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previous year question of Continuity of functions for NDA

Previous year NDA question of continuity

See questions
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 2166523475

A function `f(x)` is defined as follows

` f(x) = { tt ( ( x + pi , x in [ - pi ,0) ),(pi cos x , x in [ 0 , pi/2] ),( (x - pi/2)^2 ,x in ( pi/2 , pi] ))`

Consider the following statements
1. The function `f(x)` is continuous at `x = 0`.
2. The function `f( x)` is continuous at `x = pi/2`.

Which of the above statements is/are correct?
NDA Paper 1 2016
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

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

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

`= pi cos (0) = pi`

Also, `f(0) = pi cos (0) = pi`

Hence, `f(x)` is continuous at `x = 0`.

Statement `1` is correct.

Now, for `x = pi/2`.

`LHL = lim_(x -> pi//2^-) f(x) = lim_(x -> pi//2) pi cos x = pi cos (pi/2) = 0`

`RHL = lim_(x -> pi//2^+) f(x) = lim_(x -> pi//2) (x - pi/2)^2 = ( pi/2 - pi/2)^2 = 0`

Also, `f ( pi/2) = pi cos pi/2 = 0`

Hence, `f(x)` is continuous at `x = pi/2`.

Statement `2` is correct.
Correct Answer is `=>` (C) Both `1` and `2`
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 1618291100

`f(x) = tt { (( -1 , if , x <= 0) ,( ax + b , if , 0 < x < 1) ,( 1 , if , x >= 1))`
where `a, b` are constants.
The function is continuous everywhere.

What is the value of `b?`
NDA Paper 1 2015
(A)

`-1`

(B)

`1`

(C)

`0`

(D)

`2`

Solution:

We have, `f(x) = tt { (( -1 , if , x <= 0) ,( ax + b , if , 0 < x < 1) ,( 1 , if , x >= 1))`

where `a` and `b` are constants and `f(x)` is continuous

everywhere.

As `f(x)` is continuous everywhere, therefore `f` is

continuous at `x = 0` and `x = 1` also.

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

and `lim_(x -> 0^+) f(x) = lim_(x -> 0^-) f(x) = f(1)`

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

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

`=> lim_(h -> 0) a(0 + h)+ b = -1`
Correct Answer is `=>` (A) `-1`
Q 1638291102

`f(x) = tt { (( -1 , if , x <= 0) ,( ax + b , if , 0 < x < 1) ,( 1 , if , x >= 1))`
where `a, b` are constants.
The function is continuous everywhere.

What is the value of `a?`

NDA Paper 1 2015
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

We have, `f(x) = tt { (( -1 , if , x <= 0) ,( ax + b , if , 0 < x < 1) ,( 1 , if , x >= 1))`

where `a` and `b` are constants and `f(x)` is continuous

everywhere.

As `f(x)` is continuous everywhere, therefore `f` is

continuous at `x = 0` then

`:. b = -1`

and `lim_(x -> 1) f(x) = f(1)`

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

`=> lim_(h -> 0) a(1 -h)+ b = 1 => a + b = 1`

` :. a = 2`
Correct Answer is `=>` (D) `2`
Q 1679701616

Consider the function
`f(x) = {tt ((( alpha cos x)/(pi-2x) ,if , x != pi/2),(3 ,if , x = pi/2) )`
which is continuous at `x = pi/2`. where

What is the value of `alpha?`
NDA Paper 1 2015
(A)

`6`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

We have ,

`f(x) = {tt (( alpha cos x)/(pi-2x) ,if , x != pi/2),(3 ,if , x = pi/2) )`

is continuous at `x = pi/ 2` , where a is a constant

`=> lim _ (x -> pi/2) f(x) = f(pi/2) = 3` ..........(1)

Now, consider ` lim _ (x -> pi/2) = lim _ (x -> pi/2) ( alpha cos alpha)/(pi-2x) = alpha lim _ (x -> pi/2) (cos x) /(pi - 2x)`

Now, put `y = x - pi/2 ` then as `x -> pi/2 => y -> 0`

`:. lim _ (x -> pi/2) f(x) = lim _ (y -> 0) ( cos ( y + pi/2) )/( pi -2 ( pi/2 - y)) = alpha lim _ (y -> 0) (( -siny)/(-2y))`


