RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS-2

RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS

Function Exercise 3D, 3E, 3F

Exercise 3D

1. Consider the real function f: R → R: f(x) = x + 5 for all x ∈ R.
Find its domain and range.
Draw the graph of this function.

Solution: Given: f(x) = x + 5 for all x ∈ R
∴ The domain of the given function is all real numbers.
      D_f(x) = R
As f(x) is a polynomial function,
∴ The range of the given function is all real numbers.
Therefore, D_f(x) = R

Graph:

On a graph paper plot (-5, 0), (0, 5) and join these point successively to get the required graph.

GRAPH:

2. Consider the function f: R → R, defined by

  f(x) = {1 - x, when x < 0
          x, when x = 0
          x + 1, when x > 0 

Write its domain and range. Also, draw the graph of f(x).

Solution: Given:
f(x) = 1 – x, when x < 0
x, when x = 0
x + 1, when x > 0

CASE I
f(x) = 1 – x, when x < 0
Here f(x) is defined for any negative value of x. 
∴ Domain of f(x) = (-∞, 0)
As -∞ < x < 0
 ∴ 0 < -x < ∞
 ⇒ 1 + 0 < 1 – x < 1 + ∞
 ⇒ 1 < f(x) < ∞
∴ Range of f(x) = (1, ∞)

CASE II
f(x) = x, when x = 0
Here f(x) is defined for any negative value of x.
 ∴ Domain of f(x) = {0}
∴ Range of f(x) = {0}

CASE III
f(x) = x + 1, when x > 0
Here f(x) is defined for any positive value of x.
 ∴ Domain of f(x) = (0, ∞)
As 0 < x < ∞
∴ 0 +1 < x + 1 < ∞ +1
⇒ 1 ∞  < x + 1 < ∞
 ⇒ 1 < f(x) < ∞
∴ Range of f(x) = (1, ∞)

As a whole,
∴ Domain of f(x) = (-∞, ∞)
∴ Range of f(x) = 0 ∪ (1, ∞)
Graph:
when x < 0,
f(x) = 1 – x,

Exercise 3D

when x = 0,
y = f(x) = 1 ,

when x > 0,
f(x) = x + 1,

On a graph paper plot (-1, 2), (-2, 3), (0, 1), (1, 2), (2, 3) and join these point successively to get the required graph.

GRAPH:

3. Find the domain and the range of the square root function,
f: R⁺ U {0} → R: f(x) = √x for all non-negative real numbers.
Also, draw its graph.

Solution: Given:
f(x) = √x for all non-negative real numbers.
√x is defined for the set of all positive real numbers including 0.
Therefore, Domain of f(x) = [0, ∞)
For 0 ≤ x < ∞
The value of √x varies from 0 to ∞.
∴ Range of f(x) = [0, ∞)
Graph:
Let y = f(x) = √x

On a graph paper plot (0, 0), (1, 1), (2.25, 1.5), (4, 2), (6.25, 2.5), (9, 3) and join these point successively to get the required graph.

GRAPH:

4. Find the domain and the range of the cube root function, f: R → R: f(x) = x^⅓ for all x ∈ R.
Also, draw its graph.

Solution: Given: f(x) = x ^1/₃ for all x ∈ R.
Cube root function is defined for all real numbers.
So Domain of the function is (-∞, ∞).
Cube root of a real number is real number.
So Range of the  function is = (-∞, ∞)
Graph:
Let y = f(x) = x ^1/₃ for all x ∈ R.

On a graph paper plot (-8, -2), (-5, -1.7), (-3, -1.4), (-1, -1), (0, 0), (1,-1),  (3, 1.4), (5, -1.7), (8, 2) and join these point successively to get the required graph.

GRAPH:

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Exercise 3E

1. (i) Let f : R → R : f(x) = x + 1 and g : R → R : g(x) = 2x – 3. Find: (f + g)(x)

Solution: Given:
f(x) = x + 1 and g(x) = 2x -3
∴ (f + g)(x)
= f(x) + g(x)
⇒ (x + 1) + (2x -3)
= x + 1 + 2x – 3
= 3x – 2
Therefore,  (f + g)(x) = 3x -2

1. (ii) Let f : R → R : f(x) = x + 1 and g : R → R : g(x) = 2x – 3. Find: (f – g)(x)

Solution: Given: f(x) = x + 1 and g(x) = 2x -3
∴ (f – g)(x)
= f(x) – g(x)
= (x + 1) – (2x -3)
⇒ x + 1 – 2x + 3
= 4 – x
Therefore,(f – g)(x) = 4 – x

