sets relations and functions jee mains questions | JEE Main Mathematics Sets Relations And Functions

Q 11 :

Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps $f : A \to B$ , such that $f (1) + f (3) = 14$ , is: [2024]

180
480
120
240

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(4)

A = {1, 3, 7, 9, 11}, B = {2, 4, 5, 7, 8, 10, 12}

$f : A \to B$ is one-one such that

$f (1) + f (3) = 14$

$(f (1), f (3)) = {(2, 12), (4, 10), (12, 2), (10, 4)}$

Since $f$ is one-one so $f (1) \neq f (3)$ so we cannot take (7, 7).

$∴$ So, for $f (1)$ we have 4 choices and for $f (3)$ we have 4 choices and remaining 3 elements have 5! choices for mapping to be one-one.

Total number of ways = $4 \times 5! / 2$

$= \frac{480}{2} = 240$ [ $∵$ Pair (2, 12) and (12, 2) will be considered same]

Q 12 :

The function $f (x) = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}, x \in R$ is [2024]

both one-one and onto.
neither one-one nor onto.
onto but not one-one.
one-one but not onto.

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2

3

4

(2)

$We have, f (x) = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}$

$f (x) = \frac{(x + 5) (x - 3)}{x^{2} - 4 x + 9}$

$f (x) has two roots as x = - 5, 3$

$So f (- 5) = 0 and f (3) = 0$

$So f cannot be one - one .$

$Now consider, x^{2} - 4 x + 9$

$D = 16 - 36 = - 20 < 0$

$Now, let y = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}$

$\Rightarrow y x^{2} - 4 x y + 9 y = x^{2} + 2 x - 15$

$\Rightarrow x^{2} (y - 1) - 2 x (2 y + 1) + (9 y + 15) = 0$

$D = 4 {(2 y + 1)}^{2} - 4 (y - 1) (9 y + 15) \geq 0$ $[∵ x \in R]$

$\Rightarrow 4 (4 y^{2} + 1 + 4 y) - (36 y^{2} - 36 y + 60 y - 60) \geq 0$

$\Rightarrow 16 y^{2} + 4 + 16 y - 36 y^{2} - 24 y + 60 \geq 0$

$\Rightarrow - 20 y^{2} - 8 y + 64 \geq 0$

$\Rightarrow 5 y^{2} + 2 y - 16 \leq 0$

$\Rightarrow (5 y - 8) (y + 2) \leq 0$

$\Rightarrow y \in [- 2, \frac{8}{5}] which is the range of function f$

$So, f is not onto if f : R \to R .$

$Here in given question co - domain is not defined .$

Q 13 :

Let $f (x) = \frac{1}{7 - \sin 5 x}$ be a function defined on R. Then the range of the function $f (x)$ is equal to : [2024]

$[\frac{1}{7}, \frac{1}{5}]$
$[\frac{1}{7}, \frac{1}{6}]$
$[\frac{1}{8}, \frac{1}{6}]$
$[\frac{1}{8}, \frac{1}{5}]$

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(3)

Since, $- 1 \leq sinθ \leq 1$ $\forall θ$ , so we have

$- 1 \leq \sin$ $5 x \leq 1$

$\Rightarrow - 1 \leq - \sin 5 x \leq 1$

$\Rightarrow - 1 + 7 \leq 7 - \sin 5 x \leq 1 + 7$

$\Rightarrow \frac{1}{8} \leq \frac{1}{7 - \sin 5 x} \leq \frac{1}{6}$

$\Rightarrow Range f (x) = [\frac{1}{8}, \frac{1}{6}]$

Q 14 :

Let [ $t$ ] be the greatest integer less than or equal to $t$ . Let A be the set of all prime factors of 2310 and $f : A \to Z be the function f (x) = [\log_{2} (x^{2} + [\frac{x^{3}}{5}])] .$ The number of one-to-one functions from A to the range of $f$ is [2024]

25
24
20
120

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(4)

$2310 = 2 \times 3 \times 5 \times 7 \times 11$

$∴ A = {2, 3, 5, 7, 11}$

$f : A \to Z is a function such that$

$f (x) = [\log_{2} (x^{2} + [\frac{x^{3}}{5}])]$

$f (2) = [\log_{2} (4 + [1.6])] = [\log_{2} (4 + 1)] = [\log_{2} 5] = [\log_{2} (2^{2} + 1)] = 2$

