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#### Question 1:

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

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

Hence, $x=-1,\frac{2}{3}$.

#### Question 2:

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

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

#### Question 3:

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

Given:
$f\left(x\right)=\frac{ax-b}{bx-a}$
Let y = f (x) .
y( bx $-$ a) = ax b
xybay = axb
xybax = ayb
x(bya) = ayb
$⇒x=\frac{ay-b}{by-a}$
x = f (y)
Hence proved.

#### Question 4:

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

Given:
$f\left(x\right)=\frac{1}{1-x}\phantom{\rule{0ex}{0ex}}$
Thus,
$f\left\{f\left(x\right)\right\}=f\left\{\frac{1}{1-x}\right\}\phantom{\rule{0ex}{0ex}}$

$=\frac{1}{1-\frac{1}{1-x}}\phantom{\rule{0ex}{0ex}}$

$=\frac{1}{\frac{1-x-1}{1-x}}\phantom{\rule{0ex}{0ex}}=\frac{1-x}{-x}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{x}$
Again,
$f\left[f\left\{f\left(x\right)\right\}\right]=f\left[\frac{x-1}{x}\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{1-\left(\frac{x-1}{x}\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{x-x+1}{x}}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}\phantom{\rule{0ex}{0ex}}=x$

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

#### Question 5:

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

Given:
$f\left(x\right)=\frac{x+1}{x-1}$
Therefore,
$f\left[f\left\{\left(x\right)\right\}\right]=f\left(\frac{x+1}{x-1}\right)\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(\frac{x+1}{x-1}\right)+1}{\left(\frac{x+1}{x-1}\right)-1}\phantom{\rule{0ex}{0ex}}$
$=\frac{\frac{x+1+x-1}{x-1}}{\frac{x+1-x+1}{x-1}}=\frac{\frac{2x}{x-1}}{\frac{2}{x-1}}=\frac{2x}{2}=x$
Thus,
f [ f {(x)}] = x
Hence proved.

#### Question 6:

If
find: (a) f(1/2), (b) f(−2), (c) f(1), (d) $f\left(\sqrt{3}\right)$ and (e) $f\left(\sqrt{-3}\right)$.

Given:

Now,
(a) $f\left(\frac{1}{2}\right)=\frac{1}{2}$                    [ Using f (x) = x, 0 ≤ x < 1]
(b) f ($-$2) = ( $-$ 2)2 = 4
(c) $f\left(1\right)=\frac{1}{1}=1$
(d) $f\left(\sqrt{3}\right)=\frac{1}{\sqrt{3}}$
(e) $f\left(\sqrt{-3}\right)$
Since x is not defined in R, $f\left(\sqrt{-3}\right)$ does not exist.

#### Question 7:

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

Given:
$f\left(x\right)={x}^{3}-\frac{1}{{x}^{3}}\phantom{\rule{0ex}{0ex}}$  ...(i)
Thus,
$f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^{3}-\frac{1}{{\left(\frac{1}{x}\right)}^{3}}\phantom{\rule{0ex}{0ex}}$

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

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

$f\left(x\right)+f\left(\frac{1}{x}\right)=\left({x}^{3}-\frac{1}{{x}^{3}}\right)+\left(\frac{1}{{x}^{3}}-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}$

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

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

#### Question 8:

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

Given:
$f\left(x\right)=\frac{2x}{1+{x}^{2}}$
Thus,
$f\left(\mathrm{tan}\theta \right)=\frac{2\left(\mathrm{tan}\theta \right)}{1+{\mathrm{tan}}^{2}\theta }\phantom{\rule{0ex}{0ex}}$

Hence,  f (tan θ) = sin 2θ.

#### Question 9:

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

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

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

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

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

#### Question 10:

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

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

Thus, f(f(x)) = x.
Hence proved.

#### Question 11:

If for non-zero x, af(x) + bf $\left(\frac{1}{x}\right)=\frac{1}{x}-5$, where ab, then find f(x).

