Maths M.C.Q (1 Marks)

1. $\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$

Solution:

If we solve it by the trial-and-error method, we can see that (a) satisfies the given condition.

From (a):

$\text{f(x)}=\sin^2\text{x}$ and $\text{g(x)}=\sqrt{\text{x}}$

$\Rightarrow\ \text{f(g(x))}=\text{f}(\sqrt{\text{x}})=\sin^2\sqrt{\text{x}}$

$=(\sin\sqrt{\text{x}})^2$

4. {1, 2, 3}

Solution:

We know that

$7-\text{x}>0;\ \text{x}-3\geq0$ and $7-\text{x}\geq\text{x}-3$

$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $2\text{x}\leq10$

$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $\text{x}\leq5$

Therefore, x = 3, 4, 5

Range of $\text{f}=\Big\{^{(7-3)}\text{P}_{(3-3)},\ ^{(7-4)}\text{P}_{(4-3)},\ ^{(5-3)}\text{P}_{(7-5)}\Big\}$

= {4P₀, 3P₁, 2P₂}

= {1, 3, 2}

= {1, 2, 3}

3. x for all $\text{x}\in\text{R}-\{0,1\}$

Solution:

Domain of f: $1-\text{x}\neq0$

$\Rightarrow\ \text{x}\neq1$

Domain of f = R - {1}

Range of f: $\text{y}=\frac{1}{1-\text{x}}$

$\Rightarrow\ 1-\text{x}=\frac{1}{\text{y}}$

$\Rightarrow\ \text{x}=1-\frac{1}{\text{y}}$

$\Rightarrow\ \text{y}\neq0$

Range of f = R - {0}

So, f : R - {1} → R - {0} and f : R - {1} → R - {0}

Range of f is not a subset of the domain of f.

Domain (fof) = {x : $\text{x}\in$ domain of f and $\text{f(x)}\in$ domain of f}

Domain (fof) $=\big\{\text{x}:\text{x}\in\text{R}-\{1\}\text{ and }\frac{1}{1-\text{x}}\in\text{R}-\{1\}\big\}$

Domain (fof) $=\big\{\text{x}:\text{x}\neq1\text{ and }\frac{1}{1-\text{x}}\neq1\big\}$

Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }1-\text{x}\neq1\}$

Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }\text{x}\neq0\}$

Domain (fof) = R - {0, 1}

(fof)(x) = f(f(x))

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

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

For range of fof, $\text{x}\neq0$

Now, fof : R → {0, 1} → R - {0} and f : R - {1} → R - {0}

Range of fof is not a subset of domain of f.

Domain (fo(fof)) $=\{\text{x}:\text{x}\in$ domain of fof and (fof)(x) $\in$ domain of f$\}$

Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\in\text{R}-\{0,1\}\text{ and }\frac{\text{x}-1}{\text{x}}\in\text{R}-\{1\}\Big\}$

Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\neq0,1\text{ and }\frac{\text{x}-1}{\text{x}}\neq1\Big\}$

Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}-1\neq\text{x}\}$

Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}\in\text{R}\}$

Domain (fo(fof)) = R - {0, 1}

​​​​​​​Domain (fo(fof)) = f((fof)(x))

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

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

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

$=\text{x}$

So, (fo(fof))(x) = x, where $\text{x}\neq0,1$

2. One-one but not onto.

Solution:

Given function is $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ on $\text{f}:[0,\infty)\rightarrow\ \text{R}$

If f(x) = f(y)

$\Rightarrow\ \frac{\text{x}}{\text{x}+1}=\frac{\text{y}}{\text{y}+1}$

⇒ xy + x = xy + y

⇒ x = y

Hence, f is one-one.

If y = f(x)

$\text{y}=\frac{\text{x}}{\text{x}+1}$

⇒ xy + y = x

⇒ xy - x = -y

x(y - 1) = -y

$\text{x}=\frac{-\text{y}}{\text{y}-1}\neq\text{f(x)}$

It is not onto.

4. One-one and onto both.

Solution:

Injectivity: Let x and y be any two elements in the domain (N).

Case-1: Both x and y are even.