$f (3) = [\log_{2} (9 + 5)] = [\log_{2} (2^{3} + 6)] = 3$

$f (5) = [\log_{2} (25 + 25)] = [\log_{2} (2^{5} + 18)] = 5$

$f (7) = [\log_{2} (49 + 68)] = [\log_{2} (2^{6} + 53)] = 6$

$f (11) = [\log_{2} (121 + 266)] = [\log_{2} (2^{8} + 131)] = 8$

$Range of f = {2, 3, 5, 6, 8}$

$Number of one-one functions = 5! = 120$

Q 15 :

Let $f (x) = {\begin{matrix} - a if - a \leq x \leq 0 \\ x + a if 0 < x \leq a \end{matrix}$ where $a > 0$ and $g (x) = \frac{(f (| x |) - | f (x) |)}{2}$ . Then the function $g :$ $[- a, a] \to [- a, a]$ is [2024]

onto.
both one-one and onto.
one-one.
neither one-one nor onto.

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2

3

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(4)

$y = f (x) = {\begin{matrix} - a, & - a \leq x \leq 0 \\ x + a, & 0 < x \leq a \end{matrix}$

$y = f | x | = {\begin{matrix} - a, & - a \leq | x | \leq 0 \\ | x | + a, & 0 < | x | \leq a \end{matrix}$

but $| x | < 0$ is not possible.

$f | x | = {\begin{matrix} - x + a, & - a \leq x < 0 \\ x + a, & 0 < x \leq a \end{matrix}$

$y = | f (x) | = {\begin{matrix} a, & - a \leq x \leq 0 \\ x + a, & 0 < x \leq a \end{matrix}$

$g (x) = \frac{f | x | - | f (x) |}{2} = {\begin{matrix} \frac{- x + a - a}{2} = - \frac{x}{2} if - a \leq x \leq 0 \\ \frac{x + a - x - a}{2} = 0 if 0 < x \leq a \end{matrix} \begin{matrix} \end{matrix}$

$g : [- a, a] \to [- a, a]$ is neither one-one nor onto as set $[0, a]$ has only one image i.e. 0.

Q 16 :

If the domain of the function $f (x) = \frac{\sqrt{x^{2} - 25}}{(4 - x^{2})} + \log_{10} (x^{2} + 2 x - 15)$ is $(- \infty, α) \cup [β, \infty),$ then $α^{2} + β^{3}$ is equal to: [2024]

140
175
125
150

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(4)

$For domain, 4 - x^{2} \neq 0 \Rightarrow x \neq \pm 2$

$x^{2} - 25 \geq 0; x^{2} \geq 25 \Rightarrow x \in (- \infty, - 5] \cup [5, \infty)$

$Also, x^{2} + 2 x - 15 > 0 \Rightarrow (x + 5) (x - 3) > 0$

$\Rightarrow x \in (- \infty, - 5) \cup (3, \infty)$ $∴ D_{f} = (- \infty, - 5) \cup [5, \infty)$

$So, α = - 5 and β = 5$ $∴ α^{2} + β^{3} = 25 + 125 = 150$

Q 17 :

Let $f (x) = {\begin{matrix} x - 1, & x is even, \\ 2 x, & x is odd, \end{matrix} x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $\lim_{x \to a^{-}} {\frac{| x |^{3}}{a} - [\frac{x}{a}]},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is equal to: [2024]

169
121
225
144

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(4)

There are two cases arise:

Case I : Let $a$ is even $∴$ $f (a) = a - 1$ (odd)

$\Rightarrow f (f (a)) = f (a - 1) = 2 a - 2$ (even)

$\Rightarrow f (f (f (a))) = f (2 a - 2) = 2 a - 2 - 1 = 2 a - 3$

$\Rightarrow 21 = 2 a - 3 \Rightarrow a = 12$

$So, \lim_{x \to a^{-}} {\frac{| x |^{3}}{a} - [\frac{x}{a}]} = \lim_{x \to 12^{-}} {\frac{| x^{3} |}{12} - [\frac{x}{12}]}$