` = alpha /2 lim _ (y -> 0) (siny)/y = alpha /2 ` ...........(2)

From Eqs. (i) and (ii), we get` alpha /2 = 3 => alpha = 6`
Correct Answer is `=>` (A) `6`
Q 1629801711

Consider the function
`f(x) = tt { (( alpha cos x)/(pi-2x) ,if , x != pi/2),(3 ,if , x = pi/2) )`
which is continuous at `x = pi/2`. where

What is `lim_(x -> 0) f(x)` equal to?
NDA Paper 1 2015
(A)

`0`

(B)

`3`

(C)

`3/pi`

(D)

`6/pi`

Solution:

We have ,

`f(x) = tt { (( alpha cos x)/(pi-2x) ,if , x != pi/2),(3 ,if , x = pi/2) )`

is continuous at `x = pi/ 2` , where a is a constant

`=> lim _ (x -> pi/2) f(x) = f(pi/2) = 3` ..........(1)

Now, consider ` lim _ (x -> pi/2) = lim _ (x -> pi/2) ( alpha cos alpha)/(pi-2x) = alpha lim _ (x -> pi/2) (cos x) /(pi - 2x)`

Now, put `y = x - pi/2 ` then as `x -> pi/2 => y -> 0`

`:. lim _ (x -> pi/2) f(x) = lim _ (y -> 0) ( cos ( y + pi/2) )/( pi -2 ( pi/2 - y)) = alpha lim _ (y -> 0) (( -siny)/(-2y))`


` = alpha /2 lim _ (y -> 0) (siny)/y = alpha /2 ` ...........(2)

From Eqs. (i) and (ii), we get` alpha /2 = 3 => alpha = 6`

Consider,

`lim_(x -> 0) f(x) = lim_(x -> 0) (6 cos x)/(pi - 2x) = 6 lim_(x -> 0) ( cos x)/(pi - 2x)`

` = 6 ( ( cos 0)/( pi - 2 (0))) = 6/pi`
Correct Answer is `=>` (D) `6/pi`
Q 2271491326

` f(x) = { tt ((-2 sin x , text(if) , x <= - pi/2 ) , ( A sin x + B , text(if) , - pi/2 < x < pi/2 ),( cos x , text(if) , x >= pi/2 ))` which is continuous

The value of `A` is
NDA Paper 1 2015
(A)

`1`

(B)

`0`

(C)

`-1`

(D)

`-2`

Solution:

Given, ` f(x) = { tt ((-2 sin x , text(if) , x <= - pi/2 ) , ( A sin x + B , text(if) , - pi/2 < x < pi/2 ),( cos x , text(if) , x >= pi/2 ))`

Case `I` For continuity at `x = pi/2,`

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

` => lim_( h -> 0) f( pi/2 - h) = lim_( h -> 0) f( pi/2 + h) = f (pi/2)`

`=> lim_( h -> 0) A sin ( pi/2 - h) + B = lim_( h -> 0) cos ( pi/2 + h) = cos (pi/2)`

` => A + B = 0` .........(i)

Case `II` For continuity at `x = - pi/2`,

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

` => lim_( h -> 0) f ( - pi/2 - h) = lim_(h -> 0) ( - pi/2 + h) = f( - pi /2)`

` => lim_(h -> 0) [ - 2 sin ( - pi/2 - h) ]`

` = lim_(h -> 0) [ A sin ( - pi/2 + h) + B ]`

` = -2 sin ( - pi/2) => 2 = - A + B = 2`

` => A - B = 0` ..........(ii)

On solving Eqs. (i) and (ii), we get

`A = -1`
Correct Answer is `=>` (C) `-1`
Q 2221591421

` f(x) = { tt ((-2 sin x , text(if) , x <= - pi/2 ) , ( A sin x + B , text(if) , - pi/2 < x < pi/2 ),( cos x , text(if) , x >= pi/2 ))` which is continuous

The value of `B` is
NDA Paper 1 2015
(A)

`1`

(B)

`0`

(C)