Let f(x) = f(y)

$\Rightarrow\ \frac{-\text{x}}{2}=\frac{-\text{y}}{2}$

$\Rightarrow-\text{x}=-\text{y}$

$\Rightarrow\ \text{x}=\text{y}$

Case-2: Both x and y are odd.

Let f(x) = f(y)

$\Rightarrow\ \frac{\text{x}-1}{2}=\frac{\text{y}-1}{2}$

$\Rightarrow\ \text{x}-1=\text{y}-1$

$\Rightarrow\ \text{x}=\text{y}$

Case-3: Let x be even and y be odd.

Then, $\text{f(x)}=\frac{-\text{x}}{2}$ and $\text{f(y)}=\frac{\text{y}-1}{2}$

Then, clearly

$\text{x}\neq\text{y}$

$\Rightarrow\ \text{f(x)}\neq\text{f(y)}$

From all the cases, f is one-one.

Surjectivity: Co-domain of f = Z = {......, -3, -2, -1, 0, 1, 2, 3, ......}

Range of $\text{f}=\Big\{....,\ \frac{-3-1}{2},\ \frac{-(-2)}{2},\ \frac{-1-1}{2},\ \frac{0}{2},\ \frac{1-1}{2},\ \frac{-2}{2},\ \frac{3-1}{2},\ ....\Big\}$

⇒ Range of f = {....., -2, 1, -1, 0, 0, -1, 1, .....}

⇒ Range of f = {....., -2, -1, 0, 1, 2, ......}

⇒ Co-domain of f = Range of f

⇒ f is onto.

4. Onto but not one-one.

Solution:

$\text{M}=\begin{Bmatrix}\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}:\text{a, b, c, d}\in\text{R}\end{Bmatrix}$

f : M → R is given by f(A) = |A|

Injectivity: $\text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}0&0\\0&0\end{vmatrix}=0$

and $\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}1&0\\0&0\end{vmatrix}=0$

$\Rightarrow\ \text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=0$

So, f is not one-one.

Surjectivity: Let y be an element of the co-domain, such that

$\text{f(A)}=-\text{y},\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}$

$\Rightarrow\ \begin{vmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{vmatrix}=\text{y}$

$\Rightarrow\ \text{ad}-\text{bc}=\text{y}$

$\Rightarrow\ \text{a, b, c, d}\in\text{R}$

$\Rightarrow\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}\in\text{M}$

⇒ f is onto.

4. One-one and onto.

Solution:

Given function is,

$\text{f(n)}=\frac{\text{n}-1}{2}$ for n is odd

$=-\frac{\text{n}}{2}$ for n is even

For n is odd,

If f(n) = f(m) then

$\frac{\text{n}-1}{2}=\frac{\text{m}-1}{2}$

⇒ n = m

Also, for n is even if f(n) = f(m) then n = m

Hence, f is one-one.

Also, each element of y is associated with at least one element of x,

f is onto.

1. Bijection.

Solution:

$\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3)$

$\Rightarrow\ \text{f(x)}=3\sin^{-1}\text{x}$

Injectivity: Let x and y be two elements in the domain $\Big[\frac{-1}{2},\frac{1}{2}\Big],$ such that

f(x) = f(y)

$\Rightarrow\ 3\sin^{-1}\text{x}=3\sin^{-1}\text{y}$

$\Rightarrow\ \sin^{-1}\text{x}=\sin^{-1}\text{y}$

$\Rightarrow\ \text{x}=\text{y}$

So, f is one-one.

Surjectivity: Let y be any element in the co-domain, such that

f(x) = y

$\Rightarrow\ 3\sin^{-1}\text{x}=\text{y}$

$\Rightarrow\ \sin^{-1}\text{x}=\frac{\text{y}}{3}$

$\Rightarrow\ \text{x}=\sin\frac{\text{y}}{3}\in\Big[\frac{-1}{2},\frac{1}{2}\Big]$

⇒ f is onto.

⇒ f is a bijection.

1. Onto but not one-one.

Solution:

Given function is

$\text{f(x)}=\frac{\text{x}}{2}$ if x is even

= 0 if x is odd

For f(3) = 0 and f(4) = 0

⇒ f(3) = f(4)

But, $3\neq4$

Hence, it is not one-one.