Given:
$af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5\phantom{\rule{0ex}{0ex}}$              ...(i)
$⇒af\left(\frac{1}{x}\right)+bf\left(x\right)=\frac{1}{\frac{1}{x}}-5\phantom{\rule{0ex}{0ex}}$
$⇒af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$        ...(ii)

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

$af\left(x\right)+bf\left(x\right)+bf\left(\frac{1}{x}\right)+af\left(\frac{1}{x}\right)=\frac{1}{x}-5+x-5\phantom{\rule{0ex}{0ex}}$
$⇒\left(a+b\right)f\left(x\right)+\left(a+b\right)f\left(\frac{1}{x}\right)=\frac{1}{x}+x-10\phantom{\rule{0ex}{0ex}}$
$⇒f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{\left(a+b\right)}\left[\frac{1}{x}+x-10\right]$             ...(iii)

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

$af\left(x\right)-bf\left(x\right)+bf\left(\frac{1}{x}\right)-af\left(\frac{1}{x}\right)=\frac{1}{x}-5-x+5\phantom{\rule{0ex}{0ex}}$
$⇒\left(a-b\right)f\left(x\right)-f\left(\frac{1}{x}\right)\left(a-b\right)=\frac{1}{x}-x\phantom{\rule{0ex}{0ex}}$
$⇒f\left(x\right)-f\left(\frac{1}{x}\right)=\frac{1}{\left(a-b\right)}\left[\frac{1}{x}-x\right]$                   ...(iv)

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

$2f\left(x\right)=\frac{1}{a+b}\left[\frac{1}{x}+x-10\right]+\frac{1}{a-b}\left[\frac{1}{x}-x\right]\phantom{\rule{0ex}{0ex}}$
$⇒2f\left(x\right)=\frac{\left(a-b\right)\left[\frac{1}{x}+x-10\right]+\left(a+b\right)\left[\frac{1}{x}-x\right]}{\left(a+b\right)\left(a-b\right)}\phantom{\rule{0ex}{0ex}}$
$⇒2f\left(x\right)=\frac{\frac{a}{x}+ax-10a-\frac{b}{x}-bx+10b+\frac{a}{x}-ax+\frac{b}{x}-bx}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}$
$⇒2f\left(x\right)=\frac{\frac{2a}{x}-10a+10b-2bx}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}$
$⇒f\left(x\right)=\frac{1}{{a}^{2}-{b}^{2}}×\frac{1}{2}\left[\frac{2a}{x}-10a+10b-2bx\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-5a+5b-bx\right]$

Therefore,
$f\left(x\right)=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx-5a+5b\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5\left(a-b\right)}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5}{\left(a+b\right)}$

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

#### Question 1:

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

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

(i) Given: $f\left(x\right)=\frac{1}{x}$
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: $f\left(x\right)=\frac{1}{\left(x-7\right)}$
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: $f\left(x\right)=\frac{3x-2}{\left(x+1\right)}$
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: $f\left(x\right)=\frac{2x+1}{{x}^{2}-9}$
Domain of f :
Clearly,  f (x) is defined for all  xR except for x2 $-$ 9 ≠  0, i.e. x = ± 3.
At x = $-$3, 3,  f (x) takes the intermediate form $\frac{1}{0}.$
Hence, domain ( f ) = R $-$$-$ 3, 3 }.

(v) Given:

$=\frac{{x}^{2}+2x+1}{{x}^{2}-6x-2x+12}\phantom{\rule{0ex}{0ex}}$

$=\frac{{x}^{2}+2x+1}{x\left(x-6\right)-2\left(x-6\right)}\phantom{\rule{0ex}{0ex}}$

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

#### Question 2:

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

(i) $f\left(x\right)=\sqrt{x-2}$
(ii) $f\left(x\right)=\frac{1}{\sqrt{{x}^{2}-1}}$
(iii) $f\left(x\right)=\sqrt{9-{x}^{2}}$
(iv) $f\left(x\right)=\frac{\sqrt{x-2}}{3-x}$

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

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

(iv) Given: $f\left(x\right)=\sqrt{\frac{x-2}{3-x}}$
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) .