`-1`

(D)

`-2`

Solution:

Given, ` f(x) = { tt ((-2 sin x , text(if) , x <= - pi/2 ) , ( A sin x + B , text(if) , - pi/2 < x < pi/2 ),( cos x , text(if) , x >= pi/2 ))`

Case `I` For continuity at `x = pi/2,`

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

` => lim_( h -> 0) f( pi/2 - h) = lim_( h -> 0) f( pi/2 + h) = f (pi/2)`

`=> lim_( h -> 0) A sin ( pi/2 - h) + B = lim_( h -> 0) cos ( pi/2 + h) = cos (pi/2)`

` => A sin pi/2 + = cos pi/2 = 0`

` => A + B = 0` .........(i)

Case `II` For continuity at `x = - pi/2`,

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

` => lim_( h -> 0) f ( - pi/2 - h) = lim_(h -> 0) ( - pi/2 + h) = f( - pi /2)`

` => lim_(h -> 0) [ - 2 sin ( - pi/2 - h) ]`

` = lim_(h -> 0) [ A sin ( - pi/2 + h) + B ]`

` = -2 sin ( - pi/2) => 2 = - A + B = 2`

` => A - B = 0` ..........(ii)

On solving Eqs. (i) and (ii), we get

` B = 1`
Correct Answer is `=>` (A) `1`
Q 1659167014

Consider the function `f(x) = tt { ( (x^2 - 5 , x <= 3) ,( sqrt(x + 13) , x > 3) )`

Consider the following statements
1. The function is discontinuous at `x = 3`.
2. The function is not differentiable at `x =0`.
Which of the above statement (s) is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

1. For continuous,

`lim_(x -> 3^-) f(x) = lim_(x -> 3^+) f(x) = f(3)`

`:. lim_(x -> 3^-) f(x) = lim_(x -> 3^+) f(x) = 4`

Hence, `f(x)` is continuous at `x = 4`.

2. We have, `f(x) = x^2 - 5, x <= 3`

` => f'(x)=2x => f'(0)= 0`

Hence, `f(x)` is differentiable at `x = 0`.

So, neither Statement `1` nor `2` is correct.
Correct Answer is `=>` (D) Neither 1 nor 2
Q 1609367218

Consider the function

`f(x) = {tt (((tan kx)/x , x < 0),(3x + 2k^2, x >= 0) )`

What is the non-zero value of `k` for which the function is continuous at `x = 0`?
NDA Paper 1 2014
(A)

`1/4`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

We have,

`f(x) = {tt (((tan kx)/x , x < 0),(3x + 2k^2, x >= 0) )`

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

`:. lim_(x -> 0^-) f(x) = lim f(x)_(x -> 0^+) = f (0)`

`=> lim_(x -> 0^-) ((tan kx)/x) = lim_(x -> 0^+) (3x + 2k^2) = 3(0) + 2k^2`

`=> lim_(x -> 0-h) [(tan k (0 - h))/(0-h)]= lim_(x -> 0 + h) + 2k^2 ] = 2k^2`

` => lim_(h -> 0) ((tan kh)/h) = 2k^2`

`=> lim_(h -> 0) ((k tan kh)/h) = 2k^2`

` => k = 2k^2`

` :. k = 1/2 `
Correct Answer is `=>` (B) `1/2`
Q 2460101915

Consider the following functions

I. `f(x) =e^x`, where `x > 0`

II. `g(x)= | x-3 |`

Which of the above functions is/arc continuous?
NDA Paper 1 2013
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. Given function, `f(x) =e^x`, Wilere `x > 0`

Graph of `f(x)` is

We observe from the graph that the graph of `e^x` is not breaking
when `x > 0`. So, the graph is continuous at `x > 0`.

II. Given function, `g (x) = | x - 3 |`

Graph of `g (x)` is


We observe from the graph that, the graph of ` | x - 3|` is not
breaking anywhere but have sharp turn at `x = 3`. So, `g(x)` is
continuous but not differentiable at `x = 3`.
Correct Answer is `=>` (C) Both I and II
Q 2460012815

Consider the following statements in respect of a
function` f (x)`


I. `f(x)` is continuous at `x =a` iff `lim_(x->a) f(x)` exist.

II. If `f(x)` is continuous at a point , then `1/(f(x))` is also continuous at the point.

Which of the above statements is/are correct?
NDA Paper 1 2013
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. We know that, If `f{a)= lim_(x->a) f(x)`, then `f(x)` is continuous

at `x =a` , while both hand must be exist.

