Jee 2014 Booklet5 Hwt Solutions Differential Calculus 1

Vidyamandir Classes

Solutions to Home Practice Test-5/Mathematics Differential Calculus - 1 1.

HWT - 1

For statement 1  x   2  0   x   2  x  2 ,  1

Also

4  x2 0  x   2

For x  2, 4  x 2  0

 x   2  0 

and



x  2 is in domain of the function.

For

1  x  2 4  x2 0  x   2

4  x 2  0 and  x   2  0  For



x   1, 2  is in domain of the function.



Domain of the function is x    ,  2    1, 2 

x2 4  x2 0  x   2

4  x 2  0 and  x   2  0  

4  x2 0  x   2

x > 2 is not in domain of the function

For statement 2 2

 

 x   4  x   3  0 

y 2  4 y  3  0 , where y = [x]

 y  1  y  3  0 or x    , 2   3,  

y    , 1  3,   or  x     , 1  3,  

For statement 3 sin log 2 log3 x  0

or 4.

7.

log3 x   22 n , 2 

 tan x  y   tan y  x

 

log 2 log3 x   2n ,  2n  1  

or

 2 n  1   2 n x  32 , 32 ,n I  

y log  tan x   xlog  tan y 





dy y x dy log  tan x   sec 2 x  log  tan y   sec 2 y dx tan x tan y dx



2y log  tan y   dy 2 sin x cos x log  tan y   2 y cos ec 2 x   2x dx log tan x  log  tan x   2 x cos ec 2 y   2 sin y cos y

x  a cos 3 



dx  3 a cos 2  sin d

y  a sin3 



dy  3a sin 2  cos  d



dy   tan  dx

d2y dx 2



d  dy  d 1 1   sec 2    sec 2    dx  dx  dx 3a cos 2  sin 3a cos 4  sin

  dy  2  1       dx   d2y dx 2

10.

2 n  1 



lim

x

3x  x 7x  5 x

32

1  tan    2

3

2

1

 3a cos 4  sin  sec3   3a sin cos 

3a cos 4  sin

3x  x 4x  lim 2 7 x  5 x 2 x x x

 lim

VMC/Solutions/Differential Calculus - 1

11

HWT-5/Mathematics

Vidyamandir Classes

Differential Calculus - 1 1.

 x1 3  a1 3 y  tan 1   1  x1 3 a1 3 

   tan 1 x1 3  tan 1 a1 3  

 

 

dy 1 1 1    dx 1  x 2 3 3 x 2 3 3 x 2 3 1  x 2 3



HWT - 2





2

3.

1 1  1  f  x    x2  2   x    2 x x    x Put x 

4.

5.

lim

1  y , to get f  y   y 2  2 x

a x  log 1  x   sin x  cos 2 x ex 1

x0

f  x   x2  2

or



1  0  0 1 0 11

Note that f (x) is always +ve.

f  x  f



 x   f  x  f  x 

 1  1   x      x   1  x  0 2  2   1 1      x  x  0  x 1   2   2  2 f  x

7.

LHL = lim

5 sin cos x 

RHL = lim x

8.

9.

lim

n



cos x   2

 2

x

5 sin  1 1  2





5 sin  0  02

0

 5 sin 1

2

yn  xn



1n

1n

 xn   lim y  1  n  n  y  

 0.5log3 log1 3  x 2  

4  2 4   1  log3 log1 3  x  5   0 5     17  2 4  2 4 1  x2   0 log1 3  x    1   x    15 5 5 3   17 15

or

x2 

Also,

4  log1 3  x 2    0 or 5 

and

x2 

or

4 0 5

  17   17 x    ,  ,     15 15     x2 

4 1 5

or

x2 

9 5

 3 3  x ,  5  5

or

  2   2 x    , ,   5 5      17 3   3 17  x  , ,    15  5  5  15

x2 

or

Combining all these results, we get :

10.

  x  y  As  1 y  

4 5

or

 1  x  1     sin  2 x  1   6   1  x  1          lim x  sin    6   xlim 1 x    2 x  1    x Using LH rule, we get :

1

  lim

x

 x 1  1    2x 1 

2



 2 x  1   x  1  2   2 x  12 1 x2

VMC/Solutions/Differential Calculus - 1

 lim

x2

x

 2 x  1

12

3x  2 x 2

 lim

x

x

 2 x  1

2 3 x



1 2 3

HWT-5/Mathematics

Vidyamandir Classes

Differential Calculus - 1 1.