#### Question 3:

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

(i) $f\left(x\right)=\frac{ax+b}{bx-a}$

(ii) $f\left(x\right)=\frac{ax-b}{cx-d}$

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

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

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

(vi) $f\left(x\right)=\left|x-1\right|$

(vii) $f\left(x\right)=-\left|x\right|$

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

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

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

(i)
Given:
$f\left(x\right)=\frac{ax+b}{bx-a}$
Domain of f : Clearly,  f (x) is a rational function of x as $\frac{ax+b}{bx-a}$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.
$⇒x=\frac{a}{b}$
Hence, domain ( f ) = $R-\left\{\frac{a}{b}\right\}$
Range of f :
Let f (x) = y
$⇒\frac{ax+b}{bx-a}=y\phantom{\rule{0ex}{0ex}}$

⇒ (ax + b) = y (bx $-$ a)
⇒ (ax + b) = (bxy $-$ ay)
b + ay = bxy $-$ ax
b + ay = x(by $-$ a)
$⇒x=\frac{b+ay}{by-a}$
Clearly, f (x) assumes real values for all x except for all those values of x for which ( by $-$ a) = 0, i.e. by = a.
$⇒y=\frac{a}{b}$.
Hence, range ( f ) = $R-\left\{\frac{a}{b}\right\}$
(ii)
Given:
$f\left(x\right)=\frac{ax-b}{cx-d}$
Domain of f : Clearly,  f (x) is a rational function of x as $\frac{ax-b}{cx-d}$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.
$⇒x=\frac{d}{c}$.
Hence, domain ( f ) = $R-\left\{\frac{d}{c}\right\}$
Range of f :
Let f (x) = y
$⇒\frac{ax-b}{cx-d}=y\phantom{\rule{0ex}{0ex}}$

⇒ (ax $-$ b) = y( cx $-$ d)
⇒ (ax $-$ b) = (cxy $-$ dy)
dy $-$ b = cxy $-$ ax
dy $-$ b = x(cy $-$ a)
$⇒x=\frac{dy-b}{cy-a}$
Clearly, f (x) assumes real values for all x except for all those values of x for which ( cy $-$ a) = 0, i.e. cy = a.
$⇒y=\frac{a}{c}$.
Hence, range ( f ) = $R-\left\{\frac{a}{c}\right\}$ .

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

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

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

$\frac{x-2}{2-x}=y$
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 xR, (x – 1) assumes all real numbers.
Thus, the range of f is the set of all non-negative real numbers.
Hence, range of f = [0, ∞) .

(vii)
f (x) = – | x |, xR
We know that
$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x& x<0\end{array}\right\\phantom{\rule{0ex}{0ex}}$
$\therefore f\left(x\right)=-\left|x\right|=\left\{\begin{array}{ll}-x,& x\ge 0\\ x,& x<0\end{array}\right\$
Since f(x) is defined for xR, domain of f  = R.
It can be observed that the range of f (x) = – | x | is all real numbers except positive real numbers.
∴ The range of f is (– ∞, 0).

(viii) Given:
$f\left(x\right)=\sqrt{9-{x}^{2}}$

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

$\frac{1}{\sqrt{16-{x}^{2}}}$ 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

Hence, range ( f ) = [).

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

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

For x ≤ – 4, we have:
x2 $-$ 16 ≥ 0
$⇒\sqrt{{x}^{2}-16}\ge 0$
f (x) ≥ 0

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

#### Question 1:

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

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

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

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

(b) Given:
$f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$
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) = $\sqrt{x-1}×\sqrt{x+1}=\sqrt{{x}^{2}-1}$ .

#### Question 2:

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

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.
.

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

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

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

(iv) Given:
g(x) = x2 + x
g(x) = 0 ⇒ x2 + x = 0 = x(x+ 1) = 0
x = 0 or (x + 1) = 0
x = 0 or x$-$ 1
Now,
.
Hence, $\text{d}\mathrm{omain}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\mathrm{R}-\left\{-1,0\right\}$.

#### Question 3:

If f(x) be defined on [−2, 2] and is given by and g(x) $=f\left(\left|x\right|\right)+\left|f\left(x\right)\right|$, find g(x).

