RD Sharma XI 2020 Volume 1 Solutions for Class 11 Humanities Math Chapter 3 - Functions

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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
⇒ 4x² – x² – 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
$\Rightarrow x = - 1 o r x = \frac{2}{3}$
Hence, $x = - 1, \frac{2}{3}$ .

Page No 3.11:

Question 2:

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

Answer:

Given:
f (x) = (x – a)²(x – b)²
Thus,
f (a + b) = (a + b – a)²(a + b – b)²
= b²a²
Hence, f (a + b) = a²b² .

Page No 3.11:

Question 3:

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

Answer:

Given:
$f (x) = \frac{a x - b}{b x - a}$
Let y = f (x) .
⇒ y( bx $-$ a) = ax – b
⇒ xyb – ay = ax – b
⇒ xyb – ax = ay – b
⇒ x(by – a) = ay – b
$\Rightarrow x = \frac{a y - b}{b y - a}$
⇒ x = f (y)
Hence proved.

Page No 3.11:

Question 4:

If $f (x) = \frac{1}{1 - x}$ , show that f[f[f(x)]] = x.

Answer:

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

$= \frac{1}{1 - \frac{1}{1 - x}}$

$= \frac{1}{\frac{1 - x - 1}{1 - x}} = \frac{1 - x}{- x} = \frac{x - 1}{x}$
Again,
$f [f \{f (x)\}] = f [\frac{x - 1}{x}]$
$= \frac{1}{1 - (\frac{x - 1}{x})} = \frac{1}{\frac{x - x + 1}{x}} = \frac{x}{1} = x$

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

Page No 3.11:

Question 5:

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

Answer:

Given:
$f (x) = \frac{x + 1}{x - 1}$
Therefore,
$f [f \{(x)\}] = f (\frac{x + 1}{x - 1})$
$= \frac{(\frac{x + 1}{x - 1}) + 1}{(\frac{x + 1}{x - 1}) - 1}$
$= \frac{\frac{x + 1 + x - 1}{x - 1}}{\frac{x + 1 - x + 1}{x - 1}} = \frac{\frac{2 x}{x - 1}}{\frac{2}{x - 1}} = \frac{2 x}{2} = x$
Thus,
f [ f {(x)}] = x
Hence proved.

Page No 3.11:

Question 6:

If $f (x) = \{\begin{matrix} x^{2}, & when x < 0 \\ x, & when 0 \leq x < 1 \\ \frac{1}{x}, & when x \geq 1 \end{matrix}$
find: (a) f(1/2), (b) f(−2), (c) f(1), (d) $f (\sqrt{3})$ and (e) $f (\sqrt{- 3})$ .

Answer:

Given:
$f (x) = \{\begin{cases} x^{2}, & when x < 0 \\ x, & when 0 \leq x < 1 \\ \frac{1}{x}, & when x \geq 1 \end{cases}$
Now,
(a) $f (\frac{1}{2}) = \frac{1}{2}$ [ Using f (x) = x, 0 ≤ x < 1]
(b) f ( $-$ 2) = ( $-$ 2)² = 4
(c) $f (1) = \frac{1}{1} = 1$
(d) $f (\sqrt{3}) = \frac{1}{\sqrt{3}}$
(e) $f (\sqrt{- 3})$
Since x is not defined in R, $f (\sqrt{- 3})$ does not exist.

Page No 3.11:

Question 7:

If $f (x) = x^{3} - \frac{1}{x^{3}}$ , show that $f (x) + f (\frac{1}{x}) = 0 .$

Answer:

Given:
$f (x) = x^{3} - \frac{1}{x^{3}}$ ...(i)
Thus,
$f (\frac{1}{x}) = {(\frac{1}{x})}^{3} - \frac{1}{{(\frac{1}{x})}^{3}}$

$= \frac{1}{x^{3}} - \frac{1}{\frac{1}{x^{3}}}$

$∴ f (\frac{1}{x}) = \frac{1}{x^{3}} - x^{3}$ ...(ii)

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

$= x^{3} - \frac{1}{x^{3}} + \frac{1}{x^{3}} - x^{3} = 0$

Hence, $f (x) + f (\frac{1}{x}) = 0$ .

Page No 3.11:

Question 8:

If $f (x) = \frac{2 x}{1 + x^{2}}$ , show that f(tan θ) = sin 2θ.

Answer:

Given:
$f (x) = \frac{2 x}{1 + x^{2}}$
Thus,
$f (\tan θ) = \frac{2 (\tan θ)}{1 + \tan^{2} θ}$
$= \frac{2 \times \frac{\sin θ}{\cos θ}}{1 + (\frac{\sin^{2} θ}{\cos^{2} θ})} = \frac{2 \sin θ}{\cos θ} \times \frac{\cos^{2} θ}{\cos^{2} θ + \sin^{2} θ} = \frac{2 \sin θ \cos θ}{1} [∵ \cos^{2} θ + \sin^{2} θ = 1] = \sin 2 θ [∵ 2 \sin θ \cos θ = \sin 2 θ]$

Hence, f (tan θ) = sin 2θ.

Page No 3.12:

Question 9:

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

(i) Replacing x by $\frac{1}{x}$ in (1), we get
$f (\frac{1}{x}) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} = \frac{1 - x}{1 + x} = - \frac{x - 1}{x + 1} = - f (x)$

(ii) Replacing x by $- \frac{1}{x}$ in (1), we get
$f (- \frac{1}{x}) = \frac{- \frac{1}{x} - 1}{- \frac{1}{x} + 1} = \frac{- 1 - x}{- 1 + x} = - \frac{x + 1}{x - 1} = - \frac{1}{(\frac{x - 1}{x + 1})} = - \frac{1}{f (x)}$

Page No 3.12:

Question 10:

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

Answer:

Given:
f(x) = (a − xⁿ)^1/n, a > 0
Now,
f{ f (x)} = f (a − xⁿ)^1/n
             = [a – {(a – xⁿ)^1/n}ⁿ]^1/n
             = [ a – (a – xⁿ)]^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 $(\frac{1}{x}) = \frac{1}{x} - 5$ , where a ≠ b, then find f(x).

Answer:

Given:
$a f (x) + b f (\frac{1}{x}) = \frac{1}{x} - 5$               ...(i)
$\Rightarrow a f (\frac{1}{x}) + b f (x) = \frac{1}{\frac{1}{x}} - 5$
$\Rightarrow a f (\frac{1}{x}) + b f (x) = x - 5$         ...(ii)

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

$a f (x) + b f (x) + b f (\frac{1}{x}) + a f (\frac{1}{x}) = \frac{1}{x} - 5 + x - 5$
$\Rightarrow (a + b) f (x) + (a + b) f (\frac{1}{x}) = \frac{1}{x} + x - 10$
$\Rightarrow f (x) + f (\frac{1}{x}) = \frac{1}{(a + b)} [\frac{1}{x} + x - 10]$              ...(iii)

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

$a f (x) - b f (x) + b f (\frac{1}{x}) - a f (\frac{1}{x}) = \frac{1}{x} - 5 - x + 5$
$\Rightarrow (a - b) f (x) - f (\frac{1}{x}) (a - b) = \frac{1}{x} - x$
$\Rightarrow f (x) - f (\frac{1}{x}) = \frac{1}{(a - b)} [\frac{1}{x} - x]$                    ...(iv)

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

$2 f (x) = \frac{1}{a + b} [\frac{1}{x} + x - 10] + \frac{1}{a - b} [\frac{1}{x} - x]$
$\Rightarrow 2 f (x) = \frac{(a - b) [\frac{1}{x} + x - 10] + (a + b) [\frac{1}{x} - x]}{(a + b) (a - b)}$
$\Rightarrow 2 f (x) = \frac{\frac{a}{x} + a x - 10 a - \frac{b}{x} - b x + 10 b + \frac{a}{x} - a x + \frac{b}{x} - b x}{a^{2} - b^{2}}$
$\Rightarrow 2 f (x) = \frac{\frac{2 a}{x} - 10 a + 10 b - 2 b x}{a^{2} - b^{2}}$
$\Rightarrow f (x) = \frac{1}{a^{2} - b^{2}} \times \frac{1}{2} [\frac{2 a}{x} - 10 a + 10 b - 2 b x]$
            $= \frac{1}{a^{2} - b^{2}} [\frac{a}{x} - 5 a + 5 b - b x]$

Therefore,
$f (x) = \frac{1}{a^{2} - b^{2}} [\frac{a}{x} - b x - 5 a + 5 b]$
      $= \frac{1}{a^{2} - b^{2}} [\frac{a}{x} - b x] - \frac{5 (a - b)}{a^{2} - b^{2}}$
      $= \frac{1}{a^{2} - b^{2}} [\frac{a}{x} - b x] - \frac{5 (a - b)}{(a - b) (a + b)}$
     $= \frac{1}{a^{2} - b^{2}} [\frac{a}{x} - b x] - \frac{5}{(a + b)}$

Hence,
$f (x) = \frac{1}{a^{2} - b^{2}} [\frac{a}{x} - b x] - \frac{5}{(a + b)}$

Page No 3.18:

Question 1:

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

(i) $f (x) = \frac{1}{x}$
(ii) $f (x) = \frac{1}{x - 7}$
(iii) $f (x) = \frac{3 x - 2}{x + 1}$
(iv) $f (x) = \frac{2 x + 1}{x^{2} - 9}$
(v) $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12}$

Answer:

(i) Given: $f (x) = \frac{1}{x}$
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 $\frac{1}{0} .$
Hence, domain ( f ) = R $-$ { 0 }

(ii) Given: $f (x) = \frac{1}{(x - 7)}$
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 $\frac{1}{0} .$
Hence, domain ( f ) = R $-$ { 7 }.

(iii) Given: $f (x) = \frac{3 x - 2}{(x + 1)}$
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 $\frac{1}{0} .$
Hence, domain ( f ) = R $-$ { $-$ 1 }.

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

(v) Given: $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12}$

                     $= \frac{x^{2} + 2 x + 1}{x^{2} - 6 x - 2 x + 12}$

                     $= \frac{x^{2} + 2 x + 1}{x (x - 6) - 2 (x - 6)}$

                    $= \frac{x^{2} + 2 x + 1}{(x - 6) (x - 2)}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12}$ 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) $f (x) = \sqrt{x - 2}$
(ii) $f (x) = \frac{1}{\sqrt{x^{2} - 1}}$
(iii) $f (x) = \sqrt{9 - x^{2}}$
(iv) $f (x) = \frac{\sqrt{x - 2}}{3 - x}$

Answer:

(i) Given: $f (x) = \sqrt{x - 2}$
Clearly, f (x) assumes real values if x $-$ 2 ≥ 0.
⇒ x ≥ 2
⇒ x ∈ [2, ∞)
Hence, domain (f) = [2, ∞) .

