Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 2 - Relations And Functions

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Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 2 Relations And Functions are provided here with simple step-by-step explanations. These solutions for Relations And Functions are extremely popular among class 11 Science students for Maths Relations And Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 11 Science Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 27:

Question 1:

Let A = {–1, 2, 3} and B = {1, 3}. Determine
(i) A × B
(ii) B × A
(iii) B × B
(iv) A × A

Answer:

We have, A = {–1, 2, 3} and B = {1, 3}

(i) A × B = {–1, 2, 3} × {1, 3} = {(–1, 1), (–1, 3), (2, 1), (2, 3), (3, 1), (3, 3)}
(ii) B × A = {1, 3} × {–1, 2, 3} = {(1, –1), (1, 2), (1, 3), (3, –1), (3, 2), (3, 3)}
(iii) B × B = {1, 3} × {1, 3} = {(1, 1), (1, 3), (3, 1), (3, 3)}
(iv) A × A = {–1, 2, 3} × {–1, 2, 3} = {(–1, –1), (–1, 2), (–1, 3), (2, –1), (2, 2), (2, 3), (3, –1), (3, 2), (3, 3)}

Page No 28:

Question 2:

If P = {x : x < 3, x ∈ N}, Q = {x : x ≤ 2, x ∈ W}. Find (P â‹ƒ Q) × (P â‹‚ Q), where W is the set of whole numbers.

Answer:

We have, P = {x : x < 3, x ∈ N} and Q = {x : x ≤ 2, x ∈ W}

(P â‹ƒ Q) = {0, 1, 2} and (P â‹‚ Q) = {1, 2}
(P â‹ƒ Q) × (P â‹‚ Q) = {0, 1, 2} × {1, 2} = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}

Page No 28:

Question 3:

If A = {x : x ∈ W, x < 2} B = {x : x ∈ N, 1 < x < 5} C = {3, 5} find
(i) A × (B â‹‚ C)
(ii) A × (B â‹ƒ C)

Answer:

A = {0, 1}
B = {2, 3, 4}
C = {3, 5}
Now, (B â‹‚ C) = {3} and (B â‹ƒ C) = {2, 3, 4, 5}

(i) A × (B â‹‚ C) = {0, 1} × {3} = {(0, 3), (1, 3)}
(ii) A × (B â‹ƒ C) = {0, 1} × {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

Page No 28:

Question 4:

In each of the following cases, find a and b.

(i) (2a + b, a – b) = (8, 3)

(ii) $(\frac{a}{4}, a - 2 b) = (0, 6 + b)$

Answer:

Two ordered pairs can be equal iff their corresponding coordinates are equal, thus

(i) 2a + b = 8.....(1)
a – b = 3.....(2)
Solving (1) and (2), we get, $a = \frac{11}{3} and b = \frac{2}{3}$
(ii) $\frac{a}{4} = 0$ .....(1)
$a - 2 b = 6 + b$ .....(2)
Solving (1) and (2), we get, a = 0 and b = $-$ 2

Page No 28:

Question 5:

Given A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈ A, y ∈ A}. Find the ordered pairs which satisfy the conditions given below:
(i) x + y = 5
(ii) x + y < 5
(iii) x + y > 8

Answer:

(i) The set of ordered pairs satisfying x + y = 5 are {(1, 4), (2, 3), (3, 2), (4, 1)}
(ii) The set of ordered pairs satisfying x + y < 5 are {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}
(iii) The set of ordered pairs satisfying x + y > 8 are {(4, 5), (5, 4), (5, 5)}

Page No 28:

Question 6:

Given R = {(x, y) : x, y ∈ W, x² + y² = 25}. Find the domain and Range of R.

Answer:

The set of ordered pairs satisfying R is {(0, 5), (5, 0), (3, 4), (4, 3)}

Hence, the Domain of R = {0, 3, 4, 5} and Range of R = {0, 3, 4, 5}

Page No 28:

Question 7:

If R₁ = {(x, y) | y = 2x + 7, where x ∈ R and –5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R₁.

