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Page No 3.11:

Question 1:

If f(x) = x2 − 3x + 4, then find the values of x satisfying the equation f(x) = f(2x + 1).

Answer:

Given:
f (x) = x2 – 3x + 4
Therefore,
f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4
                = 4x2 + 1 + 4x – 6x – 3 + 4
                = 4x2 – 2x + 2
Now,
f (x) = f (2x + 1)
x2 – 3x + 4 = 4x2 – 2x + 2
⇒ 4x2x2 – 2x + 3x + 2 – 4 = 0
⇒ 3x2 + x – 2 = 0
⇒ 3x2 + 3x – 2x – 2 = 0
⇒ 3x(x + 1) – 2(x +1) = 0
⇒ (3x – 2)(x +1) = 0
⇒ (x + 1) = 0  or  ( 3x – 2) = 0
x=-1 or x=23
Hence, x=-1,23.

Page No 3.11:

Question 2:

If f(x) = (xa)2 (xb)2, find f(a + b).

Answer:

Given:
f (x) = (xa)2(xb)2
Thus,
f (a + b) = (a + ba)2(a + bb)2
             = b2a2
Hence, f (a + b) = a2b2 .

Page No 3.11:

Question 3:

If y=fx=ax-bbx-a, show that x = f(y).

Answer:

Given:
fx=ax-bbx-a
Let y = f (x) .
y( bx - a) = ax b
xybay = axb
xybax = ayb
x(bya) = ayb
x=ay-bby-a
x = f (y)      
Hence proved.

Page No 3.11:

Question 4:

If fx=11-x, show that f[f[f(x)]] = x.

Answer:

Given:
fx=11-x
Thus,
ffx=f11-x

=11-11-x

=11-x-11-x=1-x-x=x-1x
Again,
fffx=fx-1x
=11-x-1x=1x-x+1x=x1=x

Therefore,  f[f{f(x)}] = x.
Hence proved.

Page No 3.11:

Question 5:

If fx=x+1x-1, show that f[f[(x)]] = x.

Answer:

Given:
fx=x+1x-1
Therefore,
ffx=fx+1x-1
=x+1x-1+1x+1x-1-1
=x+1+x-1x-1x+1-x+1x-1=2xx-12x-1=2x2=x
Thus,
f [ f {(x)}] = x
Hence proved.

Page No 3.11:

Question 6:

If fx=x2,when x <0x,when 0x<11x,when x 1
find: (a) f(1/2), (b) f(−2), (c) f(1), (d) f3 and (e) f-3.

Answer:

Given:
fx=x2,when x<0x,when 0x<11x,when x1
Now,
(a) f12=12                    [ Using f (x) = x, 0 ≤ x < 1]
(b) f (-2) = ( - 2)2 = 4  
(c) f1=11=1
(d) f3=13
(e) f-3
Since x is not defined in R, f-3 does not exist.

Page No 3.11:

Question 7:

If fx=x3-1x3, show that fx+f1x=0.

Answer:

Given:
fx=x3-1x3  ...(i)
Thus,
f1x=1x3-11x3

          =1x3-11x3

f1x=1x3-x3 ...(ii)

fx+f1x=x3-1x3+1x3-x3

=x3-1x3+1x3-x3=0

Hence, fx+f1x=0 .

Page No 3.11:

Question 8:

If fx=2x1+x2, show that f(tan θ) = sin 2θ.

Answer:

Given:
fx=2x1+x2
Thus,
ftanθ=2tanθ1+tan2θ
            =2×sin θcos θ1+sin2θcos2θ= 2 sin θ cos θ×cos2θcos2θ+sin2θ= 2 sin θ cos θ1    cos2θ+sin2θ =1=sin 2θ        2 sin θ cos θ = sin 2θ
                            
Hence,  f (tan θ) = sin 2θ.



Page No 3.12:

Question 9:

If fx=x-1x+1, then show that

(i) f1x=-fx                                     (ii) f-1x=-1fx

Answer:

Given: fx=x-1x+1                  .....(1)

(i) Replacing  x by 1x in (1), we get
f1x=1x-11x+1          =1-x1+x          =-x-1x+1          =-fx

(ii) Replacing  x by -1x in (1), we get
f-1x=-1x-1-1x+1             =-1-x-1+x             =-x+1x-1             =-1x-1x+1             =-1fx

Page No 3.12:

Question 10:

If f(x) = (axn)1/n, a > 0 and n ∈ N, then prove that f(f(x)) = x for all x.

Answer:

Given:
f(x) = (axn)1/n, a > 0
Now,
f{ f (x)} = f (axn)1/n
             = [a – {(a xn)1/n}n]1/n
             = [ a – (axn)]1/n
             = [ a a + xn)]1/n = (xn)1/n = x(n × 1/n) = x

Thus, f(f(x)) = x.
Hence proved.

Page No 3.12:

Question 11:

If for non-zero x, af(x) + bf 1x=1x-5, where ab, then find f(x).

Answer:

Given:
afx+bf1x=1x-5              ...(i)
af1x+bfx=11x-5
af1x+bfx=x-5        ...(ii)

On adding equations (i) and (ii), we get:

afx+bfx+bf1x+af1x=1x-5+x-5
a+bfx+a+bf1x=1x+x-10
fx+f1x=1a+b1x+x-10             ...(iii)

On subtracting (ii) from (i), we get:

afx-bfx+bf1x-af1x=1x-5-x+5
a-bfx-f1xa-b=1x-x
fx-f1x=1a-b1x-x                   ...(iv)

On adding equations (iii) and (iv), we get:

2fx=1a+b1x+x-10+1a-b1x-x
2fx=a-b1x+x-10+a+b1x-xa+ba-b
2fx=ax+ax-10a-bx-bx+10b+ax-ax+bx-bxa2-b2
2fx=2ax-10a+10b-2bxa2-b2
fx=1a2-b2×122ax-10a+10b-2bx
            =1a2-b2ax-5a+5b-bx

Therefore,
fx=1a2-b2ax-bx-5a+5b
      =1a2-b2ax-bx-5a-ba2-b2
      =1a2-b2ax-bx-5a-ba-ba+b
     =1a2-b2ax-bx-5a+b

Hence,
fx=1a2-b2ax-bx-5a+b



Page No 3.18:

Question 1:

Find the domain of each of the following real valued functions of real variable:

(i) fx=1x
(ii) fx=1x-7
(iii) fx=3x-2x+1
(iv) fx=2x+1x2-9
(v) fx=x2+2x+1x2-8x+12

Answer:

(i) Given: fx=1x
Domain of f :
We observe that f (x) is defined for all x except at x = 0.
At x = 0, f (x) takes the intermediate form 10.
Hence, domain ( f ) = R -{ 0 }

(ii) Given: fx=1x-7
Domain of f :
Clearly,  f (x) is not defined for all (x - 7)  = 0 i.e. x = 7.
At x = 7,  f (x) takes the intermediate form 10.
Hence, domain ( f ) = R - { 7 }.

(iii) Given: fx=3x-2x+1
Domain of f :
Clearly,  f (x) is not defined for all (x + 1)  = 0, i.e. x- 1.
At x = -1,  f (x) takes the intermediate form 10.
Hence, domain ( f ) = R --1 }.

(iv) Given: fx=2x+1x2-9
Domain of f :
Clearly,  f (x) is defined for all  xR except for x2 - 9 ≠  0, i.e. x = ± 3.
At x = -3, 3,  f (x) takes the intermediate form 10.
Hence, domain ( f ) = R -- 3, 3 }.

(v) Given: fx=x2 +2x+1x2-8x+12

                     =x2+2x+1x2-6x-2x+12

                     =x2+2x+1xx-6-2x-6

                    =x2+2x+1x-6x-2
Domain of f : Clearly,  f (x) is a rational function of x as x2+2x+1x2-8x+12 is a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which x2 - 8x + 12 = 0, i.e. x = 2, 6.
Hence, domain ( f ) = R - {2,6}.

Page No 3.18:

Question 2:

Find the domain of each of the following real valued functions of real variable:

(i) fx=x-2
(ii) fx=1x2-1
(iii) fx=9-x2
(iv) fx=x-23-x

Answer:

(i) Given: fx=x-2
Clearly, f (x) assumes real values if x - 2 ≥ 0.
x ≥ 2
x ∈ [2, ∞)
Hence, domain (f) = [2, ∞) .

(ii) Given: fx=1x2-1
Clearly, f (x) is defined for x2 - 1 > 0 .
(x + 1)(x - 1) > 0     [ Since a2 - b2 = ( a + b)(a - b)]
x-1 and  x > 1
x ∈ (-∞ , - 1) ∪ (1, ∞)
Hence, domain (f) = (- ∞ , - 1) ∪ (1, ∞)

(iii) Given: fx=9-x2
We observe that f (x) is defined for all satisfying
- x2 ≥ 0 .
x2 - 9 ≤ 0
⇒ (x + 3)(x - 3) ≤ 0
-3 ≤ x ≤ 3
x ∈ [ - 3, 3]
Hence, domain ( f ) = [ -3, 3]

(iv) Given: fx=x-23-x
Clearly, f (x) assumes real values if
x - 2 ≥ 0 and 3 - x > 0
x ≥ 2 and 3 > x
x ∈ [2, 3)
Hence, domain ( f ) = [2, 3) .