(ii) Given: $f (x) = \frac{1}{\sqrt{x^{2} - 1}}$
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: $f (x) = \sqrt{9 - x^{2}}$
We observe that f (x) is defined for all satisfying
9 $-$ x² ≥ 0 .
⇒ x² $-$ 9 ≤ 0
⇒ (x + 3)(x $-$ 3) ≤ 0
⇒ $-$ 3 ≤ x ≤ 3
x ∈ [ $-$ 3, 3]
Hence, domain ( f ) = [ $-$ 3, 3]

(iv) Given: $f (x) = \sqrt{\frac{x - 2}{3 - x}}$
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) $f (x) = \frac{a x + b}{b x - a}$

(ii) $f (x) = \frac{a x - b}{c x - d}$

(iii) $f (x) = \sqrt{x - 1}$

(iv) $f (x) = \sqrt{x - 3}$

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

(vi) $f (x) = |x - 1|$

(vii) $f (x) = - |x|$

(viii) $f (x) = \sqrt{9 - x^{2}}$

(ix) $f (x) = \frac{1}{\sqrt{16 - x^{2}}}$

(x) $f (x) = \sqrt{x^{2} - 16}$

Answer:

(i)
Given:
$f (x) = \frac{a x + b}{b x - a}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{a x + b}{b x - a}$ is 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.
$\Rightarrow x = \frac{a}{b}$
Hence, domain ( f ) = $R - \{\frac{a}{b}\}$
Range of f :
Let f (x) = y
$\Rightarrow \frac{a x + b}{b x - a} = y$
⇒ (ax + b) = y (bx $-$ a)
⇒ (ax + b) = (bxy $-$ ay)
⇒ b + ay = bxy $-$ ax
⇒ b + ay = x(by $-$ a)
$\Rightarrow x = \frac{b + a y}{b y - a}$
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.
$\Rightarrow y = \frac{a}{b}$ .
Hence, range ( f ) = $R - \{\frac{a}{b}\}$
(ii)
Given:
$f (x) = \frac{a x - b}{c x - d}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{a x - b}{c x - d}$ is 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.
$\Rightarrow x = \frac{d}{c}$ .
Hence, domain ( f ) = $R - \{\frac{d}{c}\}$
Range of f :
Let f (x) = y
$\Rightarrow \frac{a x - b}{c x - d} = y$
⇒ (ax $-$ b) = y( cx $-$ d)
⇒ (ax $-$ b) = (cxy $-$ dy)
⇒ dy $-$ b = cxy $-$ ax
⇒ dy $-$ b = x(cy $-$ a)
$\Rightarrow x = \frac{d y - b}{c y - a}$
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.
$\Rightarrow y = \frac{a}{c}$ .
Hence, range ( f ) = $R - \{\frac{a}{c}\}$ .

(iii)
Given: $f (x) = \sqrt{x - 1}$
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
$\Rightarrow \sqrt{x - 1} \geq 0$
⇒ f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .

(iv)
Given: $f (x) = \sqrt{x - 3}$
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
$\Rightarrow \sqrt{x - 3} \geq 0$
⇒ f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .

(v)
Given:
$f (x) = \frac{x - 2}{2 - x}$
Domain ( f ) :
Clearly, f (x) is defined for all x satisfying: if 2 $-$ x ≠ 0 ⇒ x ≠ 2.
Hence, domain ( f ) = R $-$ {2}.
Range of f :
Let f (x) = y
⇒ $\frac{x - 2}{2 - x} = y$
⇒ 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 x ∈ R, (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 |, x ∈ R
We know that
$|x| = \{\begin{cases} x, & x \geq 0 \\ - x & x < 0 \end{cases}$
$∴ f (x) = - |x| = \{\begin{cases} - x, & x \geq 0 \\ x, & x < 0 \end{cases}$
Since f(x) is defined for x ∈ R, 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:
$f (x) = \sqrt{9 - x^{2}}$
$(9 - x^{2}) \geq 0 \Rightarrow 9 \geq x^{2} \Rightarrow x \in [- 3, 3]$

$\sqrt{9 - x^{2}}$ 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:
$f (x) = \frac{1}{\sqrt{16 - x^{2}}}$

$(16 - x^{2}) > 0 \Rightarrow 16 > x^{2} \Rightarrow x \in (- 4, 4)$

$\frac{1}{\sqrt{16 - x^{2}}}$ 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
$\Rightarrow \frac{1}{\sqrt{16 - x^{2}}} = y \Rightarrow \frac{1}{16 - x^{2}} = y^{2} \Rightarrow \frac{1}{y^{2}} = 16 - x^{2} \Rightarrow x^{2} = 16 - \frac{1}{y^{2}} Since, - 4 < x < 4 \Rightarrow 0 \leq x^{2} < 16 \Rightarrow 0 \leq 16 - \frac{1}{y^{2}} < 16 \Rightarrow - 16 \leq - \frac{1}{y^{2}} < 0 \Rightarrow 16 \geq \frac{1}{y^{2}} > 0 \Rightarrow \frac{1}{16} \leq y^{2} < \infty \Rightarrow \frac{1}{4} \leq y < \infty (∵ y \geq 0)$

Hence, range ( f ) = [ $\frac{1}{4}, \infty$ ).

(x) Given:
$f (x) = \sqrt{x^{2} - 16}$

$(x^{2} - 16) \geq 0 \Rightarrow x^{2} \geq 16 \Rightarrow x \in (- \infty, - 4] \cup [4, \infty)$

$\sqrt{x^{2} - 16}$ 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
$\Rightarrow \sqrt{x^{2} - 16} \geq 0$
⇒ f (x) ≥ 0

For x ≤ – 4, we have:
x² $-$ 16 ≥ 0
$\Rightarrow \sqrt{x^{2} - 16} \geq 0$
⇒ 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, f − g, cf (c ∈ R, c ≠ 0), fg, $\frac{1}{f} and \frac{f}{g}$ in each of the following:

(a) If f(x) = x³ + 1 and g(x) = x + 1
(b) If $f (x) = \sqrt{x - 1}$ and $g (x) = \sqrt{x + 1}$

Answer:

(a) Given:
f (x) = x³ + 1 and g (x) = x + 1
Thus,
(f + g) (x) : R → R is given by (f + g) (x) = f (x) + g (x) = x³ + 1 + x + 1 = x³ + x + 2.
(f $-$ g) (x) : R → R is given by (f $-$ g) (x) = f (x) $-$ g (x) = (x³ + 1) $-$ (x + 1 ) = x³ + 1 $-$ x $-$ 1 = x³ $-$ x.
cf : R → R is given by (cf) (x) = c(x³ + 1).
(fg) (x) : R → R 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).
$\frac{1}{f} : R - \{- 1\} \to R is given by (\frac{1}{f}) (x) = \frac{1}{f (x)} = \frac{1}{(x^{3} + 1)} .$
$\frac{f}{g} : R - \{- 1\} \to R is given by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{(x^{3} + 1)}{(x + 1)} = \frac{(x + 1) (x^{2} - x + 1)}{(x + 1)} = (x^{2} - x + 1) .$

Note that : (x³ + 1) = (x + 1) (x² $-$ x + 1)]

(b) Given:
$f (x) = \sqrt{x - 1}$ and $g (x) = \sqrt{x + 1}$
Thus,
(f + g) ) : [1, ∞) → R is defined by (f + g) (x) = f (x) + g (x) = $\sqrt{x - 1} + \sqrt{x + 1}$ .
(f   $-$ g) ) : [1, ∞) → R is defined by (f $-$ g) (x) = f (x) $-$ g (x) = $\sqrt{x - 1} - \sqrt{x + 1}$ .
cf : [1, ∞) → R is defined by (cf) (x) = $c \sqrt{x - 1}$ .
(fg) : [1, ∞) → R is defined by (fg) (x) = f(x).g(x) = $\sqrt{x - 1} \times \sqrt{x + 1} = \sqrt{x^{2} - 1}$ .
$\frac{1}{f} : (1, \infty) \to R is defined by (\frac{1}{f}) (x) = \frac{1}{f (x)} = \frac{1}{\sqrt{x - 1}} .$
$\frac{f}{g} : [1, \infty) \to R is defined by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt{x - 1}}{\sqrt{x + 1}} = \sqrt{\frac{x - 1}{x + 1}} .$

Page No 3.38:

Question 2:

Let f(x) = 2x + 5 and g(x) = x² + x. Describe (i) f + g (ii) f − g (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.
$∴ D (f) \cap D (g) = R$ .

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

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

(iii) (fg) : R → R 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,
$\frac{f}{g} : R - \{- 1, 0\} \to R is given by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{2 x + 5}{x^{2} + x}$ .
Hence, $d omain (\frac{f}{g}) = R - \{- 1, 0\}$ .

Page No 3.38:

Question 3:

If f(x) be defined on [−2, 2] and is given by $f (x) = \{\begin{cases} - 1, & - 2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{cases}$ and g(x) $= f (|x|) + |f (x)|$ , find g(x).

Answer:

Given:
$f (x) = \{\begin{cases} - 1, & - 2 ⩽ x ⩽ 0 \\ x - 1, & 0 < x ⩽ 2 \end{cases}$
Thus,
$g (x) = f (|x|) + |f (x)|$
$= \{\begin{cases} x - 1 + 1, & - 2 ⩽ x ⩽ 0 \\ x - 1 + (- x + 1), & 0 < x < 1 \\ x - 1 + x - 1, & 1 \leq x \leq 2 \end{cases} = \{\begin{cases} x, & - 2 ⩽ x ⩽ 0 \\ 0, & 0 < x < 1 \\ 2 x - 2, & 1 \leq x \leq 2 \end{cases}$

Page No 3.38:

Question 4:

Let f and g be two real functions defined by $f (x) = \sqrt{x + 1}$ and $g (x) = \sqrt{9 - x^{2}}$ . Then, describe each of the following functions:
(i) f + g
(ii) g − f
(iii) f g
(iv) $\frac{f}{g}$
(v) $\frac{g}{f}$
(vi) $2 f - \sqrt{5} g$
(vii) f² + 7f
(viii) $\frac{5}{8}$

Answer:

Given:
$f (x) = \sqrt{x + 1} and g (x) = \sqrt{9 - x^{2}}$
Clearly, $f (x) = \sqrt{x + 1}$ is defined for all x ≥ $-$ 1.
Thus, domain (f) = [ $-$ 1, ∞]
Again,

$g (x) = \sqrt{9 - x^{2}}$ is defined for
9 $-$ x² ≥ 0 ⇒ x² $-$ 9 ≤ 0
⇒ x² $-$ 3² ≤ 0
⇒ (x + 3)(x $-$ 3) ≤ 0
⇒ $x \in [- 3, 3]$
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) = $\sqrt{x + 1} + \sqrt{9 - x^{2}}$ .

(ii) ( g $-$ f ) : [ $-$ 1 , 3] → R is given by ( g $-$ f ) (x) = g (x) $-$ f (x) = $\sqrt{9 - x^{2}} - \sqrt{x + 1}$ .

(iii) (fg) : [ $-$ 1 , 3] → R is given by (fg) (x) = f(x).g(x) = $\sqrt{x + 1} . \sqrt{9 - x^{2}} = \sqrt{(x + 1) (9 - x^{2})} = \sqrt{9 + 9 x - x^{2} - x^{3}}$ .

(iv) $\frac{f}{g} : [- 1, 3] \to R is given by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt{x + 1}}{\sqrt{9 - x^{2}}} = \sqrt{\frac{x + 1}{9 - x^{2}}}$ .

(v) $\frac{g}{f} : [- 1, 3] \to R is given by (\frac{g}{f}) (x) = \frac{g (x)}{f (x)} = \frac{\sqrt{9 - x^{2}}}{\sqrt{x + 1}} = \sqrt{\frac{9 - x^{2}}{x + 1}}$ .

(vi) $2 f - \sqrt{5} g : [- 1, 3] \to R is given by (2 f - \sqrt{5} g) (x) = 2 \sqrt{x + 1} - \sqrt{5} (\sqrt{9 - x^{2}})$
                                                                              $= 2 \sqrt{x + 1} - \sqrt{45 - 5 x^{2}}$ .