$\text{x}\in\text{R}\Rightarrow\ \text{y}\in\text{R}$

Here, Domain = range of f

Hence, it is onto.

2. y

Solution:

Given that f be an injective map with domain {x, y, z} and range {1, 2, 3}

f(x) = 1, $\text{f(y)}\neq1,\ \text{f(z)}\neq2$

As f(x) = 1 ⇒ f^-1(1) = y

2. $1-\sqrt{1-\text{x}}$

Solution:

Let y be the element in the co-domain R such that f^-1(x) = y ......(1)

⇒ f(y) = x and $\text{y}\leq1$

⇒ y(2 - y) = x

⇒ 2y - y² = x

⇒ y² - 2y + x = 0

⇒ y² - 2y = -x

⇒ y² - 2y + 1 = 1 - x

$\Rightarrow\ (\text{y}-1)^2=\sqrt{1-\text{x}}$

$\Rightarrow\ \text{y}-1=\pm\sqrt{1-\text{x}}$​​​​​​​

$\Rightarrow\ \text{y}=1\pm\sqrt{1-\text{x}}$

$\Rightarrow\ \text{y}=1-\sqrt{1-\text{x}}$ $(\because\ \text{y}\leq1)$

3. f(x) = -x - 1, g(x) = x - 1

Solution:

​​​​​​​Since f is invertible, range of f = co-domain of f = x

So, we need to find the range of f to find X.

For finding the range, let f(x) = y

⇒ 4x - x² = y

⇒ x² - 4x = -y

⇒ x² - 4x + 4 = 4 - y

⇒ (x - 2)² = 4 - y

$\Rightarrow\ \text{x}-2=\pm4-\text{y}$

$\Rightarrow\ \text{x}=2\pm4-\text{y}$

This is defined only when $4-\text{y}\geq0$

$\Rightarrow\ \text{y}\leq4,$

X = Range of f $=(-\infty,4]$

4. $\text{None of these}$

Solution:

Given function is f : R → R be given by f(x) = x² - 3.

$\text{y} = \text{x}^2 - 3$

$\text{y} + 3 = \text{x}^2$

$\text{x}=\pm\sqrt{\text{y}+3}$

$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+3}$

1. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$

Solution:

f(x) = -x² + 6x - 8, is a polynomial function and the domain of polynomial function is real number.

$\therefore\ \text{x}\in\text{R}$

f(x) = -x² + 6x - 8

= -(x² - 6x + 8)

= -(x² - 6x + 9 - 1)

= -(x - 3)² + 1

Maximum value of -(x - 3)² woud be 0

$\therefore$ Maximum value of -(x - 3)² + 1 woud be 1

$\therefore\ \text{f(x)}\in(-\infty,1]$

We can see from the given graph that function is symmetrical about x = 3 and the given function is bijective.

So, x would be either $(-\infty,3]\text{ or }[3,\infty)$

The correct option which satisfy A and B both is:

$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$

2. is given by $\frac{\text{x}+5}{3}$

Solution:

Given function is f : R → R is given by f(x) = 3x - 5

To find f^-1(x)

y = f(x)

⇒ y = 3x - 5

⇒ y + 5 = 3x

$\Rightarrow\ \text{y}=\frac{\text{y}+5}{3}$

Hence, $\text{f}^{-1}(\text{x})=\frac{\text{x}+5}{3}$

4. [0, 1)

Solution:

As f is surjective, range of f = co-domain of f

⇒ A = range of f

$=\frac{\text{x}^2}{\text{x}^2+1},$

$\text{y}=\frac{\text{x}^2}{\text{x}^2+1}$

$\Rightarrow\ \text{y}(\text{x}^2+1)$

$\Rightarrow\ \text{x}^2=\frac{-\text{y}}{(\text{y}-1)}$

$\Rightarrow\ \text{x}=\sqrt{\frac{\text{y}}{(1-\text{y})}}$

$\Rightarrow\ \frac{\text{y}}{(1-\text{y})}\geq0$

$\Rightarrow\ \text{y}\in[0,1)$

⇒ Range of f = [0, 1)

⇒ A = [0, 1)

1. S defines a function from A to B.

Solution:

Given that $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}$

$\text{x}^2+\text{y}^2=1$

$\Rightarrow\ \text{y}^2=1-\text{x}^2$

$\Rightarrow\ \text{y}=\sqrt{1-\text{x}^2}$

$\text{y}\in\text{B}$

Hence, S defines a function from A to B.