Page No 3.18:

Question 3:

Find the domain and range of each of the following real valued functions:

(i) fx=ax+bbx-a

(ii) fx=ax-bcx-d

(iii) fx=x-1

(iv) fx=x-3

(v) fx=x-22-x

(vi) fx=x-1

(vii) fx=-x

(viii) fx=9-x2

(ix) fx=116-x2

(x) fx=x2-16

Answer:

(i)
Given:
fx=ax+bbx-a
Domain of f : Clearly,  f (x) is a rational function of x as ax+bbx-ais a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which ( bx - a) = 0, i.e. bx = a.
x=ab
Hence, domain ( f ) = R-ab
Range of f :
Let f (x) = y
ax+bbx-a=y

⇒ (ax + b) = y (bx - a)
⇒ (ax + b) = (bxy - ay)
b + ay = bxy - ax
b + ay = x(by - a)
x=b+ayby-a
Clearly, f (x) assumes real values for all x except for all those values of x for which ( by - a) = 0, i.e. by = a.
y=ab.
Hence, range ( f ) = R-ab
(ii)
Given:
fx=ax-bcx-d
Domain of f : Clearly,  f (x) is a rational function of x as ax-bcx-dis a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which ( cx  - d) = 0, i.e. cx = d.
x=dc.
Hence, domain ( f ) = R-dc
Range of f :
Let f (x) = y
ax-bcx-d=y

⇒ (ax - b) = y( cx - d)
⇒ (ax - b) = (cxy - dy)
dy - b = cxy - ax
dy - b = x(cy - a)
x=dy-bcy-a
Clearly, f (x) assumes real values for all x except for all those values of x for which ( cy - a) = 0, i.e. cy = a.
y=ac.
Hence, range ( f ) = R-ac .

(iii) 
Given: fx=x-1
Domain ( f ) : Clearly, f (x) assumes real values if x - 1 ≥ 0 ⇒ x ≥ 1 ⇒ x ∈ [1, ∞) .
Hence, domain (f) = [1, ∞)
Range of f : For x ≥  1, we have:
x - 1 ≥ 0
x-10
f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .

(iv)
Given: fx=x-3
Domain ( f ) : Clearly, f (x) assumes real values if x - 3 ≥ 0 ⇒ x ≥ 3 ⇒ x ∈ [3, ∞) .
Hence, domain ( f ) = [3, ∞)
Range of f : For x ≥  3, we have:
x - 3 ≥ 0
x-30
f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .

(v)
Given:
fx=x-22-x
Domain ( f ) :
Clearly,  f (x) is defined for all x satisfying: if 2 - x  ≠ 0 ⇒ ≠ 2.
Hence, domain ( f ) = R - {2}.
Range of f :
Let f (x) = y

x-22-x=y
x - 2 = y (2 - x)
x - 2 = - y (x - 2)
y- 1
Hence, range ( f ) = {- 1}.

(vi)
The given real function is f (x) = |x – 1|.
It is clear that |x – 1| is defined for all real numbers.
Hence, domain of f = R.
Also, for xR, (x – 1) assumes all real numbers.
Thus, the range of f is the set of all non-negative real numbers.
Hence, range of f = [0, ∞) .

(vii)
f (x) = – | x |, xR
We know that
x=x,x0-xx<0
fx=-x=-x,x0x,x<0
Since f(x) is defined for xR, domain of f  = R.
It can be observed that the range of f (x) = – | x | is all real numbers except positive real numbers.
∴ The range of f is (– ∞, 0).

(viii) Given:
fx=9-x2
(9-x2) 09 x2 x -3,3

9-x2 is defined for all real numbers that are greater than or equal to – 3 and less than or equal to 3.
Thus, domain of f (x) is {x : – 3 ≤ x ≤ 3} or [– 3, 3].
For any value of x such that – 3 ≤ x ≤ 3, the value of f (x) will lie between 0 and 3.
Hence, the range of f (x) is {x: 0 ≤ x ≤ 3} or [0, 3].

(ix) Given:
fx=116-x2

(16-x2) >016 >x2 x -4,4

116-x2 is defined for all real numbers that are greater than  – 4 and less than 4.
Thus, domain of f (x) is {x : – 4 < x < 4} or (– 4, 4).

Range of f :
Let f (x) = y
116-x2=y116-x2=y21y2=16-x2x2=16-1y2Since, -4<x<40x2<16016-1y2<16-16-1y2<0161y2>0116y2<14y<        y0


Hence, range ( f ) = [14, ).


(x) Given:
fx=x2-16

(x2-16) 0x216x (-,-4]  [4, )

x2-16 is defined for all real numbers that are greater than or equal to 4 and less than or equal to –4.
Thus, domain of f (x) is {x : x ≤ – 4 or x ≥ 4} or (–∞, –4] ∪ [4, ∞).

Range of f :
For
x ≥ 4, we have:
x2 - 16 ≥ 0
x2-160
f (x) ≥ 0

For x ≤ – 4, we have:
x2 - 16 ≥ 0
x2-160
f (x) ≥ 0

Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞).



Page No 3.38:

Question 1:

Find f + g, fg, cf (c ∈ R, c ≠ 0), fg, 1f and fgin each of the following:

(a) If f(x) = x3 + 1 and g(x) = x + 1
(b) If fx=x-1 and gx=x+1

Answer:

(a) Given:
f (x)  = x3  + 1 and g (x) = x + 1
Thus,
(f + g) (x) : RR is given by (f + g) (x) = f (x) + g (x) = x3 + 1 + x + 1 = x3 + x + 2.
(- g) (x) : RR is given by (f - g) (x) = f (x- g (x) = (x3 + 1) - (x + 1 ) = x3 + 1 - x - 1 = x3 - x.
cf : RR is given by (cf) (x) = c(x3  + 1).
(fg) (x) : RR is given by (fg) (x) = f(x).g(x) = (x3 + 1) (x + 1) = (x + 1) (x2 - x + 1) (x + 1) = (x + 1)2 (x2 - x + 1).
1f:R--1R is given by1fx=1fx=1x3+1.
fg:R--1R is given byfgx=fxgx=x3+1x+1=x+1x2-x+1x+1=x2-x+1 .  

Note that : (x3 + 1) = (x + 1) (x2 - x + 1)]
                                                                               
(b) Given:
fx=x-1 and gx=x+1
Thus,
(f + g) ) : [1, ∞) → R is defined by (f + g) (x) = f (x) + g (x) = x-1+x+1.
(f  - g) ) : [1, ∞) → R is defined by (f - g) (x) = f (x- g (x) = x-1-x+1 .
cf : [1, ∞) → R is defined by (cf) (x) = cx-1 .
(fg) : [1, ∞) → R is defined by (fg) (x) = f(x).g(x) = x-1×x+1=x2-1 .
1f:1,R is defined by 1fx=1fx=1x-1.
fg:[1,)R is defined by fgx=fxgx=x-1x+1=x-1x+1.

Page No 3.38:

Question 2:

Let f(x) = 2x + 5 and g(x) = x2 + x. Describe (i) f + g (ii) fg (iii) fg (iv) f/g. Find the domain in each case.

Answer:

Given:
f(x) = 2x + 5 and g(x) = x2 + x
Clearly, f (x) and g (x) assume real values for all x.
Hence,
domain (f) = R and domain (g) = R.
  DfDg=R .

Now,
(i) (f + g) : RR is given by (f + g) (x) = f (x) + g (x) = 2x + 5 + x2 + x = x2 + 3x + 5.
 Hence, domain ( f + g) = R .

(ii) (- g) : RR is given by (f - g) (x) = f (x- g (x) = (2x + 5) - (x2 + x) = 5 + x - x2
 Hence, domain ( f - g) = R.

(iii) (fg) : RR is given by (fg) (x) = f(x).g(x) = (2x + 5)(x2 + x)
                                                                     = 2x3 + 2x2 + 5x2 + 5x
                                                                     = 2x3 + 7x2 + 5x
Hence, domain ( f.g) = R .

(iv) Given:
 g(x) = x2 + x
g(x) = 0 ⇒ x2 + x = 0 = x(x+ 1) = 0
x = 0 or (x + 1) = 0
x = 0 or x- 1
Now,
fg:R--1,0R is given by fgx=fxgx=2x+5x2+x .
Hence, domainfg=R--1,0.

Page No 3.38:

Question 3:

If f(x) be defined on [−2, 2] and is given by fx=  -1,-2x0x-1,   0<x2and g(x) =fx+fx, find g(x).