Given:
$f\left(x\right)=\left\{\begin{array}{ll}-1,& -2⩽x⩽0\\ x-1,& 0
Thus,
$g\left(x\right)=f\left(\left|x\right|\right)+\left|f\left(x\right)\right|$

#### Question 4:

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

Given:

Clearly, $f\left(x\right)=\sqrt{x+1}$ is defined for all x$-$1.
Thus, domain (f) = [ $-$1, ∞]
Again,

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

(iv) .

(v) .

(vi)
$=2\sqrt{x+1}-\sqrt{45-5{x}^{2}}$ .

(vii)            {Since domain(f) = [$-$ 1, ∞]}
$={\left(\sqrt{x+1}\right)}^{2}+7\left(\sqrt{x+1}\right)=x+1+7\sqrt{x+1}\phantom{\rule{0ex}{0ex}}$

(viii)            {Since domain(g) = [$-$ 3, 3]}

#### Question 5:

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

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

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

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

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

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

Thus,
$\mathrm{domain}\left(\frac{f}{g}\right)=\mathrm{domain}\left(f\right)\cap \mathrm{domain}\left(g\right)-\left\{x:g\left(x\right)=0\right\}$

(iv) Given:
f(x) = loge (1 − x)
$⇒\frac{1}{f\left(x\right)}=\frac{1}{{\mathrm{log}}_{e}\left(1-x\right)}$
$\frac{1}{f\left(x\right)}$ is defined if loge( 1$-$x) is defined and loge(1 – x) ≠ 0.
⇒ (1$-$ x) > 0 and (1 $-$ x) ≠ 0
x < 1 and x ≠ 0
x ∈ ( $-$ ∞, 0)∪ (0, 1)
Thus, $\text{d}\mathrm{omain}\left(\frac{g}{f}\right)=\left(-\infty ,0\right)\cup \left(0,1\right)$ = ( $-$ ∞, 1)  .

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

(fg)(0) = loge ( 1 – 0) × [0] = 0

$\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)=\frac{\left[\frac{1}{2}\right]}{{\mathrm{log}}_{\mathrm{e}}\left(1-\frac{1}{2}\right)}=0$

#### Question 6:

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

Given:

Clearly, f (x) is defined for x + 1 ≥ 0 .
x$-$ 1
x ∈ [ $-$1, ∞]
Thus, domain ( f ) = [ $-$1, ∞] .
Clearly, g (x) is defined for x ≠ 0 .
x ∈ R – { 0} and h(x) is defined for all x such that  x ∈ R .
Thus,
domain ( f ) ∩ domain (g) ∩ domain (h) = [ $-$ 1, ∞] – { 0}.
Hence,
(2f + g h) : [ $-$ 1, ∞] – { 0} → R is given by:
(2f + gh)(x) = 2f (x) + g (x$-$ h (x)
$=2\sqrt{x+1}+\frac{1}{x}-2{x}^{2}+3\phantom{\rule{0ex}{0ex}}$

(2f + gh) (0) does not exist because 0  does not lie in the domain x [ - 1, ∞] – {0}.

#### Question 7:

The function f is defined by . Draw the graph of f(x).

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:

#### Question 8:

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

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

$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)},g\left(x\right)\ne 0,x\in \mathbit{R}\phantom{\rule{0ex}{0ex}}$

$⇒\left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{2x-3},x\ne \frac{3}{2}$

#### Question 9:

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

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

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

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

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

#### Question 10:

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

Given:
f (x)  = x2 and g (x) = 2x + 1
Clearly, D (f) =  R and D (g) = R

$D\left(fg\right)=D\left(f\right)\cap D\left(g\right)=R\cap R=R\phantom{\rule{0ex}{0ex}}$
$D\left(\frac{f}{g}\right)=D\left(f\right)\cap D\left(g\right)-\left\{x:g\left(x\right)=0\right\}=R\cap R-\left\{-\frac{1}{2}\right\}=R-\left\{-\frac{1}{2}\right\}$
Thus,
(f + g) (x) : RR is given by (f + g) (x) = f (x) + g (x) = x2 + 2x + 1= (x + 1)2 .
($-$ g) (x) : RR is given by (f $-$ g) (x) = f (x$-$ g (x) = x2 $-$ 2x $-$1.
(fg) (x) : RR is given by (fg) (x) = f(x).g(x) = x2(2x + 1) = 2x3 + x2 .

