Mathematics Tricks & Tips of differentiability for NDA Lecture - 1
Click for Only Video

Differentiability of Piecewise Function ( Function Define Differently In Different Intervals)

Following methods will be covered :

- Checking differentiability by differentiating in different intervals

- Two variable problems : use continuity and differentiability

- Differentiability At point where function is not changing

- Checking differentiability by graph

- Differentiability by definition

- Checking differentiability by differentiating in different intervals

Q 2900223118

If `f(x) = { tt(( x^2, x le 0),(ax, x > 0))` then find the values of `a` if `f(x)` differentiable at `x = 0`.

(A)

a=1

(B)

a=0

(C)

a=3

(D)

None of these

Solution:

If `f(x) = { tt(( x^2, x le 0),(ax, x > 0))`

`f'(x) = { tt(( 2x, x le 0),(a, x > 0))`

`f (x)` will be differentiable at `x = 0`

if Limit of `f'(x)` exist at x = 0

`:.` LHL = RHL

`2xx0 = a`

`a=0`
Q 2665267165

If `f(x) = { tt(( x, x le 1),(x^2+bx+c, x > 1))` then find the values of `b` and `c` if `f(x)` differentiable at `x = 1`.



Solution:

`f(x) = { tt(( x, x le 1),(x^2+bx+c, x > 1))`

` => f'(x) = { tt(( 1, x < 1),(2x+b , x > 1))`

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

Then, it must be continuous at `x = 1`

for which `Lim_(x-> 1^+) f(x) = Lim _(x-> 1^(-)) f(x)`

`=> 1+b+c=1`

`=> b+c =0`

Also, `f'(0) =f'(0^(-))`....................(i)

`Lim_(x-> 1^(+)) f'(x) =Lim_(x->1^(-)) f'(x)=> 2+b=1 => b=-1`

`=> c=1`
[from equation (i)]

- Two variable problems : use continuity and differentiability

Q 2372012836

If the derivative of the function

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

is everywhere continuous and differentiable

then the values of `a` and `b` are
BITSAT Mock
(A)

`(2, 3)`

(B)

`(3, 2)`

(C)

` (- 2, - 3)`

(D)

` (- 3, - 2)`

Solution:

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

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

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

`=> underset (x-> -1^-) (Lt) f(x) = underset (x-> -1^+) ( Lt) f(x)`

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

`=> a = 2`

`f(x)` is differentiable at `x = - 1`

`=>` LHD at `x = 1`

`=` RHD at `x = - 1`

`- 2a = - 2b + a`

`=> b = 3`
Correct Answer is `=>` (A) `(2, 3)`
Q 2811012820

Find the value of `a` and `b` if `f(x) = { tt ((ax^2+1 , x le 1) , ( x^2 + ax+b, x > 1))` is differentiable at `x = 1`.

Solution:

`a=2, b=0`
Q 1765601565

Find the values of `p` and `q`, so that `f(x) = { tt( (x^2 +3x+p , text(if) x le 1) ,(qx +2, text(if) x >1) )` is differentiable at `x=1`.
NCERT Exemplar
Solution:

We have,

`f(x) = { tt( (x^2 +3x+p , text(if) x le 1) ,(qx +2, text(if) x >1) )` is differentiable at `x=1`.

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

LHD `= lim_(x-> 1^-) ( (x^2 +3x + p) -(1+3 +p))/(x-1)`

`= lim_(h-> 0)( [ (1-h)^2 + 3 (1-h) + p] - [1+3+p])/( (1-h)-1)`

`= lim_(h->0) ([1+h^2 -2h +3 -3h +p] -[4+p ])/(-h)`

`= lim_(h->0) ( [h^2 -5h + p+4 -4 -p])/(-h) = lim_(h->0) (h [h-5]) /(-h)`

`= lim_(h->0) - [h-5] =5`

RHD ` f'(1) =lim_(x->1^+) ( f(x) - f(1))/(x-1) = lim_(x-> 1^+) ( (qx+2) - (1+3 +p) )/(x-1)`