Answer:

Given:
fx=-1,-2x0x-1,0<x2
Thus,
gx=fx+fx
       =x-1+1 ,-2x0x-1+(-x+1),0<x<1x-1+x-1,1x2=x,-2x00,0<x<12x-2,1x2 

Page No 3.38:

Question 4:

Let f and g be two real functions defined by fx=x+1and gx=9-x2. Then, describe each of the following functions:
(i) f + g
(ii) gf
(iii) f g
(iv) fg
(v) gf
(vi) 2f-5 g
(vii) f2 + 7f
(viii) 58

Answer:

Given:
fx=x+1and gx=9-x2
Clearly, fx=x+1 is defined for all x-1.
Thus, domain (f) = [ -1, ∞]
Again,

gx=9-x2 is defined for
 9 -x2 ≥ 0 ⇒ x2 - 9 ≤ 0
x2 - 32 ≤ 0
⇒ (x + 3)(x - 3) ≤ 0
x-3,3
Thus, domain (g) = [- 3, 3]
Now,
domain ( f ) ∩ domain( g ) = [ -1, ∞] ∩ [- 3, 3]
                                           = [ -1, 3]
(i) ( f + g ) : [ - 1 , 3] → R is given by ( f + g ) (x) = f (x) + g (x) = x+1+9-x2 .

(ii) ( - f ) : [ -1 , 3] → R is given by ( g - f ) (x) = g (x)- f (x) = 9-x2-x+1 .

(iii) (fg) : [ -1 , 3] → R is given by (fg) (x) = f(x).g(x) = x+1.9-x2=x+19-x2=9 +9x-x2-x3 .

(iv) fg:-1,3R is given by fgx=fxgx=x+19-x2=x+19-x2 .

(v) gf:-1,3R is given by gfx=gxfx=9-x2x+1=9-x2x+1 .

(vi) 2f-5g:-1,3R is given by 2f-5gx=2x+1-59-x2
                                                                              =2x+1-45-5x2 .

(vii) f2+7f:-1,R is given by f2+7fx=f2x+7fx           {Since domain(f) = [- 1, ∞]}
                                                                       =x+12+7x+1=x+1+7x+1

(viii) 5g:-3,3R is defined by 5gx=59-x2.           {Since domain(g) = [- 3, 3]}                                                          

Page No 3.38:

Question 5:

If f(x) = loge (1 − x) and g(x) = [x], then determine each of the following functions:

(i) f + g
(ii) fg
(iii) fg
(iv) gf
Also, find (f + g) (−1), (fg) (0), fg 12, gf 12.

Answer:

Given:
f(x) = loge (1 − x) and g(x) = [x]
Clearly, f(x) = loge (1 − x)  is defined for all ( 1 - x)  > 0.
⇒ 1 > x
x < 1
x ∈ ( -∞, 1)
Thus, domain (f ) = ( - ∞, 1)
Again,
g(x) = [x] is defined for all x ∈ R.
Thus, domain (g) = R
∴ Domain (f) ∩ Domain (g) = ( - ∞, 1) ∩ R
                                              = ( - ∞, 1)

Hence,
(i ) ( f + g ) : ( -∞, 1) → R is given by ( f + g ) (x) = f (x) + g (x) = loge (1 − x) + [ x ].

(ii) (fg) : ( - ∞, 1) → R is given by (fg) (x) = f(x).g(x) = loge (1 − x)[ x ] = [ x ]loge (1 − x).

(iii) Given:
      g(x) = [ x ]
   If  [ x ]  = 0,
 x ∈ (0, 1)

Thus,
domainfg=domainfdomaing-x:gx=0
fg:-,0R is defined by fgx=fxgx=loge1-xx.

(iv) Given:
f(x) = loge (1 − x)
1fx=1loge1-x
1fx is defined if loge( 1-x) is defined and loge(1 – x) ≠ 0.
⇒ (1- x) > 0 and (1 - x) ≠ 0
x < 1 and x ≠ 0
x ∈ ( - ∞, 0)∪ (0, 1)
Thus, domaingf=-,00,1 = ( - ∞, 1)  .
gf:-,00,1R defined by gfx=gxfx=xloge1-x

(f + g)( -1) = f( -1) + g( -1)
 = loge{1 – (- 1)}+ [ -1]
= loge  2 – 1
Hence, (f + g)( - 1) = loge  2 – 1

(fg)(0) = loge ( 1 – 0) × [0] = 0
fg12=does not exist.
gf12=12loge1-12=0

Page No 3.38:

Question 6:

If f, g and h are real functions defined by fx=x+1,  gx=1xand h(x) = 2x2 − 3, find the values of (2f + gh) (1) and (2f + gh) (0).

Answer:

Given:
fx=x+1,gx=1xand hx=2x3-3
Clearly, f (x) is defined for x + 1 ≥ 0 .
x- 1
x ∈ [ -1, ∞]
Thus, domain ( f ) = [ -1, ∞] .
Clearly, g (x) is defined for x ≠ 0 .
x ∈ R – { 0} and h(x) is defined for all x such that  x ∈ R .
Thus,
domain ( f ) ∩ domain (g) ∩ domain (h) = [ - 1, ∞] – { 0}.
Hence,
(2f + g h) : [ - 1, ∞] – { 0} → R is given by:
(2f + gh)(x) = 2f (x) + g (x- h (x)
                      =2x+1+1x-2x2+3
                      (2f+g-h)(1) =22+1-2+3 = 22+4-2=22+2
(2f + gh) (0) does not exist because 0  does not lie in the domain x [ - 1, ∞] – {0}.

Page No 3.38:

Question 7:

The function f is defined by fx=1-x,x<01     ,x=0x+1,x>0. Draw the graph of f(x).

Answer:

Here,
f (x) = 1 – x for x < 0. So,
f (- 4) = 1 – ( - 4) = 5
f (- 3) = 1 – ( - 3) = 4
f (-2) = 1 – ( - 2) = 3
f (-1) = 1 – (- 1) = 2 etc.

Also, f(x) = 1 for x = 0.

Lastly,  f (x) = x + 1 for, x > 0.
and f (1) = 2, f (2) = 3, f (3) = 4, f (4) = 5 and so on.

Thus, the graph of f is as shown below:

Page No 3.38:

Question 8:

Let f, g : R → R be defined, respectively by f(x) = x + 1 and g(x) = 2x − 3. Find f + g, fg and fg.

Answer:

f, g : R → R is defined, respectively, by f(x) = x + 1 and g(x) = 2x − 3.
(f + g) (x) = f(x) + g(x)
                = (x + 1) + (2x – 3)
                = 3x - 2
∴ (f + g) (x) = 3x – 2

(f - g)(x) = f(x- g(x)
              = (x + 1) - (2x – 3)
              = x + 1 – 2x + 3 
              = -x + 4
∴ (f - g) (x) = - x + 4

fgx=fxgx,gx0,xR

fgx=x+12x-3,2x-30 or 2x3

fgx=x+12x-3,x32

Page No 3.38:

Question 9:

Let f : [0, ∞) → R and g : R → R be defined by fx=xand g(x) = x. Find f + g, fg, fg and fg.

Answer:

It is given that f : [0, ∞) → R and g : RR such that fx=x and g(x) = x .
Df+g=[0,)R=[0,)
So, f + g : [0, ∞) → R is given by
f+gx=fx+gx=x+x

Df-g=DfDg=[0,)R=[0,)
So, f - g : [0, ∞) → R is given by
f-gx=fx-gx=x-x

Dfg=DfDg=[0,)R=[0,)
So, fg : [0, ∞) → R is given by
fgx=fxgx=x.x=x32

Dfg=DfDg-x:gx=0=0,
So,fg:0,R is given by
fgx=fxgx=xx=1x

Page No 3.38:

Question 10:

Let f(x) = x2 and g(x) = 2x+ 1 be two real functions. Find (f + g) (x), (fg) (x), (fg) (x) and fg x.

Answer:

Given:
 f (x)  = x2 and g (x) = 2x + 1
Clearly, D (f) =  R and D (g) = R
  Df±g=DfDg=RR=R
Dfg=DfDg=RR=R
Dfg=DfDg-x:gx=0=RR--12=R--12
Thus,
(f + g) (x) : RR is given by (f + g) (x) = f (x) + g (x) = x2 + 2x + 1= (x + 1)2 .
(- g) (x) : RR is given by (f - g) (x) = f (x- g (x) = x2 - 2x -1.
(fg) (x) : RR is given by (fg) (x) = f(x).g(x) = x2(2x + 1) = 2x3 + x2 .

fg:R--12R is given byfgx=fxgx=x22x+1.



Page No 3.41:

Question 1:

Write the range of the real function f(x) = |x|.

Answer:

Given:
f (x) =  | x |, xR
We know that
x=x,x0-xx<0
It can be observed that the range of f (x) =  | x | is all real numbers except negative real numbers.
∴ The range of f is [0, ∞) .

Page No 3.41:

Question 2:

If f is a real function satisfying fx+1x=x2+1x2for all x ∈ R − {0}, then write the expression for f(x).

Answer:

Given:
fx+1x=x2+1x2
                =x2+1x2+2-2
                =x+1x2-2
Thus,
fx+1x=x+1x2-2
Hence,
f (x)  = x2 - 2 , where | x | ≥  2.

Page No 3.41:

Question 3:

Write the range of the function f(x) = sin [x], where -π4xπ4.

Answer:

Given : f(x) = sin [x], where -π4xπ4.
-π4xπ4-7.85  x  0.785 x-7.85, 7.85Or x = {0,1}Thus, range of f(x) = sin [x] is{sin 0, sin1} = {0, sin 1}

Page No 3.41:

Question 4:

If f(x) = cos [π2]x + cos [−π2] x, where [x] denotes the greatest integer less than or equal to x, then write the value of f(π).