HWT - 3

Check formula in option (B). When 0  x  3, f  x   x  3  2 x   3  x   2 x  x  3 When x  3, f  x   x  3  2 x   x  3  2 x  3 x  3

3.

f (x) is discontinuous when x 2  x  1 is an integer. Check the graph of y  x 2  x  1 in the internal x   0 , 2  The only points of discontinuity are when x 2  x  1  1 and x 2  x  1  2

x2  x  1  1

and

x2  x  1  2



x2  x  0

and

x2  x  1  0



x  0, 1

and

x

So, we get in the desired interval x  1,

1 5 2

1 5 2 14

5.

2    x  2 2 x 1 f  x   9 x  27 3  219  3       

 9 x  9 

x  2

 219  9

14 x 1 



14

 73    . 9 x  219   81 

7.

Now,

73 x . 9  219  0 or 81

73.9 x  219 81

or

9 x  243

or

32 x  35

2x  5

or

x

or

5 2

 1 x 2 x x    2  x log 1    log 1   log  log     1  x 2   2  x  2 2 f  x    x x x

Now

2  x 1 2  x    2  x  1  2 x   log   2 x  2  x 2 2 x   lim  lim x 1 x 0 x 0

 lim

x 0

4 4  x2

(Using LH rule)

1

As f (x) is continuous at x = 0, lim f  x   f  0   a  1 x 0

Differential Calculus - 1 2.

 x  2  0 

3.

4.

x  2  0 , 1



x  2 ,  1

x    ,  2    1,  

x 2  xy  y 2 





HWT - 4

7 4



2x  y  x

dy dy  2y 0 dx dx

1   2   dy   2 x  y  2  5     dx x  2y 11 4

y  sin x  sin x  sin x  . . . .  sin x  y 

y 2  sin x  y



2y

VMC/Solutions/Differential Calculus - 1

dy dy  cos x  dx dx



13

dy cos x  dx 2 y  1

HWT-5/Mathematics

Vidyamandir Classes 5.

  2 x  3   2x  3  f  x   sin  log x  , y  f   sin log     3  2x    3  2 x    2 x  3   3  2 x 2  3  2 x    2 x  3 2  dy  cos log     dx   3  2 x  2 x  3  3  2 x 2

  2 x  3  12  cos log      3  2 x    2 x  3  3  2 x  dy 12 cos log 5 at x = 1 is equal to dx 5



6.

lim

x 3   k  4  x  2k x3

x3

 7.

 lim

x3

23  k  8

3x 2   k  4  1



 23  k

k  15

    y  sin 1  cos x   cos 1  sin x     x     x     2 x 2  2  dy  2 dx

Differential Calculus - 1

1.

HWT - 5

    tan x   log    1 2      lim       4x    x sin x   4  4        1  tan  t   4    log     2     1  Put x   t to get :   lim   4 4t sint  t 0     

 sec 2  t   4   2 1 1 1   lim    lim Apply LH rule to get :   lim  4 2 t  0 sint t  0  1  tan  t   4  t  0 sint    Hence  doesn’t exist. 2.

f  x   1  tan x

f (x) is not differentiable when tan x  0 or x  n , n  Z . Also, the function is discontinuous and non-differentiable when cos x = 0 or x   2n  1 2

4.

lim

x 1

x  x4  x9  . . . . .  xn  n 1  4 x 3  9 x8  . . . .  n 2 x n  lim x 1 x 1 1





 1  4  9  . . . .  n2 

5.

 sin x    lim   x0  x 



1x

 log     lim

x0

 ,nZ . 2

2 1

(Using LH rule)

n  n  1  2n  1 6

1  sin x  log   x  x

  x 3 x5   . . . . x 2 4   1 3 5   lim 1 log 1  x  x  . . .  log     lim log    x0 x x0 x x 3 5      

   x2 x4    . . . .  1        x2 3 5   x   1 x4 x3    lim log     . . . .  lim    . . . .  0 or       2 4 x0 x x  0 3 5 3 5    x  x  . . . .          3  5    

VMC/Solutions/Differential Calculus - 1

14

  e0  1

HWT-5/Mathematics

Vidyamandir Classes 6.

