Jee 2014 Booklet3 Hwt Solutions Functions

Vidyamandir Classes

Solutions to Home Work Test/Mathematics Functions 1.(B)

HWT - 1

f  x   6 x  7   cos  x  6 x  6 x   7  cos  x  6 x





 cos  x  7  6 x  6 x   cos  x  7  6 x Period of cos  x  Period of 6 x 

2 2 

1 6

Period of f (x) = LCM of 2 and 2.(B)

For f (x) to be defined





1 =2 6

log10 1  x3  0

and

1  x3  0



1  x3  1

or

x3  1

or

x3  0

or

x  1

or

x>0

or

22 x  1

and

y

or

f  x, y  

Combining both of these results, we get : x   0 ,   5.(C)

 2 x  2 x Let f  x   sin  x  2  2 x   2 x  2 x f   x   sin   x  2  2 x 



   

  2 x  2 x    sin  x   2  2 x  

    f  x  

f  x  is an odd function.

For f (x) to be defined or

2 x  2 x  0

or

2 x  2 x

2x  0

or

x0

So, domain of f (x) is x  R  0 6.(A)

8.(D)

Put x  2 y  a and x  2 y  b 

2x  a  b



f  a, b  

or

x

ab 2

 a  b  .  a  b   a 2  b2 2 4

8

a b 4 x2  y 2 8

 x2  e  f  x   ln  2   x 1   

As x increases from 0 to 

 x2  e  decreases or ln  2  decreases  x 1  x2  1  

x2  e

f  0   ln  e   1 and f (x) approaches 0 as x approaches .

 9.(D)

Range of f (x) is f (x)  (0, 1].

See in Module.

VMC/Functions

32

HWT-Solutions/Mathematics

Vidyamandir Classes

Functions 1.(A)

  x , x  0 f  x    x , x  0



HWT - 2

even function.

  x   x , x  0 f  x   , x0   x

2.(C)

See in Module.

3.(A)

h x  f  x  f x g  x  g x













h   x   f   x   f  x  g   x   g  x   h  x   5.(D)

odd function.

As f (x) is even f   x   f  x  As f (x) is odd f   x    f  x  

f  x   f  x

2 f  x  0

or

or

f  x  0

f  3  f  2   0  0  0

9.(B)

For f (x) to be defined x>0 and

log e x  0



x>0

and

log e x  0



x>0

and

x 1

Functions 4 x  x   x 

1.(B)

3x  x   5

or

f  x 

2.(A)

f x 

x e 1 x



x e

x



1



HWT - 3





4 x   x   x   x 



x = 0 is the only solution

or

3 x  5 x 

x 1 2 

x  xe x x xe x x 1   1  x  1 x 2 2 1 e e 1 2



x  e x  1  1   x x x   x  1  x  1  f  x 2 2 e 1 e 1  3.(C)

f (x) is an even function.

Period of sin x  2 and period of {x} = 1  period of f (x) = LCM of 2 and 1 = not defined So, the function f (x) is non periodic.

VMC/Functions

33

HWT-Solutions/Mathematics

Vidyamandir Classes 4.(A)

f  x   2 f 1  x   x 2  2

. . . .(i)

Put x  1  x to get : f 1  x   2 f  x   1  x   2 2

. . . .(ii)

(i) – 2(ii) gives



 



2 3 f  x    x 2  2   2 1  x   2   x 2  2  2 x 2  2 x  3   x 2  4 x  4    

or

f  x 

x2  4 x  4  x  2  3 3 

2

x  e x which is not possible as seen from the graphs of y = x and y = ex.

8.(A)

log x  x

9.(A)

2 x  4 x  52 is always positive as 2 x , 4 x and 5 x all are always positive.

10.(B)

2 x  4 x  5 x  0 gives 2 x  4 x  5 x 2 x  4 x increases from 0 to  as x increases from –  to .

5 x decreases from  to 0 as x increases from –  to . 

The graphs of 2 x  4 x and 5 x will intersect at a point.

Functions 1.(B)

Check from graph

2.(A)

Discontinuous at e x  1  0 i.e.

3.(CD)

ex  1

1 f  x f    f  x  x



x=0

1 f   x

Now

  3  1  28



f  x   x3  1



f 1  13  1  2



n

0  f 0

VMC/Functions

HWT - 4



f  x    xn  1 n=3

i.e.

x3  1  0



x3  1



x  1 only (for real solution)

34

HWT-Solutions/Mathematics

Vidyamandir Classes 4.(B)

Graph is

6.(A)

If we observe 

9.(C)

x 1  x2

lies between 1 and 1

only integer in range is 0.

Draw graph for it.