Answer:

f(x) = cos [π2]x + cos [−π2] x
Thus, f(π) =  cosπ2 π+cos -π2 πf(π) =cos 9.8π+ cos-9.8πf(π) =cos 10π+ cos 9πf(π) =1+(-1) = 0

Page No 3.41:

Question 5:

Write the range of the function f(x) = cos [x], where -π2<x<π2.

Answer:

Since f(x) = cos [x], where -π2<x<π2,
-π2<x<π2-1.57< x < 1.57[x]  {-1,0,1,2}Thus, cos [x] = {cos (-1), cos 0, cos1, cos 2 } .Range of f(x) = {cos 1, 1, cos 2} .

Page No 3.41:

Question 6:

Write the range of the function f(x) = ex[x], x ∈ R.

Answer:

f(x) = ex[x], x ∈ R

We know that x-[x]  ={x}, which is the fractional part of any number x.Thus, f(x) = e{x}Also, 0  {x} <1e0 e{x} <e11f(x) < eThus range of f(x) is [1, e) .

Page No 3.41:

Question 7:

Let fx=αxx+1, x-1. Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.

Answer:

Given:
fx=αxx+1, x-1

Since f(f(x))  = x,ααxx+1αxx+1+1  =xα2xαx+x+1  =xα2x -αx2 -(x2+x) = 0Solving the quadratic equation in α:α = x2±x4+4x(x2+x)2x  α = x+1 or -1Since, α x+1, α = -1.



Page No 3.42:

Question 8:

If fx=1-1x, then write the value of ff1x.

Answer:

Given:
fx=1-1x
Now,
f1x=1-11x=1-x
ff1x=f1-x
Again,
 If fx=1-1x
Thus,
f1-x=1-11-x

=1-x-11-x=-x1-x=-x-x-1=xx-1

Page No 3.42:

Question 9:

Write the domain and range of the function fx=x-22-x.

Answer:

Given:
fx=x-22-x
Domain ( f ) :
Clearly,  f (x) is defined for all x satisfying: if 2 -x  ≠ 0 ⇒ ≠ 2.
Hence, domain ( f ) = R - {2}
Range of f :
Let f (x) = y

x-22-x=y
x - 2 = y (2 - x)
x - 2 = - y (x - 2)
y- 1
Hence, range ( f ) = {- 1}.

Page No 3.42:

Question 10:

If f(x) =  4xx2, x ∈ R, then write the value of f(a + 1) −f(a − 1).

Answer:

Given:
f(x) =  4xx2, x ∈ R
Now,
f(a + 1) = 4(a + 1) - (a + 1)2
             = 4a + 4 - (a2 + 1 + 2a)
             = 4a + 4 - a2  -- 2a
             = 2a - a2 + 3
f(a - 1) = 4(a - 1) - (a - 1)2
             = 4a -- (a2 + 1 - 2a)
             = 4a -- a2  - 1 + 2a
             = 6a - a2  - 5
Thus,
f(a + 1) − f(a − 1) = ( 2a - a2 + 3) - (6a - a2  - 5)
                             = 2a - a2 + 3 - 6a + a2 + 5
                             =  8 - 4a
                             = 4(2 - a)

Page No 3.42:

Question 11:

If f, g, h are real functions given by f(x) = x2, g(x) = tan x and h(x) = loge x, then write the value of (hogof) π4.

Answer:

Given : f(x) = x2, g(x) = tan x and h(x) = loge x.
(hogof) π4 = hgfπ4 
= hgπ4 2= hgπ4= htan π4= h1= loge 1 =0 

Page No 3.42:

Question 12:

Write the domain and range of function f(x) given by fx=1x-x.

Answer:

Given:
fx=1x-x

We know that
x=x,if x0-x,if x<0
x-x=x-x=0,if x0x+x=2x,if x<0
x - | x| ≤ 0 for all x.
1x-x does not take any real values for any x ∈ R.
f (x) is not defined for any x ∈ R.

Hence,
domain ( f ) = Φ and range ( f ) = Φ .

Page No 3.42:

Question 13:

Write the domain and range of fx=x-x.

Answer:

fx=x-x
Since f(x) is defined for all values of x,  xR.Or dom (f(x)) = RSince, x-[x] = {x}, which is the fractional part of any real number x, f(x) = {x}   .....(1)We know that 0  {x}<10 {x} <10  f(x) <1   {from(1)}Thus, range of f(x) is [0,1).

Page No 3.42:

Question 14:

Write the domain and range of function f(x) given by fx=x-x.

Answer:

fx=x-x

We know that[x] - x =-{x}, which is the fractional part of any real number x.Thus, f(x)  =-{x} .Since {x} is always a positive number, f(x) is not defined for any x.Or dom (f) = φThus, range (f)  = φ.

Page No 3.42:

Question 15:

Let A and B be two sets such that n(A) = p and n(B) = q, write the number of functions from A to B.

Answer:

It is given that A and B are two sets such that n(A) = p and n(B) = q.

Now, any element of set A, say ai (1 ≤ i ≤ p), is related with an element of set B in q ways. Similarly, other elements of set A are related with an element of set B in q ways.

Thus, every element of set A is related with every element of set B in q ways.

∴ Total number of functions from A to B = q × × q × ... × q (p times) = qp

Page No 3.42:

Question 16:

Let f and g be two functions given by

f = {(2, 4), (5, 6), (8, −1), (10, −3)} and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, −5)}.

Find the domain of f + g.

Answer:

It is given that f and g are two functions such that

f = {(2, 4), (5, 6), (8, −1), (10, −3)}

and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, −5)}

Now,

Domain of f = Df = {2, 5, 8, 10}

Domain of g = Dg = {2, 7, 8, 10, 11}

∴ Domain of f + g = Df Dg = {2, 8, 10}

Page No 3.42:

Question 17:

Find the set of values of x for which the functions f(x) = 3x2 − 1 and g(x) = 3 + x are equal.

Answer:

It is given that the functions f(x) = 3x2 − 1 and g(x) = 3 + x are equal.

fx=gx3x2-1=3+x3x2-x-4=0x+13x-4=0
x+1=0 or 3x-4=0x=-1 or x=43

Hence, the set of values of x for which the given functions are equal is -1,43.

Page No 3.42:

Question 18:

Let f and g be two real functions given by

f = {(0, 1), (2, 0), (3, −4), (4, 2), (5, 1)} and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}

Find the domain of fg.

Answer:

It is given that f and g are two real functions such that

f = {(0, 1), (2, 0), (3, −4), (4, 2), (5, 1)}

and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}

Now,

Domain of f = Df = {0, 2, 3, 4, 5}

Domain of g = Dg = {1, 2, 3, 4, 5}

∴ Domain of fg = Df Dg = {2, 3, 4, 5}

Page No 3.42:

Question 1:

Let A = {1, 2, 3} and B = {2, 3, 4}. Then which of the following is a function from A to B?

(a) {(1, 2), (1, 3), (2, 3), (3, 3)}
(b) [(1, 3), (2, 4)]
(c) {(1, 3), (2, 2), (3, 3)}
(d) {(1, 2), (2, 3), (3, 2), (3, 4)}

Answer:

(c) {(1, 3), (2, 2), (3, 3)}
We have
R = {(1, 3), (2, 2), (3, 3)}
We observe that each element of the given set has appeared as first component in one and only one ordered pair of R.
So, R = {(1, 3), (2, 2), (3, 3)} is a function.

Page No 3.42:

Question 2:

If f : Q → Q is defined as f(x) = x2, then f−1 (9) is equal to

(a) 3
(b) −3
(c) {−3, 3}
(d) ϕ

Answer:

(c) {−3, 3}
If f : AB, such that yB, then f-1{ y }={xA: f (x) = y}.
In other words, f-1{ y} is the set of pre-images of  y.
Let f-1{9} = x
Then, f (x) = 9
x2  = 9
x = ± 3
f-1{9} = {- 3, 3}.

Page No 3.42:

Question 3:

Which one of the following is not a function?

(a) {(x, y) : x, y ∈ R, x2 = y}
(b) {(x, y) : x, y ∈, R, y2 = x}
(c) {(x, y) : x, y ∈ R, x2 = y3}
(d) {(x, y) : x, y ∈, R, y = x3}

Answer:

(b) {(x, y) : x, y ∈, R, y2 = x}

y2  =x gives two values of y for a value of x.i.e. there are two images for a value of x.For example: (2)2  =4 and (-2)2  =4Thus, it is not a function.