 sin  1  h    sin  h  1  lim f  x   lim    lim    sin1 1  h  h  0  h  1  h  0  x  0  sin  h   lim f  x   lim   1  h  0 x0  h 

7.

lim

x p

x p

 9.

g  x f  p  g  p f  x

lim f  x  does not exist.



x0

g x f  p  g  p f  x

 lim

x p

1

g p f  p   g  p  f  p   0

g p

or

g  p

We need to ensure that the function is continuous at x 



1 

lim x

 2

cos x



 

g p f  p  g  p f  p 1 f  p  f  p



0

6 3  2 1

 2 p

p cos x

q

      lim  1  cos   h   cos  2  h  h0  2 

or

 ep  q

and

lim e x

      cot   x   cot m  x    2 2    



tan  mh  tan  h  

h0

10.

p

cot   h  cot m  h  

lim e 

or

h0

2

 lim e  

q

m 

 lim e mh h  q

em

or

h0



q

q  em 

and

For inverse we have x

e y  e y e y  e y



e2 y  1 e2 y  1



x 1  e2 y 1 x



Domain is x   1, 1



log

x 1  2y 1 x

And there is no point in domain where function is discontinuous.

Differential Calculus - 1 1.

HWT - 6

f  x   3 x 2  1  3 x 2   1     Critical points are 3x 2  n , n Z

4.

n 3



x





f (x) is not continuous at x 

x  0,

1 , 3

2 ,1 3

1 2 , 3 3

 1  k  k  1 2  k  2    tk  cot 1    2  k  1      k  1  k  2    k  k  1  2  k  1   tan 1     tan 1  2  1  k  k  1  k  2    1  k  k  12  k  2      







tk  tan 1  k  1 k  2   tan 1 k  k  1 n



t

k













 tan 1  n  1  n  2   tan 1 11  1



k 1

 tan 1  n  1  n  2   tan 1  2 

VMC/Solutions/Differential Calculus - 1

15

HWT-5/Mathematics

Vidyamandir Classes 5.

6.

9 9 , 0 , are the critical points for mod function. 2 2  f   x  changes sign here. x

e   4 eh  4  lim h  3 h  h h  0 h h h

lim f  x   lim  h  h0

x  0

lim f  x   lim   h  h0

x  0

 7.

h  h

f (x) is discontinuous at x = 0

u  3x12 , v  x 6 

8.

e   4 e 1  4  lim  h 0 h0 1   h 1

u  3v 2

du  6v  6 x 6 dv



 3 2  3 f  1  f     1    2 2  1 

3

27 8

35 8 2

27  3  3 f   1  f     2  1  3     2  2 2

9.

Let



f  x   3 x  log 1  3x  2 x 2



f  0  0

f  x  3 

 4 x  3 1  3x  2 x 2

f  0  0

f   x  

8 x 2  12 x  5

1  3x  2 x  2

2

f   x   0

 Now

f  x  0  x  0

Let

f  x  0  x  0







g  x   log 1  3x  2 x 2  3x 

5x 2 2

g  0  0

g x 

4x  3

1  3x  2 x  2

 3  5x

g  0  0

g   x  

 8 x 2  12 x  5   5 2 1  3x  2 x 2





g   0   0

g   x   

32 x 3  72 x 2  60 x  18



1  3x  2 x 2



3

0  x 0

g   x  is increasing  x  0

VMC/Solutions/Differential Calculus - 1

16

HWT-5/Mathematics

Vidyamandir Classes

Differential Calculus - 1 2.

1 

6  3x 1 4



4  6  3 x  4



x

x 1  1 2 x  1

 

x 1    ,  1  1,   2

and

x

2 3

x 1 1 2 x3

or or

 10  x   3,   3



3.

10 3

and

HWT - 7

 x cos    2  0 form lim  x 1 1 x 0

Apply LH rule

   sin  x   2  2  2 sin       22 1   2 x



4.

lim

x3

x 5  243 x x  aa



will be 0 if a  3

a3

5.

So f (x) is continuous and differentiable at x = 0 6.

8.

It has removable discontinuity as limit exists but it is not equal to f  0  (at (x = 0))

dy dy  dt dx dx dt dy 2  dt  2t  12  1 2 dx  2te t dt

9.