$= 144$ $(∵ x < 12)$

Case II : Let $a$ is odd

$∴ f (a) = 2 a (even) \Rightarrow f (f (a)) = f (2 a) = 2 a - 1 (odd)$

$\Rightarrow f (f (f (a))) = f (2 a - 1) = 2 (2 a - 1)$

$\Rightarrow 21 = 4 a - 2 \Rightarrow a = \frac{23}{4} \notin N$

Q 18 :

The function $f : N - {1} \to N$ ; defined by $f (n)$ = the highest prime factor of $n$ , is: [2024]

one-one only
both one-one and onto
neither one-one nor onto
onto only

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2

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(3)

Given, $f : N - {1} \to N$

$f (n) =$ the highest prime factor of $n$

For one-one: If $n = 4, f (n) = 2 .$

If $n = 8, f (n) = 2$ $∴$ $f$ is not one-one.

For onto : Range = All prime numbers

Co-domain = Set of natural numbers

$\Rightarrow$ Range $\neq$ Co-domain $∴$ $f$ is not onto.

Q 19 :

Let $f : R - {\frac{- 1}{2}} \to R$ and $g : R - {\frac{- 5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{| x | + 1}{2 x + 5}$ . Then, the domain of the function $f o g$ is [2024]

$R - {- \frac{5}{2}, - \frac{7}{4}}$
$R - {- \frac{7}{4}}$
$R$
$R - {- \frac{5}{2}}$

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(4)

Q 20 :

If $f (x) = {\begin{matrix} 2 + 2 x, & - 1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{matrix}; g (x) = {\begin{matrix} - x, & - 3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{matrix},$ then range of $(f o g) (x)$ is [2024]

[0, 1)
[0, 3)
(0, 1]
[0, 1]

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(4)

Given,

$f (x) = {\begin{matrix} 2 + 2 x, & - 1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{matrix}$ and $g (x) = {\begin{matrix} - x, & - 3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{matrix}$

$f o g (- 3) = f (3) = 0; f o g (- 2) = f (2) = \frac{1}{3}; f o g (- 1) = f (1) = \frac{2}{3}$

$f o g (0) = f (0) = 1; f o g (1) = f (1) = \frac{2}{3}$

$∴$ $Range of f o g (x) = [0, 1]$

Q 21 :

If $f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3}$ and $(f o f) (x) = g (x)$ , where $g : R - {\frac{2}{3}} \to R - {\frac{2}{3}},$ then $(g o g o g) (4)$ is equal to [2024]

$\frac{19}{20}$
$- \frac{19}{20}$
$- 4$
$4$

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4

(4)

Here, $f (x) = \frac{4 x + 3}{6 x - 4}$

$f [f (x)] = \frac{4 f (x) + 3}{6 f (x) - 4} = \frac{4 (\frac{4 x + 3}{6 x - 4}) + 3}{6 (\frac{4 x + 3}{6 x - 4}) - 4}$

$= \frac{16 x + 12 + 18 x - 12}{6 x - 4} = \frac{34 x}{34} = x$

$∴$ $f [f (x)] = x \Rightarrow g (x) = x$

Now, $(g o g o g) (4) = g o g [g (4)] = g o g (4) = g (4) = 4$

Q 22 :

Consider the function $f : R \to R$ defined by $f (x) = \frac{2 x}{\sqrt{1 + 9 x^{2}}} .$ If the composition of $\underset{10 times}{f, \underset{⏟}{(f o f o f o . . . o f)}} (x)$ $= \frac{2^{10} x}{\sqrt{1 + 9 α x^{2}}},$ then the value of $\sqrt{3 α + 1}$ is equal to _________ . [2024]

(1024)

We have, $f (x) = \frac{2 x}{\sqrt{1 + 9 x^{2}}}$

$∴$ $(f o f) (x) = \frac{2 f (x)}{\sqrt{1 + 9 (f {(x))}^{2}}} = \frac{\frac{4 x}{\sqrt{1 + 9 x^{2}}}}{\sqrt{1 + 9 \times \frac{4 x^{2}}{1 + 9 x^{2}}}} = \frac{4 x}{\sqrt{1 + 45 x^{2}}} = \frac{2^{2} x}{\sqrt{1 + 5 \times 9 x^{2}}}$