Page No 3.42:

Question 4:

If f(x) = cos (log x), then the value of f(x2) f(y2) −12fx2y2+fx2y2is
(a) −2
(b) −1
(c) 1/2
(d) None of these

Answer:

(d) None of these

Given:
fx=coslogx
fx2=coslogx2fx2=cos2logx

Similarly,
fy2=cos2logy

Now,
fx2y2=coslogx2y2=coslogx2-logy2
and
fx2y2=coslogx2y2=coslogx2+logy2

fx2y2+fx2y2=cos2logx-2logy+cos2logx+2logyfx2y2+fx2y2=2cos2logxcos2logy12fx2y2+fx2y2=cos2logxcos2logy
fx2fy2-12fx2y2+fx2y2=cos2logxcos2logy-cos2logxcos2logy=0



Page No 3.43:

Question 5:

If f(x) = cos (log x), then the value of f(x) f(y) −12fxy+fxy is
(a) −1
(b) 1/2
(c) −2
(d) None of these

Answer:

(d) None of these

Given:
fx=coslogx
fy=coslogy

Now,
fxy=coslogxy=coslogx-logy
and
fxy=coslogxy=coslogx+logy

fxy+fxy=coslogx-logy+coslogx+logyfxy+fxy=2coslogxcoslogy12fxy+fxy=coslogxcoslogy
fxfy-12fxy+fxy=coslogxcoslogy-coslogxcoslogy=0

Page No 3.43:

Question 6:

Let f(x) = |x − 1|. Then,

(a) f(x2) = [f(x)]2
(b) f(x + y) = f(x) f(y)
(c) f(|x| = |f(x)|
(d) None of these

Answer:

(d) None of these

f(x)  =x-1Since, x2-1 x-1 2 ,f(x2) (f(x)) 2Thus, (i) is wrong.Since, x+y-1 x-1y-1,f(x+y) f(x) f(y)Thus, (ii) is wrong.Since x-1  x-1 = x-1,fx  f(x)Thus, (iii) is wrong.Hence, none of the given options is the answer.

Page No 3.43:

Question 7:

The range of f(x) = cos [x], for π/2 < x < π/2 is

(a) {−1, 1, 0}
(b) {cos 1, cos 2, 1}
(c) {cos 1, −cos 1, 1}
(d) [−1, 1]

Answer:

(b) {cos 1, cos 2, 1}

Since, f(x) = cos [x], where -π2<x<π2,
-π2<x<π2-1.57< x < 1.57[x]  {-1,0,1,2}Thus, cos [x] = {cos (-1), cos 0, cos1, cos 2 }Range of f(x) = {cos 1, 1, cos 2}

Page No 3.43:

Question 8:

Which of the following are functions?

(a) {(x, y) : y2 = x, x, y ∈ R}
(b) {(x, y) : y = |x|, x, y ∈ R}
(c) {(x, y) : x2 + y2 = 1, x, y ∈ R}
(d) {(x, y) : x2y2 = 1, x, y ∈ R}

Answer:

(b) {(x, y) : y = |x|, x, y ∈ R}

For every value of x ∈ R, there is a unique value y∈ R.
i.e. there is a unique image for all values of x ∈ R.
Also, values of x occur only once in the ordered pairs.
Thus, it is a function.

Page No 3.43:

Question 9:

If fx=log 1+x1-x and gx=3x+x31+3x2, then f(g(x)) is equal to
(a) f(3x)
(b) {f(x)}3
(c) 3f(x)
(d) −f(x)

Answer:

(c) 3f(x)

fx=log 1+x1-x and gx=3x+x31+3x2
Now,1+g(x)1-g(x) = 1+3x+x31+3x21-3x+x31+3x2=1+3x2+3x+x31+3x2-3x-x3=(1+x)3(1-x)3Then, f(g(x)) = log 1+g(x)1-g(x)=log 1+x1-x3=3 log 1+x1-x=3f(x)) 

Page No 3.43:

Question 10:

If A = {1, 2, 3} and B = {x, y}, then the number of functions that can be defined from A into B is

(a) 12
(b) 8
(c) 6
(d) 3

Answer:

(b) 8

Given:
Number of elements in set A = 3
Number of elements in set B = 2
Therefore, the number of functions that can be defined from A into B is = 23 = 8.

Page No 3.43:

Question 11:

If fx=log 1+x1-x, then f2x1+x2 is equal to
(a) {f(x)}2
(b) {f(x)}3
(c) 2f(x)
(d) 3f(x)

Answer:

(c) 2f(x)

fx=log 1+x1-x
Then, f2x1+x2 = log 1+2x1+x21-2x1+x2= log 1+x2+2x1+x21+x2-2x1+x2= log (1+x)2(1-x)2= 2 log 1+x1-x= 2 (f(x)) 

Page No 3.43:

Question 12:

If f(x) = cos (log x), then value offx f4-12 fx4+f4xis
(a) 1
(b) −1
(c) 0
(d) ±1

Answer:

(c) 0

Given : f(x) = cos (log x)
Then,  fx f4-12 fx4+f4x
= cos (log x)cos(log 4)-12cos logx4+coslog4x= 12coslogx+log4+cos logx-log4-12cos logx4+coslog4x= 12cos (log 4x)+cos log x4-cos log x4-cos log 4x= 12×0 = 0

Page No 3.43:

Question 13:

If fx=2x+2-x2, then f(x + y) f(xy) is equal to
(a) 12f2x+f2y
(b) 12f2x-f2y
(c) 14f2x+f2y
(d) 14f2x-f2y

Answer:

(a) 12f2x+f2y

Given:
fx=2x+2-x2
Now,
f(x + y) f(xy) = 2x+y+2-x-y22x-y+2-x+y2
⇒ f(x + y) f(xy) = 1422x+2-2y+22y+2-2x
⇒ f(x + y) f(xy) = 1222x+2-2x2+22y+2-2y2
⇒ f(x + y) f(xy) = 12f2x+f2y

Page No 3.43:

Question 14:

If 2f (x) − 3f1x=x2 (x ≠ 0), then f(2) is equal to

(a) -74
(b) 52
(c) −1
(d) None of these

Answer:

(a) -74
 
2f (x) − 3f1x=x2         (x ≠ 0)                  ....(1)
Replacing x by 1x:2f1x-3f(x) = 1x2            ...(2) Solving equations (1) & (2) -5f(x) = 3x2+2x2f(x)  = -15 3x2+2x2Thus, f(2) = -15 34+2×4= -15 3+324 = -74

Page No 3.43:

Question 15:

Let f : R → R be defined by f(x) = 2x + |x|. Then f(2x) + f(−x) − f(x) =

(a) 2x
(b) 2|x|
(c) −2x
(d) −2|x|

Answer:

(b) 2|x|

f(x) = 2x + |x|
Then, f(2x) + f(−x) − f(x)
=22x+2x+-2x+-x -2x+x=4x-2x-2x+2x+-x-x=0+2x+x-x=2x

Page No 3.43:

Question 16:

The range of the function fx=x2-xx2+2x is
(a) R
(b) R − {1}
(c) R − {−1/2, 1}
(d) None of these

Answer:

(c) R − {-1/2,1}

fx=x2-xx2+2x
Let y = x2-xx2+2x    Also, x0y = x(x-1)x(x+2)y = (x-1)(x+2)xy+2y = x-1x = 2y+11-yHere, 1-y  0.or, y  1 .Also, x02y+11-y0y-12Thus, range (f) = R-{-12,1} .

Page No 3.43:

Question 17:

If x ≠ 1 and fx=x+1x-1is a real function, then f(f(f(2))) is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(c) 3
fx=x+1x-1
f(f(f(2))) = ff2+12-1=ff(3)=f3+13-1=f(2)  = 3

Page No 3.43:

Question 18:

If f(x) = cos (loge x), then f1xf1y-12fxy+fxyis equal to
(a) cos (xy)
(b) log (cos (xy))
(c) 1
(d) cos (x + y)

Answer:

Given:
fx=coslogex
f1x=cosloge1xf1x=cos-logexf1x=coslogex

Similarly,
f1y=coslogey

Now,
fxy=coslogexy=coslogex+logey
and
fxy=coslogexy=coslogex-logey

fxy+fxy=coslogex-logey+coslogex+logeyfxy+fxy=2coslogexcoslogey12fxy+fxy=coslogexcoslogey
f1xf1y-12fxy+fxy=coslogexcoslogey-coslogexcoslogey=0
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.



Page No 3.44:

Question 19:

Let f(x) = x, gx=1x and h(x) = f(x) g(x). Then, h(x) = 1
(a) x ∈ R
(b) x ∈ Q
(c) x ∈ R − Q
(d) x ∈ R, x ≠ 0

Answer:

(d) x ∈ R, x ≠ 0

Given:
f(x) = x, gx=1x and h(x) = f(x) g(x)
Now,
h(x)=x×1x=1
We observe that the domain of f is  and the domain of g is -0.
∴ Domain of h = Domain of f ⋂ Domain of g = -0=-0
x ∈ R, x ≠ 0

Page No 3.44:

Question 20:

If fx=sin4 x+cos2 xsin2 x+cos4 xfor x ∈ R, then f (2002) =
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1

Given:
fx=sin4 x+cos2 xsin2 x+cos4 x

On dividing the numerator and denominator by cos4 x, we get
fx=tan4 x+sec2 x1+tan2 xsec2 x=1+tan4 x+tan2 x1+tan2 x1+tan2 x=1+tan4 x+tan2 x1+tan4 x+tan2 x=1 (For every x ∈ R)

For x = 2002, we have
f (2002) = 1

Page No 3.44:

Question 21:

The function f : R → R is defined by f(x) = cos2 x + sin4 x. Then, f(R) =

(a) [3/4, 1)
(b) (3/4, 1]
(c) [3/4, 1]
(d) (3/4, 1)

Answer:

(c) (3/4, 1)

Given:
f(x) = cos2x + sin4x
fx=1-sin2x+sin4x
fx=sin2x-122+34

The minimum value of fx is 34.
Also,
sin2x1sin2x-1212sin2x-12214sin2x-122+3414+34fx1

The maximum value of fx is 1.

