FUNCTION EX 1 PROBLEMS WITH SOLUTION UPTO JEE LEVEL

FUNCTIONS
1) If f = {(4,5), (5,6), (6,-4)} and g = { (4,-4), (6,5), (8,5)} then
find i) f+g
fdom = {4,5,6},
Solution
fdomgdom
gdom = {4,6,8}
= {4,6} = x (say)
Before performing any
algebraic operations,
we find fdomgdom
To find f+g, f-g, fg, f/g we take
first elements 4,6 and
corresponding operations
performed on 2nd coordinates

FUNCTIONS
f+g
f-g=
i) f+g
Solution
= {(4,1),(6,1)}
ii) f-g
Solution
= {(4,9),(6,-9)}
f={(4,5), (5,6), (6, -4)}
g={(4,-4), (6,5), (8,5)}
4,5-4
= 6,-4+5
4,5+4 6,-4-5
,
,

FUNCTIONS
-8+20
10-16
2f 2nd ordinates of f multiplied by 2
4g2nd ordinates of g multiplied by 4
2f={(4,10), (5, 12), (6, -8)}
iii) 2f+4g
Solution
and 4g = {(4,-16), (6,20), (8,20)}
Now 2f+4g=
f={(4,5), (5,6), (6, -4)}
g={(4,-4), (6,5), (8,5)}
4, 6,
,
={(4,-6), (6,12)}

FUNCTIONS
f+4
f+4
fg
fg
iv) f+4
Solution
 Add 4 to 2nd coordinates of f
= {(4,5+4), (5, 6+4), (6, -4+4)}
= {(4,9), (5,10),(6,0)}
v) fg
Solution
multiply 2nd coordinates of f and g, when 1st
coordinates are same
={(4, 5-4), (6, -4 5)}
= {(4, -20), (6, -20)}
f={(4,5), (5,6), (6, -4)}
g={(4,-4), (6,5), (8,5)}

FUNCTIONS
vi)
f
g
Solution
vii) |f|Apply || to all 2nd coordinates in f
Solution
f = { (4,5), (5,6), (6, -4) }
g = { (4,-4), (6,5), (8,5) }
f
g
= 4,
−5
4
, 6,
−4
5
f =
4, 5 , 5, 6 , 6, −4
= 4,5 , 5,6 , 6,4

FUNCTIONS
ix) f2square the 2nd coordinates in f
viii) fApply to all 2nd coordinates in f
Solution
Solution
Does not
exsist
f={(4,5), (5,6), (6,-4)}
f = 4, 5 , 5, 6 , 6, −4
= 4, 5 , 5, 6
f2 = 4,52 , 5,62 , 6,(−4)2
= 4,25 , 5,36 , 6,16

FUNCTIONS
f3 =
x) f3cube the 2nd coordinates in f
Solution
{(4, 53), (5,63),(6,(-4)3}
= {(4,125), (5,216),(6,-64)}
f={(4,5),(5,6),(6, -4)}

FUNCTIONS
2. Is g={(1,1), (2,3), (3,5), (4,7)} is a function from
A={1,2,3,4}, to B={1,3,5,7}? If this is given by the
formula g(x)=ax+b then find a and b.
All the elements in the domain A have unique images,
Hence g is a function
Given that
Now,
Solution
g(x) = ax+b
g(1) = 1
a+b=1(1)

FUNCTIONS
From (1) and (2) we get
2a+(1–a) = 3
From (1)
a+1 = 3
a = 2
b = 1 - a
b = -1
g(2) = 3
 2a+b=3(2)
= 1 - 2
g={(1,1), (2,3), (3,5), (4,7)}
a+b=1(1)

FUNCTIONS
2. If f(x)=
cos2x+sin4x
sin2x+cos4x
xR then show that f(2012)=1
Solution:
=
1-sin2x+sin4x
1-cos2x+cos4x
cos2x can be
written as
sin2x can
be written
as
=
1-sin2x(1-sin2x)
1-cos2x(1-cos2x)
Take ‘-sin2x’
as common
Take ‘-cos2x’
as common
=
1-sin2xcos2x
1-sin2xcos2x
f(x)=
cos2x+sin4x
sin2x+cos4x
= 1
f(2012)=1