$(f o f o f) (x) = \frac{4 \times \frac{2 x}{\sqrt{1 + 9 x^{2}}}}{\sqrt{1 + 45 \times \frac{4 x^{2}}{1 + 9 x^{2}}}} = \frac{2^{3} x}{\sqrt{1 + 21 \times 9 x^{2}}}$

$(f o f o f o f) (x) = \frac{2^{4} x}{\sqrt{1 \times 85 \times 9 x^{2}}}$

$∴$ $\underset{n times}{\underset{⏟}{(f o f o f o . . . . . o f)}} (x)$ $= \frac{2^{n} x}{\sqrt{1 + 9 (\frac{4^{n} - 1}{3}) x^{2}}}$

$∴$ $\underset{10 times}{\underset{⏟}{(f o f o f o . . . . . o f)}} (x)$ $= \frac{2^{10} x}{\sqrt{1 + 9 (\frac{4^{10} - 1}{3}) x^{2}}} = \frac{2^{10} x}{\sqrt{1 + 9 α x^{2}}}$

( $∵$ Given)

On comparing, we get

$α = \frac{4^{10} - 1}{3} \Rightarrow 3 α + 1 = 4^{10}$

$\Rightarrow \sqrt{3 α + 1} = \sqrt{4^{10}} = 4^{5} = 1024 .$

Q 23 :

If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m,$ $n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq {(2022)}^{2}$ is equal to ___________ . [2024]

(1010)

We have, $f (m + n) = f (m) + f (n)$

So, $f (x) = k x$

$∵$ $f (1) = 1 \Rightarrow k = 1$

Hence, $f (x) = x$

Now, $\sum_{k = 1}^{2022} f (λ + k) = \sum_{k = 1}^{2022} (λ + k)$

$= \underset{2022}{\underset{⏟}{λ + λ + . . . + λ}}$ $+ (1 + 2 + . . . . + 2022)$

$= 2022 λ + \frac{2022 \times 2023}{2} \leq {(2022)}^{2}$ (Given)

$\Rightarrow λ \leq \frac{2021}{2}$

So, largest $λ = 1010$

Q 24 :

Let $A = {(x, y) : 2 x + 3 y = 23, x, y \in N}$ and $B = {x : (x, y) \in A}$ . Then the number of one-one functions from A to B is equal to ________ . [2024]

(24)

We have, $A = {(x, y) : 2 x + 3 y = 23, x, y \in N}$

$B = {x : (x, y) \in A}$

$∴ A = {(1, 7), (4, 5), (7, 3), (10, 1)}$

and $B = {1, 4, 7, 10}$

Total number of one-one functions from A to B = 4! = 24

Q 25 :

Let A = {1, 2, 3, ..., 7} and let P(A) denote the power set of A. If the number of functions $f : A \to P (A)$ such that $a \in f (a)$ , $\forall$ $a \in A$ is $m^{n}$ , $m$ and $n \in N$ and $m$ is least, then $m + n$ is equal to _____. [2024]

(44)

Given, $f : A \to P (A) \Rightarrow a \in f (a)$

It means $' a'$ will connect with subset which contain element $a .$

Total options for 1 will be $2^{6}$ ( $∵$ $2^{6}$ subsets contains 1)

Similarly, for every other element

Now, number of functions from A to P(A) = ${(2^{6})}^{7} = 2^{42}$

i.e., $m + n = 2 + 42 = 44$

Q 26 :

Let $f, g : R \to R$ be defined as: [2024]

$f (x) = | x - 1 |$ and $g (x) =$ ${\begin{matrix} e^{x}, & x \geq 0 \\ x + 1, & x \leq 0 \end{matrix} .$

Then the function $f (g (x))$ is

onto but not one-one.
neither one-one nor onto.
both one-one and onto.
one-one but not onto.