(vii) $f^{2} + 7 f : [- 1, \infty] \to R is given by (f^{2} + 7 f) (x) = f^{2} (x) + 7 f (x)$            {Since domain(f) = [ $-$ 1, ∞]}
                                                                       $= {(\sqrt{x + 1})}^{2} + 7 (\sqrt{x + 1}) = x + 1 + 7 \sqrt{x + 1}$

(viii) $\frac{5}{g} : [- 3, 3] \to R is defined by (\frac{5}{g}) (x) = \frac{5}{\sqrt{9 - x^{2}}} .$            {Since domain(g) = [ $-$ 3, 3]}

Page No 3.38:

Question 5:

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

(i) f + g
(ii) fg
(iii) $\frac{f}{g}$
(iv) $\frac{g}{f}$
Also, find (f + g) (−1), (fg) (0), $(\frac{f}{g}) (\frac{1}{2}), (\frac{g}{f}) (\frac{1}{2})$ .

Answer:

Given:
f(x) = log_e (1 − x) and g(x) = [x]
Clearly, f(x) = log_e (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) = log_e (1 − x) + [ x ].

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

(iii) Given:
      g(x) = [ x ]
   If [ x ] = 0,
x ∈ (0, 1)
Thus,
$domain (\frac{f}{g}) = domain (f) \cap domain (g) - \{x : g (x) = 0\}$
$\frac{f}{g} : (- \infty, 0) \to R is defined by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\log_{e} (1 - x)}{[x]} .$

(iv) Given:
f(x) = log_e (1 − x)
$\Rightarrow \frac{1}{f (x)} = \frac{1}{\log_{e} (1 - x)}$
$\frac{1}{f (x)}$ is defined if log_e( 1 $-$ x) is defined and log_e(1 – x) ≠ 0.
⇒ (1 $-$ x) > 0 and (1 $-$ x) ≠ 0
⇒ x < 1 and x ≠ 0
⇒ x ∈ ( $-$ ∞, 0)∪ (0, 1)
Thus, $d omain (\frac{g}{f}) = (- \infty, 0) \cup (0, 1)$ = ( $-$ ∞, 1) .
$\frac{g}{f} : (- \infty, 0) \cup (0, 1) \to R defined by (\frac{g}{f}) (x) = \frac{g (x)}{f (x)} = \frac{[x]}{\log_{e} (1 - x)}$

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

(fg)(0) = log_e ( 1 – 0) × [0] = 0
$(\frac{f}{g}) (\frac{1}{2}) = does not exist .$
$(\frac{g}{f}) (\frac{1}{2}) = \frac{[\frac{1}{2}]}{\log_{e} (1 - \frac{1}{2})} = 0$

Page No 3.38:

Question 6:

If f, g and h are real functions defined by $f (x) = \sqrt{x + 1}, g (x) = \frac{1}{x}$ and h(x) = 2x² − 3, find the values of (2f + g − h) (1) and (2f + g − h) (0).

Answer:

Given:
$f (x) = \sqrt{x + 1}, g (x) = \frac{1}{x} and h (x) = 2 x^{3} - 3$
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 + g – h)(x) = 2f (x) + g (x) $-$ h (x)
$= 2 \sqrt{x + 1} + \frac{1}{x} - 2 x^{2} + 3$
$(2 f + g - h) (1) = 2 \sqrt{2} + 1 - 2 + 3 = 2 \sqrt{2} + 4 - 2 = 2 \sqrt{2} + 2$
(2f + g – h) (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 $f (x) = \{\begin{matrix} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{matrix}$ . 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, f − g and $\frac{f}{g}$ .

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

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}, g (x) \neq 0, x \in R$

$\Rightarrow (\frac{f}{g}) (x) = \frac{x + 1}{2 x - 3}, 2 x - 3 \neq 0 o r 2 x \neq 3$

$\Rightarrow (\frac{f}{g}) (x) = \frac{x + 1}{2 x - 3}, x \neq \frac{3}{2}$

Page No 3.38:

Question 9:

Let f : [0, ∞) → R and g : R → R be defined by $f (x) = \sqrt{x}$ and g(x) = x. Find f + g, f − g, fg and $\frac{f}{g}$ .

Answer:

It is given that f : [0, ∞) → R and g : R → R such that $f (x) = \sqrt{x}$ and g(x) = x .
$D (f + g) = [0, \infty) \cap R = [0, \infty)$
So, f + g : [0, ∞) → R is given by
$(f + g) (x) = f (x) + g (x) = \sqrt{x} + x$

$D (f - g) = D (f) \cap D (g) = [0, \infty) \cap R = [0, \infty)$
So, f - g : [0, ∞) → R is given by
$(f - g) (x) = f (x) - g (x) = \sqrt{x} - x$

$D (f g) = D (f) \cap D (g) = [0, \infty) \cap R = [0, \infty)$
So, fg : [0, ∞) → R is given by
$(f g) (x) = f (x) g (x) = \sqrt{x} . x = x^{\frac{3}{2}}$

$D (\frac{f}{g}) = [D (f) \cap D (g) - \{x : g (x) = 0\}] = (0, \infty)$
So, $\frac{f}{g} : (0, \infty) \to R$ is given by
$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$

Page No 3.38:

Question 10:

Let f(x) = x² and g(x) = 2x+ 1 be two real functions. Find (f + g) (x), (f − g) (x), (fg) (x) and $(\frac{f}{g}) (x)$ .

Answer:

Given:
f (x) = x² and g (x) = 2x + 1
Clearly, D (f) = R and D (g) = R
$∴ D (f \pm g) = D (f) \cap D (g) = R \cap R = R$
$D (f g) = D (f) \cap D (g) = R \cap R = R$
$D (\frac{f}{g}) = D (f) \cap D (g) - \{x : g (x) = 0\} = R \cap R - \{- \frac{1}{2}\} = R - \{- \frac{1}{2}\}$
Thus,
(f + g) (x) : R → R is given by (f + g) (x) = f (x) + g (x) = x² + 2x + 1= (x + 1)² .
(f $-$ g) (x) : R → R is given by (f $-$ g) (x) = f (x) $-$ g (x) = x² $-$ 2x $-$ 1.
(fg) (x) : R → R is given by (fg) (x) = f(x).g(x) = x²(2x + 1) = 2x³ + x² .

$(\frac{f}{g}) : R - \{- \frac{1}{2}\} \to R is given by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{x^{2}}{2 x + 1}$ .

Page No 3.41:

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 3.41:

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) Ï•

Answer:

(c) {−3, 3}
If f : A → B, such that y ∈ B, then $f^{- 1}$ { y }={x ∈ A: 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 3.41:

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}

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

Page No 3.41:

Question 4:

If f(x) = cos (log x), then the value of f(x²) f(y²) − $\frac{1}{2} \{f (\frac{x^{2}}{y^{2}}) + f (x^{2} y^{2})\}$ is
(a) −2
(b) −1
(c) 1/2
(d) None of these

Answer:

(d) None of these

Given:
$f (x) = \cos (\log x)$
$\Rightarrow f (x^{2}) = \cos (\log (x^{2})) \Rightarrow f (x^{2}) = \cos (2 \log (x))$

Similarly,
$f (y^{2}) = \cos (2 \log (y))$

Now,
$f (\frac{x^{2}}{y^{2}}) = \cos (\log (\frac{x^{2}}{y^{2}})) = \cos (\log x^{2} - \log y^{2})$
and
$f (x^{2} y^{2}) = \cos (\log x^{2} y^{2}) = \cos (\log x^{2} + \log y^{2})$

$\Rightarrow f (\frac{x^{2}}{y^{2}}) + f (x^{2} y^{2}) = \cos ((2 \log x - 2 \log y)) + \cos ((2 \log x + 2 \log y)) \Rightarrow f (\frac{x^{2}}{y^{2}}) + f (x^{2} y^{2}) = 2 \cos (2 \log x) \cos (2 \log y) \Rightarrow \frac{1}{2} [f (\frac{x^{2}}{y^{2}}) + f (x^{2} y^{2})] = \cos (2 \log x) \cos (2 \log y)$
$\Rightarrow f (x^{2}) f (y^{2}) - \frac{1}{2} \{f (x^{2} y^{2}) + f (\frac{x^{2}}{y^{2}})\} = \cos (2 \log x) \cos (2 \log y) - \cos (2 \log x) \cos (2 \log y) = 0$

Page No 3.42:

Question 5:

If f(x) = cos (log x), then the value of f(x) f(y) − $\frac{1}{2} \{f (\frac{x}{y}) + f (x y)\}$ is
(a) −1
(b) 1/2
(c) −2
(d) None of these

Answer:

(d) None of these

Given:
$f (x) = \cos (\log x)$
∴ $f (y) = \cos (\log y)$

Now,
$f (\frac{x}{y}) = \cos (\log (\frac{x}{y})) = \cos (\log x - \log y)$
and
$f (x y) = \cos (\log x y) = \cos (\log x + \log y)$

$\Rightarrow f (\frac{x}{y}) + f (x y) = \cos (\log x - \log y) + \cos (\log x + \log y) \Rightarrow f (\frac{x}{y}) + f (x y) = 2 \cos (\log x) \cos (\log y) \Rightarrow \frac{1}{2} [f (\frac{x}{y}) + f (x y)] = \cos (\log x) \cos (\log y)$
$\Rightarrow f (x) f (y) - \frac{1}{2} \{f (x y) + f (\frac{x}{y})\} = \cos (\log x) \cos (\log y) - \cos (\log x) \cos (\log y) = 0$

Page No 3.42:

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

$f (x) = |x - 1| Since, |x^{2} - 1| \neq |x - 1|^{2}, f (x^{2}) \neq (f (x))^{2} Thus, (i) is wrong . Since, |x + y - 1| \neq |x - 1| |y - 1|, f (x + y) \neq f (x) f (y) Thus, (ii) is wrong . Since ||x| - 1| \neq ||x - 1|| = |x - 1|, f (|x|) \neq |f (x)| Thus, (iii) is wrong . Hence, none of the given options is the answer.$

Page No 3.42:

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 $\frac{- π}{2} < x < \frac{π}{2}$ ,
$- \frac{π}{2} < x < \frac{π}{2} \Rightarrow - 1.57 < x < 1.57 \Rightarrow [x] \in {- 1, 0, 1, 2} Thus, \cos [x] = {\cos (- 1), \cos 0, \cos 1, \cos 2} Range of f (x) = {\cos 1, 1, \cos 2}$

Page No 3.42:

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) : x² − y² = 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 3.42:

Question 9:

If $f (x) = \log (\frac{1 + x}{1 - x}) and g (x) = \frac{3 x + x^{3}}{1 + 3 x^{2}}$ , then f(g(x)) is equal to
(a) f(3x)
(b) {f(x)}³
(c) 3f(x)
(d) −f(x)

Answer:

(c) 3f(x)

$f (x) = \log (\frac{1 + x}{1 - x}) and g (x) = \frac{3 x + x^{3}}{1 + 3 x^{2}}$
$Now, \frac{1 + g (x)}{1 - g (x)} = \frac{1 + \frac{3 x + x^{3}}{1 + 3 x^{2}}}{1 - \frac{3 x + x^{3}}{1 + 3 x^{2}}} = \frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}} = \frac{(1 + x)^{3}}{(1 - x)^{3}} Then, f (g (x)) = \log (\frac{1 + g (x)}{1 - g (x)}) = \log {(\frac{1 + x}{1 - x})}^{3} = 3 \log (\frac{1 + x}{1 - x}) = 3 f (x))$

Page No 3.42:

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 3.42:

Question 11:

If $f (x) = \log (\frac{1 + x}{1 - x})$ , then $f (\frac{2 x}{1 + x^{2}})$ is equal to
(a) {f(x)}²
(b) {f(x)}³
(c) 2f(x)
(d) 3f(x)

Answer:

(c) 2f(x)