∴ f(R) = (3/4, 1)

Page No 3.44:

Question 22:

Let A = {x ∈ R : x ≠ 0, −4 ≤ x ≤ 4} and f : A ∈ R be defined by fx=xxfor x ∈ A.
Then th (is

(a) [1, −1]
(b) [x : 0 ≤ x ≤ 4]
(c) {1}
(d) {x : −4 ≤ x ≤ 0}

Answer:

As, x=x, x0-x<0So, f(x)=xxWhen x<0 i.e. x[-4,0)f(x)=x-x=-1and when x>0 i.e. x(0,4]f(x)=xx=1 So, range(f)=-1,1
Disclaimer: The question in the book has some error. The solution is created according to the question given in the book.

Page No 3.44:

Question 23:

If f : R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the values of x such that g(f(x)) = 8 are

(a) 1, 2
(b) −1, 2
(c) −1, −2
(d) 1, −2

Answer:

(c) −1, −2
f(x) = 2x + 3 and g(x) = x2 + 7

g(f(x)) = 8f(x)2+7 = 8(2x+3)2+7 = 8x2+3x+2 = 0(x+2)(x+1) = 0x = -1,-2

Page No 3.44:

Question 24:

If f : [−2, 2] → R is defined by fx=  -1,for -2x0x-1,for 0x2, then
{x ∈ [−2, 2] : x ≤ 0 and f (|x|) = x} =

(a) {−1}
(b) {0}
(c) -12
(d) ϕ

Answer:

(c) -12

Given:
fx=  -1,for -2x0x-1,for 0x2
We know,  
x0
fx=x-1             ...(1)

Also,
If x0, then x=-x    ...(2)

∴ {x ∈ [−2, 2]: x ≤ 0 and f (|x|) = x}
= x: x-1=x                    [Using (1)]
= x: -x-1=x                  [Using (2)]
= x: 2x=-12
= x: x=-12
= -12

Page No 3.44:

Question 25:

If efx=10+x10-x, x ∈ (−10, 10) and fx=kf200 x100+x2, then k =
(a) 0.5
(b) 0.6
(c) 0.7
(d) 0.8

Answer:

(a) 0.5

efx=10+x10-x
f(x) = log e10+x10-x   ...(1)
fx=kf200 x100+x2
log e10+x10-x = k loge10+200x100+x210-200x100+x2   {from (1)}log e10+x10-x = k loge1000+10x2+200x1000+10x2-200xlog e10+x10-x = k logex+102x-102log e10+x10-x = 2k  logex+10x-101 = 2kk = 1/2=0.5

Page No 3.44:

Question 26:

f is a real valued function given by fx=27x3+1x3and α, β are roots of 3x+1x=12. Then,
(a) f(α) ≠ f(β)
(b) f(α) = 10
(c) f(β) = −10
(d) None of these

Answer:

(d) None of these

Given:
fx=27x3+1x3
fx=3x+1x9x2+1x2-3
fx=3x+1x3x+1x2-9
fα=3α+1α3α+1α2-9
Since α and β are the roots of 3x+1x=12,
3α+1α=12 and 3β+1β=12
fα=12122-9 and fβ=12122-9
fα=fβ=12122-9

Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.

Page No 3.44:

Question 27:

If fx=64x3+1x3and α, β are the roots of 4x+1x=3. Then,
(a) f(α) = f(β) = −9
(b) f(α) = f(β) = 63
(c) f(α) ≠ f(β)
(d) none of these

Answer:

(a) fα=fβ=-9

Given:
fx=64x3+1x3
fx=4x+1x16x2+1x2-4
fx=4x+1x4x+1x2-12

fα=4α+1α4α+1α2-12 and fβ=4β+1β4β+1β2-12
Since α and β are the roots of 4x+1x=3,
 4α+1α=3 and 4β+1β=3
fα=332-12=-9 and fβ=332-12=-9
fα=fβ=-9

Page No 3.44:

Question 28:

If 3fx+5f1x=1x-3 for all non-zero x, then f(x) =

(a) 1143x+5x-6
(b) 114-3x+5x-6
(c) 114-3x+5x+6
(d) None of these

Answer:

(d) None of these

3fx+5f1x=1x-3   ...(1)

Multiplying (1) by 3:15 f1x +9 f(x) = 3x-9   .....(2)Replacing x by 1xin (1):3 f1x +5 f(x) = x-3  Multiplying by 5:15 f1x +25 f(x) = 5x-15  ....(3)Solving (2) and (3):-16 f(x) = 3x-5x+6f(x) = 116-3x+5x-6
Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.

Page No 3.44:

Question 29:

If f : R → R be given by for all fx=4x4x+2 x ∈ R, then

(a) f(x) = f(1 − x)
(b) f(x) + f(1 − x) = 0
(c) f(x) + f(1 − x) = 1
(d) f(x) + f(x − 1) = 1

Answer:

(c) f(x) + f(1 − x) = 1

fx=4x4x+2x ∈ R
f (1-x) = 41-x41-x+2= 42×4x+4=24x+2f(x) +f(1-x) = 4x4x+2+24x+2 = 4x+24x+2 =1

Page No 3.44:

Question 30:

If f(x) = sin [π2] x + sin [−π]2 x, where [x] denotes the greatest integer less than or equal to x, then

(a) f(π/2) = 1
(b) f(π) = 2
(c) f(π/4) = −1
(d) None of these

Answer:

(a) f(π/2) = 1

f(x) = sin [π2] x + sin [−π2]x
f(x) =  sin 9.8x+ sin -9.8xf(x) = sin 9x - sin 10xfπ2 =  sin 9×π2 - sin 10×π2fπ2 = 1- 0 =1



Page No 3.45:

Question 31:

The domain of the function fx=2-2x-x2 is

(a) -3, 3
(b) -1-3, -1+3
(c) [−2, 2]
(d) -2-3, -2+3

Answer:

(b) -1-3, -1+3

fx=2-2x-x2
Since, 2-2x-x20, x2+2x-2 0x2-2x-2+1-10x-12-320x--1-3x--1+3 0-1-3 x -1+3Thus, dom(f) = -1-3,-1+3  .

Page No 3.45:

Question 32:

The domain of definition of fx=x+32-x x-5is

(a) (−∞, −3] ∪ (2, 5)
(b) (−∞, −3) ∪ (2, 5)
(c) (−∞, −3) ∪ [2, 5]
(d) None of these

Answer:

(a) (−∞, −3] ∪ (2, 5)

fx=x+32-x x-5
For f(x) to be defined ,2-xx-50x2, 5            ....(1)Also, x+32-xx-50x+32-xx-52-x2x-520x+3x-2x-50x(-,-3](2,5)   ....(2)From (1) and (2),x(-,-3] (2,5)

Page No 3.45:

Question 33:

The domain of the function fx=x+1 x-3x-2is
(a) [−1, 2) ∪ [3, ∞)
(b) (−1, 2) ∪ [3, ∞)
(c) [−1, 2] ∪ [3, ∞)
(d) None of these

Answer:

(a) [−1, 2) ∪ [3, ∞)

fx=x+1 x-3x-2
For f(x) to be defined,(x-2) 0x  2           ...(1)Also,(x+1)(x-3)(x-2)0(x+1)(x-3)(x-2)(x-2)20(x+1)(x-3)(x-2)0x  [-1,2)[3,)   .....(2)From (1) and (2),x  [-1,2)  [3,)  

Page No 3.45:

Question 34:

The domain of definition of the function fx=x-1+3-x is

(a) [1, ∞)
(b) (−∞, 3)
(c) (1, 3)
(d) [1, 3]

Answer:

(d) [1, 3]

fx=x-1+3-x
For f(x) to be defined, x-10x1       ...(1)and 3-x 03 x      ...(2)From (1) and (2),x[1,3]

Page No 3.45:

Question 35:

The domain of definition of the function fx=x-2x+2+1-x1+xis
(a) (−∞, −2] ∪ [2, ∞)
(b) [−1, 1]
(c) ϕ
(d) None of these

Answer:

(c) ϕ

fx=x-2x+2+1-x1+x
For f(x) to be defined,x+2 0x -2  ...(1)And 1+x 0 x  -1  ....(2)Also, x-2x+2 0(x-2)(x+2)(x+2)2  0(x-2)(x+2)0x  (-,-2)  [2,)  ...(3)And 1-x1+x 0(1-x)(1+x)(1+x)2 0(1-x)(1+x) 0 x (-,-1)  [1,)   ...(4)From (1), (2), (3) and (4), we get,x ϕ .Thus, dom (f(x)) = ϕ.