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2

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4

(2)

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

$=$ ${\begin{matrix} | e^{x} - 1 |, & x \geq 0 \\ | x + 1 - 1 |, & x \leq 0 \end{matrix}$

$=$ ${\begin{matrix} e^{x} - 1, & x \geq 0 \\ - x, & x \leq 0 \end{matrix}$

$f (g (x))$ is neither one-one nor onto as negative numbers have no pre-image.

Q 27 :

Let $f : R \to R$ and $g : R \to R$ be defined as:

$f (x)$ $= {\begin{matrix} \log_{e} x, & x > 0 \\ e^{- x}, & x \leq 0 \end{matrix}$ and $g (x)$ $= {\begin{matrix} x, & x \geq 0 \\ e^{x}, & x < 0 \end{matrix} .$

Then, $g o f : R \to R$ is [2024]

neither one-one nor onto
onto but not one-one
one-one but not onto
both one-one and onto

1

2

3

4

(1)

$g (f (x))$ $=$ ${\begin{matrix} f (x), & f (x) \geq 0 \\ e^{f (x)}, & f (x) < 0 \end{matrix}$

$=$ ${\begin{matrix} \log_{e} x, & x \geq 1 \\ e^{- x}, & x \leq 0 \\ e^{\log_{e} x} = x, & 0 < x < 1 \end{matrix}$

$g o f (x) =$ ${\begin{matrix} e^{- x}, & x \leq 0 \\ x, & 0 < x < 1 \\ \log_{e} x, & x \geq 1 \end{matrix}$

Now, range of $g o f = [0, \infty) \neq R$

Hence, not one-one and not onto.

Q 28 :

If the function $f : (- \infty, - 1] \to (a, b]$ defined by $f (x) = e^{x^{3} - 3 x + 1}$ is one-one and onto, then the distance of the point $P (2 b + 4, a + 2)$ from the line $x + e^{- 3} y = 4$ is : [2024]

$2 \sqrt{1 + e^{6}}$
$4 \sqrt{1 + e^{6}}$
$3 \sqrt{1 + e^{6}}$
$\sqrt{1 + e^{6}}$

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(1)

Given $f (x) = e^{x^{3} - 3 x + 1}$

$∴$ $f' (x) = e^{x^{3} - 3 x + 1} [3 (x - 1) (x + 1)] \geq 0$

As, $e^{x^{3} - 3 x + 1}$ is always positive.

$∴$ $3 (x - 1) (x + 1) \geq 0$

$\Rightarrow x \in (- \infty, - 1] \cup [1, \infty)$

For onto function, Range = Co-domain

$∴$ $a = f (- \infty) = e^{- \infty} = 0$

and $b = f (- 1) = e^{- 1 + 3 + 1} = e^{3}$

$∴$ $P (2 b + 4, a + 2) = P (2 e^{3} + 4, 2)$

Now, distance of the point $P (2 e^{3} + 4, 2)$ from the line $x + e^{- 3} y - 4 = 0$

$= \frac{2 e^{3} + 4 + 2 e^{- 3} - 4}{\sqrt{1 + e^{- 6}}} = \frac{2 (e^{3} + e^{- 3})}{\sqrt{1 + e^{- 6}}} = 2 (\frac{e^{6} + 1}{e^{3}}) \times \frac{1}{\sqrt{1 + e^{6}}} \times e^{3}$

$= 2 \sqrt{1 + e^{6}}$

Q 29 :

If the domain of the function

$f (x) = \frac{1}{\sqrt{10 + 3 x - x^{2}}} + \frac{1}{\sqrt{x + | x |}}$ is (a, b), then

${(1 + a)}^{2} + b^{2}$ is equal to: [2025]

29
26
30
25

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2

3

4

(2)

We have, $f (x) = \frac{1}{\sqrt{10 + 3 x - x^{2}}} + \frac{1}{\sqrt{x + | x |}}$

For f(x) to be defined, we must have $10 + 3 x - x^{2} > 0$

$\Rightarrow x^{2} - 3 x - 10 < 0 \Rightarrow (x - 5) (x + 2) < 0 \Rightarrow - 2 < x < 5$

Also, x + |x| > 0

Now, if x > 0, then x + |x| > 0

If x < 0, then |x| = –x $\Rightarrow$ x + |x| = x – x = 0

$∴$ The domain of $\frac{1}{\sqrt{x + | x |}}$ is x > 0

$∴$ Domain of f(x) is 0 < x < 5 i.e., (0, 5)

$∴$ a = 0 and b = 5

$\Rightarrow {(1 + a)}^{2} + b^{2} = 1^{2} + 5^{2} = 26$ .