1. (iii) Let f : R → R : f(x) = x + 1 and g : R → R : g(x) = 2x – 3. Find: (fg)(x)

Solution: Given:f(x) = x + 1 and g(x) = 2x -3
∴ (fg)(x)
= f(x). g(x)
= (x+1) (2x-3)
⇒ 2x² – 3x + 2x -3
= 2x² – x -3
Therefore,(fg)(x) = 2x² – x -3

1. (iv) Let f : R → R : f(x) = x + 1 and g : R → R : g(x) = 2x – 3. Find:  (^f/_g)(x)

Solution: Given: f(x) = x + 1 and g(x) = 2x -3
∴ (^f/_g)(x)
= f(x)/g(x)
= ^{x + 1}/_{2x -3}
Therefore, (^f/_g)(x) = ^{x + 1}/_{2x -3}

2. (i) Let f : R → R : f(x) = 2x + 5 and g : R → R : g(x) = x² + x.Find: (f + g)(x)

Solution: Given: f(x) = 2x + 5 and g(x) = x² + x
∴ (f + g)(x)
= f(x) + g(x)
= (2x + 5) + (x² + x)
⇒ 2x + 5 + x² + x
= x² + 3x + 5
Therefore,
(f + g)(x) = x² + 3x + 5

2. (ii) Let f : R → R : f(x) = 2x + 5 and g : R → R : g(x) = x² + x.Find: (f – g)(x)

Solution: Given: f(x) = 2x + 5 and g(x) = x² + x
∴ (f – g)(x)
= f(x) – g(x)
= (2x + 5) – (x² + x)
⇒ 2x + 5 – x² – x
= -x² + x + 5
Therefore, (f + g)(x) = -x² + x + 5

2. (iii) Let f : R → R : f(x) = 2x + 5 and g : R → R : g(x) = x² + x.Find: (fg)(x)

Solution: Given: f(x) = 2x + 5 and g(x) = x² + x
∴ (fg)(x) = f(x).g(x)
= (2x + 5).(x² + x)
⇒ 2x³ + 2x² + 5x² + 5x
= 2x³ + 7x² + 5x
Therefore, (fg)(x) = 2x³ + 7x² + 5x

2. (iv) Let f : R → R : f(x) = 2x + 5 and g : R → R : g(x) = x² + x.Find: (^f/_g)(x)

Solution: Given:f(x) = 2x + 5 and g(x) = x² + x
∴ (^f/_g)(x)
= ^f(x)/_g(x)
= ^{2x + 5}/_{x² + x}
Therefore, (^f/_g)(x) = ^{2x + 5}/_{x² + x}

3. (i) Let f: R → R: f(x) = x³ + 1 and g: R → R: g(x) = (x + 1). Find: (f + g)(x)

Solution: Given: f(x) = x³ + 1 and g(x) = x + 1
∴ (f + g)(x)
= f(x) + g(x)
= (x³ + 1) + (x + 1)
⇒ x³ + 1 + x + 1
= x³ + x + 2
Therefore, (f + g)(x) = x³ + x + 2

3. (ii) Let f: R → R: f(x) = x³ + 1 and g: R → R: g(x) = (x + 1). Find: (f – g)(x)

Solution: Given: f(x) = x³ + 1 and g(x) = x + 1
∴ (f – g)(x)
= f(x) – g(x)
= (x³ + 1) – (x + 1)
⇒ x³ + 1 – x – 1
= x³ – x
Therefore, (f – g)(x) = x³ – x

3. (iii) Let f: R → R: f(x) = x³ + 1 and g: R → R: g(x) = (x + 1). Find: (¹/_f)(x)

Solution: Given: f(x) = x³ + 1 and g(x) = x + 1

3. (iv) Let f: R → R: f(x) = x³ + 1 and g: R → R: g(x) = (x + 1). Find: (^f/_g)(x) 

Solution: Given: f(x) = x³ + 1 and g(x) = x + 1

4. (i) Let f: R → R; f(x) = ^x/_c, where c is a constant. Find: (cf)(x)

Solution: Given: f(x) = ^x/_c, where c is a constant.
∴ (cf)(x)
= c.f(x)
= c. ^x/_c = x
Therefore, (cf)(x) = x

4. (ii) Let f: R → R; f(x) = ^x/_c, where c is a constant. Find: (c²f)(x)

Solution: Given: f(x) = ^x/_c, where c is a constant.
∴ (c²f)(x)
= c².f(x)
= c². ^x/_c = cx
Therefore, (c²f)(x) = cx