.

#### Question 1:

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

Given:
f (x) =  | x |, xR
We know that
$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x& x<0\end{array}\right\\phantom{\rule{0ex}{0ex}}$
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, ∞) .

#### Question 2:

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

Given:
$f\left(x+\frac{1}{x}\right)={x}^{2}+\frac{1}{{x}^{2}}$
$={x}^{2}+\frac{1}{{x}^{2}}+2-2\phantom{\rule{0ex}{0ex}}$
$={\left(x+\frac{1}{x}\right)}^{2}-2$
Thus,
$f\left(x+\frac{1}{x}\right)={\left(x+\frac{1}{x}\right)}^{2}-2$
Hence,
f (x)  = x2 $-$ 2 , where | x | ≥  2.

#### Question 3:

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

Given : f(x) = sin [x], where $\frac{-\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{4}$.

#### 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).

f(x) = cos [π2]x + cos [−π2] x

#### Question 5:

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

Since f(x) = cos [x], where $\frac{-\mathrm{\pi }}{2},

#### Question 6:

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

f(x) = ex[x], x ∈ R

#### Question 7:

Let . Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.

Given:

#### Question 8:

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

Given:
$f\left(x\right)=1-\frac{1}{x}\phantom{\rule{0ex}{0ex}}$
Now,
$f\left(\frac{1}{x}\right)=1-\frac{1}{\frac{1}{x}}=1-x$
$⇒f\left(f\left(\frac{1}{x}\right)\right)=f\left(1-x\right)$
Again,
If $f\left(x\right)=1-\frac{1}{x}\phantom{\rule{0ex}{0ex}}$
Thus,
$f\left(1-x\right)=1-\frac{1}{1-x}\phantom{\rule{0ex}{0ex}}$

$=\frac{1-x-1}{1-x}\phantom{\rule{0ex}{0ex}}=\frac{-x}{1-x}\phantom{\rule{0ex}{0ex}}=\frac{-x}{-\left(x-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{x}{x-1}$

#### Question 9:

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

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

$\frac{x-2}{2-x}=y$
x $-$ 2 = y (2 $-$ x)
x $-$ 2 = $-$ y (x $-$ 2)
y$-$ 1
Hence, range ( f ) = {$-$ 1}.

#### Question 10:

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

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

#### Question 11:

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

Given : f(x) = x2, g(x) = tan x and h(x) = loge x.
(hogof) $\left(\sqrt{\frac{\mathrm{\pi }}{4}}\right)$ =

#### Question 12:

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

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

We know that

x $-$ | x| ≤ 0 for all x.
$⇒\frac{1}{\sqrt{x-\left|x\right|}}$ 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 .

#### Question 13:

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

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

#### Question 14:

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

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

#### 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.

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

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

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

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

#### 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.

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

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

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

Now,

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

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

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

#### Question 17:

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

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

$\therefore f\left(x\right)=g\left(x\right)\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-1=3+x\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-x-4=0\phantom{\rule{0ex}{0ex}}⇒\left(x+1\right)\left(3x-4\right)=0$

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

#### 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.

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

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

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

Now,

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

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

∴ Domain of fg = Df Dg = {2, 3, 4, 5}

#### 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)}

(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.

#### Question 2:

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

(a) 3
(b) −3
(c) {−3, 3}
(d) ϕ

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

#### 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}

(b) {(x, y) : x, y ∈, R, y2 = x}

#### Question 4:

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

(d) None of these

Given:
$f\left(x\right)=\mathrm{cos}\left(\mathrm{log}x\right)$
$⇒f\left({x}^{2}\right)=\mathrm{cos}\left(\mathrm{log}\left({x}^{2}\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left({x}^{2}\right)=\mathrm{cos}\left(2\mathrm{log}\left(x\right)\right)$

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

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

$⇒f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)=\mathrm{cos}\left(\left(2\mathrm{log}x-2\mathrm{log}y\right)\right)+\mathrm{cos}\left(\left(2\mathrm{log}x+2\mathrm{log}y\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)=2\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)\right]=\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)$
$⇒f\left({x}^{2}\right)f\left({y}^{2}\right)-\frac{1}{2}\left\{f\left({x}^{2}{y}^{2}\right)+f\left(\frac{{x}^{2}}{{y}^{2}}\right)\right\}=\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)-\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)=0$