$f (x) = \log (\frac{1 + x}{1 - x})$
$Then, f (\frac{2 x}{1 + x^{2}}) = \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}}) = \log (\frac{\frac{1 + x^{2} + 2 x}{1 + x^{2}}}{\frac{1 + x^{2} - 2 x}{1 + x^{2}}}) = \log (\frac{(1 + x)^{2}}{(1 - x)^{2}}) = 2 \log (\frac{1 + x}{1 - x}) = 2 (f (x))$

Page No 3.42:

Question 12:

If f(x) = cos (log x), then value of $f (x) f (4) - \frac{1}{2} \{f (\frac{x}{4}) + f (4 x)\}$ is
(a) 1
(b) −1
(c) 0
(d) ±1

Answer:

(c) 0

Given : f(x) = cos (log x)
Then, $f (x) f (4) - \frac{1}{2} \{f (\frac{x}{4}) + f (4 x)\}$
$= \cos (\log x) \cos (\log 4) - \frac{1}{2} \{\cos (\log \frac{x}{4}) + \cos (\log 4 x)\} = \frac{1}{2} [\cos (\log x + \log 4) + \cos (\log x - \log 4)] - \frac{1}{2} \{\cos (\log \frac{x}{4}) + \cos (\log 4 x)\} = \frac{1}{2} \{\cos (\log 4 x) + \cos (\log \frac{x}{4}) - \cos (\log \frac{x}{4}) - \cos (\log 4 x)\} = \frac{1}{2} \times 0 = 0$

Page No 3.42:

Question 13:

If $f (x) = \frac{2^{x} + 2^{- x}}{2}$ , then f(x + y) f(x − y) is equal to
(a) $\frac{1}{2} [f (2 x) + f (2 y)]$
(b) $\frac{1}{2} [f (2 x) - f (2 y)]$
(c) $\frac{1}{4} [f (2 x) + f (2 y)]$
(d) $\frac{1}{4} [f (2 x) - f (2 y)]$

Answer:

(a) $\frac{1}{2} [f (2 x) + f (2 y)]$

Given:
$f (x) = \frac{2^{x} + 2^{- x}}{2}$
Now,
f(x + y) f(x − y) = $(\frac{2^{x + y} + 2^{- x - y}}{2}) (\frac{2^{x - y} + 2^{- x + y}}{2})$
⇒ f(x + y) f(x − y) = $\frac{1}{4} (2^{2 x} + 2^{- 2 y} + 2^{2 y} + 2^{- 2 x})$
⇒ f(x + y) f(x − y) = $\frac{1}{2} (\frac{2^{2 x} + 2^{- 2 x}}{2} + \frac{2^{2 y} + 2^{- 2 y}}{2})$
⇒ f(x + y) f(x − y) = $\frac{1}{2} [f (2 x) + f (2 y)]$

Page No 3.42:

Question 14:

If 2f (x) − $3 f (\frac{1}{x}) = x^{2}$ (x ≠ 0), then f(2) is equal to

(a) $- \frac{7}{4}$
(b) $\frac{5}{2}$
(c) −1
(d) None of these

Answer:

(a) $- \frac{7}{4}$

2f (x) − $3 f (\frac{1}{x}) = x^{2}$ (x ≠ 0) ....(1)
$Replacing x by \frac{1}{x} : 2 f (\frac{1}{x}) - 3 f (x) = \frac{1}{x^{2}} . . . (2) Solving equations (1) & (2) - 5 f (x) = \frac{3}{x^{2}} + 2 x^{2} \Rightarrow f (x) = \frac{- 1}{5} (\frac{3}{x^{2}} + 2 x^{2}) Thus, f (2) = \frac{- 1}{5} (\frac{3}{4} + 2 \times 4) = \frac{- 1}{5} (\frac{3 + 32}{4}) = \frac{- 7}{4}$

Page No 3.42:

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)
$= 2 (2 x) + 2 |x| + (- 2 x) + |- x| - 2 x + |x| = 4 x - 2 x - 2 x + 2 |x| + |- x| - |x| = 0 + 2 |x| + |x| - |x| = 2 |x|$

Page No 3.42:

Question 16:

The range of the function $f (x) = \frac{x^{2} - x}{x^{2} + 2 x}$ is
(a) R
(b) R − {1}
(c) R − {−1/2, 1}
(d) None of these

Answer:

(c) R − {-1/2,1}

$f (x) = \frac{x^{2} - x}{x^{2} + 2 x}$
$Let y = \frac{x^{2} - x}{x^{2} + 2 x} [Also, x \neq 0] \Rightarrow y = \frac{x (x - 1)}{x (x + 2)} \Rightarrow y = \frac{(x - 1)}{(x + 2)} \Rightarrow x y + 2 y = x - 1 \Rightarrow x = \frac{2 y + 1}{1 - y} Here, 1 - y \neq 0 . or, y \neq 1 . Also, x \neq 0 \Rightarrow \frac{2 y + 1}{1 - y} \neq 0 \Rightarrow y \neq - \frac{1}{2} Thus, range (f) = R - {- \frac{1}{2}, 1} .$

Page No 3.42:

Question 17:

If x ≠ 1 and $f (x) = \frac{x + 1}{x - 1}$ is a real function, then f(f(f(2))) is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

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

Page No 3.42:

Question 18:

If f(x) = cos (log_ex), then $f (\frac{1}{x}) f (\frac{1}{y}) - \frac{1}{2} \{f (x y) + f (\frac{x}{y})\}$ is equal to
(a) cos (x − y)
(b) log (cos (x − y))
(c) 1
(d) cos (x + y)

Answer:

Given:
$f (x) = \cos (\log_{e} x)$
$\Rightarrow f (\frac{1}{x}) = \cos (\log_{e} (\frac{1}{x})) \Rightarrow f (\frac{1}{x}) = \cos (- \log_{e} (x)) \Rightarrow f (\frac{1}{x}) = \cos (\log_{e} (x))$

Similarly,
$f (\frac{1}{y}) = \cos (\log_{e} y)$

Now,
$f (x y) = \cos (\log_{e} x y) = \cos (\log_{e} x + \log_{e} y)$
and
$f (\frac{x}{y}) = \cos (\log_{e} \frac{x}{y}) = \cos (\log_{e} x - \log_{e} y)$

$\Rightarrow f (\frac{x}{y}) + f (x y) = \cos (\log_{e} x - \log_{e} y) + \cos (\log_{e} x + \log_{e} y) \Rightarrow f (\frac{x}{y}) + f (x y) = 2 \cos (\log_{e} x) \cos (\log_{e} y) \Rightarrow \frac{1}{2} [f (\frac{x}{y}) + f (x y)] = \cos (\log_{e} x) \cos (\log_{e} y)$
$\Rightarrow f (\frac{1}{x}) f (\frac{1}{y}) - \frac{1}{2} \{f (x y) + f (\frac{x}{y})\} = \cos (\log_{e} x) \cos (\log_{e} y) - \cos (\log_{e} x) \cos (\log_{e} y) = 0$
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 3.43:

Question 19:

Let f(x) = x, $g (x) = \frac{1}{x}$ 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, $g (x) = \frac{1}{x}$ and h(x) = f(x) g(x)
Now,
$h (x) = x \times \frac{1}{x} = 1$
We observe that the domain of f is $ℝ$ and the domain of g is $ℝ - \{0\}$ .
∴ Domain of h = Domain of f â‹‚ Domain of g = $ℝ \cap [ℝ - \{0\}] = ℝ - \{0\}$
$\Rightarrow$ x ∈ R, x ≠ 0

Page No 3.43:

Question 20:

If $f (x) = \frac{\sin^{4} x + \cos^{2} x}{\sin^{2} x + \cos^{4} x}$ for x ∈ R, then f (2002) =
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1

Given:
$f (x) = \frac{\sin^{4} x + \cos^{2} x}{\sin^{2} x + \cos^{4} x}$

On dividing the numerator and denominator by $\cos^{4} x$ , we get
$f (x) = \frac{\tan^{4} x + \sec^{2} x}{1 + \tan^{2} x \sec^{2} x} = \frac{1 + \tan^{4} x + \tan^{2} x}{1 + \tan^{2} x (1 + \tan^{2} x)} = \frac{1 + \tan^{4} x + \tan^{2} x}{1 + \tan^{4} x + \tan^{2} x} = 1$ (For every x ∈ R)

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

Page No 3.43:

Question 21:

The function f : R → R is defined by f(x) = cos²x + sin⁴x. 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
$\Rightarrow f (x) = 1 - \sin^{2} x + \sin^{4} x$
$\Rightarrow f (x) = {(\sin^{2} x - \frac{1}{2})}^{2} + \frac{3}{4}$
The minimum value of $f (x)$ is $\frac{3}{4}$ .
Also,
$\sin^{2} x \leq 1 \Rightarrow \sin^{2} x - \frac{1}{2} \leq \frac{1}{2} \Rightarrow {(\sin^{2} x - \frac{1}{2})}^{2} \leq \frac{1}{4} \Rightarrow {(\sin^{2} x - \frac{1}{2})}^{2} + \frac{3}{4} \leq \frac{1}{4} + \frac{3}{4} \Rightarrow f (x) \leq 1$

The maximum value of $f (x)$ is 1.

∴ f(R) = (3/4, 1)

Page No 3.43:

Question 22:

Let A = {x ∈ R : x ≠ 0, −4 ≤ x ≤ 4} and f : A ∈ R be defined by $f (x) = \frac{|x|}{x}$ for x ∈ A.
Then th (is

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

Answer:

$As, |x| = \{\begin{cases} x, x \geq 0 \\ - x < 0 \end{cases} So, f (x) = \frac{x}{|x|} When x < 0 i . e . x \in [- 4, 0) f (x) = \frac{x}{- x} = - 1 and when x > 0 i . e . x \in (0, 4] f (x) = \frac{x}{x} = 1 So, range (f) = \{- 1, 1\}$
Disclaimer: The question in the book has some error. The solution is created according to the question given in the book.

Page No 3.43:

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

$g (f (x)) = 8 \Rightarrow {(f (x))}^{2} + 7 = 8 \Rightarrow (2 x + 3)^{2} + 7 = 8 \Rightarrow x^{2} + 3 x + 2 = 0 \Rightarrow (x + 2) (x + 1) = 0 \Rightarrow x = - 1, - 2$

Page No 3.43:

Question 24:

If f : [−2, 2] → R is defined by $f (x) = \{\begin{cases} - 1, & for - 2 \leq x \leq 0 \\ x - 1, & for 0 \leq x \leq 2 \end{cases}$ , then
{x ∈ [−2, 2] : x ≤ 0 and f (|x|) = x} =

(a) {−1}
(b) {0}
(c) $\{- \frac{1}{2}\}$
(d) Ï•

Answer:

(c) $\{- \frac{1}{2}\}$

Given:
$f (x) = \{\begin{cases} - 1, & for - 2 \leq x \leq 0 \\ x - 1, & for 0 \leq x \leq 2 \end{cases}$
We know,
$|x| \geq 0$
⇒ $f (|x|) = |x| - 1$ ...(1)

Also,
If $x \leq 0$ , then $|x| = - x$ ...(2)

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

Page No 3.43:

Question 25:

If $e^{f (x)} = \frac{10 + x}{10 - x}$ , x ∈ (−10, 10) and $f (x) = k f (\frac{200 x}{100 + x^{2}})$ , then k =
(a) 0.5
(b) 0.6
(c) 0.7
(d) 0.8

Answer:

(a) 0.5

$e^{f (x)} = \frac{10 + x}{10 - x}$
$\Rightarrow f (x) = \log_{e} (\frac{10 + x}{10 - x})$ ...(1)
$f (x) = k f (\frac{200 x}{100 + x^{2}})$
$\Rightarrow \log_{e} (\frac{10 + x}{10 - x}) = k \log_{e} (\frac{10 + \frac{200 x}{100 + x^{2}}}{10 - \frac{200 x}{100 + x^{2}}}) {from (1)} \Rightarrow \log_{e} (\frac{10 + x}{10 - x}) = k l {og}_{e} (\frac{1000 + 10 x^{2} + 200 x}{1000 + 10 x^{2} - 200 x}) \Rightarrow \log_{e} (\frac{10 + x}{10 - x}) = k l {og}_{e} (\frac{{(x + 10)}^{2}}{{(x - 10)}^{2}}) \Rightarrow \log_{e} (\frac{10 + x}{10 - x}) = 2 k l {og}_{e} \frac{(x + 10)}{(x - 10)} \Rightarrow 1 = 2 k \Rightarrow k = 1 / 2 = 0.5$

Page No 3.43:

Question 26:

f is a real valued function given by $f (x) = 27 x^{3} + \frac{1}{x^{3}}$ and α, β are roots of $3 x + \frac{1}{x} = 12$ . Then,
(a) f(α) ≠ f(β)
(b) f(α) = 10
(c) f(β) = −10
(d) None of these

Answer:

(d) None of these

Given:
$f (x) = 27 x^{3} + \frac{1}{x^{3}}$
$\Rightarrow f (x) = (3 x + \frac{1}{x}) (9 x^{2} + \frac{1}{x^{2}} - 3)$
$\Rightarrow f (x) = (3 x + \frac{1}{x}) ({(3 x + \frac{1}{x})}^{2} - 9)$
$\Rightarrow f (α) = (3 α + \frac{1}{α}) ({(3 α + \frac{1}{α})}^{2} - 9)$
Since α and β are the roots of $3 x + \frac{1}{x} = 12$ ,
$3 α + \frac{1}{α} = 12 and 3 β + \frac{1}{β} = 12$
$\Rightarrow f (α) = 12 ({(12)}^{2} - 9)$ and $f (β) = 12 ({(12)}^{2} - 9)$
$\Rightarrow f (α) = f (β) = 12 ({(12)}^{2} - 9)$

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 3.43:

Question 27:

If $f (x) = 64 x^{3} + \frac{1}{x^{3}}$ and α, β are the roots of $4 x + \frac{1}{x} = 3$ . Then,
(a) f(α) = f(β) = −9
(b) f(α) = f(β) = 63
(c) f(α) ≠ f(β)
(d) none of these

Answer:

(a) $f (α) = f (β) = - 9$

Given:
$f (x) = 64 x^{3} + \frac{1}{x^{3}}$
$\Rightarrow f (x) = (4 x + \frac{1}{x}) (16 x^{2} + \frac{1}{x^{2}} - 4)$
$\Rightarrow f (x) = (4 x + \frac{1}{x}) ({(4 x + \frac{1}{x})}^{2} - 12)$

$\Rightarrow f (α) = (4 α + \frac{1}{α}) ({(4 α + \frac{1}{α})}^{2} - 12) and f (β) = (4 β + \frac{1}{β}) ({(4 β + \frac{1}{β})}^{2} - 12)$
Since α and β are the roots of $4 x + \frac{1}{x} = 3$ ,
$4 α + \frac{1}{α} = 3 and 4 β + \frac{1}{β} = 3$
$\Rightarrow f (α) = 3 ({(3)}^{2} - 12) = - 9$ and $f (β) = 3 ({(3)}^{2} - 12) = - 9$
$\Rightarrow f (α) = f (β) = - 9$

Page No 3.43:

Question 28:

If $3 f (x) + 5 f (\frac{1}{x}) = \frac{1}{x} - 3$ for all non-zero x, then f(x) =

(a) $\frac{1}{14} (\frac{3}{x} + 5 x - 6)$
(b) $\frac{1}{14} (- \frac{3}{x} + 5 x - 6)$
(c) $\frac{1}{14} (- \frac{3}{x} + 5 x + 6)$
(d) None of these

Answer:

(d) None of these

$3 f (x) + 5 f (\frac{1}{x}) = \frac{1}{x} - 3$ ...(1)

$Multiplying (1) by 3 : 15 f (\frac{1}{x}) + 9 f (x) = \frac{3}{x} - 9 . . . . . (2) Replacing x by \frac{1}{x} in (1) : 3 f (\frac{1}{x}) + 5 f (x) = x - 3 Multiplying by 5 : 15 f (\frac{1}{x}) + 25 f (x) = 5 x - 15 . . . . (3) Solving (2) and (3) : - 16 f (x) = \frac{3}{x} - 5 x + 6 \Rightarrow f (x) = \frac{1}{16} (- \frac{3}{x} + 5 x - 6)$
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 3.43:

Question 29:

If f : R → R be given by for all $f (x) = \frac{4^{x}}{4^{x} + 2}$ 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

$f (x) = \frac{4^{x}}{4^{x} + 2}$ ; x ∈ R
$f (1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{4}{2 \times 4^{x} + 4} = \frac{2}{4^{x} + 2} f (x) + f (1 - x) = \frac{4^{x}}{4^{x} + 2} + \frac{2}{4^{x} + 2} = \frac{4^{x} + 2}{4^{x} + 2} = 1$

Page No 3.43:

Question 30:

If f(x) = sin [π²] x + sin [−π]²x, 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
$\Rightarrow f (x) = \sin [9.8] x + \sin [- 9.8] x \Rightarrow f (x) = \sin 9 x - \sin 10 x f (\frac{π}{2}) = \sin 9 \times \frac{π}{2} - \sin 10 \times \frac{π}{2} \Rightarrow f (\frac{π}{2}) = 1 - 0 = 1$

Page No 3.44:

Question 31:

The domain of the function $f (x) = \sqrt{2 - 2 x - x^{2}}$ is

(a) $[- \sqrt{3}, \sqrt{3}]$
(b) $[- 1 - \sqrt{3}, - 1 + \sqrt{3}]$
(c) [−2, 2]
(d) $[- 2 - \sqrt{3}, - 2 + \sqrt{3}]$

Answer:

(b) $[- 1 - \sqrt{3}, - 1 + \sqrt{3}]$

$f (x) = \sqrt{2 - 2 x - x^{2}}$
$Since, 2 - 2 x - x^{2} \geq 0, x^{2} + 2 x - 2 \leq 0 \Rightarrow x^{2} - 2 x - 2 + 1 - 1 \leq 0 \Rightarrow {(x - 1)}^{2} - {(\sqrt{3})}^{2} \leq 0 \Rightarrow [x - (- 1 - \sqrt{3})] [x - (- 1 + \sqrt{3})] \leq 0 \Rightarrow (- 1 - \sqrt{3}) \leq x \leq (- 1 + \sqrt{3}) Thus, dom (f) = [- 1 - \sqrt{3}, - 1 + \sqrt{3}] .$

Page No 3.44:

Question 32:

The domain of definition of $f (x) = \sqrt{\frac{x + 3}{(2 - x) (x - 5)}}$ is

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

Answer:

(a) (−∞, −3] ∪ (2, 5)

$f (x) = \sqrt{\frac{x + 3}{(2 - x) (x - 5)}}$
$For f (x) to be defined, (2 - x) (x - 5) \neq 0 \Rightarrow x \neq 2, 5 . . . . (1) Also, \frac{(x + 3)}{(2 - x) (x - 5)} \geq 0 \Rightarrow \frac{(x + 3) (2 - x) (x - 5)}{{(2 - x)}^{2} {(x - 5)}^{2}} \geq 0 \Rightarrow (x + 3) (x - 2) (x - 5) \leq 0 \Rightarrow x \in (- \infty, - 3] \cup (2, 5) . . . . (2) From (1) and (2), x \in (- \infty, - 3] \cup (2, 5)$

Page No 3.44:

Question 33:

The domain of the function $f (x) = \sqrt{\frac{(x + 1) (x - 3)}{x - 2}}$ is
(a) [−1, 2) ∪ [3, ∞)
(b) (−1, 2) ∪ [3, ∞)
(c) [−1, 2] ∪ [3, ∞)
(d) None of these

Answer:

(a) [−1, 2) ∪ [3, ∞)

$f (x) = \sqrt{\frac{(x + 1) (x - 3)}{x - 2}}$
$For f (x) to be defined, (x - 2) \neq 0 \Rightarrow x \neq 2 . . . (1) Also, \frac{(x + 1) (x - 3)}{(x - 2)} \geq 0 \Rightarrow \frac{(x + 1) (x - 3) (x - 2)}{(x - 2)^{2}} \geq 0 \Rightarrow (x + 1) (x - 3) (x - 2) \geq 0 \Rightarrow x \in [- 1, 2) \cup [3, \infty) . . . . . (2) From (1) and (2), x \in [- 1, 2) \cup [3, \infty)$

Page No 3.44:

Question 34:

The domain of definition of the function $f (x) = \sqrt{x - 1} + \sqrt{3 - x}$ is

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

Answer:

(d) [1, 3]

$f (x) = \sqrt{x - 1} + \sqrt{3 - x}$
$For f (x) to be defined, (x - 1) \geq 0 \Rightarrow x \geq 1 . . . (1) and (3 - x) \geq 0 \Rightarrow 3 \leq x . . . (2) From (1) and (2), x \in [1, 3]$

Page No 3.44:

Question 35:

The domain of definition of the function $f (x) = \sqrt{\frac{x - 2}{x + 2}} + \sqrt{\frac{1 - x}{1 + x}}$ is
(a) (−∞, −2] ∪ [2, ∞)
(b) [−1, 1]
(c) Ï•
(d) None of these

Answer:

(c) Ï•

$f (x) = \sqrt{\frac{x - 2}{x + 2}} + \sqrt{\frac{1 - x}{1 + x}}$
$For f (x) to be defined, x + 2 \neq 0 \Rightarrow x \neq - 2 . . . (1) And 1 + x \neq 0 \Rightarrow x \neq - 1 . . . . (2) Also, \frac{x - 2}{x + 2} \geq 0 \Rightarrow \frac{(x - 2) (x + 2)}{(x + 2)^{2}} \geq 0 \Rightarrow (x - 2) (x + 2) \geq 0 \Rightarrow x \in (- \infty, - 2) \cup [2, \infty) . . . (3) And \frac{1 - x}{1 + x} \geq 0 \Rightarrow \frac{(1 - x) (1 + x)}{(1 + x)^{2}} \geq 0 \Rightarrow (1 - x) (1 + x) \geq 0 \Rightarrow x \in (- \infty, - 1) \cup [1, \infty) . . . (4) From (1), (2), (3) and (4), we get, x \in ϕ . Thus, dom (f (x)) = ϕ .$

Page No 3.44:

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

Page No 3.44:

Question 37:

The domain of definition of $f (x) = \sqrt{4 x - x^{2}}$ is

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

Answer:

(d) [0, 4]

Given:
$f (x) = \sqrt{4 x - x^{2}}$
Clearly, f (x) assumes real values if
4x $-$ x² ≥ 0
⇒ x(4 $-$ x) ≥ 0
⇒ $-$ x(x $-$ 4) ≥ 0
⇒ x(x $-$ 4) ≤ 0
⇒ x ∈ [0, 4]

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

Page No 3.44:

Question 38:

The domain of definition of $f (x) = \sqrt{x - 3 - 2 \sqrt{x - 4}} - \sqrt{x - 3 + 2 \sqrt{x - 4}}$ is

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

Answer:

(a) [4, ∞)