II. If `f(x)` is continuous at a point, then it is not necessary that
`1/(f(x))` is also continuous at that point.

e.g.,

(i) `f(x) =x` is continuous at `x =0` but `f(x)=1/x` is not
continuous at `x = 0`.

(ii) `f(x) = e^x` is continuous at `x = 0` and `f(x) = e^(-x) ` is also
continuous at `x = 0`.
Correct Answer is `=>` (D) Neither I nor II
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
Q 2470145016

Consider the following in respect of the function

`f(x) = | x-3 |`

I. `f(x)` is continuous at `x = 3`.

II. `f(x)` is differentiable at `x = 0`.

Which of the above statements is/are correct ?
NDA Paper 1 2012
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Given function, `f (x) = | x - 3 |`

I. `LHL` at `x= 3, f(3-0) = lim_(x->3^-) f(3)`

`= lim_(h->0) f(3-h) = lim_(x->0) h =0`

`= lim_(x->0) | -h | = lim_(x->0) h =0`

RHL at `x = 3, f (3 + 0)`

`lim_(x->3^+) = lim_(h->0) f(3+h)`

`= lim_(h->0) |3+h -3 | = lim_(h->0) | h | =0`

and `f(3) = |3-3 | =0`

Here, `f(x)` is continuous at `x = 3`.


II. `Lf' (0) = lim_(x->0) (f(0-h) -f (0) )/(-h) = lim_(h->0) (|-h -3 | - |-3 |)/(-h)`

`= lim_(h->0) (h+3-3)/(-h) = lim_(h->0) h/(-h) = -1`


`Rf' (0) = lim_(h->0) (f (0+h) -f (0) )/h = lim_(h->0) ( | h-3| - |-3 | )/h`

`= lim_(h->0) (-h +3 -3)/h = lim_(h->0) (-h)/h`

`= -1`

`:. Lf' (0) = Rf' (0)`

Hence, `f(x)` is differentiable at `x = 0`.
Correct Answer is `=>` (C) Both I and II
Q 2400645518

What is the value of `k` for which the following
function `f(x)` is continuous for all `x`?

`f(x) = { tt ( ((x^3 -3x +2)/( (x-1)^2) , x ne 1 ), (k, x =1) )`
NDA Paper 1 2011
(A)

`3`

(B)

`2`

(C)

`1`

(D)

`-1`

Solution:

` :. f(x) = { tt ( ( (x^3 -3x +2)/( (x-1)^2) , AA x ne 1 ) , (k, AA x =1) )`

and `f(x)` is continuous for all `x`.

`:. lim_(x->1) f(x) = f(1)`

`=> lim_(x->1) (x^3 -3x +2)/( (x-1)^2 ) = k` (`0/0` form )

By L' Hospital rule,

`k = lim_(x->1) (3x^2 -3)/(2 (x-1) )` (`0/0` form)

By L' Hospital rule,

`=lim_(x->1) (6x)/2 =3 * (1)`

`=3`
Correct Answer is `=>` (A) `3`
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 2450356214

Consider the following function `f : R -> R` such
that `f(x) = x`, if `x ge 0` and `f(x) =- x^2` , if `x < 0`. Then,
which one of the following is correct?
NDA Paper 1 2009
(A)

`f(x)` is continuous at every `x in R`

(B)

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

(C)

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

(D)

`f(x)` is discontinuous at every `x in R`

Solution:

Given, `f(x) = { tt ( (x, x ge 0), (-x^2 , x < 0) )`

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

`=> = f (0-h) = lim_(h->oo) - (0-h)^2`

`=> lim_(h->0) -h^2 = 0`

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

and `f(0) = 0`

`:. LHL= RHL =f(0)`

Since, it is continuous at `x = 0`.

