Jee 2014 Booklet5 Hwt Solutions Differential Calculus 1

Solutions to Home Practice Test-5/Mathematics

1. For statement 1



2 0 2 2 1 x    x    x  ,           Also 2 4 0 2 x x  _      For x 2 4, x20 and 2 4 2 0 0 2 x x x          _{ }    x  is in domain of the function.2

For   1 x 2 2 4x  and0 2 4 2 0 0 2 x x x          _{ }

   x 



1 2,



is in domain of the function.

For x2 2 4x  and0 2 4 2 0 0 2 x x x          _{ }  

 x > 2 is not in domain of the function  Domain of the function is x  



, 2



 



1 2,



For statement 2 2 4 3 0 x  x            y24y  , where y = [x]3 0 



y1

 

y 3



0 or y 



,1 _{ } 3,



or  _{ }x  



,1 _{ } 3,



 x 



,2



3,



For statement 3



2 3



0

sin log log x   log log x2 3 2n ,



2n1



 or log x₃  22n,22n1_   or 2 1 2 2 2 3 n 3 n x_ ,   _, nI  

4.



tan x

 

y tan y



x  y log tan x





x log tan y





 dy





y 2





x 2 dy

log tan x sec x log tan y sec y

dx tan x  tan y dx 

















2 2 2 2 2 2 2 2  _     y log tan y

log tan y y cos ec x

dy sin x cos x

x

dx _{log tan x} log tan x x cos ec y sin y cos y

7. xa cos3  dx 3a cos2sin d  3

ya sin   dy3a sin2cos d 

 dy tan dx   2 2 2 2 2 4 1 1 3 3 d y d dy d sec sec dx dx dx

dx a cos sin a cos sin

          _{ }        





3 2 2 3 2 2 4 3 2 4 2 1 1 3 3 1 3 dy tan dx

a cos sin sec a sin cos d y a cos sin dx          _{ }   _{  }         _{   } 10. 3 3 4 2 7 5 7 5 2 x x x x x x x x

lim lim lim

x x x x x

     

 _  _{ }

 

1.

 

 

1 3 1 3 1 1 1 3 1 1 3 1 3 1 3 1 x a

y tan tan x tan a

x a        _ _     





2 3 2 3 2 3 2 3 1 1 1 1 3 3 1 dy dx x x x x      3. 2 2 2 1 1 1 2 f x x x x x x  _ _{  } _  _         Put x 1 y x   , to get

 

2 2   f y y or f x

 

x22 4.





0 1 2 1 0 0 1 0 1 1 1 x x x

a log x sin x cos x lim e      _    _  

5. Note that f (x) is always +ve.



 

 

 

 

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

 

2 f x  7. LHL =

 

2 5 5 0 0 2 0 2 x

sin cos x sin lim cos x      _{  }       RHL =

 

 

2 5 1 5 1 1 2 x sin lim sin        8.





1 1 1 As 1           _  _         n n n n n n n n x x lim y x lim y y y y 9.

 

3 1 3 2 3 1 3 2 4 4 0 5 1 0 5 5 log log . _x  _  log log _x  _      2 1 3 4 1 5 log _x  _    2 4 1 5 3 x  _ _      2 17 0 15 x   or 2 17 15 x  or 17 17 15 15 x  _ , __ ,_       Also, _{1 3} 2 4 0 5 log _x  _   or 2 4 1 5 x   or 2 9 5 x  or 3 3 5 5 x  , _   and 2 4 0 5 x   or 2 4 5 x  or 2 2 5 5 x _ ,   _{ } ,_    

Combining all these results, we get : 17 3 3 17

15 5 5 15 x , _ _ ,        10. 1 1 1 2 1 6 1 1 2 1 6          _{ }    _             _{  } _       x x x sin x x

lim x sin lim

x

x

 

Using LH rule, we get :



 

  









2 2 2 2 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 3 2 x x x x x x x x lim lim x x x x              _{ } _  _    





1 2 2 3 2 1 3       x x lim x x

Differential Calculus - 1

HWT - 2

1. Check formula in option (B).

When 0 x 3, f x

 

 x 3 2x 



3 x



2x x 3 When x3, f x

 

 x 3 2x



x 3



2x3x3 3. f (x) is discontinuous when x2  is an integer.x 1

Check the graph of yx2  in the internalx 1 x

 

0 2,

The only points of discontinuity are when x2   andx 1 1 x2  x 1 2 2 1 1 x   x and x2  x 1 2  2 0 x  x and x2  x 1 0  x0 1, and 1 5 2 x  So, we get in the desired interval 11 5

2 x ,  5.

 

        1 4 2 _{1 4} 2 _{2 1 2 1} 3 9 27 219 3 9 9 219 9 x _{x x x} x x f x  _{  }           _    _       1 4 73 9 219 81 x .   _  _   Now, 73 9 219 0 81   x . or 73 9 219 81 x . _ or 9x243 or 32x35 or 2x5 or 5 2 x 7.