1. f^-1(x) = f(x)

Solution:

Given function is $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}$

fof(x) = f(f(x))

$=\text{f}\Big(\frac{3\text{x}+2}{5\text{x}-3}\Big)$

$=\frac{3\big(\frac{3\text{x}+2}{5\text{x}-3}\big)+2}{5\big(\frac{3\text{x}+2}{5\text{x}-3}\big)-3}$

After solving you will get

f(f(x)) = x

Also, f^-1(x) = f(x) you can check.

4. None of these.

Solution:

$\text{f}:\text{A}\rightarrow\text{B}$

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

$\Rightarrow\ 3^{\text{f(x)}}=4-2^{-\text{x}}$

Taking log on both the sides,

$\text{f(x)}\log3=\log(4-2^{-\text{x}})$

$\Rightarrow\ \text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$

Logaritmic function will only be defined if $4-2^{-\text{x}}>0$

$\Rightarrow\ 4>2^{-\text{x}}$

$\Rightarrow\ 2^2>2^{-\text{x}}$

$\Rightarrow\ 2>-\text{x}$

$\Rightarrow-2<\text{x}$

$\Rightarrow\ \text{x}\in(-2,\infty)$

That means $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\}$

As we know that, $\text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$

We take $\text{x}=0\in(-2,\infty)$

⇒ f(x) = 1 which does not belong to any of the options.

3. $(-\infty,4]$

Solution:

Since f is invertible, range of f = co-domain of f = X

So, we need to find the range of f to find X.

For finding the range, let

f(x) = y

⇒ 4x - x² = y

⇒ x² - 4x = -y

⇒ x² - 4x + 4 = 4 - y

⇒ (x - 2)² = 4 - y

$\Rightarrow\ \text{x}-2=\pm\sqrt{4-\text{y}}$

$\Rightarrow\ \text{x}=2\pm\sqrt{4-\text{y}}$

This is defined only when

$4-\text{y}\geq0$

$\Rightarrow\ \text{y}\leq4$

X = Range of $\text{f}=(-\infty,4]$

1. $\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$

Solution:

Let f^-1(x) = y .....(1)

$\Rightarrow\ \text{f(y)}=\text{x}$

$\Rightarrow\ \frac{\text{e}^{\text{y}}-\text{e}^{-\text{y}}}{\text{e}^{\text{y}}+\text{e}^{-\text{y}}}=\text{x}$

$\Rightarrow\ \frac{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}-1)}{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}+1)}=\text{x}$

$\Rightarrow\ (\text{e}^{2\text{y}}-1)=\text{x}(\text{e}^{2\text{y}}+1)$

$\Rightarrow\ \text{e}^{2\text{y}}-1=\text{xe}^{2\text{y}}+\text{x}$

$\Rightarrow\ \text{e}^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$

$\Rightarrow\ 2\text{y}=\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

$\Rightarrow\ \text{y}=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ [From (1)]

1. 9

Solution:

We have,

$\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}$

Now,

f(-1) + f(2) + f(4)

= 3(-1) + 2² + 2(4)

= -3 + 4 + 8

= 9

2. Many-one and into.

Solution:

f : R → R

= [x²] + [x + 1] - 3

It is many one function because in this case for two different values of x we would get the same value of f(x).

For $\text{x}=1.1,\ 1.2\in\text{R}$

f(1.1) = (1.1)² + [1.1 + 1] - 3

= [1.21] + [2.1] - 3

= 1 + 2 + 3 = 0

f(1.1) = [1.2]² + [1.2 + 1] - 3

= [1.44] + [2.2] - 3

= 1 + 2 - 3

= 0

It is into function because for the given domain we would only get the integral values of f(x).