Q 30 :

If the domain of the function

$f (x) = \log_{e} (\frac{2 x - 3}{5 + 4 x}) + \sin^{- 1} (\frac{4 + 3 x}{2 - x}) is [α, β), then α^{2} + 4 β$

is equal to [2025]

7
5
3
4

1

2

3

4

(4)

We have, $f (x) = \log_{e} (\frac{2 x - 3}{5 + 4 x}) + \sin^{- 1} (\frac{4 + 3 x}{2 - x})$

For f(x) to be defined we have,

$\frac{2 x - 3}{5 + 4 x} > 0 and | \frac{4 + 3 x}{2 - x} | \leq 1$

Now, $\frac{2 x - 3}{5 + 4 x} > 0$

Case I : 2x – 3 > 0 and 5 + 4x > 0

$\Rightarrow$ x > 3/2 and x > – 5/4

$\Rightarrow x \in (3 / 2, \infty)$ ... (i)

Case II : 2x – 3 < 0 and 5 + 4x < 0

$\Rightarrow$ x < 3/2 and x < – 5/4

$\Rightarrow x \in (- \infty, - 5 / 4)$ ... (ii)

From (i) and (ii), we get

$x \in (- \infty, - \frac{5}{4}) \cup (\frac{3}{2}, \infty)$ ... (iii)

Also, $| \frac{4 + 3 x}{2 - x} | \leq 1$

$\Rightarrow - 1 \leq \frac{4 + 3 x}{2 - x} \leq 1 \Rightarrow - 1 \leq \frac{4 + 3 x}{2 - x} and \frac{4 + 3 x}{2 - x} \leq 1$

$\Rightarrow 0 \leq \frac{4 + 3 x}{2 - x} + 1 and \frac{4 + 3 x}{2 - x} - 1 \leq 0$

$\Rightarrow 0 \leq \frac{6 + 2 x}{2 - x} and \frac{2 + 4 x}{2 - x} \leq 0$

$\Rightarrow - 3 \leq x and x \leq - \frac{1}{2}, x \neq 2$

$\Rightarrow x \in [- 3, - \frac{1}{2}]$ ... (iv)

From (iii) and (iv), we get

$x \in [- 3, - \frac{5}{4})$

$∴ α = - 3 and β = - \frac{5}{4}$

Thus, $α^{2} + 4 β = 9 - 5 = 4 .$ .

Q 31 :

If the domain of the function $f (x) = \log_{7} (1 - \log_{4} (x^{2} - 9 x + 18))$ is $(α, β) \cup (γ, δ)$ , then $α + β + γ + δ$ is equal to [2025]

18
15
16
17

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2

3

4

(1)

For f(x) to be defined we have,

$1 - \log_{4} (x^{2} - 9 x + 18) > 0 i . e ., x^{2} - 9 x + 18 < 4$

Also, $x^{2} - 9 x + 18 > 0$

$\Rightarrow (x - 3) (x - 6) > 0$

$\Rightarrow x \in (- \infty, 3) \cup (6, \infty)$ ... (i)

Now, $x^{2} - 9 x + 18 < 4$

$\Rightarrow x^{2} - 9 x + 14 < 0$

$\Rightarrow (x - 2) (x - 7) < 0$

$\Rightarrow x \in (2, 7)$ ... (ii)

From equation (i) & (ii), we get

$x \in (2, 3) \cup (6, 7) = (α, β) \cup (γ, δ)$ [Given]

Hence, $α + β + γ + δ = 2 + 3 + 6 + 7 = 18$ .

Q 32 :

Let f be a function such that $f (x) + 3 f (\frac{24}{x}) = 4 x, x \neq 0$ . Then f(3) + f(8) is equal to [2025]

13
12
10
11

1

2

3

4

(4)

We have, $f (x) + 3 f (\frac{24}{x}) = 4 x$

Put x = 3, f(3) + 3f(8) = 12 ... (i)

Put x = 8, f(8) + 3f(3) = 32 ... (ii)

Adding (i) and (ii), we get f(3) + f(8) = 11.