Also, `f(x)` is continuous in the given interval. Hence, `f(x) ` is

continuous in every `x in R`.
Correct Answer is `=>` (A) `f(x)` is continuous at every `x in R`
Q 2420556411

If `f(x) = { tt ( (3x-4, 0 le x le 2), (2x+ lambda , 2< x le 3) )`

is continuous at `x = 2`, then what is the value of `lambda` ?
NDA Paper 1 2009
(A)

`1`

(B)

`-1`

(C)

`2`

(D)

`-2`

Solution:

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

Also, `f(x)` is continuous at `x = 2`.

Now, `LHL = f(2-0) = lim_(h->0) f (2-h)`

`= lim_(h->0) 3 (2-h) -4 = lim_(h->0) 2-3h = 2`

`RHL = f(2+0) = lim_(h->0) f (2+h)`

`= lim_(h->0) 2 (2+h) + lambda = 4 +lambda`

and `f(2) = 2(2) +lambda = 4+ lambda`

Now, `LHL = RHL = f(2)`

`:. 2= 4+ lambda => lambda = -2`
Correct Answer is `=>` (D) `-2`
Q 2400167918

Let `f : R -> R` be defined as

`f(x) =sin( | x |)`

Which one of the following is correct?
NDA Paper 1 2008
(A)

`f` is not differentiable only at `0`

(B)

`f` is differentiable at `0` only

(C)

`f` is differentiable everywhere

(D)

`f` is non-differentiable at many points

Solution:

`f(x) = sin | x |`

`= { tt ( (sin x , x ge 0 ), (sin(-x) , x < 0) )`

`= { tt ((sin(x) , x ge 0), (-sin x , x < 0) )`

`=> f'(x) = { tt ( (cos x , x ge 0), (- cos x, x < 0) )`


From above it is clear that `f(x)` is not differentiabrd at `x = 0` and
`f(x)` is a periodic function. Thus, `f(x)` is non-differertiable at many
points.
Correct Answer is `=>` (D) `f` is non-differentiable at many points
Q 2480178017

A function `f` is defined as follows

`f(x) = x^p cos (1/x) , x ne 0 , f(0) = 0`

What conditions should be imposed on `p`, so that `f`
may be continuous at `x = 0`?
NDA Paper 1 2007
(A)

`p = 0`

(B)

`p> 0`

(C)

`p < 0`

(D)

No value of `p`

Solution:

`:. f(x) = { tt ( x^p cos (1/x) , x ne 0) , (0, x = 0) )`

`lim_(x->0) f(x) = lim_(x->0) x^p cos (1/x)`

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

`:. lim_(x->0) x^p cos (1/x) = f(0) = 0`

which is possible only, if `p> 0`.
Correct Answer is `=>` (B) `p> 0`
Q 2410378210

If the derivative of the function

`f(x) = { tt ( (ax^2 +b , x < -1), (bx^2 +ax +4 , x ge -1) )`

is everywhere continuous, then what are the
values of `a` and `b`?
NDA Paper 1 2007
(A)

`a =2, b = 3`

(B)

`a= 3, b = 2`

(C)

`a = -2, b = - 3`

(D)

`a=- 3, b = -2`

Solution:

Given that,

`f(x) = { tt ( (ax^2 +b , x < -1) , (bx^2 +ax +4, x ge -1) )`

Since `f'(x)` is continuous everywhere, so `f(x)` is also continuous
everywhere.

So, `f(-1) = lim_(x->1^-) f(x) = lim_(h->0) a (-1-h)^2 + b`

`=> b-a +4 =a+b`

`=> a= 2`

Now, `f'(x) = { tt ( (2ax , x < -1), (2bx +a , x ge -1) )`

Now, given that `f '(x)` is everywhere continuous ..

`:. f(-1) = lim_(x->1^-) f(x) = lim_(h->0) (2a) (-1-h) = -2a`

`=> 2b (-1) +a = -2a`

`=> 3a -2b = 0`

i.e., `a=1, b =3`

Here, we can take the value of `a = 2`, because at `a = 2` the
function `f(x)` is continuous everywhere.
Correct Answer is `=>` (A) `a =2, b = 3`
 
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