#### Question 5:

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

(d) None of these

Given:
$f\left(x\right)=\mathrm{cos}\left(\mathrm{log}x\right)$
$f\left(y\right)=\mathrm{cos}\left(\mathrm{log}y\right)$

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

$⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=\mathrm{cos}\left(\mathrm{log}x-\mathrm{log}y\right)+\mathrm{cos}\left(\mathrm{log}x+\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=2\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]=\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)$
$⇒f\left(x\right)f\left(y\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}=\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)-\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)=0$

#### 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

(d) None of these

#### 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]

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

Since, f(x) = cos [x], where $\frac{-\mathrm{\pi }}{2},

#### Question 8:

Which of the following are functions?

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

(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.

#### Question 9:

If , then f(g(x)) is equal to
(a) f(3x)
(b) {f(x)}3
(c) 3f(x)
(d) −f(x)

(c) 3f(x)

#### 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

(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.

#### Question 11:

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

(c) 2f(x)

#### Question 12:

If f(x) = cos (log x), then value ofis
(a) 1
(b) −1
(c) 0
(d) ±1

(c) 0

Given : f(x) = cos (log x)
Then,

#### Question 13:

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

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

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

#### Question 14:

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

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

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

2f (x) − $3f\left(\frac{1}{x}\right)={x}^{2}$         (x ≠ 0)                  ....(1)

#### 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|

(b) 2|x|

f(x) = 2x + |x|
Then, f(2x) + f(−x) − f(x)

#### Question 16:

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

(c) R − {-1/2,1}

$f\left(x\right)=\frac{{x}^{2}-x}{{x}^{2}+2x}$

#### Question 17:

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

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

#### Question 18:

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

Given:
$f\left(x\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x\right)$
$⇒f\left(\frac{1}{x}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}\left(\frac{1}{x}\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{1}{x}\right)=\mathrm{cos}\left(-{\mathrm{log}}_{e}\left(x\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{1}{x}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}\left(x\right)\right)$

Similarly,
$f\left(\frac{1}{y}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}y\right)$

Now,
$f\left(xy\right)=\mathrm{cos}\left({\mathrm{log}}_{e}xy\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x+{\mathrm{log}}_{e}y\right)$
and
$f\left(\frac{x}{y}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}\frac{x}{y}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x-{\mathrm{log}}_{e}y\right)$

$⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x-{\mathrm{log}}_{e}y\right)+\mathrm{cos}\left({\mathrm{log}}_{e}x+{\mathrm{log}}_{e}y\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=2\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]=\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)$
$⇒f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}=\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)-\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)=0$
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.

#### Question 19:

Let f(x) = x, $g\left(x\right)=\frac{1}{x}$ 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

(d) x ∈ R, x ≠ 0

Given:
f(x) = x, $g\left(x\right)=\frac{1}{x}$ and h(x) = f(x) g(x)
Now,
$h\left(x\right)=x×\frac{1}{x}=1$
We observe that the domain of f is $\mathrm{ℝ}$ and the domain of g is $\mathrm{ℝ}-\left\{0\right\}$.
∴ Domain of h = Domain of f ⋂ Domain of g = $\mathrm{ℝ}\cap \left[\mathrm{ℝ}-\left\{0\right\}\right]=\mathrm{ℝ}-\left\{0\right\}$
$⇒$x ∈ R, x ≠ 0

#### Question 20:

If for x ∈ R, then f (2002) =
(a) 1
(b) 2
(c) 3
(d) 4

(a) 1

Given:

On dividing the numerator and denominator by , we get
(For every x ∈ R)

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

#### Question 21:

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

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

(c) (3/4, 1)

Given:
f(x) = cos2x + sin4x
$⇒f\left(x\right)=1-{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x$
$⇒f\left(x\right)={\left({\mathrm{sin}}^{2}x-\frac{1}{2}\right)}^{2}+\frac{3}{4}$

The minimum value of $f\left(x\right)$ is $\frac{3}{4}$.
Also,
${\mathrm{sin}}^{2}x\le 1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}x-\frac{1}{2}\le \frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{\left({\mathrm{sin}}^{2}x-\frac{1}{2}\right)}^{2}\le \frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒{\left({\mathrm{sin}}^{2}x-\frac{1}{2}\right)}^{2}+\frac{3}{4}\le \frac{1}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)\le 1$

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

∴ f(R) = (3/4, 1)

#### Question 22:

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

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

Disclaimer: The question in the book has some error. The solution is created according to the question given in the book.