Q 33 :

Let $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f o g : [2, 4] \to R$ is $[α, β]$ then $\frac{1}{β - α}$ is equal to [2025]

56
68
29
2

1

2

3

4

(1)

Given $f, g : (1, \infty) \to R$ , $f (x) = \frac{2 x + 3}{5 x + 2}$ , $g (x) = \frac{2 - 3 x}{1 - x}$

also, we have $f o g : [2, 4] \to R$

Now, $g (2) = \frac{2 - 6}{1 - 2} = 4, g (4) = \frac{2 - 12}{1 - 4} = \frac{10}{3}$

$∴ f (g (2)) = \frac{8 + 3}{20 + 2} = \frac{1}{2}, f (g (4)) = \frac{20 + 9}{50 + 6} = \frac{29}{56}$

i.e., $α = \frac{1}{2} and β = \frac{29}{56}$

Then, $\frac{1}{β - α} = \frac{1}{\frac{29}{56} - \frac{1}{2}} = \frac{1}{\frac{1}{56}} = 56$ .

Q 34 :

Consider the sets $A = {(x, y) \in R \times R : x^{2} + y^{2} = 25}$ , $B = {(x, y) \in R \times R : x^{2} + 9 y^{2} = 144}$ , $C = {(x, y) \in Z \times Z : x^{2} + y^{2} \leq 4}$ and $D = A \cap B$ . The total number of one-one functions from the set D to the set C is: [2025]

18290
15120
17160
19320

1

2

3

4

(3)

We Have, $A : x^{2} + y^{2} = 25$ ... (i)

$B : \frac{x^{2}}{144} + \frac{y^{2}}{16} = 1$ ... (ii)

$C : x^{2} + y^{2} \leq 4$ ... (iii)

Solving (i) and (ii), we get

$x^{2} + 9 (25 - x^{2}) = 144 \Rightarrow - 8 x^{2} = - 81 \Rightarrow x = \pm \frac{9}{2 \sqrt{2}}$

From (i), $\frac{81}{8} + y^{2} = 25 \Rightarrow y^{2} = 25 - \frac{81}{8} \Rightarrow y = \pm \frac{\sqrt{119}}{2 \sqrt{2}}$

As, $D = A \cap B$

$= {(\frac{9}{2 \sqrt{2}}, \frac{\sqrt{119}}{2 \sqrt{2}}), (\frac{9}{2 \sqrt{2}}, - \frac{\sqrt{119}}{2 \sqrt{2}}), (\frac{- 9}{2 \sqrt{2}}, \frac{\sqrt{119}}{2 \sqrt{2}}), (- \frac{9}{2 \sqrt{2}}, - \frac{\sqrt{119}}{2 \sqrt{2}})}$

$\Rightarrow n (D) = 4$

Also, $C = {(x, y) \in Z \times Z : x^{2} + y^{2} \leq 4}$

={(0, 2), (0, –2), (2, 0), (–2, 0), (1, 1), (–1, –1), (–1, 1), (1, –1), (0, 1), (0, –1), (1, 0), (–1, 0), (0, 0)}.

$\Rightarrow n (C) = 13$

$∴$ Total number of one-one function from D to C = $^{13} P_{4}$ = 17160.

Q 35 :

If the range of the function $f (x) = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$ , is $(- \infty, α] \cup [β, \infty)$ , then $α^{2} + β^{2}$ is equal to : [2025]

190
194
188
192

1

2

3

4

(2)

$y = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$

$\Rightarrow y x^{2} - 3 x y + 2 y + x - 5 = 0$

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

Case I : If y = 0

$\Rightarrow$ x = 5

Case II : if y $\neq$ 0

For real solutions, $D \geq 0$

$\Rightarrow {(- 3 y + 1)}^{2} - 4 (y) (2 y - 5) \geq 0$

$\Rightarrow 9 y^{2} + 1 - 6 y - 8 y^{2} + 20 y \geq 0$

$\Rightarrow y^{2} + 14 y + 1 \geq 0$

$\Rightarrow {(y + 7)}^{2} - 48 \geq 0$

$\Rightarrow | y + 7 | \geq 4 \sqrt{3} \Rightarrow y + 7 \geq 4 \sqrt{3} or y + 7 \leq - 4 \sqrt{3}$