Page No 3.45:

Question 36:

The domain of definition of the function f(x) = log |x| is

(a) R
(b) (−∞, 0)
(c) (0, ∞)
(d) R − {0}

Answer:

(d) R − {0}

f(x) = log |x|
For f(x) to be defined,x >0, which is always true.But x  0 x  0Thus, dom (f) = R- {0}.

Page No 3.45:

Question 37:

The domain of definition of fx=4x-x2 is

(a) R − [0, 4]
(b) R − (0, 4)
(c) (0, 4)
(d) [0, 4]

Answer:

(d) [0, 4]

Given:
fx=4x-x2
Clearly, f (x) assumes real values if
4x - x2 ≥ 0
x(4 - x) ≥ 0
 ⇒ - x(- 4) ≥ 0
x(- 4) ≤ 0
x ∈ [0, 4]

Hence, domain (f )= [0, 4].

Page No 3.45:

Question 38:

The domain of definition of fx=x-3-2x-4-x-3+2x-4is

(a) [4, ∞)
(b) (−∞, 4]
(c) (4, ∞)
(d) (−∞, 4)

Answer:

(a) [4, ∞)

fx=x-3-2x-4-x-3+2x-4

For f(x) to be defined, x-4 0x-4  0 x  4               ....(1)Also, x-3-2x-4  0x-3-2x-4  0x-3  2x-4(x-3)2  2x-42x2+9-6x   4x-4 x2-10x+25  0 (x-5) 2 0, which is always true.Similarly, x-3+2x-40 is always true.Thus, dom(f(x))=[4,)

Page No 3.45:

Question 39:

The domain of the function fx=5 x-x2-6 is

(a) (−3, − 2) ∪ (2, 3)
(b) [−3, − 2) ∪ [2, 3)
(c) [−3, − 2] ∪ [2, 3]
(d) None of these

Answer:

(c) [−3, − 2] ∪ [2, 3]

fx=5 x-x2-6
For f(x) to be defined, 5x-x2-6 05x-x2-6 0x 2-5x +6  0For x>0, x=xx 2-5x +6  0(x-2)(x-3)  0x[2,3]              ........(1)For x<0, x=-xx 2+5x +6  0 (x+2)(x+3)  0x [-3,-2]         .......(2)From (1) and (2),x[-3,-2][2,3]  or, dom(f) =[-3,-2][2,3]   

Page No 3.45:

Question 40:

The range of the function fx=xxis

(a) R − {0}
(b) R − {−1, 1}
(c) {−1, 1}
(d) None of these

Answer:

(c) {−1, 1}

fx=xx
Let y = xxFor x>0, x=xy = xx = 1For x <0, =-xy = x-x  =-1Thus, range of f(x) is {-1,1}.

Page No 3.45:

Question 41:

The range of the function fx=x+2x+2, x ≠ −2 is
(a) {−1, 1}
(b) {−1, 0, 1}
(c) {1}
(d) (0, ∞)

Answer:

(a) {−1, 1}

fx=x+2x+2 , x ≠ −2
Let y= x+2x+2For x+2 >0,or x>-2 ,y = x+2x+2 = 1For x+2 < 0,or x<-2,y = x+2-(x+2) = -1Thus, y = {-1,1}or range f(x)= {-1,1}.

Page No 3.45:

Question 42:

The range of the function f(x) = |x − 1| is

(a) (−∞, 0)
(b) [0, ∞)
(c) (0, ∞)
(d) R

Answer:

(b) [0, ∞)

f(x)=x-10 xRThus, range=[0,0

Page No 3.45:

Question 43:

Let fx=x2+1. Then, which of the following is correct?

(a) fxy=fxfy                     (b) fxyfxfy                     (c) fxyfxfy                     (d) none of these                        

Answer:

Given: fx=x2+1              .....(1)

Replacing x by y in (1), we get

fy=y2+1

fxfy=x2+1y2+1                 =x2+1y2+1                 =x2y2+x2+y2+1

Also, replacing x by xy in (1), we get

fxy=x2y2+1

Now,

x2y2+1x2y2+x2+y2+1x2y2+1x2y2+x2+y2+1fxyfxfy

Hence, the correct answer is option (c).

Page No 3.45:

Question 44:

If x2-5x+6=0, where [.] denotes the greatest integer function, then

(a) x ∈ [3, 4]                           (b) x ∈ (2, 3]                           (c) x ∈ [2, 3]                           (d) x ∈ [2, 4)                           

Answer:

The given equation is x2-5x+6=0.

x2-5x+6=0x2-3x-2x+6=0xx-3-2x-3=0x-2x-3=0
x-2=0 or x-3=0x=2 or x=3
x ∈ [2, 3) or x ∈ [3, 4)
x ∈ [2, 4)

Hence, the correct answer is option (d).

Page No 3.45:

Question 45:

The range of fx=11-2cosx is

(a) [1/3, 1]                        (b) [−1, 1/3]                        (c) (−∞, −1) ∪ [1/3, ∞)                        (d) [−1/3, 1]                       

Answer:

We know that −1 ≤ cosx ≤ 1 for all x ∈ R.

Now,

-1cosx1-1-cosx1-2-2cosx2-11-2cosx3                Adding 1 to each term

But,

cosx12
1-2cosx-1,3-011-2cosx(-,-1][13,)

∴ Range of f(x) = (−∞, −1] ∪[13, ∞)

Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as "(−∞, −1] ∪ [1/3, ∞)".



Page No 3.7:

Question 1:

Define a function as a set of ordered pairs.

Answer:

A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components.
Sometimes we say that a function is a rule (correspondence) that assigns to each element of one set, X, only one element of another set, Y.

The elements of set X are often called inputs and the elements of set Y are called outputs.
The domain of a function is the set of all first components, x, in the ordered pairs.
The range of a function is the set of all second components, y, in the ordered pairs.

A function can be defined by a set of ordered pairs.

Example: {(1,a), (2, b), (3, c), (4,a)} is a function, since there are no two pairs with the same first component.
The domain is then the set {1,2,3,4} and the range is the set {a, b, c}.

Page No 3.7:

Question 2:

Define a function as a correspondence between two sets.

Answer:

A function is a correspondence between two sets of elements, such that for each element in the first set there is only one corresponding element in the second set.
The first set is called the domain and the set of all corresponding elements in the second set is called the range.
Let A = {1, 2, 3} and  B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Let f : A B be the correspondence which assigns to each element in A its square.
Hence,
f (1) = 12 = 1
f (2) = 22 = 4
f (3) = 32 = 9
Since for each element (1 or 2 or 3) of A, there is exactly one element of B, so f is a function.
In this case, every element of B is not an image of some element of A.

Page No 3.7:

Question 3:

What is the fundamental difference between a relation and a function? Is every relation a function?

Answer:

Differences between relation and function

  1. If R is a relation from A to B, then domain of R may be a subset of A. But if f is a function from A to B, then domain f is equal to A.
  2. In a relation from A to B, an element of A may be related to more than one element in B. But in a function from A to B, each element of A must be associated to one and only one element of B.
Thus, every function is a relation, but every relation is not necessarily a function.

Page No 3.7:

Question 4:

Let A = {−2, −1, 0, 1, 2} and f : A → Z be a function defined by f(x) = x2 − 2x − 3. Find:

(a) range of f, i.e. f(A).
(b) pre-images of 6, −3 and 5.

Answer:

(a) Given:
f (x) = x2 − 2x − 3
f (−2) = (− 2)2 − 2(− 2) − 3
          = 4 + 4 – 3
          = 8 − 3 = 5
f (−1) = (−1)2 − 2(−1) − 3
          = 1+ 2 − 3
          = 3 − 3 = 0
f (0) = (0)2 − 2(0) − 3
        = 0 − 0 − 3
        = − 3
f (1) = (1)2 − 2(1) − 3
        = 1 − 2 − 3
        =1 − 5 = − 4
f (2) = (2)2 – 2(2) − 3
        = 4 − 4 – 3
        = 4 – 7 = − 3
Thus, range of f(A) = (− 4, − 3, 0, 5).

(b) Let x be the pre-image of 6.
Then,
f(6) = x2 − 2x − 3 = 6
x2 − 2x − 9 = 0
x=1±10
Since x=1±10A, there is no pre-image of 6.

Let x be the pre-image of -3. Then,
f(− 3) ⇒ x2 − 2x − 3 = − 3
x2 − 2x  = 0
x = 0, 2
Clearly 0,2A. So, 0 and 2 are pre-images of  −3.

Let x be the pre-image of  5. Then,
f(5) ⇒ x2 − 2x − 3 = 5
x2 − 2x − 8 = 0
⇒ (x − 4) (x + 2) = 0 ⇒ x = 4, − 2
Since -2A, − 2 is the pre-image of 5.
Hence,
pre-images of 6, − 3 and 5 are ϕ,0,2,,-2 respectively.

Page No 3.7:

Question 5:

If a function f : R → R be defined by
fx=3x-2,x<0;1,x=0;4x+1,x>0.
find: f(1), f(−1), f(0) and f(2).

Answer:

f (1) = 4 × 1 + 1 = 5          [By using f (x) = 4x + 1, x > 0]
f (- 1) = 3 × (-1) - 2          [By using f (x) = 3x - 2, x < 0]
          = -- 2 = - 5
f (0) = 1                             [By using f (x) = 1, x = 0]
f (2) = 4 × 2 + 1                 [By using f (x) = 4x + 1, x > 0]
        = 9
Hence,
f (1) = 5, f (-1) = - 5, f (0) = 1 and f (2) = 9.