 

1 2 2 1 1 1 2 2 2 2 x x x x log

log log log

x x f x x x x      _ _  _          _               Now



 

 





2 0 0 1 2 2 1 2 2 2 ₂ 2 1 x x x x x x log _x x x lim lim x             _   _  _{  (Using LH rule)} 2 0 4 1 4 x lim x     As f (x) is continuous at x = 0,

 

 

0 0 1     x lim f x f a 2.



x2



0  x  2 0 1,



 x   2, 1



 x  



, 2



  



1,



3. 2 2 7 4 x xyy   2x y x dy2y dy 0 dx dx 





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

4. y sin x sin x sin x. . . .  sin xy

 2 y sin xy  2y dy cos x dy dx dx  2 1 dy cos x dx y

Differential Calculus - 1

HWT - 3

Differential Calculus - 1

HWT - 4

5.

 





2 3 2 3

3 2 3 2

x x

f x sin log x , y f sin log

x x           _{ } _{  }       



 

 





2 2 3 2 2 3 2 2 3 3 2 3 2 2 3 _{3 2} x x dy x x cos log dx x x _x           _{  }        



 



2 3 12 3 2 2 3 3 2 x cos log x x x      _{  }         dy dx at x = 1 is equal to 12 5 5 cos log  6.









3 2 3 3 4 2 3 4 23 3 1            x x x k x k x k lim lim k x  23 k 8  k15 7. 1





1





2 2 2

ysin cos x cos sin x _ x _{ }  x_  x

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





0 1 4 2 ₁ 4 t tan t log lim t sin t          _ _   _{ }   _  _         

Apply LH rule to get :









2 0 0 0 4 2 1 1 1 1 4 4 2 t t t sec t

lim lim lim

tan t sint sint

      _               

Hence  doesn’t exist. 2. f x

 

 1 tan x

f (x) is not differentiable when tan x or x n , n Z0    .

Also, the function is discontinuous and non-differentiable when cos x = 0 or



2 1



2 x n  , n .Z 4. 2 2 4 9 3 8 2 1 1 1 1 4 9 1 1         _      n n x x x x x . . . . . x n x x . . . . n x lim lim x (Using LH rule)



2





1

 

2 1



1 4 9 6 n n n . . . . n         5.

 

1 0 0 1 x x x sin x sin x

lim log lim log

x x x        _{ }   _{ }       

 

3 5 2 4 0 0 1 3 5 1 1 3 5          _{ }    _{ }                  x x x x x . . . . x x

log lim log lim log . . .

x x x 2 4 2 4 3 2 4 0 0 1 3 5 1 0 3 5 3 5 3 5    _ _{ }   _ _           _{  }   _ _   _ _   _{   }                x x x x . . . . x x x x

lim log . . . . lim . . . .

x _{x x}

. . . .

or e01

6.

 









0 0 0 1 1 1 1 1 h h x sin h sin h

lim f x lim lim sin

h h              _{ }   _         

 

 

0 0 1 h x sin h lim f x lim h              x0

 

lim f x does not exist.

7.

   

   

   

   

   

   

0 1 1              x p x p g x f p g p f x g x f p g p f x g p f p g p f p lim lim x p  g

       

p f p g p f p 0 or

 

 

 

 

62 31 g p f p g p f p     

9. We need to ensure that the function is continuous at 2 x 





2 1 p cos x x lim cos x q     or ₂ 0 1 2  _        _  _   _ _   p cos h hlim cos h   p e q   and 2 2 2 cot x cot m x x lim e q      _   _   _ _ _ _      or     0 cot h cot m h h lim e        0 0 tan mh tan h _{mh h} hlim e hlim e q            _or m e q  pm  and m qe 

10. For inverse we have 2 2 1 1 y y y y y y e e e x e e e          1 2 1 y x e x     1 2 1 x log y x  _   Domain is x 



1 1,



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

1. f x

 

_3x2 1_{ } 3x2_1 Critical points are 3x2 n , nZ

 3 n x  0 1 2 1 3 3 x , , ,  f (x) is not continuous at 1 2 3 3 x , 4.



 







2 1 1 1 2 2 1 k k k k t cot k  _    _      







 



1 2 2 1 1 1 2 k tan k k k  _  _   _{  }   



 

  





 



1 2 1 2 1 1 1 2 k k k k tan k k k  _     _   _{  }   



















1 1 1 2 1 k t tan k k tan k k  1





 





1



 



1 1 2 1 1 1 n k k t tan n n tan      





 







 

1 1 1 2 2 tan n n tan    

Differential Calculus - 1

HWT - 6

5. 9 0 9

2 2

x , , are the critical points for mod function.

 f

 

x changes sign here.

6.

 

 

 

 

0 0 0 4 4 3 h h h h h x e e

lim f x lim h lim h

h h h            

 

 

 

1 0 0 0 4 4 0 1 1 h h h h x e e

lim f x lim h lim h

h                    f (x) is discontinuous at x = 0 7. u3x12, vx6  2 3  u v  du 6v 6x6 dv  8.