#### 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

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

#### Question 24:

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

(a) {−1}
(b) {0}
(c) $\left\{-\frac{1}{2}\right\}$
(d) ϕ

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

Given:

We know,
$\left|x\right|\ge 0$
$f\left(\left|x\right|\right)=\left|x\right|-1$             ...(1)

Also,
If $x\le 0$, then $\left|x\right|=-x$    ...(2)

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

#### Question 25:

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

(a) 0.5

${e}^{f\left(x\right)}=\frac{10+x}{10-x}$
...(1)

#### Question 26:

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

(d) None of these

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

$⇒f\left(\alpha \right)=12\left({\left(12\right)}^{2}-9\right)$ and $f\left(\beta \right)=12\left({\left(12\right)}^{2}-9\right)$
$⇒f\left(\alpha \right)=f\left(\beta \right)=12\left({\left(12\right)}^{2}-9\right)$

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.

#### Question 27:

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

(a) $f\left(\alpha \right)=f\left(\beta \right)=-9$

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

Since α and β are the roots of $4x+\frac{1}{x}=3$,

$⇒f\left(\alpha \right)=3\left({\left(3\right)}^{2}-12\right)=-9$ and $f\left(\beta \right)=3\left({\left(3\right)}^{2}-12\right)=-9$
$⇒f\left(\alpha \right)=f\left(\beta \right)=-9$

#### Question 28:

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

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

(d) None of these

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

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.

#### Question 29:

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

(c) f(x) + f(1 − x) = 1

$f\left(x\right)=\frac{{4}^{x}}{{4}^{x}+2}$x ∈ R

#### Question 30:

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

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

(a) f(π/2) = 1

f(x) = sin [π2] x + sin [−π2]x

#### Question 31:

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

(a)
(b)
(c) [−2, 2]
(d)

(b)

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

#### Question 32:

The domain of definition of is

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

(a) (−∞, −3] ∪ (2, 5)

#### Question 33:

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

(a) [−1, 2) ∪ [3, ∞)

#### Question 34:

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

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

(d) [1, 3]

$f\left(x\right)=\sqrt{x-1}+\sqrt{3-x}$

#### Question 35:

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

(c) ϕ

$f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$

#### Question 36:

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

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

(d) R − {0}

f(x) = log |x|

#### Question 37:

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

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

(d) [0, 4]

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

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

#### Question 38:

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

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

(a) [4, ∞)

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

#### Question 39:

The domain of the function is

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

(c) [−3, − 2] ∪ [2, 3]

#### Question 40:

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

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

(c) {−1, 1}

$f\left(x\right)=\frac{x}{\left|x\right|}$

#### Question 41:

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

(a) {−1, 1}

$f\left(x\right)=\frac{x+2}{\left|x+2\right|}$ , x ≠ −2

#### Question 42:

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

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

(b) [0, ∞)

#### Question 43:

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

(a) $f\left(xy\right)=f\left(x\right)f\left(y\right)$                     (b) $f\left(xy\right)\ge f\left(x\right)f\left(y\right)$                     (c) $f\left(xy\right)\le f\left(x\right)f\left(y\right)$                     (d) none of these

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

Replacing x by y in (1), we get

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

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

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

Now,

${x}^{2}{y}^{2}+1\le {x}^{2}{y}^{2}+{x}^{2}+{y}^{2}+1\phantom{\rule{0ex}{0ex}}⇒\sqrt{{x}^{2}{y}^{2}+1}\le \sqrt{{x}^{2}{y}^{2}+{x}^{2}+{y}^{2}+1}\phantom{\rule{0ex}{0ex}}⇒f\left(xy\right)\le f\left(x\right)f\left(y\right)$

Hence, the correct answer is option (c).