But R is the co-domain of the given function.

That means, $\text{Co-domain}\neq\text{Range}$

Hence, the given function is into function.

Therefore, f(x) is many one and into.

2. Onto but not one-one.

Solution:

Given function is f(x) = (x - 1)(x - 2)(x - 3)

If f(x) = f(y) then

(x - 1)(x - 2)(x - 3) = (y - 1)(y - 2)(y - 3)

⇒ f(1) = f(2) = f(3) = 0

It is not one-one.

y = f(x)

$\text{x}\in\text{R}$ also $\text{y}\in\text{R}$ hence f is onto.

4. Neither one-one nor onto.

Solution:

We have,

$\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$

Here, $-2,2\in\text{R}$

Now, $2\neq-2$

But, f(2) = f(-2)

Therefore, function is not one-one.

And,

The minimum value of the function is 0 and maximum value is 1.

That is range of the function is [0, 1] but the co-domain of the function is given R.

Therefore, function is not onto.

$\therefore$ function is neither one-one nor onto.

2. $\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$

Solution:

Given function is $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$

The inverse of the function f : A → A given by $\text{f(x)}=2^{\text{x}(\text{x}-1)}$

$\text{f(x)}=\text{y}$

$2^{\text{x}(\text{x}-1)}=\text{y}$

$\text{x}(\text{x}-1)=\log_2\text{y}$

$\text{x}^2+\text{x}=\log_2\text{y}$

$\text{x}^2+\text{x}+\frac{1}{4}=\log_2\text{y}+\frac{1}{4}$

$\Big(\text{x}-\frac{1}{2}\Big)^2=\frac{4\log_2\text{y}+1}{4}$

$\text{x}-\frac{1}{2}=\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$

$\text{x}=\frac{1}{2}\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$

$\text{x}=\frac{1}{2}+\sqrt{\frac{4\log_2\text{y}+1}{4}}$

$\text{f}^{-1}(\text{x})=\frac{1+\sqrt{4\log_2\text{y}+1}}{2}$

2. f(x) = x + 2

Solution:

1. f is not because for $\text{y}=3\in\text{Co-domain (Z)},$ there is no value of $\text{x}\in\text{Domain (Z)}$

$\text{x}^3=3$

$\Rightarrow\ \text{x}=\sqrt[3]{3}\notin\text{Z}$

⇒ f is not onto.

So, f is not a bijection.

2. Injectivity: Let x and y be two elements of the domain (Z), such that

x + 2 = y + 2

⇒ x = y

So, f is one-one.

Surjectivity: Let y be an element in the co-domain (Z), such that

y = f(x)

⇒ y = x + 2

$\Rightarrow\ \text{x}=\text{y}-2\in\text{Z}$ (Domain)

⇒ f is onto.

So, f is a bijection.

3. f(x) = 2x + 1 is not onto because if we take $4\in\text{Z}$ (co domain), then 4 = f(x)

$\Rightarrow4 = 2\text{x} + 1$

$\Rightarrow 2\text{x} = 3$

$\Rightarrow\ \text{x}=\frac{3}{2}\notin\text{Z}$

So, f is not a bijection.

4. f(0) = 0² + 0 = 0

⇒ and f(-1) = (-1)² + (-1) = 1 - 1 = 0

⇒ 0 and -1 have the same image.

⇒ f is not one-one.

So, f is not a bijection.

3. $(\text{x}-3)^\frac{1}{3}$

Solution:

Let f^-1(x) = y

f(y) = x

⇒ y³ + 3 = x

⇒ y³ = x - 3

⇒ y = (x - 3)³

$\Rightarrow\ \text{y}=(\text{x}-3)^\frac{1}{3}$

3. 0

Solution:

As, the number of bijection from A into B can only be possible when provided $\frac{7}{(\text{A})}>\frac{7}{(\text{B})}$

But here n(A) < n(B)

So, the number of bijection.

i.e. one-one and onto mapping from A to B.

1. 2x - 3

Solution:

We will solve this problem by the trial-and-error method.