`= lim_(h-> 0) ( [q(1+h) +2 ] - (4+p)) /(1+ h-1)`

`= lim_(h->0) ( [ q +qh +2 -4 -p])/h = lim_(h->0) ( qh + (q-2-p)) /h`

`=> q-2 -p =0 => p -q =-2` ......................(i)

`=> lim_(h-> 0) (qh +0)/h =q` [for existing the limit]

If `L H D f'(1) = R H D f'(1) `, then `5 =q`

`=> p-5 = -2 => p =3`

`:. p =3` and `q =5`

Differentiability At point where function is not changing

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 1752391234

Let `f(x)` be a function defined in `1 <= x < oo` by

` f(x) ={tt {(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

Consider the following statements
I. The function is continuous at every point in the interval `[1, oo)`,
II. The function is differentiable at `x = 1.5`.

Which of the above statements is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given function, `f(x) = {tt {(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

and whole function defined in `1 <= x < oo`.

I. Since, the function is polynomial, so it is continuous

as well as differentiable in its domain `[1, oo) - {2}`.

Now, we check the continuity of the function at `x = 2`.

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

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

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

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

` = lim_(h -> 0) 3(2 + h) - (2 + h)^2`

` = 3(2 + 0) - (2 + 0)^2`

` = 6 - 4 = 2`

and `f(2) = 2 - 2 = 0`

`∵ f(2) = LHL != RHL`

So, the function is not continuous at every point in the interval

`[1, oo)` i.e., not continuous at `x = 2`.

II. We also check the differentiability of the function at

`x=1.5`.

`Rf'(1.5) = lim_(h -> 0) (f(1.5 + h) - 0 f(1.5))/h`

` lim_(h -> 0)( 2- (1.5+ h)- (2 - 1.5))h`

` = lim_(h -> 0)( 0.5- h - 0.5)/h`

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

`Lf' (1.5) = lim_(h -> 0) (f(1. 5- h) f(1.5))/(-h)`

` = lim_(h -> 0) (2-(1.5- h)-(2 - 1.5))/(-h)`

` = lim_(h -> 0) ( 0.5 + h - 0.5)/(-h)`

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

`∵ Lf' (1.5) = Rf' (1.5)`

So, the function is differentiable at `x = 1.5`
Correct Answer is `=>` (B) Only 2
Q 1702491338

Let `f(x)` be a function defined in `1 <= x < oo` by

` f(x) =tt ({(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

What is the differentiable coefficient of `f(x)` at `x = 3`?

NDA Paper 1 2014
(A)

`1`

(B)

`2`

(C)

`-1`

(D)

`-3`

Solution:

Given function, `f(x) = tt ({(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

and whole function defined in `1 <= x < oo`.

`∵ f(x) = tt ({(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

` => f'(x) = tt ({(-1, text(for) 1<= x <= 2),( 3 - 2x, text(for) x > 2))`...........(1)

So, the differentiable coefficient of `f(x)` at `x = 3` is

`f ' (3) = 3 - 2(3) = 3- 6 = - 3 quad [∵ f' (x) = 3 - 2x` for `x > 2]`
Correct Answer is `=>` (D) `-3`

- Checking differentiability by graph

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`

- Differentiability by definition

Q 2960412315

lf `f(x) = { tt (( (e^(1//x) -1)/(e^(1//x)+1) text(,), x ne 0),(0 text(,), x =0))` then `f(x)` is

(A)

continuous as well as differentiable at `x = 0`

(B)

continuous but not differentiable at `x = 0`

(C)

diffi:rentiable but not continuous at `x = 0`

(D)

None of these

Solution:

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

and `lim_(h->0) f(0+h) =lim _(h->0) (e^(1//h)-1)/(e^(1//h)+1)= lim_(h->0) (1- e^(-1//h))/(1+ e^(-1//h))`

`=(1-0)/(1+0) =1`

Since `lim _(h->0) f(0-h) ne lim_(h->0) f(0 +h),` therefore, `f(x)` is not continuous and hence not differentiable at `x =0`.
Correct Answer is `=>` (D) None of these

 
SiteLock