Answer:

We have, R₁ = {(x, y)|y = 2x + 7, where x∈ R and -5 ≤ x ≤ 5}
Domain of R₁ = {-5 ≤ x ≤ 5, x ∈ R} = [-5, 5]
x ∈ [-5, 5]
=> 2x ∈ [-10, 10]
=>2x + 7 ∈ [-3, 17]
Range is [-3, 17]

Page No 28:

Question 8:

If R₂ = {(x, y) | x and y are integers and x² + y² = 64} is a relation. Then find R₂.

Answer:

The set of ordered pairs satisfying the above relation is {(0, 8), (8, 0), (0, −8), (−8, 0)}

Hence, R₂= {(0, 8), (8, 0), (0,− 8), (−8, 0)}

Page No 28:

Question 9:

If R₃ = {(x, |x|) | x is a real number} is a relation. Then find domain and range of R₃.

Answer:

Given that: R₃ = {(x, |x|) | x is a real number}
The domain of R₃ = R.
As |x| is always positive, so the range will be R⁺
Range of R₃ = (0, ∞)

Page No 28:

Question 10:

Is the given relation a function? Give reasons for your answer.
(i) h = {(4, 6), (3, 9), (–11, 6), (3, 11)}
(ii) f = {(x, x) |x is a real number}
(iii) $g = n, \frac{1}{n}| n$ is a positive integer
(iv) s = {(n, n²) |n is a positive integer}
(v) t = {(x, 3) |x is a real number.

Answer:

(i) Given: h = {(4, 6), (3, 9), (–11, 6), (3, 11)}
The two ordered pairs (3, 9) and (3, 11) clearly show that the element 3 has two different images. So h is not a function.

(ii) Given: f = {(x, x) | x is a real number}
For every element, there will be a unique image that will be the same as the element. So f is a function.

(iii)Given: $g = {n, \frac{1}{n}| n$ is a positive integer}
For every element, there will be a unique image that will be equal to the reciprocal of the element. So g is a function.

(iv)Given: s = {(n, n²) | n is a positive integer}
The square of any number is a unique number. So s is a function.

(v)Given: t = {(x, 3) | x is a real number}
For all the different real values of x, the image will be constant 3. So t is a constant function.

Page No 28:

Question 11:

If f and g are real functions defined by f(x) = x² + 7 and g(x) = 3x + 5, find each of the following

(a) f(3) + g (– 5)

(b) $f (\frac{1}{2}) \times g (14)$

(c) f(–2) + g (–1)

(d) f(t) – f(–2)

(e) $\frac{f (t) - f (5)}{t - 5}, if t \neq 5$

Answer:

Given: f(x) = x² + 7 and g(x) = 3x + 5

$(a) f (3) = {(3)}^{2} + 7 = 9 + 7 = 16 g (- 5) = 3 (- 5) + 5 = - 15 + 5 = - 10 ∴ f (3) + g (- 5) = 16 + (- 10) = 6 (b) f (\frac{1}{2}) = {(\frac{1}{2})}^{2} + 7 = \frac{1}{4} + 7 = \frac{29}{4} g (14) = 3 (14) + 5 = 42 + 5 = 47 ∴ f (\frac{1}{2}) + g (14) = \frac{29}{4} + 47 = \frac{217}{4} (c) f (- 2) = {(- 2)}^{2} + 7 = 4 + 7 = 11 g (- 1) = 3 (- 1) + 5 = - 3 + 5 = 2 ∴ f (- 2) + g (- 1) = 11 + 2 = 13 (d) f (t) = {(t)}^{2} + 7 = t^{2} + 7 f (- 2) = {(- 2)}^{2} + 7 = 4 + 7 = 11 ∴ f (t) - f (- 2) = {(t)}^{2} + 7 - 11 = t^{2} - 4 (e) f (t) = {(t)}^{2} + 7 = t^{2} + 7 f (5) = {(5)}^{2} + 7 = 25 + 7 = 32 ∴ \frac{f (t) - f (5)}{t - 5} = \frac{t^{2} + 7 - 32}{t - 5} = \frac{t^{2} - 25}{t - 5}, t \neq 5 = t + 5$

Page No 29:

Question 12:

Let f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x – 7.
(a) For what real numbers x, f(x) = g(x)?
(b) For what real numbers x, f(x) < g(x)?