Let us check option (a) first.

If f(x) = 2x - 3

$\frac{1}{2}(\text{gof})(x)=\text{g(f(x))}$

$=\frac{1}{2}\text{g}(2\text{x}-3)$

$=\frac{1}{2}\big[(2\text{x}-3)^2+(2\text{x}-3)-2\big]$

$=\frac{1}{2}[4\text{x}^2+9-12\text{x}+2\text{x}-3-2]$

$=\frac{1}{2}[4\text{x}^2-10\text{x}+4]$

$=2\text{x}^2-5\text{x}+2$

The given condition is satisfied by (a).

2. 1

Solution:

When, -1 < x < 0

Then, g(x) = 1 + x - [x]

= 1 + x - (-1) = 2 + x

$\therefore$ f(g(x)) = 1

When, x = 0

Then, g(x) = 1 + x - [x]

= 1 + x - 0 = 1 + x

$\therefore$ f(g(x)) = 1

When, x > 1

Then, g(x) = 1 + x - [x]

= 1 + x - 1 = x

$\therefore$ f(g(x)) = 1

Therefore, for each interval f(g(x)) = 1

2. $-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$

Solution:

Given function is f : R → (-1, 1) is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2}$

Here, for mod function we will have to consider three cases as,

x < 0, x = 0, x > 0

x < 0 ⇒ |x| = -x

$\text{f(|x|)}=\frac{-\text{x}(-\text{x})}{1+\text{x}^2}$

$\text{y}=\frac{\text{x}^2}{1+\text{x}^2}$

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

$\text{y}+\text{yx}^2=\text{x}^2$

$\text{y}=\text{x}^2-\text{yx}^2$

$\text{y}=(1-\text{y})\text{x}^2$

$\text{x}^2=\frac{\text{y}}{1-\text{y}}$

$\text{x}=-\sqrt{\frac{\text{y}}{1-\text{y}}}$

$\Rightarrow\ \text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}\ \text{x} < 0$ 

Also you can check for the cases x = 0 and x > 0 that $\text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}$

$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$

4. -1

Solution: Given function is $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1$

Also f(f(x)) = x

$\text{f}\Big(\frac{\alpha\text{x}}{\text{x}+1}\Big)=\text{x}$

$\frac{\alpha\big(\frac{\alpha\text{x}}{\text{x}+1}\big)}{\frac{\alpha\text{x}}{\text{x}+1}+1}=\text{x}$

$\frac{\alpha^2\text{x}}{\alpha\text{x}+\text{x}+1}=\text{x}$

$\alpha^2=\alpha\text{x}+\text{x}+1$

$\alpha^2=(\alpha+1)\text{x}+1$

Comparing on both sides,

$\alpha+1=0\Rightarrow\ \alpha=-1$​​​​​​​

2. $2^\text{n}-2$

Solution:

The number of functions from a set with n number of elements into a set with 2 number of elements = 2ⁿ

But two functions can be many-one into function.

4. Many one and into.

Solution:

Graph of the given function is as follows:

A line parallel to X-axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y.

That means it is many one function.

From the given graph we can see that the range is $[2,\infty)$ and R is the co-domain of the given function.

Hence, Co-dornain = Range

Therefore, the given function is into.

4. f is neither an injection nor a surjection.

Solution:

f : R → R

$\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

For x = -2 and -3 $\in\text{R}$

$\text{f(-2)}=\frac{\text{e}^{|-2|}-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$

$=\frac{\text{e}^2-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$

$=0$

Hence, for different values of x we are getting same values of f(x)

That means, the given function is many one.

Therefore, this function is not injective.

For x < 0

f(x) = 0

For x > 0

$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

$=\frac{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{}e^{-\text{x}}}-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

$=1-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

The value of $\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is always positive.

Therefore, the value of f(x) is always less than 1.

Numbers more than 1 are not included in the range but they are included in co-domain.

As the codomain is R.

$\therefore\ \text{Co-domain}\neq\text{Range}$

Hence, the given function is not onto.

Therefore, this function is not surjective.