#### Question 44:

If ${\left[x\right]}^{2}-5\left[x\right]+6=0$, where [.] denotes the greatest integer function, then

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

The given equation is ${\left[x\right]}^{2}-5\left[x\right]+6=0$.

${\left[x\right]}^{2}-5\left[x\right]+6=0\phantom{\rule{0ex}{0ex}}⇒{\left[x\right]}^{2}-3\left[x\right]-2\left[x\right]+6=0\phantom{\rule{0ex}{0ex}}⇒\left[x\right]\left(\left[x\right]-3\right)-2\left(\left[x\right]-3\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\left[x\right]-2\right)\left(\left[x\right]-3\right)=0$

x ∈ [2, 3) or x ∈ [3, 4)
x ∈ [2, 4)

Hence, the correct answer is option (d).

#### Question 45:

The range of $f\left(x\right)=\frac{1}{1-2\mathrm{cos}x}$ is

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

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

Now,

But,

$\mathrm{cos}x\ne \frac{1}{2}$
$⇒1-2\mathrm{cos}x\in \left[-1,3\right]-\left\{0\right\}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{1-2\mathrm{cos}x}\in \left(-\infty ,-1\right]\cup \left[\frac{1}{3},\infty \right)$

∴ 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, ∞)".

#### Question 1:

Define a function as a set of ordered pairs.

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}.

#### Question 2:

Define a function as a correspondence between two sets.

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.

#### Question 3:

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

Differences between relation and function

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

#### 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.

(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
$x=1±\sqrt{10}$
Since $x=1±\sqrt{10}\notin A$, 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 $\varphi ,\left\{0,2,\right\},-2$ respectively.

#### Question 5:

If a function f : R → R be defined by
$f\left(x\right)=\left\{\begin{array}{cc}3x-2,& x<0;\\ 1,& x=0;\\ 4x+1,& x>0.\end{array}\right\$
find: f(1), f(−1), f(0) and f(2).

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

#### 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].

(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

#### Question 7:

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

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

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

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

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

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

#### 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]}

(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.

#### 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?

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.

#### 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.

(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}

#### 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)}

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

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

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

#### 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.

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

#### Question 13:

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

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

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

#### 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)].

(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.

#### 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.

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

#### Question 16:

The function f is defined by$f\left(x\right)=\left\{\begin{array}{cc}{x}^{2},& 0\le x\le 3\\ 3x,& 3\le x\le 10\end{array}\right\$
The relation g is defined by $g\left(x\right)=\left\{\begin{array}{ll}{x}^{2},& 0\le x\le 2\\ 3x,& 2\le x\le 10\end{array}\right\$
Show that f is a function and g is not a function.

The function f is defined by $f\left(x\right)=\left\{\begin{array}{ll}{x}^{2}& 0⩽x⩽3\\ 3x& 3⩽x⩽10\end{array}\right\$
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\left(x\right)=\left\{\begin{array}{ll}{x}^{2},& 0⩽x⩽2\\ 3x,& 2⩽x⩽10\end{array}\right\$
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.

#### Question 17:

If f(x) = x2, find $\frac{f\left(1.1\right)-f\left(1\right)}{\left(1.1\right)-1}$.

Given:
f(x) = x2

Therefore,

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

#### Question 18:

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

$f\left(-1\right)={\left(-1\right)}^{3}+1=0\phantom{\rule{0ex}{0ex}}f\left(0\right)={0}^{3}+1=1\phantom{\rule{0ex}{0ex}}f\left(3\right)={3}^{3}+1=28\phantom{\rule{0ex}{0ex}}f\left(7\right)={7}^{3}+1=344\phantom{\rule{0ex}{0ex}}f\left(9\right)={9}^{3}+1=730$
So, $f=\left\{\left(x,f\left(x\right)\right):x\in X\right\}=\left\{\left(-1,0\right),\left(0,1\right),\left(3,28\right),\left(7,344\right),\left(9,730\right)\right\}$