Answer:

Given: f(x) = 2x + 1 and g(x) = 4x – 7

$(a) f (x) = g (x) 2 x + 1 = 4 x - 7 - 2 x = - 8 x = 4 Hence, for x = 4 f (x) = g (x) (b) f (x) < g (x) 2 x + 1 < 4 x - 7 - 2 x < - 8 x > 4 Hence, for x > 4 f (x) > g (x)$

Page No 29:

Question 13:

If f and g are two real valued functions defined as f(x) = 2x + 1, g(x) = x² + 1, then find.

(i) f + g

(ii) f – g

(iii) fg

(iv) $\frac{f}{g}$

Answer:

Given: f(x) = 2x + 1, g(x) = x² + 1

$(i) f + g = f (x) + g (x) = (2 x + 1) + (x^{2} + 1) = x^{2} + 2 x + 1 (ii) f - g = f (x) - g (x) = (2 x + 1) - (x^{2} + 1) = - x^{2} + 2 x (iii) f g = f (x) \times g (x) = (2 x + 1) \times (x^{2} + 1) = 2 x^{3} + x^{2} + 2 x + 1 (iv) \frac{f}{g} = \frac{f (x)}{g (x)} = \frac{(2 x + 1)}{(x^{2} + 1)}$

Page No 29:

Question 14:

Express the following functions as set of ordered pairs and determine their range.
f : X → R, f(x) = x³ + 1, where X = {–1, 0, 3, 9, 7}

Answer:

Given: f(x) = x³ + 1

$When x = - 1, f (- 1) = {(- 1)}^{3} + 1 = - 1 + 1 = 0 When x = 0, f (0) = {(0)}^{3} + 1 = 0 + 1 = 1 When x = 3, f (3) = {(3)}^{3} + 1 = 27 + 1 = 28 When x = 9, f (9) = {(9)}^{3} + 1 = 729 + 1 = 730 When x = 7, f (7) = {(7)}^{3} + 1 = 343 + 1 = 344$

Hence, The set of ordered pairs satisfying f(x) are ${(- 1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}$

Range of f(x) is {0, 1, 28, 730, 344}

Page No 29:

Question 15:

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

Answer:

Given: f(x) = g(x)

$3 x^{2} - 1 = 3 + x \Rightarrow 3 x^{2} - x - 4 = 0 \Rightarrow 3 x^{2} + 3 x - 4 x - 4 = 0 \Rightarrow 3 x (x + 1) - 4 (x + 1) = 0 \Rightarrow (3 x - 4) (x + 1) = 0 \Rightarrow x = - 1 and x = \frac{4}{3}$

Page No 29:

Question 16:

Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g(x) = αx + β, then what values should be assigned to α and β?

Answer:

Given: g(x) = αx + β

The set of ordered pairs satisfying g is {(1. 1), (2, 3), (3, 5), (4, 7)}
Here each element has a unique image so g is a function.

$for (1, 1) g (1) = 1 \Rightarrow α (1) + β = 1 \Rightarrow α + β = 1 . . . . . (1) for (2, 3) g (2) = 1 \Rightarrow α (2) + β = 3 \Rightarrow 2 α + β = 3 . . . . . (2)$

Solving (1) and (2) we will get $α = 2 and β = - 1$

Page No 29:

Question 17:

Find the domain of each of the following functions given by

(i) $f (x) = \frac{1}{\sqrt{1 - \cos x}}$

(ii) $f (x) = \frac{1}{\sqrt{x + |x|}}$

(iii) $f (x) = x |x|$

(iv) $f (x) = \frac{x^{3} - x + 3}{x^{2} - 1}$

(v) $f (x) = \frac{3 x}{2 x - 8}$

Answer:

$(i) f (x) = \frac{1}{\sqrt{1 - \cos x}} For f (x) to be defined, 1 - \cos x > 0 1 > \cos x Range of \cos x is [- 1, 1] Here \cos x can take all the values except \cos x = 1 Now, \cos x = 1 when x = 2 n π, n \in Z Hence, the domain of f (x) is R - {2 n π, n \in Z}$

$(ii) f (x) = \frac{1}{\sqrt{x + |x|}} for f (x) to be defined, x + |x| > 0 The sum of a positive number and its modulus will always be positve . The sum of a negative number and its modulus will always be zero . Hence, the domain of f (x) i s R^{+} .$

$(iii) f (x) = x |x| it is clear that f (x) is defined for all x \in R Hence, the domain of f (x) is R .$