2. $\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$

Solution:

We have, f : R → R is given by

$\text{f(x)}=\tan\text{x}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\tan^{-1}\text{x}$

$\therefore\ \text{f}^{-1}(1)=\tan^{-1}1=\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$

4. Neither injective nor surjective.

Solution:

Given function is f : R → R, f(x) = x²

If f(x) = f(y) then

x² = y²

$\Rightarrow\ \text{x}\pm\text{y}$

Hence, it is not one-one or injective.

f(x) = y

y = x²

$\text{x}=\pm\sqrt{\text{y}}$

But co-domain is R.

Hence, it is not onto or surjective.

2. $\text{g(x)}=\sin\big(\frac{\pi\text{x}}{2}\big)$

Solution:

1. Range of $\text{f}=\Big[\frac{-1}{2},\frac{1}{2}\Big]\neq\text{A}$

So, f is not a bijection.

2. Range $=\Big[\sin\Big(\frac{-\pi}{2}\Big),\ \sin\Big(\frac{\pi}{2}\Big)\Big]=[-1,1]=\text{A}$

So, g is a bijection.

3. h(-1) = |-1| = 1

And h(1) = |1| = 1

⇒ -1 and 1 have the same images.

So, h is not a bijection.

4. k(-1) = (-1)² = 1

And k(1) = (1)² = 1

⇒ -1 and 1 have the same images.

So, k is not a bijection.

1. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

​​​​​​​Solution:

Let f^-1(x) = y

​​​​​​​$\Rightarrow\ \text{f(y)} = \text{x}$

$\Rightarrow\ \text{y}+\frac{1}{\text{y}}=\text{x}$

$\Rightarrow\ \text{y}^2 + 1 = \text{xy}$

$\Rightarrow\ \text{y}^2 - \text{xy} + 1 = 0$

$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2-\big(\frac{\text{x}}{2}\big)^2+1=0$

$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2=\frac{\text{x}^2-1}{4}$

$\Rightarrow\ \Big(\text{y}-\frac{\text{x}}{2}\Big)^2=\frac{\text{x}^2-1}{4}$

$\Rightarrow\ \text{y}-\frac{\text{x}}{2}=\frac{\sqrt{\text{x}^2-4}}{2}$

$\Rightarrow\ \text{y}=\frac{\text{x}}{2}+\frac{\sqrt{\text{x}^2-4}}{2}$

$\Rightarrow\ \text{y}=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

4. None of these.

Solution:

$\text{f(x)}=\text{x}+\sqrt{\text{x}^2}=\text{x}\pm\text{x}=0\text{ or }2\text{x}$

⇒ Each element of the domain has 2 images.

f is not a function.

1. A bijection.

Solution:

Given function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and f : A → A such that $\text{f(x)}=\text{x}|\text{x}|$

For the mod function we have to check three cases as x < 0, x = 0, x > 0.

For example,

x < 0

f(x) = x|x| < 0

|x| = -x

y = -x²

$\text{x}=-\sqrt{-\text{y}}$ which is not possible for x > 0

Hence, f is onto.

⇒ f is bijection.

2. f is one-one into.

Solution:

Given function f : R - {n} → R be a function defined by $\text{f(x)}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}},\ \text{m}\neq\text{n}$

If f(x) = f(y) then

$\frac{\text{x}-\text{m}}{\text{x}-\text{n}}=\frac{\text{y}-\text{m}}{\text{y}-\text{n}}$

⇒ (x - m)(y - n) = (y - m)(x - n)

After solving this we will get x = y

Hence, it is one-one.

$\text{f(x)}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}},\ \text{m}\neq\text{n}$

$\text{y}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}}$

⇒ y(x - n) = x - m

⇒ yx - yn = x - m

⇒ yx - x = ny - m

⇒ x(y - 1) = ny - m

$\Rightarrow\ \text{x}=\frac{\text{ny}-\text{m}}{\text{y}-1}$

Here, for y = 1 we can not define x.

Hence, it is not onto.