$f (x) = \sqrt{x - 3 - 2 \sqrt{x - 4}} - \sqrt{x - 3 + 2 \sqrt{x - 4}}$

$For f (x) to be defined, x - 4 \geq 0 \Rightarrow x - 4 \geq 0 \Rightarrow x \geq 4 . . . . (1) Also, x - 3 - 2 \sqrt{x - 4} \geq 0 \Rightarrow x - 3 - 2 \sqrt{x - 4} \geq 0 \Rightarrow x - 3 \geq 2 \sqrt{x - 4} \Rightarrow (x - 3)^{2} \geq {(2 \sqrt{x - 4})}^{2} \Rightarrow x^{2} + 9 - 6 x \geq 4 (x - 4) \Rightarrow x^{2} - 10 x + 25 \geq 0 \Rightarrow (x - 5)^{2} \geq 0, which is always true . Similarly, x - 3 + 2 \sqrt{x - 4} \geq 0 is always true . Thus, dom (f (x)) = [4, \infty)$

Page No 3.44:

Question 39:

The domain of the function $f (x) = \sqrt{5 |x| - x^{2} - 6}$ 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]

$f (x) = \sqrt{5 |x| - x^{2} - 6}$
$For f (x) to be defined, 5 |x| - x^{2} - 6 \geq 0 \Rightarrow 5 |x| - x^{2} - 6 \geq 0 \Rightarrow x^{2} - 5 |x| + 6 \leq 0 For x > 0, |x| = x \Rightarrow x^{2} - 5 x + 6 \leq 0 \Rightarrow (x - 2) (x - 3) \leq 0 \Rightarrow x \in [2, 3] . . . . . . . . (1) For x < 0, |x| = - x \Rightarrow x^{2} + 5 x + 6 \leq 0 \Rightarrow (x + 2) (x + 3) \leq 0 \Rightarrow x \in [- 3, - 2] . . . . . . . (2) From (1) and (2), x \in [- 3, - 2] \cup [2, 3] or, dom (f) = [- 3, - 2] \cup [2, 3]$

Page No 3.44:

Question 40:

The range of the function $f (x) = \frac{x}{|x|}$ is

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

Answer:

(c) {−1, 1}

$f (x) = \frac{x}{|x|}$
$Let y = \frac{x}{|x|} For x > 0, |x| = x \Rightarrow y = \frac{x}{x} = 1 For x < 0, = - x \Rightarrow y = \frac{x}{- x} = - 1 Thus, r ange of f (x) is {- 1, 1} .$

Page No 3.44:

Question 41:

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

Answer:

(a) {−1, 1}

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

Page No 3.44:

Question 42:

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

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

Answer:

(b) [0, ∞)

$f (x) = |x - 1| \geq 0 \forall x \in R Thus, range = [0, \infty 0$

Page No 3.44:

Question 43:

Let $f (x) = \sqrt{x^{2} + 1}$ . Then, which of the following is correct?

(a) $f (x y) = f (x) f (y)$ (b) $f (x y) \geq f (x) f (y)$ (c) $f (x y) \leq f (x) f (y)$ (d) none of these

Answer:

Given: $f (x) = \sqrt{x^{2} + 1}$ .....(1)

Replacing x by y in (1), we get

$f (y) = \sqrt{y^{2} + 1}$

$∴ f (x) f (y) = \sqrt{x^{2} + 1} \sqrt{y^{2} + 1} = \sqrt{(x^{2} + 1) (y^{2} + 1)} = \sqrt{x^{2} y^{2} + x^{2} + y^{2} + 1}$

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

$f (x y) = \sqrt{x^{2} y^{2} + 1}$

Now,

$x^{2} y^{2} + 1 \leq x^{2} y^{2} + x^{2} + y^{2} + 1 \Rightarrow \sqrt{x^{2} y^{2} + 1} \leq \sqrt{x^{2} y^{2} + x^{2} + y^{2} + 1} \Rightarrow f (x y) \leq f (x) f (y)$

Hence, the correct answer is option (c).

Page No 3.44:

Question 44:

If ${[x]}^{2} - 5 [x] + 6 = 0$ , 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 ${[x]}^{2} - 5 [x] + 6 = 0$ .

${[x]}^{2} - 5 [x] + 6 = 0 \Rightarrow {[x]}^{2} - 3 [x] - 2 [x] + 6 = 0 \Rightarrow [x] ([x] - 3) - 2 ([x] - 3) = 0 \Rightarrow ([x] - 2) ([x] - 3) = 0$
$\Rightarrow [x] - 2 = 0 or [x] - 3 = 0 \Rightarrow [x] = 2 or [x] = 3$
⇒ x ∈ [2, 3) or x ∈ [3, 4)
⇒ x ∈ [2, 4)

Hence, the correct answer is option (d).

Page No 3.44:

Question 45:

The range of $f (x) = \frac{1}{1 - 2 \cos x}$ 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,

$- 1 \leq \cos x \leq 1 \Rightarrow - 1 \leq - \cos x \leq 1 \Rightarrow - 2 \leq - 2 \cos x \leq 2 \Rightarrow - 1 \leq 1 - 2 \cos x \leq 3 (Adding 1 to each term)$

But,

$\cos x \neq \frac{1}{2}$
$\Rightarrow 1 - 2 \cos x \in [- 1, 3] - \{0\} \Rightarrow \frac{1}{1 - 2 \cos x} \in (- \infty, - 1] \cup [\frac{1}{3}, \infty)$

∴ Range of f(x) = (−∞, −1] ∪[ $\frac{1}{3}$ , ∞)

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 3.45:

Question 46:

The domain of the function $f (x) \sqrt{4 - x} + \frac{1}{\sqrt{x^{2} - 1}}$ is equal to
(a) (−∞, −1) ∪ (1, 4)
(b) (−∞, −1] ∪ (1, 4]
(c) (−∞, −1) ∪ [1, 4]
(d) (−∞, −1) ∪ [1, 4)

Answer:

$f (x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^{2} - 1}}$
for $\sqrt{4 - x}$
4 − x ≥ 0
i.e 4 ≥ x
i.e x ≤ 4
and $\frac{1}{\sqrt{x^{2} - 1}}$
Since x² − 1 > 0 $(∵ x^{2} - 1 = 0 is not defined for \frac{1}{\sqrt{x^{2} - 1}})$
i.e x² > 1
i.e x < − 1 or x > 1

for $\sqrt{4 - x} + \frac{1}{\sqrt{x^{2} - 1}},$ Domain is (–∞, –1) ∪ (1, 4].

Page No 3.45:

Question 47:

Domain of $f (x) = \sqrt{a^{2} - x^{2}}, a > 0$ is
(a) (−a, a)
(b) [−a, a]
(c) [0, a]
(d) (−a, 0]

Answer:

f(x) = $\sqrt{a^{2} - x^{2}}$ ; a > 0
Since a² − x² ≥ 0
i.e −x² ≥ −a²
i.e x² ≤ a²
i.e |x| ≤ a
i.e −a ≤ x ≤ a
i.e x∈ [−a, a]
$∴$ Domain of f (x) is [−a, a]
Hence, the correct answer is option B.

Page No 3.45:

Question 48:

If f(x) = ax + b, where, a and b are integers, f(−1) = −5 and f(x) = 3, then a and b are equal
(a) a = −3, b = −1
(b) a = 2, b = −3
(c) a = 0, b = 2
(d) a = 2, b = 3

Answer:

f(x) = ax + b
Given f(−1) = −5 and f(x) = 3

(correction)

Page No 3.45:

Question 49:

The domain and range of the real function of 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) = $\frac{4 - x}{x - 4}$
Domain :- for f(x) = $\frac{4 - x}{x - 4}$
x − 4 $\neq$ 0
i.e x $\neq$ 4
∴ Domain is R −{4}
Range:- Since f(x) = $\frac{- (x - 4)}{x - 4} = - 1$
∴ Range is {−1}
Hence, the correct answer is option C.

Page No 3.45:

Question 50:

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) = $\sqrt{x - 1}$
Since x −1 ≥ 0
i.e x ≥ 1
$∴$ Domain of f(x) is [1, ∞)
and for x∈ [1, ∞)
f(x) ≥ 0
⇒ Range of f(x) is [0, ∞)
Hence, the correct answer is option C.

Page No 3.45:

Question 51:

The domain of the function f given by $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6}$
(a) R −{−2, 3}
(b) R −{−3, 2}
(c) R −[−2, 3]
(d) R −(−2, 3)

Answer:

f(x) = $\frac{x^{2} + 2 x + 1}{x^{2} - x - 6}$
f(x) is not defined for x² − x − 6 = 0
              i.e x² − 3x + 2x – 6 = 0
              i.e x(x − 3) + 2 (x − 3) = 0
              i.e (x − 3) (x + 2) = 0
i.e x = 3 or  x = −2
$∴$ Domain for f (x) is R −{−2, 3}
Hence the correct answer is option A.

Page No 3.45:

Question 52:

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|
Since f(x) is defined for every x∈R
$∴$ Domain is R
also |x − 5| ≥ 0
i.e −|x − 5| ≤ 0
i.e 2 − |x − 5| ≤ 2
$∴$ f(x) ≤ 2
i.e Range for f (x) is (−∞, 2]
Hence, the correct answer is option B.

Page No 3.45:

Question 53:

If $f (x) = x^{3} - \frac{1}{x^{3}},$ then $f (x) + f (\frac{1}{x})$ is equal to

(a) 2x³

(b) $\frac{2}{x^{3}}$

(c) 0

(d) 1

Answer:

$f (x) = x^{3} - \frac{1}{x^{3}} \Rightarrow f (\frac{1}{x}) = {(\frac{1}{x})}^{3} - {(\frac{1}{\frac{1}{x}})}^{3} f (\frac{1}{x}) = \frac{1}{x^{3}} - x^{3} ∴ f (x) + f (\frac{1}{x}) = x^{3} - \frac{1}{x^{3}} + \frac{1}{x^{3}} - x^{3} i . e f (x) + f (\frac{1}{x}) = 0$
Hence, the correct answer is option C.

Page No 3.45:

Question 54:

The domain of the function f defined by $f (x) = \frac{1}{\sqrt{x - |x|}}$ is
(a) R₀
(b) R⁺
(c) R⁻
(d) none of these

Answer:

f(x) = $f (x) = \frac{1}{\sqrt{x - |x|}}$
f(x) is defined if x − |x| > 0
i.e x > |x|
i.e |x| < x
Since no such real number x exist such that |x| < x
$∴$ Domain of f(x) is empty set.
Hence, the correct answer is option D.

Page No 3.46:

Question 1:

Let A and B be any two sets such that n(A) = p and n(B) = q, then the total functions from A to B is equal to __________ .

Answer:

n(A) = p, n(B) = q.
here any element of set A, can be connected with elements of B in q ways.

and there are p such elements in A.
$∴$ Total function possible is $\frac{q \times q \times q . . . . . . . \times q}{p times}$
i.e q^p

$∴$ Total functions from A to B q^p i.e n(B) ^n(A).

Page No 3.46:

Question 2:

If $f (x) = \frac{x}{x - 1} = \frac{1}{y}, then f (y) =$ __________ .

Answer:

f(x) = $\frac{x}{x - 1}$
given $\frac{x}{x - 1} = \frac{1}{y}$
xy = x − 1 i.e y = $\frac{x - 1}{x}$
$\begin{array}{rcl} i . e f (y) & = & \frac{y}{y - 1} \\ = & \frac{\frac{x - 1}{x}}{\frac{x - 1}{x} - 1} \\ = & \frac{\frac{x - 1}{x}}{\frac{x - 1 - x}{x}} \\ = & \frac{x - 1}{x} \times \frac{x}{- 1} \\ = & - (x - 1) \\ ∴ f (y) & = & 1 - x \end{array}$

Page No 3.46:

Question 3:

If $y = f (x) = \frac{a x + b}{c x - d}, then f (y) =$ __________ .