FUNCTIONS
3. If f:R R, g:R R are defined by f(x)=3x-1 and g(x)=x2+1,
then find
Solution
i) (fog)(2)
Given that
f(x)=3x-1
and g(x)=x2+1
i) (fog)(2) =f(g(2))
=f(22+1)
=f(5)
=3(5)-1
Substitute
x=2 in this eq.
Substitute
x=5 in this eq.
=14

FUNCTIONS
ii)(fof)(x2+1) = f(f(x2+1))
= f(3(x2+1)-1)
= f(3x2+3-1)
= f(3x2+2)
= (3(3x2+2)-1)
= (9x2+6-1)
=9x2+5
f(x)=3x-1

FUNCTIONS
4. If f (x)=
x + 1
x − 1
, x  1 then find (fofof)(x)
Solution Firstly, we calculate
(fof)(x)
(fof)(x) = f(f(x)) = f
x+1
x−1
x+1
x−1
+1
x+1
x−1
−1
=
x+1+x−1
x−1
x+1−(x−1)
x−1
=
Substitute
x+1
x−1
in
place of x
= x
=
2x
2

FUNCTIONS
(fofof)(x) = f(fof(x))
Thus (fofof)(x) =
x+1
x−1
= f(x)
=
x+1
x−1
We know that
fof(x) = x

FUNCTIONS
5. If A={-2,-1,0,1,2} and f:AB is a surjection defined by
f(x) = x2+x+1 then find B
Solution
Given that f:AB is surjection
 Codomain of f =Range of f
Domain A ={-2,-1,0,1,2}
 Range f(A) = {f(-2), f(-1), f(0), f(1), f(2)}
It is nothing
but on-to

FUNCTIONS
Given f(x)=x2+x+1
f(-2)=(-2)2-2+1=4-2+1=3
f(-1)=(-1)2-1+1 =1
f(0) =02+0+1=1
f(1)=12+1+1=3
f(2) =22+2+1=7
B = f(A) = {3,1,1,3,7} ={3,1,7}

FUNCTIONS
6. If A = 0,
π
6
,
π
4
,
π
3
,
π
2
and f:AB is a surjection defined by
f(x)=cosx then find B
Solution Given that f:AB is surjection
 Codomain B =
Given f(x) =
f (0) =
cosx
cos (0) = 1
f
π
6
= cos
π
6
=
3
2
f
π
4
= cos
π
4
=
1
2
f
π
3
= cos
π
3
=
1
2
Range of f

FUNCTIONS
 B = f (A)
= {1,
3
2
,
1
2
,
1
2
,0}
f
π
2
= cos
π
2
= 0

FUNCTIONS
 f is one-one
7. If Q is the set of all rational numbers, and f:QQ is
defined by f (x)= 5x+4, xQ, show that f is a bijection.
Solution
To show that f is one one
When a
function is
said to be
bijection?
If it is both
one-one and
onto function
So, first we
have to show
‘f’ is one-one
Domain
Let x1, x2Q be such that
f(x1) = f(x2)
5x1+4 = 5x2+4
5x1 = 5x2
x1 = x2

FUNCTIONS
such that f(x)=y
To show that f is onto:
 f is onto
Thus, f is both one-one and onto, hence bijection.
Codomain
Let yQ
Let f(x) = y
5x+4 = y
5x = y-4
x =
y −4
5
Q
 yQ xQ
Domain

FUNCTIONS
8. Determine whether the function f(x)=log(x+ x2+1 ) is
even or odd
Solution
We know that
log a + log b = log ab
f(x) = log(x+ x𝟐+1)
f(-x) = log(-x+ −x 𝟐+1)
Now f(x)+f(-x)
= log(x+ x𝟐+1) (-x+ x𝟐+1)
= log( x𝟐+1+x) ( x𝟐+1-x)
= log(x+ x𝟐+1)+log(-x+ x𝟐+1)

FUNCTIONS
Hence, f(x) is an odd function.
= log1
f(x)+f(-x)=0
f(-x)= -f(x)
= log(x2+1-x2)
= 0

FUNCTIONS
9. Determine whether the function f(x)=x
e𝐱
−1
e𝐱
+1
is even or
odd.
Solution
f(x)=x
e𝐱
−1
e𝐱
+1
f(-x)=-x
e−𝐱
−1
e−𝐱
+1
1
e𝐱
−1
1
e𝐱
+1
=-x
=−x
1−e𝐱
1+e𝐱