10.(CD) Similar question in module.

Functions 1.(B)

HWT - 5

  f  x   sin  log  x  x 2  1        f   x   sin  log   x  x 2  1         1  f   x   sin  log   x  x2  1          sin   log  x  x 2  1     sinlog  x  x 2  1    f  x       

2.(C)

f  x 

f  x   sinlog  x  x 2  1  is an odd function.   1  sin x sin x   2  cos x cos x

  tan x,   0,  f  x    tan x,    0, 

   

0 x

 2

  x  2 3  x 2 3  x  2 2

Hence f (x) is periodic with ‘2’ (Draw graph for better view) 8.(C)

1  2  cos x  3

9.(C)

Again

f  x    x n  1 here n =3

and

f  x   xn  1

VMC/Functions

35

HWT-Solutions/Mathematics

Vidyamandir Classes

Functions 1.(B)

2.(B)

1  t to get the answer x 1 And we know that x     ,  2  2 ,   x

Take x 

f  x  k   f  x  0

. . . .(i)

x xk  f  x  2k   f  x  k   0 (i) – (ii)

4.(B) 5.(D)

HWT - 6

f  x   f  x  2k   0

x 2  3 x  10  0

. . . .(ii) 

and

periodic with periodic 2k.

2x  5  1

and

  As we know tan x has domain x  R   2n  1  2      and  x  2  2 and  x  2 2  

2x  5  0

 x  2   2



x should not lie between –3 and –4.

 (x + 2) should not lies between 2 and 3  x  2  2 x should not lie between 0 and 1.

Again  x  2  2 

x should not lie between –2 and 1

8.(B)

Draw graph for this function

10.(C)

Similar question in objective worksheet.



Domain is x    ,  4    3,  2    1, 0   1,  

Functions 1.(C)

Replace x  3 y  a and x  3 y  b

 a b   a b  2 2 f  a1 b   12     a b  2  6 

 2.(D) 4.(D)

HWT - 7



f  x, y   x 2  y 2

and

  x  g  x  0

It is clearly non-periodic 

x x0

x x

Now take idea from graph. 10.(A)

  x   f  x      x   g  x    0 2

2

is possible only iff :   x  f  x  0



  x  f  x



  x  g  x

See from graph to get :   x   x  x  0 ,  

VMC/Functions

36

HWT-Solutions/Mathematics

Vidyamandir Classes

Functions 1.(A)

For f (x) domain is x   , 0  For f (x) x  x  0

5.(D)

HWT - 8

See that this situation is never possible



f  x   g  x   f  x   g  x  only if both f (x), g(x) are either positive or negative simultaneously. 

both negative in x    ,  1 and both positive in x  2 ,  

9.(D)

y is always –ve and domain has both positive and negative values of x.

10.(C)

 Period of sin x  cos x is 2 4

4

  2   period of f (x) is    . 3 6

Functions 1.(C)

HWT - 9

Since f (x) is even function we have two cases  x 1   x 1  and x x      x2  x2 

x2  x  1  0



1  5 2 4 values of x.

x 2  3x  1  0

x

4.(D)

x 

x

f  x   sin 2 x  cos 2 x  1  1



3  5 2 Period but with no fundamental period.

5.(D)

Draw graph of f  x   x 2  x and then do transformations to get graph of f

7.(C)

log x 3  2 x  3

Can also be written as



1  log3 x 2x  3



x

Graph is (x > 0)

2 solutions (provided x  1 , x > 0)

VMC/Functions

37

HWT-Solutions/Mathematics

Vidyamandir Classes 10.(A)

Graph is like (for x 2  2  2  x  )



Solution is when [x] = 2



x 6

Functions 1.(A)

f  x 



4x 4 2 x

and

f 1  x  

f  x   f 1  x   1

HWT - 10 4

4  2 .4



x



2 2  4x

 1  f   1997 

 1996  f  1  1997 

Similarly all pairs give 1 2.(D)

f (x) will be maximum when sin x is maximum and cos x is minimum i.e. x  

 2

f  x max  3 1  2 0  3

f (x) will be min when sin x is minimum and cos x is maximum i.e. x = 0  4.(C)

f  x min  3 0  2 1  2

x12  x9  x 4  x  1  0 (i) Now for x < 0 it is true (ii) (iii)

For 1 > x > 0 For 1 < x

x 4  x 9  0 and x12  1  x  0

x12  x9 and x 4  x 

x12  x9  x 4  x  1  0  x  R

6.(D)

Check for domains of the given functions.

7.(AC)

Odd functions have symmetrical graph about origin.

VMC/Functions

38

HWT-Solutions/Mathematics