4. (iii) Let f: R → R; f(x) = ^x/_c, where c is a constant. Find: (¹/_cf)(x)

Solution: Given: f(x) = ^x/_c, where c is a constant.
∴ (¹/_cf)(x)
= ¹/_c.f(x)
= ¹/_c. ^x/_c = ^x/_c²
Therefore, (1/cf)(x) = ^x/_c²

, ∴ (f + g)(x)
= f(x) + g(x)

∴ (f – g)(x)
= f(x) – g(x)

∴ (fg)(x)
= f(x).g(x)

Exercise 3F

1. Find the set of values for which the function f(x) = 1 – 3x and g(x) = 2x² – 1 are equal.

Solution: f(x) = 1 – 3x and g(x) = 2x² – 1
f(x) = g(x)
∴ 1 – 3x = 2x² – 1
⇒ 2x² – 1 – 1 + 3x = 0
⇒ 2x² + 3x – 2 = 0
= 2x² + 4x – x – 2 = 0
⇒ 2x(x + 2) – 1(x + 2) = 0
⇒ (x + 2)(2x – 1) = 0
∴ x = -2 or x = ¹/₂
∴ Set of values are { -2 , ¹/₂}.

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

Solution: f(x) = x + 3 and g(x) = 3x² – 1
    f(x) = g(x)x + 3 = 3x² – 1
⇒ 3x² – 1 – x – 3 = 0
⇒ 3x² – x – 4 = 0
3x² – 4x + 3x – 4 = 0
⇒ x(3x – 4) + 1(3x – 4) = 0
⇒ (3x – 4)(x + 1) = 0
∴ x = ⁴/₃  or x = -1
∴ Set of values are { -1 , ⁴/₃}.

3. Let X = {–1, 0, 2, 5} and f : X → R: f(x) = x³ + 1. Then, write f as a set of ordered pairs.

Solution: Given, X = {–1, 0, 2, 5} f : X → R R: f(x) = x³ + 1
 f(x) = x³ + 1
∴ f(-1) = (-1) ³ + 1 = -1 + 1 = 0
∴ f(0) = (0) ³ + 1 = 0 + 1 = 1
f(2) = (2) ³ + 1 = 8 + 1 = 9
∴ f(5) = (5) ³ + 1 = 125 + 1 = 126
f as a set of ordered pairs:
    f = {(-1, 0), (0, 1), (2, 9), (5, 126)}

4. Let A = {–2, –1, 0, 1, 2} and f : A → Z: f(x) = x² – 2x – 3. Find f(A).

Solution: Given : A = {–2, –1, 0, 2} f : A → Z: f(x) = x² – 2x – 3
  f(x) = x² – 2x – 3
∴ f(-2) = (-2) ² – 2(-2) – 3 = 4 + 4 -3 = 5
∴ f(-1) = (-1) ² – 2(-1) – 3 = 1 + 2 -3 = 0
f(0) = (0) ² – 2(0) – 3 = 0 + 0 – 3 = -3
∴ f(1) = (1) ² – 2(1) – 3 = 1 – 2 – 3 = -4
∴ f(2) = (2) ² – 2(2) – 3 = 4 – 4 -3 = -3
   f(A) = {5, 0, -3, -4}

5. Let f : R → R : f(x) = x².determine (i) range (f) (ii) {x : f(x) = 4}

solution: Given: f(x) = x²
(i) Let y = f(x)
   ⇒ y = x²
 Square of any real number is always 0 or positive real number.
∴ The value of y ≥ 0.
Hence, Range(f) is [0, ∞).

(ii) Let y = f(x)
Given f(x) = 4
Therefore,
   x² = 4
∴ x = ± 2
⇒ x = 2 or x = -2
∴ x = {2,-2}

6. Let f : R → R : f(x) = x² + 1. Find f^-1{10}.

Solution: Given:f : R → R : f(x) = x² + 1
Let y = f(x)
⇒ y = x² + 1
⇒ x² = y – 1

Substituting x = 10,

7. Let f : R⁺ → R : f(x) = log_ex Find {x : f(x) = -2}.

Solution: Given, f : R⁺ → R : f(x) = log_ex
    f(x) = -2
⇒ log_ex = -2
⇒ x = e^-2
Hence, {x : f(x) = –2} = {e^-2}.

8. Let A = {6, 10, 11, 15, 21} and let f : A → N : f(n) is the highest prime factor of n.
Find range (f).

Solution: Given, A = {6, 10, 11, 15, 21}
f : A → N : f(n) is the highest prime factor of n.
∴ f(6) = 3;              f(10) = 5;
  f(11) = 11;           f(15) = 5;
  f(21) = 7
Hence range is { 3, 5, 7, 11}.