3. {0, 2}

Solution:

Since (fog)(x) = (gof)(x),

f(g(x)) = g(f(x))

$\Rightarrow\ \text{f}(2^\text{x})=\text{g}(\text{x}^2)$

$\Rightarrow\ \big(2^{\text{x}}\big)^{2}=2^{\text{x}^2}$

$\Rightarrow\ 2^{2\text{x}}=2^{\text{x}^2}$

$\Rightarrow\ \text{x}^2=2\text{x}$

$\Rightarrow\ \text{x}^2-2\text{x}=0$

$\Rightarrow\ \text{x}(\text{x}-2)=0$

$\Rightarrow\ \text{x}=0, 2$

$\Rightarrow\ \text{x}\in\{0,2\}$

4. Neither one-one nor onto.

Solution:

Injectivity: Let x and y be two elements in the domain (R), such that

f(x) = f(y)

$\frac{\text{x}^2-8}{\text{x}^2+2}=\frac{\text{y}^2-8}{\text{y}^2+2}$

⇒ (x² - 8)(y² + 2) = (y² - 8)(x² + 2)

⇒ x²y² + 2x² - 8y² - 16 = x²y² + 2y² - 8x² - 16

⇒ 10x² = 10y²

⇒ x² = y²

$\Rightarrow\ \text{x}=\pm\text{y}$

So, f is not one-one.

Surjectivity: $\text{f}(-1)=\frac{(-1)^2-8}{(-1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$

and $\text{f(1)}=\frac{(1)^2-8}{(1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$

$\Rightarrow\ \text{f}(-1)=\text{f}(1)=\frac{-7}{3}$

⇒ f is not onto.

2. $[1,\infty)$

​​​​​​​Solution:

Since f is a bijection, co-domain of f = range of f

⇒ B = range of f

Given: f(x) = x² - 4x + 5

Let f(x) = y

⇒ y = x² - 4x + 5

⇒ x² - 4x + (5 - y) = 0

$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$

$(-4)^2-4\times1\times(5-\text{y})\geq0$

$\Rightarrow\ 16-20+4\text{y}\geq0$

$\Rightarrow\ 4\text{y}\geq4$

$\Rightarrow\ \text{y}\geq1$

$\Rightarrow\ \text{y}\in[1,\infty)$

⇒ Range of $\text{f}=[1,\infty)$

$\Rightarrow\ \text{B}=[1,\infty)$

2. $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$

Solution:

​​​​​​​It is clear that f(x) is one-one.

Range of $\text{f}=\Big[\sin\frac{\pi(-1)}{2},\sin\frac{\pi(1)}{2}\Big]=\Big[\sin\frac{-\pi}{2},\sin\frac{\pi}{2}\Big]$

= A = Co-domain of f

⇒ f is onto.

So, f is a bijection.

2. $^{10}\text{C}_7\times7!$

Solution:

As, the number of one-one functions from A to B with m and n elements, respectively $= \ ^{\text{n}}\text{P}_\text{m}=\ ^{\text{n}}\text{C}_\text{m}\times\text{m}!$

So, the number of one-one functions from A to B with 7 and 10 elements, respectively $=\ ^{10}\text{P}_7=\ ^{10}\text{C}_7\times7!$

2. $\sqrt{\text{x}}$

Solution:

Given that $\text{f(x)}=\sin^2\text{x}$ and the composite function $\text{g(f(x))}=|\sin\text{x}|$

We will do it using trial and error method.

If we take $\text{g(x)}=-\sqrt{\text{x}}$ and $\text{f(x)}=\sin^2\text{x}$

​​​​​​​$\text{g(f(x))}=\text{g}(\sin^2\text{x})$

$=-\sin\text{x}$

Which contradicts to the $\text{g(f(x))}=|\sin\text{x}|$

Hence, we take $\text{g(x)}=\sqrt{\text{x}}$

$\text{g(f(x))}=\text{g}(\sin^2\text{x})$

$=\sqrt{\sin^2\text{x}}=|\sin\text{x}|$

3. $\text{hofog}=\text{hogof}$

Solution:

hogof(x) = h(f(g(x)))

= h(f([x]))

= h(sin^-1[x])

= 2sin^-1[x]

= 2 × 0 = 0

f(x) = sin^-1x

hogof(x) = hogo(x) = 0