Answer:

$If y = f (x) = \frac{a x + b}{c x - d} f (y) = \frac{a y + b}{c y - d} = \frac{a (\frac{a x + b}{c x - d}) + b}{c (\frac{a x + b}{c x - d}) - d} f (y) = \frac{a (a x + b) + b (c x - d)}{\frac{c x - d}{\frac{c (a x + b) - d (c x - d)}{c x - d}}} = \frac{a^{2} x + a b + b c x - d b}{a c x + b c - c d x + d^{2}} f (y) = \frac{(a^{2} + b c) x + a b - b d}{(a c - c d) x + b c - d^{2}}$

Page No 3.46:

Question 4:

The domain of the function $f (x) = \frac{1}{\sqrt{|x| - x}}$ is _____________.

Answer:

f(x) = $\frac{1}{\sqrt{|x| - x}}$
f(x) is defined if |x| − x > 0
i.e |x| > x
i.e x < |x|
which is possible for negative real numbers
∴ Domain for f(x) is |Râ€‹⁻ ~ {0}

Page No 3.46:

Question 5:

The range of the function f(x) = [x] − x is __________ .

Answer:

f(x) = [x] − x
Since x ≥ [x]
Every number is greater than or equal to its greatest integral value
i.e x – [x] = {x} fractional part of x.
∴ [x] − x = −{x} fraction part only.
also [x] = x for integral value of x
hence, for non-integral values f(x) = −{x} $\in$ (−1, 0)
and for integral values f(x) = 0
Hence, Range of f(x) is (−1, 0].

Page No 3.46:

Question 6:

The range of the function $f (x) = \frac{x + 2}{|x + 2|}$ is __________ .

Answer:

$\begin{array}{rcl} f (x) & = & \frac{x + 2}{|x + 2|} \\ = & \{\begin{cases} \frac{x + 2}{x + 2}; x \geq - 2 \\ \frac{x + 2}{- (x + 2)}; x < - 2 \end{cases} \\ i . e f (x) & = & \{\begin{cases} 1; x \geq - 2 \\ - 1; x < - 2 \end{cases} \end{array}$

∴ Range of function f(x) is {−1, 1}

Page No 3.46:

Question 7:

The range of the function f(x) = log_ax, a > 0 is __________ .

Answer:

f(x) = log_ax ; a > 0
f(x) = $\frac{\log_{10} (x)}{\log_{10} (a)}$
Since a > 0 ⇒ f(x) is defined for R⁺ −{0}
∴ Range set of f(x) is R

Page No 3.46:

Question 8:

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). Then, domain of f + g is __________ .

Answer:

f = {(2, 4), (5, 6), (8, –1), (10, –3)}
g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, –5)}
Domain of f is {2, 5, 8, 10} and Domain of g = {2, 7, 8, 10, 11}
Since (f + g) (x) = f(x) + g(x)
∴ x should lie in Both Domain of f and Domain of g
∴ x ∈ Domain ( f) ∩ Domain (g)
i.e Domain of f + g is {2, 8, 10}

Page No 3.46:

Question 9:

Let f and g be two real functions given by f = {(10, 1), (2, 0), (3, −4), (4, 2), (5, 1)} and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}. Then the domain fg is given by __________ .

Answer:

f = {(10, 1),(2, 0), (3, –4), (4, 2), (5, 1)}
g = {(1, 0), (2, 2), (3, –1), (4, 4), (5, 3)}

Since (fg) (x) = f(x) g(x)
∴ Domain of fg is (Domain of f) ∩ (Domain of g)
Since Domain of f = {10, 2, 3, 4, 5}
Domain of g = {1, 2, 3, 4, 5}
⇒ Domain of f g is {2, 3, 4, 5}

Page No 3.46:

Question 10:

The domain for which the functions f(x) = 3x² −1 and g(x) = 3 + x are equal is __________ .

Answer:

f(x) = 3x² − 1
g(x) = 3 + x
Given:- f(x) = g(x)
i.e 3x² − 1 = 3 + x
3x² − x − 1 − 3 = 0
3x² − x − 4 = 0
3x²+ 3x − 4x − 4 = 0
3x(x + 1) − 4 (x + 1) = 0
(3x − 4) (x + 1) = 0
i.e x = $\frac{4}{3}$ or x = −1
∴ Domain for which f(x) and g(x) are equal is $\{- 1, \frac{4}{3}\}$

Page No 3.46:

Question 11:

The domain of the function $f (x) = \frac{x^{2} + 1}{x^{2} - 3 x + 2}$ is __________ .

Answer:

f(x) = $\frac{x^{2} + 1}{x^{2} - 3 x + 2}$
f(x) is defined if x² − 3x + 2 ≠ 0
i.e f(x) is not defined
for x² − 3x + 2 = 0
i.e x² − 2x − x + 2 = 0
(x² − 2x) −1 (x − 2) = 0
x (x − 2) −1 (x − 2) = 0
i.e (x − 1) (x − 2) = 0
i.e x = 1 or x = 2
i.e for x = 1 or x = 2, f (x) is not defined
∴ Domain of f(x) = R ~ {1,2}

Page No 3.46:

Question 12:

If $f (x) = \frac{x - 1}{x + 1},$ then $f (\frac{1}{x}) + f (x)$ is equal to __________ .

Answer:

$If f (x) = \frac{x - 1}{x + 1} f (\frac{1}{x}) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} = \frac{\frac{(1 - x)}{x}}{\frac{(1 + x)}{x}} f (\frac{1}{x}) = \frac{1 - x}{1 + x} ∴ f (\frac{1}{x}) + f (x) = \frac{1 - x}{1 + x} + \frac{x - 1}{x + 1} = \frac{1 - x + x - 1}{x + 1} ∴ f (\frac{1}{x}) + f (x) = 0$

Page No 3.46:

Question 13:

If $f (x) = \frac{x - 1}{x + 1},$ then $f (x) f (- \frac{1}{x})$ is equal to __________ .

Answer:

$If f (x) = \frac{x - 1}{x + 1} f (\frac{- 1}{x}) = \frac{\frac{- 1}{x} - 1}{\frac{- 1}{x} + 1} = \frac{\frac{(- 1 - x)}{x}}{\frac{(- 1 + x)}{x}} = - \frac{(1 + x)}{x - 1} f (\frac{- 1}{x}) = \frac{1 + x}{1 - x} ∴ f (x) f (\frac{- 1}{x}) = \frac{x - 1}{x + 1} \times \frac{1 + x}{1 - x} ∴ f (x) f (\frac{- 1}{x}) = - 1$

Page No 3.46:

Question 14:

If f(x) = [x]² − 5 [x] + 6, then the set of values of x satisfying f(x) = 0 is __________ .

Answer:

f(x) = [x]² − 5 [x] + 6
Given f(x) = 0
i.e [x]²− 5 [x] + 6 = 0
Let [x] = y
i.e y² − 5y + 6 = 0
i.e y² − 3y – 2y + 6 = 0
y(y − 3) −2 (y − 3) = 0
i.e y = 2 or y = 3
i.e [x] = 2 or [x] = 3
here, [x] = 2 if x∈ [2, 3) and [x] = 3 if x∈ [3, 4)
∴ value of x for which f(x) = 0 are [2, 3) ∪ [3, 4)
i.e x∈ [2, 4)

Page No 3.46:

Question 15:

The domain of the function $f (x) = \sqrt{9 - x} + \frac{1}{\sqrt{x^{2} - 16}}$ is equal to __________ .

Answer:

$f (x) = \sqrt{9 - x} + \frac{1}{\sqrt{x^{2} - 16}}$
here f(x) is defined for 9 − x ≥ 0
i.e 9 ≥ x
i.e x ≤ 9
and x² − 16 > 0
i.e x² > 16
i.e x > 4 or x < −4

∴ f(x) is defined for common region is (−∞, −4) ∪ (4, 9]
∴ Domain of f(x) is (−∞, −4) ∪ (4, 9]

Page No 3.46:

Question 16:

The domain and range of the function $f (x) = \frac{2 - x}{x - 2}$ are _________ and _________ respectively.

Answer:

$f (x) = \frac{2 - x}{x - 2}$
f(x) is defined for x − 2 ≠ 0
i.e x ≠ 2
∴ Domain of f(x) is R −{2}
Since
f(x) = – $\frac{(x - 2)}{x - 2}$
f(x) = −1
∴ Range of f(x) is {−1}

Page No 3.46:

Question 17:

The domain of the function $f (x) = \frac{1}{\sqrt{{[x]}^{2} - 3 [x] + 2}}$ is __________ .

Answer:

$f (x) = \frac{1}{\sqrt{{[x]}^{2} - 3 [x] + 2}}$
Since [x]² − 3[x] + 2 = 0
if [x]² − 2[x] − [x] + 2 = 0
[x] ([x] − 2) −1 ([x] − 2) = 0
i.e [x] = 2 or [x] = 1
f(x) is defined only if [x]² − 3[x] + 2 > 0
i.e ([x] −2) ([x]−1) > 0

[x] > 2 or [x] < 1
i.e [x] ≥ 3 or [x] ≤ 0
i.e x ∈[3, ∞) or x∈(−∞, 1)
i.e Domain of f(x) is [3, ∞) ∪ (−∞, 1)
i.e (−∞, 1) ∪ [3, ∞)

Page No 3.46:

Question 18:

The range of the function $f (x) = \frac{|x - 4|}{x - 4}$ is __________ .

Answer:

$f (x) = \frac{|x - 4|}{x - 4} \frac{- \frac{(x - 4)}{x - 4} \frac{(x - 4)}{x - 4}}{4} = \{\begin{cases} \frac{x - 4}{x - 4}; x \geq 4 \\ - \frac{(x - 4)}{x - 4}; x < + 4 \end{cases} f (x) = \{\begin{cases} 1; x \geq 4 \\ - 1; x < 4 \end{cases}$
Range of f(x) is {−1, 1}

Page No 3.46:

Question 19:

The domain of the function f(x) = x + [x] is __________ .

Answer:

f(x) = x + [x]
Since Domain of [x] = R
and Domain of x = R
∴ Domain of x + [x] = R ∩ R = R

Page No 3.46:

Question 20:

The range of the function $f (x) = \sqrt{1 - x^{2}}$ is __________ .

Answer:

f(x) = $\sqrt{1 - x^{2}}$

Since f(x) ≥ 0 ( Being a square root)

∴ Range of f(x) = [0, ∞)

Page No 3.46:

Question 21:

The domain of the function $f (x) = \sum_{n = 1}^{10} \frac{1}{|2 x - n|}$ is __________ .

Answer:

f(x) = $\sum_{n = 1}^{10} \frac{1}{|2 x - n|}$

f(x) is not defined if 2x − n = 0

$i . e x = \frac{n}{2}; n = 1 to 10$
∴ Domain of f(x) is $ℝ ~ \{\frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}, 4, \frac{9}{2}, 5\}$

Page No 3.47:

Question 22:

The domain of the function $f (x) = \frac{|x| - 2}{|x| - 3}$ is __________ .

Answer:

$f (x) = \frac{|x| - 2}{|x| - 3}$
here f(x) is not define if |x| − 3 = 0
i.e |x| = 3
i.e x = 3 or −3
∴ Domain of f(x) is R –{−3, 3}

Page No 3.47:

Question 23:

If f(2x + 3) = 4x² + 12x +15, then the value of f(3x + 2) is __________ .