9. Find the range of the function f(x) = sin x.

Solution: Given f(x) = sin x.
We know,
    -1 ≤ sinx ≤ 1
⇒ -1 ≤ f(x) ≤ 1
Hence, Range is [-1, + 1].

10. Find the range of the function f(x) = |x|.

Solution: Given f(x) = |x|.
  |x| = -x when  x < 0
  |x| = x when x ≥ 0
∴The value of |x| is never a negative value.
Hence range of |x| is [0, ∞).

Where [x] is the Greatest Integer Function of x.
Now
x = [x] + {x}, Where {x} is a fractional part of x
⇒ x – [x] = {x}
∴ f(x) = √{x}
Ans: Hence domain of f(x) is R.

Ans: As the root value [0, 1) lies between [0,1).
Hence the range of f(x) is [0, 1).

Solution: Given f(x) = ^{x – 5}/_{5 – x}
f(x) is undefined if denominator becomes zero
.∴ x – 5 ≠ 0
⇒ x ≠ 5
Hence the domain is (−∞, 5) U (5,  ∞)
Since x ≠ 5
∴ Range is {-1}.

13.(i) Let f = {(1, 6), (2, 5), (4, 3), (5, 2), (8, –1), (10, –3)} and g = {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)}. Find dom (f + g)

Solution: Given, f = {(1, 6), (2, 5), (4, 3), (5, 2), (8, –1), (10, –3)}
g = {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)}
∴ Domain of f = {1,2,4,5,8,10}
Domain of g = {2,3,5,7,8,10}
Domain of (f + g)
  = {x : x ∈ D_f ∩ Dg }
 = {2, 5, 8, 10}.

13.(ii) Let f = {(1, 6), (2, 5), (4, 3), (5, 2), (8, –1), (10, –3)} and g = {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)}. Find dom (^f/_g)

Solution:  Given, f = {(1, 6), (2, 5), (4, 3), (5, 2), (8, –1), (10, –3)}
g = {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)}
∴ Domain of f = {1,2,4,5,8,10}
Domain of g = {2,3,5,7,8,10}
Domain of (^f/_g)
 = {x : x ∈ D_f ∩ Dg and g (x) ≠ 0}
Domain of (^f/_g)
 = {5, 8, 10}.

Solution: Given,f(x) = ^{x – 1}/_x
Replacing x by ¹/_x

Again Replacing x by f(¹/_x)

Solution: Given,f(x) = ^kx/_x+1
Replacing x by f(x)

⇒ k²x = x(kx + x + 1)
⇒ k²x = kx² + x² + x
or, 0.x² + k²x = x²(k + 1) + x
∴ k + 1 = 0
⇒ k = -1
∴ Value of k = -1

16. Find the range of the function, f(x) = ^x/_|x|

Solution: Given f(x) = ^x/_|x|
   If x < 0 then |x| = -x
∴ f(x) = ^x/_|x| = ^x/_-x  = -1
  If x ≥ 0 then |x| = x
∴ f(x) = ^x/_|x| = ^x/_x  = 1
Hence output values of f(x) only 1 and -1.
∴ The range of the function is {1,-1}.

17. Find the domain of the function, f(x) = log |x|.

Solution: log function is not defined for x = 0.
∴ Domain of f(x) is R – {0}

Substitute (x + ¹/_x) = x,
Hence f(x) = x² – 2

    f(x) is defined if bx – a ≠ 0
⇒ bx ≠ a
⇒ x  ≠ ^a/_b
∴  Domain of the function is (-∞, ^a/_b) U (^a/_b, ∞)

⇒ y(bx – a) = ax +b
⇒ bxy – ay = ax + b
or, bxy – ax = ay + b
⇒ x(by – a) = ay + b
⇒ x = ^{ay + b}/_{by – a}
x is not defined when denominator is zero.
∴ by – a ≠ 0
⇒ y ≠ ^a/_b
Range of the function is (-∞, ^a/_b) U (^a/_b, ∞)

f(x) is defind if
    f(x) ≥ 0
⇒ x – 1 ≥ 0
⇒ x ≥  1
∴  Domain of the function is [1, ∞).
Range of the function is [0, +∞).

Q. 21. Write the domain and the range of the function, f(x) = –|x|.

Solution: |x| is defined for all real values.
Hence
-|x| is always less than or equal to 0.
∴ The domain of the function is R i.e. (-∞, +∞)
   The range of the function is ( -∞, 0].