$(iv) f (x) = \frac{x^{3} - x + 3}{x^{2} - 1} for f (x) to be defined x^{2} - 1 \neq 0 x^{2} - 1 \neq 0 (x - 1) (x + 1) \neq 0 x \neq + 1, - 1 Hence, the domain of f (x) is R - {- 1, 1}$

$(v) f (x) = \frac{3 x}{2 x - 8} for f (x) to be defined 2 x - 8 \neq 0 2 x - 8 \neq 0 2 (x - 4) \neq 0 x \neq 4 Hence, the domain of f (x) is R - {4}$

Page No 29:

Question 18:

Find the range of the following functions given by

(i) $f (x) = \frac{3}{2 - x^{2}}$

(ii)

f (x) = 1 - |x - 2|

(iii)

f (x) = |x - 3|

(iv) f(x) = 1 + 3 cos2x

Answer:

(i) We have,

$f (x) = \frac{3}{2 - x^{2}} = y (let) \Rightarrow 2 - x^{2} = \frac{3}{y} \Rightarrow x^{2} = 2 - \frac{3}{y} Since, x^{2} \geq 0, 2 - \frac{3}{y} \geq 0 \Rightarrow \frac{2 y - 3}{y} \geq 0 \Rightarrow 2 y - 3 \geq 0 and y > 0 or 2 y - 3 \leq 0 and y < 0 \Rightarrow y \geq \frac{3}{2} or y < 0 \Rightarrow y \in (- \infty, 0) \cup [\frac{3}{2}, \infty) ∴ Range of f is (- \infty, 0) \cup [\frac{3}{2}, \infty)$

(ii) We have,

f (x) = 1 - |x - 2|

|x - 2| \geq 0 \Rightarrow - |x - 2| \leq 0 \Rightarrow 1 - |x - 2| \leq 1 ∴ Range of f is (- \infty, 1]

(iii) We have,

f (x) = |x - 3|

|x - 3| \geq 0 ∴ Range of f is [0, \infty)

(iv) We have, f(x) = 1 + 3 cos2x

$- 1 \leq \cos 2 x \leq 1 \Rightarrow - 3 \leq 3 \cos 2 x \leq 3 \Rightarrow - 2 \leq 1 + 3 \cos 2 x \leq 4 \Rightarrow - 2 \leq f (x) \leq 4 ∴ Range of f is [- 2, 4] .$

Page No 29:

Question 19:

Redefine the function f(x) = |x – 2| + |2 + x|, –3 ≤ x ≤ 3

Answer:

$As we know, |x| = \{\begin{cases} x, x \geq 0 \\ - x, x < 0 \end{cases} |x - 2| = \{\begin{cases} x - 2, x \geq 2 \\ - (x - 2), x < 2 \end{cases} |2 + x| = \{\begin{cases} 2 + x, x \geq - 2 \\ - (2 + x), x < - 2 \end{cases} Now, f (x) = |x - 2| + |2 + x|, - 3 \leq x \leq 3 f (x) = \{\begin{cases} |x - 2| + |2 + x|, - 3 \leq x \leq - 2 \\ \begin{array}{l} |x - 2| + |2 + x|, - 2 \leq x \leq 2 \\ |x - 2| + |2 + x|, 2 \leq x \leq 3 \end{array} \end{cases} f (x) = \{\begin{cases} - (x - 2) - (2 + x), - 3 \leq x \leq - 2 \\ \begin{array}{l} - (x - 2) + (2 + x), - 2 \leq x \leq 2 \\ (x - 2) + (2 + x), 2 \leq x \leq 3 \end{array} \end{cases} f (x) = \{\begin{cases} - 2 x, - 3 \leq x \leq - 2 \\ \begin{array}{l} 4, - 2 \leq x \leq 2 \\ 2 x, 2 \leq x \leq 3 \end{array} \end{cases}$

Page No 30:

Question 20:

If $f (x) = \frac{x - 1}{x + 1},$ then show that

(i) $f (\frac{1}{x}) = - f (x)$

(ii) $f (- \frac{1}{x}) = \frac{- 1}{f (x)}$

Answer:

Given: $f (x) = \frac{x - 1}{x + 1},$

$(i) f (\frac{1}{x}) = \frac{(\frac{1}{x}) - 1}{(\frac{1}{x}) + 1} f (\frac{1}{x}) = \frac{\frac{1 - x}{x}}{\frac{1 + x}{x}} f (\frac{1}{x}) = \frac{1 - x}{1 + x} f (\frac{1}{x}) = - \frac{(x - 1)}{1 + x} f (\frac{1}{x}) = - f (x)$

$(ii) f (- \frac{1}{x}) = \frac{(- \frac{1}{x}) - 1}{(- \frac{1}{x}) + 1} f (- \frac{1}{x}) = \frac{\frac{- 1 - x}{x}}{\frac{- 1 + x}{x}} f (- \frac{1}{x}) = \frac{- 1 - x}{- 1 + x} f (- \frac{1}{x}) = - \frac{(x + 1)}{x - 1} f (- \frac{1}{x}) = \frac{- 1}{f (x)}$

Page No 30:

Question 21:

Let $f (x) = \sqrt{x}$ and g(x) = x be two functions defined in the domain R⁺ â‹ƒ {0}. Find

(i) (f + g) (x)

(ii) (f – g) (x)

(iii) (fg) (x)

(iv) $(\frac{f}{g}) (x)$

Answer:

Given: $f (x) = \sqrt{x}$ and g(x) = x

$(i) (f + g) (x) = f (x) + g (x) = \sqrt{x} + x (ii) (f - g) (x) = f (x) - g (x) = \sqrt{x} - x (iii) (f g) (x) = f (x) \times g (x) = \sqrt{x} \times x = x \sqrt{x} (iv) (\frac{f}{g}) (x) = \frac{f (x)}{g (x)}, g (x) \neq 0 = \frac{\sqrt{x}}{x}, x \neq 0$

Page No 30:

Question 22:

Find the domain and Range of the function $f (x) = \frac{1}{\sqrt{x - 5}} .$

Answer:

$f (x) = \frac{1}{\sqrt{x - 5}} For f (x) to be defined, x - 5 > 0 \Rightarrow x > 5 Hence, the domain of f (x) is (5, \infty) For Range, let y = \frac{1}{\sqrt{x - 5}} y^{2} = \frac{1}{x - 5} y^{2} x - 5 y^{2} = 1 x = \frac{1}{y^{2}} + 5 for x \in (5, \infty), y \in R^{+} Hence, the range of f (x) is R^{+}$

Page No 30:

Question 23:

If $f (x) = y = \frac{a x - b}{c x - a},$ then prove that f(y) = x.

Answer:

$f (x) = y = \frac{a x - b}{c x - a} Now, y = \frac{ax - b}{cx - a} y (cx - a) = ax - b cyx - ay = ax - b cyx - ax = ay - b x (cy - a) = ay - b x = \frac{ay - b}{cy - a} = f (y)$

Page No 30:

Question 24:

Choose the correct answer
Let n (A) = m, and n (B) = n. Then the total number of non-empty relations that can be defined from A to B is
(A) mⁿ
(B) n^m– 1
(C) mn – 1
(D) 2^mn – 1

Answer:

n (A) = m, and n (B) = n
The total number of elements in $A \times B = n (A \times B) = m \times n = m n$
The total number of relations from A to B will be $2^{m n}$ , out of which one will be empty.
Hence, the total number of non-empty relations that can be defined from A to B is $2^{m n} - 1$ .
Hence, the correct answer is option (D)

Page No 30:

Question 25:

Choose the correct answer
If [x]² – 5 [x] + 6 = 0, where [ . ] denote the greatest integer function, then
(A) x ∈ [3, 4]
(B) x ∈ (2, 3]
(C) x ∈ [2, 3]
(D) x ∈ [2, 4)

Answer:

${[x]}^{2} - 5 [x] - 6 = 0 {[x]}^{2} - 3 [x] - 2 [x] - 6 = 0 [x] ([x] - 3) - 2 ([x] - 3) = 0 ([x] - 3) ([x] - 2) = 0 [x] = 3 or [x] = 2 \Rightarrow 3 \leq x < 4 or 2 \leq x < 3 \Rightarrow x \in [2, 4)$

Hence, the correct answer is option (D)

Page No 30:

Question 26:

Choose the correct answer

Range of $f (x) = \frac{1}{1 - 2 \cos x}$ is

(A) $[\frac{1}{3}, 1]$

(B) $[- 1, \frac{1}{3}]$

(C) $(- \infty, - 1] \cup [\frac{1}{3}, \infty)$

(D) $[- \frac{1}{3}, 1]$

Answer:

$We know that \cos x \in [- 1, 1] \Rightarrow - 1 \leq \cos x \leq 1 \Rightarrow 2 \geq - 2 \cos x \geq - 2 \Rightarrow - 2 \leq - 2 \cos x \leq 2 \Rightarrow + 1 - 2 \leq + 1 - 2 \cos x \leq + 1 + 2 \Rightarrow - 1 \leq 1 - 2 \cos x \leq 3 \Rightarrow \frac{1}{- 1} \leq \frac{1}{1 - 2 \cos x} \leq \frac{1}{3} Hence, the Range of f (x) is [- 1, \frac{1}{3}]$
Hence, the correct answer is option (B)

Page No 31:

Question 27:

Choose the correct answer
Let $f (x) = \sqrt{1 + x^{2}},$ then

(A) f(xy) = f(x) · f(y)

(B) f(xy) ≥ f(x) · f(y)

(C) f(xy) ≤ f(x) · f(y)

(D) None of these

Answer:

$f (x) = \sqrt{1 + x^{2}} f (y) = \sqrt{1 + y^{2}} f (x y) = \sqrt{1 + {(x y)}^{2}} = \sqrt{1 + x^{2} y^{2}} f (x) \times f (y) = \sqrt{1 + x^{2}} \times \sqrt{1 + y^{2}} f (x) \times f (y) = \sqrt{(1 + x^{2}) (1 + y^{2})} f (x) \times f (y) = \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} As x^{2} and y^{2} will always be positive 1 + x^{2} + y^{2} + x^{2} y^{2} \geq 1 + x^{2} y^{2} \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} \geq \sqrt{1 + x^{2} y^{2}} f (x) \times f (y) \geq f (x y) Hence, the correct answer is option (C)$

Page No 31:

Question 28:

Choose the correct answer

Domain of $\sqrt{a^{2} - x^{2}} (a > 0)$ is

(A) (– a, a)

(B) [– a, a]

(C) [0, a]

(D) (– a, 0]

Answer:

$a^{2} - x^{2} \geq 0 (a - x) (a + x) \geq 0 (x - a) (x + a) \leq 0 x \in [- a, a] Hence, the correct answer is option (B)$

Page No 31:

Question 29:

Choose the correct answer
If f(x) = ax + b, where a and b are integers, f(–1) = –5 and f(3) = 3, then a and b are equal to
(A) a = –3, b = –1
(B) a = 2, b = –3
(C) a = 0, b = 2
(D) a = 2, b = 3

Answer:

$f (x) = a x + b f (- 1) = 5 \Rightarrow a (- 1) + b = - 5 b - a = - 5 . . . . . (1) f (3) = 5 \Rightarrow a (3) + b = 3 3 a + b = 3 . . . . . (2) solving (1) and (2) we get, a = 2 and b = - 3 Hence, the correct answer is option (B) .$

Page No 31:

Question 30:

Choose the correct answer
The domain of the function f defined by $f (x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^{2} - 1}}$ is equal to
(A) (–∞, –1) â‹ƒ (1, 4]
(B) (–∞, –1] â‹ƒ (1, 4]
(C) (–∞, –1) â‹ƒ [1, 4]
(D) (–∞, –1) â‹ƒ [1, 4)

Answer:

$f (x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^{2} - 1}} For f (x) to be defined, 4 - x \geq 0 and x^{2} - 1 > 0 4 - x \geq 0 \Rightarrow 4 \geq x . . . . . (1) x^{2} - 1 > 0 \Rightarrow (x - 1) (x + 1) \geq 0 . . . . . (2) solving (1) and (2) we get x \in (- \infty, - 1) \cup (1, 4] Hence, the correct answer is option (A)$

Page No 31:

Question 31:

Choose the correct answer
The domain and range of the real function f defined by $f (x) = \frac{4 - x}{x - 4}$ is given by
(A) Domain = R, Range = {–1, 1}
(B) Domain = R – {1}, Range = R
(C) Domain = R – {4}, Range = {–1}
(D) Domain = R – {–4}, Range = {–1, 1}