2



dy  et  dx t 4t 2  2  4t





For domain

4  x2  0 and

1 x  0



x   2 , 1

and

VMC/Solutions/Differential Calculus - 1

4  x2 0 1 x

17

HWT-5/Mathematics

Vidyamandir Classes

Differential Calculus - 1

HWT - 8

1  cos 2 x  1.

2.

sin x 2  x x Limit does not exist. 1  a2 x  

1



1  a2 x

b 1  a2 x

0

  b   sin  2    1  a x   b  1 b  b   b   2  1  a x 



3.

0 

f  x  f  x

x  0

x  0

2a  x   2 x



aR

Also f (x) should be continuous at x = 0 

f  x  f  x

x  0

 4.

b=0

Look for critical points of f (x) which are x  I and And

7.

x  0

x  2I  1

1 1 does not belong to any of these two sets  f (x) is continuous and differentiable at x  2 2

log 0.3  x  1 

1 log 0.3  x  1 2



2 log0.3  x  1  log0.3  x  1

Now,

x 1 0  x  1

Also,

 x  12  x  1

or x 2  3x  2  0

Combining above results, we get :



log0.3  x  1  log0.3  x  1



 x  1  x  2  0

2

 x    , 1   2 ,  

x   2, 

8.

f  x   0  x  R  (Roots of quadratic equation)

9.

tan 1 x is little less than x in proximity of zero



 tan 1 x   0  x 

Differential Calculus - 1 1. 2.

lim

2



2k  12k  16k 6k   3k 2 2

f (x) is discontinuous at x  11, 6 

3.

2 f   x   12 f   2 x   16 f   x 

x0

HWT - 9

2  2  f  is discontinuous at x  2  11, 6  x2

24 7 , 11 3



x



y  log x 

y x  y   log y  0 x y



y 

log x y  log y x  0



y log x  xlog y  0



y  xy log x  x 2   xy log y  y 2





VMC/Solutions/Differential Calculus - 1

 y  xlog y  y  x  y log x  x 

18

HWT-5/Mathematics

Vidyamandir Classes 4.

dy 1  1 2 dt t dx 1  1  dt t

5.

dy t 2  1  dx t 2  1



Graph is like

 7.

 1  tan x   1  sin x 

Let   lim

x0

log     lim



x0

1 log 1  tan x   log 1  sin x   sin x  x  log 1  tan x  tan x log 1  sin x  sin x  0 .  .    1 1 .1  1 . 1   0    e  1 sin x  tan x x sin x x 

 lim

x0

8.

Since 0 < a < 1

sin 2  n! 0 1 n1 a  a n

We have lim

n

n1 a  

As 9.

even function

cosec x

f  x  0

and

1 na

0

(a constant function)

As tan  I   0 I : integer 10.

f  x  1

For x  0

f  x   f  x   f  0



x  0



x  0

removable discontinuity

Differential Calculus - 1 1.

1



1 1 9y

2

dy dx

d y dx

2

d2y dx 2

2.

. . . .(i)

dy  1 9 y2 dx 2

HWT - 10



9y



1  9 y2

d2y dx 2



9y 1 9y

2

dy dx

 1 9 y2

 9y

Since g is inverse of f we have f  x   y and

g  y  x

or

f g  x  x



g   x   cos ec g  x 











  f  g  x  g  x   1 f g  y  y



VMC/Solutions/Differential Calculus - 1

19

HWT-5/Mathematics

Vidyamandir Classes 3.

sin 3  x   0  x  R

And period of tan   x is 1. 4.

f  x    sin x  cos x  now sin x  cos x Has values from 1 to 

6.

lim

2

Range of f (x) is only {1}.

sin 2 x  a sin x

x0



x3

lim

b

  2 x 3   2 x 5 . . . . .   a  x  x 3  x5 . . . . .   2x    3 5 3 5     x3

x0

  2

3

 7.

2a 0



a  2

and

3



b

a  b  b  1 3

From graph it is :

Non-differentiable when 2 sin x  1  cos x [except x = 0]

10.



4 1  cos 2 x   1  cos 2 x  2 cos x  



cos x 

f   x 

2  64 2  8 6 3    10 10 10 5



5 cos 2 x  2 cos x  3  0



3  3  x    cos 1    cos 1   5  5 

 U   x  V  x   V   x  U  x    4  1   2   2    0  u  x    2 1  V 2  x  v x     1

VMC/Solutions/Differential Calculus - 1

20

HWT-5/Mathematics