FUNCTIONS
Hence, f(x) is an even function.
 f(-x)=f(x)
= x
e𝐱
−1
e𝐱+1
=f(x)

FUNCTIONS
10. If f: Q Q is defined by f(x) = 5x + 4, find f-1.
Solution
We know
what ‘x’ is
Let f(x) = y
Now f(x) = y
x =
y − 4
5
x = f-1(y)
5x = y - 4
f-1(y) =
y − 4
5
f-1(x) =
x − 4
5
5x+4 = y
We know what
f(x) is

FUNCTIONS
11. Find the inverse function of f(x)=log2x
HINT
logax=bx=ab
Solution Let f(x) = y
Also f(x) = y
f-1(y) = 2y
f-1(x) = 2x
 x = f-1(y)
log2x = y
x = 2y
We know
what ‘x’ is
We know
what f(x) is

FUNCTIONS
12. Find the inverse function of f(x)=5x
Solution
logbam=mlogba logaa=1
Apply log5 on
both sides
Now f(x) = y
log55x = log5y
x(1) = log5y
x = log5y
f-1(x) = log5x
Let f(x) = y
x = f-1(y)
5x = y
f-1(y) = log5y
We know
what ‘x’ is
We know
what f(x) is

FUNCTIONS
1, x, x2,….. form an infinite G.P with
13. If f(x)=1+x+x2+.......for |x|<1 then show that f-1(x)=
x−1
x
Solution
a = 1, r = x
1 + x + x2 + . . . =
1
1 − x
Let f(x) = y
x = f-1(y)
f(x) =
1
1−x
Here r is
common ratio
Here a is
first term
sum of infinite
terms in G.P
S∞=
a
1 − r

FUNCTIONS
f-1(x) =
x − 1
x
f-1(y) =
y − 1
y
y =
1
1 − x
1 - x =
1
y
x = 1 −
1
y
x =
y − 1
y

FUNCTIONS
14. If f:RR, g:RR are defined by f(x)=2x-3, g(x)=x3+5,
then find (fog)-1(x)
(fog)(x) = f(g(x))
Put (fog)(x)=y
Solution
= f(x3+5)
= 2(x3+5)-3
= 2x3+7
 x=(fog)-1(y)
y = 2x3 + 7
y–7 = 2x3

FUNCTIONS
(fog)-1(y) =
y − 7
2
𝟏
𝟑
(fog)-1(x) =
x − 7
2
𝟏
𝟑
x3 =
y − 7
2
x =
y − 7
2
𝟏
𝟑
2x3 = y - 7
We know
x=(fog)-1(y)

FUNCTIONS
15. If f:RR, g:RR are defined by f(x)=3x-2, g(x)=x2+1,
then find (gof-1)(2)
Now f(x) = y
Solution
y = 3x - 2
3x = y + 2
x =
y + 2
3
Let f(x) = y
 x = f-1(y)
f-1(y) =
y + 2
3
f-1(x) =
x + 2
3

FUNCTIONS
(gof-1)(2) = g(f-1(2))
= g
2 + 2
3
=
4
3
𝟐
+1
=
16 + 9
9
= g
4
3
(gof-1)(2) =
25
9
=
25
9
f-1(x) =
x + 2
3
g(x)=x2+1

FUNCTIONS
16. If f:RR is defined as f(x) =
4x
4x + 2
, then show that
f(1-x)=1-f(x).Hence deduce the value of f
1
4
+2f
1
2
+f
3
4
Given f(x) =
4x
4x + 2
Solution
f 1 − x =
4𝟏−𝐱
4𝟏−𝐱
+ 2
=
𝟒
𝟒𝐱
𝟒
𝟒𝐱 + 𝟐

FUNCTIONS
Now 1-f(x) =1-
4
𝐱
4
𝐱
+ 2
=
4
4 + 2 4
𝐱
f 1 − x =
2
2 + 4
𝐱 (1)
=
4𝐱
+ 2 −4𝐱
4𝐱
+ 2
1−f(x)=
2
4𝐱
+ 2
(2)

FUNCTIONS
Hence proved
From (1) & (2),
f(1-x) =1-f(x)
From above f(x)+f(1-x) = 1
f(x)+f(y) = 1(3)
(let 1-x = y  x+y=1)