 

 

3 2 3 3 1 1 2 2       _{ }  _{ }     f f 27 1 8   35 8 

 

3

 

3 2 27 1 2 1 3 2 2 2 f   f       _{ }       9. Let f x

 

3x log



1 3 x2x2



 

0 0 f 

 



4 3



₂ 3 1 3 2 x f x x x      

 

0 0   f

 





2 2 2 8 12 5 1 3 2 x x f x x x      

 

0 f x   f

 

x   0 x 0  f x

 

  0 x 0 Now Let

 





2 2 5 1 3 2 3 2      x g x log x x x

 

0 0 g 

 



2



4 3 3 5 1 3 2 x g x x x x       

 

0 0 g 

 





2 2 2 8 12 5 5 1 3 2   _   _      x x g x x x

 

0 0 g 

 





3 2 3 2 32 72 60 18 0 0 1 3 2 x x x g x x x x            g

 

x is increasing  x 0

2. 1 6 3 1 4 x     and 1



1

  

1 2 x , ,          4 6 3x4  10 3 x and 2 3 x  1 1 2 x   or 1 1 2 x   x 1 or x3  310 3 x  , _   3. 1 0 2 0 1 x x cos lim x           form Apply LH rule  2 2 2 1 2 2 2 sin x sin x         _{ }     _{ }    _{ } 4. 5 3 243    x a x x lim x a will be 0 if a3  a3 5.

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

6. It has removable discontinuity as limit exists but it is not equal to f

 

0 (at (x = 0))

8.  dy dy _dt dx dx dt





2 2 2 1 1 dy dt  _{t } 2 2 t dx te dt    





2 2 4 2 4 t dy e dx _{t t t}     9. For domain 2 4x 0 and 2 4 0 1 x x  _  and 1 x 0  x 



2 1,



Differential Calculus - 1

HWT - 7

1.



1 2



2 cos x sin x x x  

Limit does not exist. 2. 1 a x 2    1₂ 0 ₂ 0 1 1 b a x  a x     2

 

2 1 1 1 b sin a x b b b b a x           _{  }    _      3.

 

 

0 0      x x f x f x

 

2a x 2x  aR

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



 

 

0 0 x x f x f x       b = 0

4. Look for critical points of f (x) which are xI and x2I1

And 1

2 does not belong to any of these two sets  f (x) is continuous and differentiable at 1 2 x 7. _{0 3}



1



1 _{0 3}



1



2 . . log x  log x  2log0 3.



x 1



log0 3.



x1













2 0 3. 1 0 3. 1 log x log x Now, x 1 0  x1 Also,



x1



2  orx 1 x23x 2 0 



x1

 

x2



0  x 



,1

 

 2,



Combining above results, we get : x



2,



8. f x

 

    (Roots of quadratic equation)0 x R 9. tan1x is little less than x in proximity of zero 

1 0 tan x x           1.

 

 

 

0 2 12 2 16 2 x f x f x f x lim       2 12 16 6 3 2 2 k k k k k      2. f (x) is discontinuous atx11 6,  2 2 f x    _    is discontinuous at 2 11 6 2 , x   24 7 11 3 x , 3. log xylog yx0

 y log xx log y0  y log x y xy log y 0 x y      



2



2 y xy log x x  xy log yy 









y x log y y y x y log x x     

Differential Calculus - 1

HWT - 8

Differential Calculus - 1

HWT - 9

4. dy 1 1₂ dt  t 1 1 dx dt  t  2 2 1 1 dy t dx _t    5. Graph is like  even function 7. Let 0 1 1 cosec x x tan x lim sin x      _{ }     

 









0 1 1 1 x

log lim log tan x log sin x sin x     _    _ 









0 1 1 x

log tan x log sin x

x tan x sin x

lim . .

sin x tan x x sin x x

            

         

0 1 1.1 1 . 1 0 e 1  _  _  _  8. Since 0 < a < 1 We have

 

2 1 ! 0 1 n a a sin n lim n n    _  As n1 a   and 1 0 a n 9. f x

 

0 (a constant function) As tan I

 

  I : integer0 10. For x0 f x

 

1 

 

 

 

0 0 0 x x f x f x f        removable discontinuity 1. 2 1 1 1 9 dy dx y   . . . .(i)  2 1 9 dy y dx   2 2 ₂ 9 1 9 d y y dy dx dx  _{ y} 2 2 2 ₂ 9 1 9 1 9 d y y y dx  _{ y}   2 2 9 d y y dx 

2. Since g is inverse of f we have f x

 

y

and g y

 

x  f g y



 



y or f g x



 



x  f



g x

 



g x

 

1

 g x

 

cos ec g x



 



3. sin3

 

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 2

 Range of f (x) is only {1}. 6. ₃ 0 2 x sin x a sin x lim b x   _ 

     

3 5 3 5 3 0 2 2 2 3 5 3 5 x x x x x x . . . . . a x . . . . . lim b x    _{ }          _{ }         2 a 0  a 2 and

 

3 2 1 3 3 a b b       7. From graph it is :

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

 2 2

4 1_ cos x_ 1 cos x2cos x  5cos x2 2cos x 3 0

 2 64 2 8 6 3 10 10 10 5 cos x       1 3 1 3 5 5 x  cos  _{ }cos _ _     10.

 

 

 

   

   

 

       

   

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