Answer:

Given f(2x + 3) = 4x² + 12x + 15
   = (2x)² + 2(2x) (3) + (3)² + 6
f(2x + 3) = (2x + 3)² + 6
i.e f(y) = y²+ 6    where y = 2x + 3
∴ f (3x + 2) = (3x + 2)² + 6
                   = 9x² + 4 + 12x + 6
Hence f(3x + 2) = 9x² + 12x + 10

Page No 3.47:

Question 24:

The number of elements of an identity function defined on a set containing four elements is __________ .

Answer:

The number of elements of an identity function defined on a set of containing.
Four elements is four only
Since identity function maps same element to same element.
i.e x → x
y → y
z → z
and w → w
∴ There are four such elements.

Page No 3.47:

Question 1:

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

Answer:

Given:
f (x) = | x |, x ∈ R
We know that
$|x| = \{\begin{cases} x, & x \geq 0 \\ - x & x < 0 \end{cases}$
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 3.47:

Question 2:

If f is a real function satisfying $f (x + \frac{1}{x}) = x^{2} + \frac{1}{x^{2}}$ for all x ∈ R − {0}, then write the expression for f(x).

Answer:

Given:
$f (x + \frac{1}{x}) = x^{2} + \frac{1}{x^{2}}$
$= x^{2} + \frac{1}{x^{2}} + 2 - 2$
$= {(x + \frac{1}{x})}^{2} - 2$
Thus,
$f (x + \frac{1}{x}) = {(x + \frac{1}{x})}^{2} - 2$
Hence,
f (x) = x² $-$ 2 , where | x | ≥ 2.

Page No 3.47:

Question 3:

Write the range of the function f(x) = sin [x], where $\frac{- π}{4} \leq x \leq \frac{π}{4}$ .

Answer:

Given : f(x) = sin [x], where $\frac{- π}{4} \leq x \leq \frac{π}{4}$ .
$- \frac{π}{4} \leq x \leq \frac{π}{4} \Rightarrow - 7.85 \leq x \leq 0.785 ∴ x \in [- 7.85, 7.85] Or x = {0, 1} Thus, range of f (x) = \sin [x] is {\sin 0, \sin 1} = {0, \sin 1}$

Page No 3.47:

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
$Thus, f (π) = \cos [π^{2}] π + \cos [- π^{2}] π \Rightarrow f (π) = c o s [9.8] π + c o s [- 9.8] π \Rightarrow f (π) = c o s 10 π + c o s 9 π \Rightarrow f (π) = 1 + (- 1) = 0$

Page No 3.47:

Question 5:

Write the range of the function f(x) = cos [x], where $\frac{- π}{2} < x < \frac{π}{2}$ .

Answer:

Since f(x) = cos [x], where $\frac{- π}{2} < x < \frac{π}{2}$ ,
$- \frac{π}{2} < x < \frac{π}{2} \Rightarrow - 1.57 < x < 1.57 \Rightarrow [x] \in {- 1, 0, 1, 2} Thus, \cos [x] = {\cos (- 1), \cos 0, \cos 1, \cos 2} . Range of f (x) = {\cos 1, 1, \cos 2} .$

Page No 3.47:

Question 6:

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

Answer:

f(x) = e^x−^[x], x ∈ R

$We know that x - [x] = {x}, which is the fractional part of any number x . Thus, f (x) = e^{{x}} Also, 0 \leq {x} < 1 \Rightarrow e^{0} \leq e^{{x}} < e^{1} \Rightarrow 1 \leq f (x) < e Thus range of f (x) is [1, e) .$

Page No 3.47:

Question 7:

Let $f (x) = \frac{α x}{x + 1}, x \neq - 1$ . Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.

Answer:

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

$Since f (f (x)) = x, \frac{α (\frac{α x}{x + 1})}{\frac{α x}{x + 1} + 1} = x \Rightarrow \frac{α^{2} x}{α x + x + 1} = x \Rightarrow α^{2} x - α x^{2} - (x^{2} + x) = 0 Solving the quadratic equation in α : α = \frac{x^{2} \pm \sqrt{x^{4} + 4 x (x^{2} + x)}}{2 x} \Rightarrow α = x + 1 or - 1 Since, α \neq x + 1, α = - 1 .$

Page No 3.47:

Question 8:

If $f (x) = 1 - \frac{1}{x}$ , then write the value of $f (f (\frac{1}{x}))$ .

Answer:

Given:
$f (x) = 1 - \frac{1}{x}$
Now,
$f (\frac{1}{x}) = 1 - \frac{1}{\frac{1}{x}} = 1 - x$
$\Rightarrow f (f (\frac{1}{x})) = f (1 - x)$
Again,
If $f (x) = 1 - \frac{1}{x}$
Thus,
$f (1 - x) = 1 - \frac{1}{1 - x}$

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

Page No 3.47:

Question 9:

Write the domain and range of the function $f (x) = \frac{x - 2}{2 - x}$ .

Answer:

Given:
$f (x) = \frac{x - 2}{2 - x}$
Domain ( f ) :
Clearly, f (x) is defined for all x satisfying: if 2 $-$ x ≠ 0 ⇒ x ≠ 2.
Hence, domain ( f ) = R $-$ {2}
Range of f :
Let f (x) = y
⇒ $\frac{x - 2}{2 - x} = y$
⇒ x $-$ 2 = y (2 $-$ x)
⇒ x $-$ 2 = $-$ y (x $-$ 2)
⇒ y = $-$ 1
Hence, range ( f ) = { $-$ 1}.

Page No 3.47:

Question 10:

If f(x) = 4x − x², x ∈ R, then write the value of f(a + 1) −f(a − 1).

Answer:

Given:
f(x) = 4x − x², x ∈ R
Now,
f(a + 1) = 4(a + 1) $-$ (a + 1)²
             = 4a + 4 $-$ (a² + 1 + 2a)
             = 4a + 4 $-$ a²   $-$ 1 $-$ 2a
             = 2a $-$ a² + 3
f(a $-$ 1) = 4(a $-$ 1) $-$ (a $-$ 1)²
             = 4a $-$ 4 $-$ (a² + 1 $-$ 2a)
             = 4a $-$ 4 $-$ 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 3.47:

Question 11:

If f, g, h are real functions given by f(x) = x², g(x) = tan x and h(x) = log_ex, then write the value of (hogof) $(\sqrt{\frac{π}{4}})$ .

Answer:

Given : f(x) = x², g(x) = tan x and h(x) = log_ex.
(hogof) $(\sqrt{\frac{π}{4}})$ = $h (g (f (\sqrt{\frac{π}{4}})))$
$= h (g ({(\sqrt{\frac{π}{4}})}^{2})) = h (g (\frac{π}{4})) = h (\tan (\frac{π}{4})) = h (1) = \log_{e} 1 = 0$

Page No 3.47:

Question 12:

Write the domain and range of function f(x) given by $f (x) = \frac{1}{\sqrt{x - |x|}}$ .

Answer:

Given:
$f (x) = \frac{1}{\sqrt{x - |x|}}$

We know that
$|x| = \{\begin{cases} x, & if x \geq 0 \\ - x, & if x < 0 \end{cases}$
$\Rightarrow x - |x| = \{\begin{cases} x - x = 0, & if x \geq 0 \\ x + x = 2 x, & if x < 0 \end{cases}$
⇒ x $-$ | x| ≤ 0 for all x.
$\Rightarrow \frac{1}{\sqrt{x - |x|}}$ 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 3.48:

Question 13:

Write the domain and range of $f (x) = \sqrt{x - [x]}$ .

Answer:

$f (x) = \sqrt{x - [x]}$
$Since f (x) is defined for all values of x, x \in R . Or dom (f (x)) = R Since, x - [x] = {x}, which is the fractional part of any real number x, f (x) = \sqrt{{x}} . . . . . (1) We know that 0 \leq {x} < 1 \Rightarrow \sqrt{0} \leq \sqrt{{x}} < \sqrt{1} \Rightarrow 0 \leq f (x) < 1 {from (1)} Thus, r ange of f (x) is [0, 1) .$

Page No 3.48:

Question 14:

Write the domain and range of function f(x) given by $f (x) = \sqrt{[x] - x}$ .

Answer:

$f (x) = \sqrt{[x] - x}$

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

Page No 3.48:

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 a_i (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 × ... × q (p times) = q^p

Page No 3.48:

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 = D_f= {2, 5, 8, 10}

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

∴ Domain of f + g = D_f∩ D_g = {2, 8, 10}

Page No 3.48:

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.

$∴ f (x) = g (x) \Rightarrow 3 x^{2} - 1 = 3 + x \Rightarrow 3 x^{2} - x - 4 = 0 \Rightarrow (x + 1) (3 x - 4) = 0$
$\Rightarrow x + 1 = 0 or 3 x - 4 = 0 \Rightarrow x = - 1 or x = \frac{4}{3}$

Hence, the set of values of x for which the given functions are equal is $\{- 1, \frac{4}{3}\}$ .

Page No 3.48:

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 = D_f= {0, 2, 3, 4, 5}

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

∴ Domain of fg = D_f∩ D_g = {2, 3, 4, 5}

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

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.
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
⇒ $x = 1 \pm \sqrt{10}$
Since $x = 1 \pm \sqrt{10} \notin A$ , 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 $0, 2 \in A$ . 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 $- 2 \in A$ , − 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
$f (x) = \{\begin{matrix} 3 x - 2, & x < 0; \\ 1, & x = 0; \\ 4 x + 1, & x > 0 . \end{matrix}$
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]
          = $-$ 3 $-$ 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) = log_ex. 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) = log_ex .............(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 :
log_ex = $-$ 2
⇒ x = $e^{- 2}$
Hence, { x : f (x) = - 2} = { e ^{– 2}} .      [Since log_ab = c ⇒ b = a^c]

(c) f (xy) = log_e(xy)      {From(i)}
         = log_ex + log_ey         [Since log_emn = log_e m + log_en]
         = 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)}
f₂is 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 → Z 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 $x \in A$ .n∈A
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 : A → B is such that y ∈ B, then $f^{- 1}$ { y }={x ∈ A: 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 = $-$ 3 $-$ 1 = $-$ 4
⇒ $x = \sqrt{- 4}$
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 : A → N 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 $n \in A$ .
Therefore,
range of f = {3, 5, 11, 13}

Page No 3.8:

Question 16:

The function f is defined by $f (x) = \{\begin{matrix} x^{2}, & 0 \leq x \leq 3 \\ 3 x, & 3 \leq x \leq 10 \end{matrix}$
The relation g is defined by $g (x) = \{\begin{cases} x^{2}, & 0 \leq x \leq 2 \\ 3 x, & 2 \leq x \leq 10 \end{cases}$
Show that f is a function and g is not a function.

Answer:

The function f is defined by $f (x) = \{\begin{cases} x^{2} & 0 ⩽ x ⩽ 3 \\ 3 x & 3 ⩽ x ⩽ 10 \end{cases}$
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 $g (x) = \{\begin{cases} x^{2}, & 0 ⩽ x ⩽ 2 \\ 3 x, & 2 ⩽ x ⩽ 10 \end{cases}$
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 $\frac{f (1.1) - f (1)}{(1.1) - 1}$ .

Answer:

Given:
f(x) = x²

Therefore,

$\frac{f (1.1) - f (1)}{(1.1) - 1} = \frac{{(1.1)}^{2} - {(1)}^{2}}{(1.1 - 1)} = \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1$

Page No 3.8:

Question 18:

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

Answer:

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

Now,

$f (- 1) = {(- 1)}^{3} + 1 = 0 f (0) = 0^{3} + 1 = 1 f (3) = 3^{3} + 1 = 28 f (7) = 7^{3} + 1 = 344 f (9) = 9^{3} + 1 = 730$

So, $f = \{(x, f (x)) : x \in X\} = \{(- 1, 0), (0, 1), (3, 28), (7, 344), (9, 730)\}$

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