FUNCTIONS
= 1
= 2
Thus f
1
4
+2f
1
2
+f
3
4
= 2
Now f
1
4
+ 2f
1
2
+ f
3
4 f
1
4
+f
3
4
+f
1
2
+ f
1
2
=
f(x)+f(y) = 1(3)
When x+y=1
+1

FUNCTIONS
(d,4)
17.f={(1,a),(2,c),(3,b),(4,d)} and g-1={(2,a),(4,b),(1,c),(3,d)} then
show that (gof)-1=f-1og-1
Given f={(1,a)(2,c)(3,b)(4,d)}
Solution
(a,1),
 f-1 = (c,2),(b,3),
Given g-1 ={(2,a),(4,b),(1,c),(3,d)}
(d,3)
(a,2),
 g = (b,4),(c,1),

FUNCTIONS
f g
Let us find gof with f taken first
and g next
gof = {(1,2),(2,1),(3,4),(4,3)}
(gof)-1 = {(2,1),(1,2),(4,3),(3,4)}(1)
Given
f={(1,a),(2,c),(3,b),(4,d)}
g={(a,2)(b,4),(c,1),(d,3)}
1 a
2 c
3 b
4 d
2
1
4
3

FUNCTIONS
Let us calculate f-1og-1 with g-1 first and f-1 next
g-1 f-1
2 a
4 b
1 c
3 d
1
3
2
4
Given
f-1={(a,1),(c,2),(b,3),(d,4)}
g-1={(2,a)(4,b),(1,c),(3,d)}
f-1og-1 = {(2,1),(4,3),(1,2),(3,4)}(2)
Now (1) & (2) gives
(gof)-1 = f-1og-1
(gof)-1 = {(2,1),(1,2),(4,3),(3,4)}(1)

FUNCTIONS
18. If f:QQ is defined by f(x)=5x+4 for all xQ, show that
f is a bijection and find f-1
Solution
When is a
function said to
be bijection?
If it is both
one-one and
on-to function
So, firstly we
prove ‘f’ is
one-one
To show that f is one-one
Let x1,x2Q be such that
domain
 f is one-one
f(x1) = f(x2)
5x1+4 = 5x2+4
5x1 = 5x2
x1 = x2

FUNCTIONS
Let yQ such that f(x)= y5x + 4 = y
 x =
y − 4
5
f is on-to
Thus, f is a bijection.
f:QQ is both one-one and on-to
Calculated as follows
f-1 exists and f-1 : QQ is
Q
codomain
domain

FUNCTIONS
Let f(x) = y
5x+4 = y
 f-1(x) =
x − 4
5
x =
y− 4
5
f(x)=5x+4
 x = f-1(y)
f-1(y) =
y− 4
5

FUNCTIONS
(1)
If 0  x  2
If 2 < x  3
Solution
then show that f[0, 3]  [0, 3] and find fof ?
19. If f : [0, 3]  [0, 3] is defined by
1 + x, 0  x  2,
3 - x, 2 < x  3
f(x) =
1  1 + x  3
-3  - x < -2
(2)
3 - 3  3 - x < 3 - 2
0  3 - x < 1
If you add 1
If you
multiply with
‘-’
If you add ‘3’
from (1) and (2) if you
substitute any
value [0,3] in f(x)
the resultant
must lie in [0,3]

FUNCTIONS
From (1) and (2) f [0,3]  [0,3]
When 0  x  1 we have
fof(x) = f(f(x))
= f(1+x)
= 2+x [∵11+x2]
= 1+1+x
When 1<x2 we have
fof(x) = f(f(x))
= f(1+x)
= 3-(1+x)
Here
f(x)=1+x
Here
f(x)=3-x
=2-x [∵2<1+x3]

FUNCTIONS
When 2<x3 we have
fof(x)=f(f(x))
=f(3-x)
=1+3-x
=4-x [∵ 03-x<1]
fof(x)=
2+x,0x<1
2-x,1<x2
4-x,2<x3

FUNCTION EX 1 PROBLEMS WITH SOLUTION UPTO JEE LEVEL

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FUNCTION EX 1 PROBLEMS WITH SOLUTION UPTO JEE LEVEL