Page No 3.7:

Question 6:

A function f : R → R is defined by f(x) = x2. Determine (a) range of f, (b) {x : f(x) = 4}, (c) [y : f(y) = −1].

Answer:

(a) Given:
f (x) = x2     
Range of f = R+     (Set of all real numbers greater than or equal to zero)

(b) Given:
f (x) = x2   
⇒ x2 = 4
x = ± 2
∴ {x : f (x) = 4 } = { - 2, 2}.

(c) { y : f (y) = -1}
f (y) = -1   
It is clear that  x2-1 but   x2 ≥ 0 .
f  (y) ≠ -1
∴ {y : f (y) = -1} = Φ

Page No 3.7:

Question 7:

Let f : R+ → R, where R+ is the set of all positive real numbers, such that f(x) = loge x. Determine

(a) the image set of the domain of f
(b) {x : f(x) = −2}
(c) whether f(xy) = f(x) : f(y) holds

Answer:

Given:
f : R+ → R
and f (x) = logex .............(i)

(a) f : R+ → R
Thus, the image set of the domain f = R .

(b) {x : f (x) = -2
f (x ) = -2    .....(ii)
From equations (i) and (ii), we get :
logex = -2
⇒ x = e-2
Hence, { x : f (x) = - 2} = { e – 2} .      [Since logab = c ⇒  b = ac]

(c) f (xy) = loge(xy)      {From(i)}
         = logex + logey         [Since logemn = loge m + logen]  
         = f (x) + f (y)
Thus, f (xy) = f (x) + f (y)
Hence, it is clear that f (xy) = f (x) + f (y) holds.



Page No 3.8:

Question 8:

Write the following relations as sets of ordered pairs and find which of them are functions:

(a) {(x, y) : y = 3x, x ∈ {1, 2, 3}, y ∈ [3,6, 9, 12]}
(b) {(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}
(c) {(x, y) : x + y = 3, x, y, ∈ [0, 1, 2, 3]}

Answer:

(a) Given:
{(x, y) : y = 3x, x ∈ {1, 2, 3}, y ∈ [3,6, 9, 12]}
On substituting x = 1, 2, 3 in x, we get :
y = 3, 6, 9, respectively.
∴ R = {(1, 3) , (2, 6), (3, 9)}
Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R . So, R is a function in the given set.

(b) Given:
{(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}
On substituting x = 1, 2 in y > x + 1, we get :
y > 2 and y > 3, respectively.
R = {(1, 4), (1, 6), (2, 4), (2, 6)}
We observe that 1 and 2 have appeared more than once as the first component of the ordered pairs. So, it is not a function.

(c) Given:
{(x, y) : x + y = 3, x, y, ∈ [0, 1, 2, 3]}
x + y = 3
y = 3 – x
On substituting x = 0,1, 2, 3 in y, we get:
y = 3, 2, 1, 0, respectively.
∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R . So, R is a function in the given set.

Page No 3.8:

Question 9:

Let f : R → R and g : C → C be two functions defined as f(x) = x2 and g(x) = x2. Are they equal functions?

Answer:

It is given that
f : R → R and g : C → C are two function defined as f (x) = x2 and g (x) = x2 .
Thus,
domain ( f ) = R and domain ( g ) = C .
Since, domain ( f ) ≠ domain ( g ),
 f (x) and g (x) are not equal functions.

Page No 3.8:

Question 10:

f, g, h are three function defined from R to R as follows:

(i) f(x) = x2
(ii) g(x) = sin x
(iii) h(x) = x2 + 1

Find the range of each function.

Answer:

(i) Given:
f (x) = x2
Range of f(x) = R+ (set of all positive integers)
                      = {y ∈ R| y ≥ 0}
(ii) Given:
g(x) = sin x
Range of g(x) = {y ∈ R : -1 ≤ y ≤ 1}

(iii) Given:
h (x) = x2 + 1
Range of h (x) = {y ∈ R : y ≥ 1}

Page No 3.8:

Question 11:

Let X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}
Determine which of the following sets are functions from X to Y.

(a) f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
(b) f2 = {(1, 1), (2, 7), (3, 5)}
(c) f3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

Answer:

(a) Given:
f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
f1 is a function from X to Y.

(b) Given:
f2 = {(1, 1), (2, 7), (3, 5)}
f2 is not a function from X to Y because 2 ∈ X has no image in Y.

(c) Given:
f3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
f3 is not a function from X to Y because 2 ∈ X has two images, 9 and 11, in Y.

Page No 3.8:

Question 12:

Let A = (12, 13, 14, 15, 16, 17) and f : A → Z be a function given by
f(x) = highest prime factor of x.
Find range of f.

Answer:

Given:
A ={12, 13, 14, 15, 16, 17}
f : A be defined by f (x) = the highest prime factor of x.
f (12) = the highest prime factor of 12 = 3
f (13) = the highest prime factor of 13 = 13
f (14) = the highest prime factor of 14 = 7
f (15) = the highest prime factor of 15 = 5
f (16) = the highest prime factor of 16 = 2
f (17) = the highest prime factor of 17 = 17
The range of f is the set of all f (x), where xA.nA
Therefore,
range  of  f  = {2, 3, 5, 7, 13, 17}.

Page No 3.8:

Question 13:

If f : R → R be defined by f(x) = x2 + 1, then find f−1 [17] and f−1 [−3].

Answer:

If f : AB is such that yB, then f-1{ y }={xA: f (x) = y}.
In other words, f -1{ y} is the set of pre - images of  y.
Let f-1{17} = x .
Then, f (x) =17 .
x2 +1 = 17
x2 = 17 - 1 = 16
x = ± 4
f-1{17} = {- 4,4}

Again,
let f-1{-3} = x .
Then, f (x) = - 3
x2 + 1 = - 3
x2-- 1 = - 4
x=-4
Clearly, no solution is available in R.
So f-1{- 3} = Φ .

Page No 3.8:

Question 14:

Let A = [p, q, r, s] and B = [1, 2, 3]. Which of the following relations from A to B is not a function?

(a) R1 = [(p, 1), (q, 2), (r, 1), (s, 2)]
(b) R2 = [(p, 1), (q, 1), (r, 1), (s, 1)]
(c) R3 = [(p, 1), (q, 2), (p, 2), (s, 3)
(d) R4 = [(p, 2), (q, 3), (r, 2), (s, 2)].

Answer:

(c) R3 = [(p, 1), (q, 2), (p, 2), (s, 3)

All the relations in (a), (b) and (d) have a unique image in B for all the elements in A.
R3 is not a function from A to B because p ∈ A has two images, 1 and 2, in B.
Hence, option (c) is not a function.

Page No 3.8:

Question 15:

Let A = [9, 10, 11, 12, 13] and let f : A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Answer:

Given:
A ={9, 10, 11, 12, 13}
f : AN  be defined by f (n) = the highest prime factor of n.
f (9) = the highest prime factor of 9 = 3
f (10) = the highest prime factor of 10 = 5
f (11) = the highest prime factor of 11 = 11
f (12) = the highest prime factor of 12 = 3
f (13) = the highest prime factor of 13 = 13
The range of f is the set of all f (n), where nA.
Therefore,
range  of  f  = {3, 5, 11, 13}

Page No 3.8:

Question 16:

The function f is defined byfx=x2,0x33x,3x10
The relation g is defined by gx=x2,0x23x,2x10
Show that f is a function and g is not a function.

Answer:

The function f is defined by fx=x20x33x3x10
It is observed that for 0 ≤ x < 3, f (x) = x2 .
3 <  x ≤ 10, f (x) = 3x
Also, at x = 3, f(x) = 32 = 9. And
f (x) = 3 × 3 = 9.
That is, at x = 3, f (x) = 9.
Therefore, for 0 ≤ x ≤ 10, the images of f (x) are unique.
Thus, the given relation is a function.
Again,
the relation g is defined as gx=x2,0x23x,2x10
It can be observed that for x = 2, g(x) = 22 = 4 and also, 
g(x) = 3 × 2 = 6.
Hence, 2 in the domain of the relation g corresponds to two different images, i.e. 4 and 6.
Hence, this relation is not a function.

Hence proved.

Page No 3.8:

Question 17:

If f(x) = x2, find f1.1-f11.1-1.

Answer:

Given:
f(x) = x2

Therefore,

f1.1-f11.1-1=1.12-121.1-1=1.21-10.1=0.210.1=2.1

Page No 3.8:

Question 18:

Express the function f : XR given by f(x) = x3 + 1 as set of ordered pairs, where X = {−1, 0, 3, 9, 7}.            [NCERT EXEMPLAR]

Answer:

The function f : XR is defined by  f(x) = x3 + 1, where X = {−1, 0, 3, 9, 7}.

Now,

f-1=-13+1=0f0=03+1=1f3=33+1=28f7=73+1=344f9=93+1=730

So, f=x,fx:xX=-1,0,0,1,3,28,7,344,9,730



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