Answer:

$f (x) = \frac{4 - x}{x - 4} for f (x) to be defined the denominator should not be equal to Zero . x - 4 \neq 0 x \neq 4 Hence, the domain of f (x) is R - {4} Option (C) is the correct answer .$

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Question 32:

Choose the correct answer
The domain and range of real function f defined by $f (x) = \sqrt{x - 1}$ is given by
(A) Domain = (1, ∞), Range = (0, ∞)
(B) Domain = [1, ∞), Range = (0, ∞)
(C) Domain = [1, ∞), Range = [0, ∞)
(D) Domain = [1, ∞), Range = [0, ∞)

Answer:

$f (x) = \sqrt{x - 1} For f (x) to be defined x - 1 \geq 0 x - 1 \geq 0 x \geq 1 Hence, the domain of f (x) is [1, \infty) for x \in [1, \infty), y \in [0, \infty) Hence, the correct answer is option (C)$

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Question 33:

Choose the correct answer
The domain of the function f given by $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6}$
(A) R – {3, –2}
(B) R – {–3, 2}
(C) R – [3, –2]
(D) R – (3, –2)

Answer:

$f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6} for f (x) to be defined the denominator should not be equal to Zero . x^{2} - x - 6 \neq 0 x^{2} - 3 x + 2 x - 6 \neq 0 x (x - 3) + 2 (x - 3) \neq 0 (x - 3) (x + 2) \neq 0 x \neq 3, - 2 Hence, the domain of f (x) is R - {3, - 2} Hence, the correct answer is Option (A) .$

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Question 34:

Choose the correct answer
The domain and range of the function f given by f(x) = 2 – |x – 5| is
(A) Domain = R⁺, Range = (–∞, 1]
(B) Domain = R, Range = (–∞, 2]
(C) Domain = R, Range = (–∞, 2)
(D) Domain = R⁺, Range = (–∞, 2]

Answer:

f(x) = 2 – |x – 5|

Modulus function is defined everywhere so the domain of f(x) will be R.

$N o w, |x| \geq 0 \Rightarrow |x - 5| \geq 0 - |x - 5| \leq 0 + 2 - |x - 5| \leq + 2 2 - |x - 5| \leq 2 Hence, the range of f (x) is (- \infty, 2], option (B) is the correct answer .$

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Question 35:

Choose the correct answer
The domain for which the functions defined by f(x) = 3x² – 1 and g(x) = 3 + x are equal is

(A) $\{- 1, \frac{4}{3}\}$

(B) $[- 1, \frac{4}{3}]$

(C) $(- 1, \frac{4}{3})$

(D) $[- 1, \frac{4}{3})$

Answer:

$f (x) = 3 x^{2} - 1 and g (x) = 3 + x f (x) = g (x) 3 x^{2} - 1 = 3 + x 3 x^{2} - x - 4 = 0 3 x^{2} - 4 x + 3 x - 4 = 0 x (3 x - 4) + 1 (3 x - 4) = 0 (3 x - 4) (x + 1) = 0 x = \{\frac{4}{3}, - 1\} Hence, the correct answer is option (A)$

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Question 36:

Fill in the blanks :
Let f and g be two real functions given by
f = {(0, 1), (2, 0), (3, –4), (4, 2), (5, 1)}
g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}
then the domain of f · g is given by _________.

Answer:

f = {(0, 1), (2, 0), (3, –4), (4, 2), (5, 1)}
Domain of f = {0, 2, 3, 4, 5}

g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}
Domain of g = {1, 2, 3, 4, 5}

Domain of f · g = (Domain of f) $\cap$ (Domain of g)
Domain of f · g = {2, 3, 4, 5}

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Question 37:

Let f = (2, 4), (5, 6), (8, – 1), (10, – 3)}

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

be two real functions. Then Match the following :

(a)	f – g	(i)	$\{(2, \frac{4}{5}), (8, \frac{- 1}{4}), (10, \frac{- 3}{13})\}$
(b)	f + g	(ii)	{(2, 20), (8, –4), (10, –39)}
(c)	f · g	(iii)	{(2, –1), (8, –5), (10, –16)}
(d)	$\frac{f}{g}$	(iv)	{(2, 9), (8, 3), (10, 10)}

