le x< 1), ( x, x ge 1) )`

`:. | f(x) | = { tt ((-x-2, x < -2 ), (x+2 , -2 le x < 0 ), ( x^2 +2 , 0 le x < 1 ) , ( x, x ge 1) )`

discontinuous at `x = 1`
`: . ` number of points of discont. `1`
Correct Answer is `=>` (A) `1`
Q 2322378231

If `f (x) = x + 2` when `x ≤ 1` and
`f (x) = 4x − 1` when `x > 1`, then :
BITSAT Mock
(A)

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

(B)

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

(C)

`Lim_( x→ 1) f (x) = 4`

(D)

None of these

Solution:

Given `f (x) = { tt ((x + 2, text(when)x ≤ 1),(4x-1, text(when)x > 1))`

`f (1) = 1 + 2 = 3` ...(1)

`lim_(x → 1^+) f (x) = lim_(h → 0) f (1 + h)`

`= lim_(h → 0) [4 (1 + h) − 1]`

`= 3` ...(2)

`lim_(x → 1^−) f (x) = lim_(h → 0) f (1 − h)`

`= lim_(h → 0) [(1 − h) + 2]`

`= 3` ...(3)

`∴ lim_(x → 1^+) f (x) = lim_(x → 1^−) f (x) = f (1)`

Hence `f (x)` is continuous at `x = 1`.
Correct Answer is `=>` (B) `f (x)` is continuous at `x = 1`
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 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))`
Q 1817634589

Find all the points of discontinuity of `f` defined by `f(x) = | x| -| x + 1 |`.
Class 12 Exercise 5.1 Q.No. 34
Solution:

`f(x) = |x| - |x +1| `, when `x< -1`,

`f(x) = -x - [- (x+1)] = -x +x +1 =1`

when `-1 le x < 0, f(x) = -x - (x+ 1) = -2x -1`,

when `x ge 0 , f(x) = x - ( x + 1 ) = -1`

`:. f(x) = { tt ( (1, text(if) x < -1), (-2x-1, text(if) -1 le x < 0), (-1, text(if) x ge 0) )`

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

`R.H.L. = lim_(x->1^+) f(x) = lim_(x->1^+) (-2x -1) = 1`

`:. f(-1) = -2 (-1) -1 =2 -1 =1`

`:. L.H.L = R.H.L. = f(-1)`

`=> f` is continuous at `x = -1`

At `x=0 ,L.H.L = lim_(x->0^-) (-2x-1) = -1`

`f(0) = -1` (given)

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

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

`:. f` is continuous at `x = 0 => `There is no point

of discontinuous. Hence `f` is continuous for all `x in R`.
Q 1837423382

Find the values of `k` so that the function `f` is continuous at tlte indicated point in

`f(x) = { tt ( ( (k cos x)/(pi-2x) , text(if) x ne pi/2), ( 3, text(if) x = pi/2) )` at `x =pi/2`
Class 12 Exercise 5.1 Q.No. 26
Solution:

`L.H.L. =lim_(x-> (pi/2)) (k cos x)/ (pi -2 x) = lim_(h->0) ( k cos ( (pi/2) -h ))/ ( pi- 2 ( (pi/2) -h) )`

(Putting `x= (pi/2) -h`)

`= lim_(h->0) (k sin h )/ (pi -pi +2h) = lim_(h->0) k/2 * (sin h)/(h) = k/2`

`R.H.L. = lim_(x-> (pi/2)^+) (k cos x)/ (pi- 2 x)`

` = lim_(h->0) (k cos ( (pi/2) +h))/ (pi- 2 ( (pi/2) +h))` (Putting `x= (pi/2) + h`)

`= lim_(h->0) (-k sin h)/ (-2h) = lim_(h->0) (k/2) (sin h )/(h ) = k/2`,

`f(pi/2) =3` (given)

`:. f` is continuous if `(k/2) =3` or `k =6`.
Q 1806291178

Is the function defined by `f(x)= {tt ( (x+5, text(if) x le 1), (x-5, text(if) x >1) )` a continuous function?
Class 12 Exercise 5.1 Q.No. 13
Solution:

At `x= 1, L.H.L. = Lim_(x-> 1^-) f(x) = Lim_(x->1^-) (x+5) =6`,

`R.H.L. = Lim_(x->1^+) f(x) = Lim_(x->1^+) (x-5) = -4`

`f(1) = 1 +5 =6`,

`f(1) = L.H.L. ne R.H.L.`

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

At `x = c < 1, Lim_(x->c) (x+5) =c+5 = f(c)`

At `x = c > 1, Lim_(x->c) (x-5) =c -5= f(c)`

`:. f` is continuous at all points `x in R` except `x = 1`.
Q 1755412364

The function `f(x) = (4- x^2) /(4x -x^3)` is
NCERT Exemplar
(A)

discontinuous at only one point

(B)

discontinuous at exactly two points

(C)

discontinuous at exactly three points

(D)

None of the above

Solution:

We have,

`f(x) = (4-x^2) /(4x -x^3) =(4-x^2)/(x (4-x^2))`

`= (4-x^2) /(x (2^2 -x^2))= (4-x^2) /(x (2+x) (2-x) )`

Clearly, `f(x)` is discontinuous at exactly three poirits `x = 0, x = -2` and `x=2`.
Correct Answer is `=>` (C) discontinuous at exactly three points
Q 1735412362

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) ne 0`.

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

which is discontinuous at `x = 0`.
Correct Answer is `=>` (D) `(g(x))/(f(x))`
Q 2814712659

Given the function,

`f(x) = [ tt ( ( x + a sqrt 2 sin x , 0 le x < pi/4 ), ( 2x cot x + b , pi/4 le x < pi/2 ), ( a cos 2x - b sin x , pi/2 le x le pi) )` is continuous

in `[ 0, pi ]` , then
Find the value of a.

(A)

`pi/6`

(B)

`-pi/6`

(C)

`pi/12`

(D)

`-pi/12`

Solution:

`f (pi/4) = b + pi/2`

`f (pi/4 +0) = b + pi/2`

`f (pi/4 -0) = pi/4 + a`

Since, f(x) is continuous in `[0, pi ]`.

`:. f (pi/4 +0) = f (pi/4) = f(pi/4 -0 ) `

`=> b + pi/2 = pi/4 +a => a-b = pi/4` ..........(i)

Now, `f (pi/2) = -a-b`

`f(pi/2 + 0) = -a -b => f (pi/2 -0) = b`

Again, `f (pi/2) = f (pi/2 +0) = f (pi/2 - 0)`

`=> -a -b = b => a = -2b` .........(ii)

From Eqs. (i) and (ii), we get

`a= pi/6 , b = - pi/12`
Correct Answer is `=>` (A) `pi/6`
 
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