Answer:

$f = (2, 4), (5, 6), (8, - 1), (10, - 3)} g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)} To perform algebric operations over two functions their domain must be same . Domain of f - g = {2, 8, 10} Domain of f + g = {2, 8, 10} Domain of f . g = {2, 8, 10} Domain of \frac{f}{g} = {2, 8, 10} f + g = f (x) + g (x) at x = 2, f (2) + g (2) = 4 + 5 = 9 at x = 8, f (8) + g (8) = - 1 + 4 = 3 at x = 10, f (10) + g (10) = - 3 + 13 = 10 The ordered pair satisfying f + g is {(2, 9), (8, 3), (10, 10)} f - g = f (x) - g (x) at x = 2, f (2) - g (2) = 4 - 5 = - 1 at x = 8, f (8) - g (8) = - 1 - 4 = - 5 at x = 10, f (10) - g (10) = - 3 - 13 = - 16 The ordered pair satisfying f + g is {(2, - 1), (8, - 5), (10, - 16)} f . g = f (x) \times g (x) at x = 2, f (2) \times g (2) = 4 \times 5 = 20 at x = 8, f (8) \times g (8) = - 1 \times 4 = - 4 at x = 10, f (10) \times g (10) = - 3 \times 13 = - 39 The ordered pair satisfying f . g is {(2, 20), (8, - 4), (10, - 39)} \frac{f}{g} = \frac{f (x)}{g (x)} at x = 2, \frac{f (2)}{g (2)} = \frac{4}{5} at x = 8, \frac{f (8)}{g (8)} = \frac{- 1}{4} at x = 10, \frac{f (10)}{g (10)} = \frac{- 3}{13} The ordered pair satisfying \frac{f}{g} is {(2, \frac{4}{5}), (8, \frac{- 1}{4}), (10, \frac{- 3}{13})}$
$Hence, the correct match is (a) \to (iii), (b) \to (iv), (c) \to (ii), (d) \to (i)$

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Question 38:

State whether the following statement is True or False.
The ordered pair (5, 2) belongs to the relation R = {(x, y) : y = x – 5, x, y ∈ Z}

Answer:

$The given relation is y = x - 5 If we use the given ordered pair (5, 2) \Rightarrow x = 5, y = 2 But, 2 \neq 5 - 5 2 \neq 0 Hence, the given statement is False$

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Question 39:

State whether the following statement is True or False.
If P = {1, 2}, then P × P × P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}

Answer:

P = {1, 2}
P × P = {1, 2}×{1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}
(P × P) × P = {(1, 1), (1, 2), (2, 1), (2, 2)}×{1, 2}
P × P × P ={(1, 1, 1), (1, 2, 1), (2, 1, 1), (2, 2, 1), (1, 1, 2), (1, 2, 2), (2, 1, 2), (2, 2, 2)}

Hence, the given statement is False.

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Question 40:

State whether the following statement is True or False.
If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, then (A × B) â‹ƒ (A × C)
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}.

Answer:

A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
A × B = {1, 2, 3} × {3, 4}
A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A × C = {1, 2, 3} × {4, 5, 6}
A × C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) â‹ƒ (A × C) = {(1, 3), (2, 3), (3, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)},

Hence, the given statement is True.

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Question 41:

State whether the following statement is True or False.

If $(x - 2, y + 5) = (- 2, \frac{1}{3})$ are two equal ordered pairs, then x = 4, $y = \frac{- 14}{3}$

Answer:

If two ordered pairs are equal then there corresponding elements are also equal.

$(x - 2, y + 5) = (- 2, \frac{1}{3}) x - 2 = - 2 \Rightarrow x = 0 y + 5 = \frac{1}{3} \Rightarrow y = \frac{- 14}{3}$

Hence, the given statement is False.

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Question 42:

State whether the following statement is True or False.
If A × B = {(a, x), (a, y), (b, x), (b, y)}, then A = {a, b}, B = {x, y}

Answer:

A = {a, b}, B = {x, y}
A × B = {a, b}×{x, y}
A × B = {(a, x), (a, y), (b, x), (b, y)}

Hence, the given statement is True.

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