$\Rightarrow y \geq 4 \sqrt{3} - 7 or y \leq - 4 \sqrt{3} - 7$

From Case I and Case II, we have

$y \in (- \infty, - 4 \sqrt{3} - 7] \cup [4 \sqrt{3} - 7, \infty)$

$∴ α = - 4 \sqrt{3} - 7 and β = 4 \sqrt{3} - 7$

$\Rightarrow α^{2} + β^{2} = {(- 4 \sqrt{3} - 7)}^{2} + {(4 \sqrt{3} - 7)}^{2} = 2 (48 + 49) = 194$ .

Q 36 :

Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16}. Then the number of many-one functions f : A $\to$ B such that $I \in f (A)$ is equal to : [2025]

163
139
151
127

1

2

3

4

(3)

Here, n(A) = 4, n(B) = 4

Total number of functions from A to B = $4^{4}$ = 256

Number of one-one functions from A to B = ${}^{4}P_{4}$ = 4! = 24

Number of many-one functions from A to B = 256 – 24 = 232

Number of many-one function for which $1 \notin f (A) = 3^{4}$ = 81

$∴$ Required number of many-one functions = 232 – 81 = 151.

Q 37 :

Let $f (x) = \log_{e} x$ and $g (x) = \frac{x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2}{2 x^{2} - 2 x + 1}$ . Then the domain of fog is [2025]

$(0, \infty)$
$[0, \infty)$
$[1, \infty)$
R

1

2

3

4

(4)

Given, $g (x) = \frac{x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2}{2 x^{2} - 2 x + 1}$

Here, $D g \in R$ [ $∵ 2 x^{2} - 2 x + 1 > 0$ ]

Also, $f (x) = \log_{e} x \Rightarrow D_{f} \in (0, \infty)$ $∴ D_{f o g} \Rightarrow g (x) > 0$

$\Rightarrow \frac{x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2}{2 x^{2} - 2 x + 1} > 0 \Rightarrow x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2 > 0$

The above expression is always positive for any value of x.

$∴ Domain of f o g \in R$ .

Q 38 :

Let $f (x) = \frac{2^{x + 2} + 16}{2^{2 x + 1} + 2^{x + 4} + 32}$ . Then the value of $8 (f (\frac{1}{15}) + f (\frac{2}{15}) + . . . + f (\frac{59}{15}))$ is equal to [2025]

118
102
92
108

1

2

3

4

(1)

$f (x) = \frac{2^{x + 2} + 16}{2^{2 x + 1} + 2^{x + 4} + 32}$

$= \frac{2^{2} (2^{x} + 4)}{2 [{(2^{x})}^{2} + 8 \times 2^{x} + 16]} = \frac{2 (2^{x} + 4)}{{(2^{x} + 4)}^{2}} = \frac{2}{2^{x} + 4}$

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

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

$∴ f (\frac{1}{15}) + f (\frac{59}{15}) = f (\frac{2}{15}) + f (\frac{58}{15})$

$= f (\frac{3}{15}) + f (\frac{57}{15}) = . . . = f (\frac{29}{15}) + f (\frac{31}{15}) = \frac{1}{2}$

$f (\frac{30}{15}) = f (2) = \frac{2}{4 + 4} = \frac{1}{4}$

$8 [f (\frac{1}{15}) + f (\frac{2}{15}) + . . . + f (\frac{59}{15})]$

$= 8 [\frac{1}{2} + \frac{1}{2} + . . . upto (29 terms) + \frac{1}{4}] = 8 (\frac{29}{2} + \frac{1}{4}) = 118$

Q 39 :

The function $f : (- \infty, \infty) \to (- \infty, 1)$ , defined by $f (x) = \frac{2^{x} - 2^{- x}}{2^{x} + 2^{- x}}$ is : [2025]

Onto but not one-one
Both one-one and onto
One-one but not onto
Neither one-one nor onto

1

2

3

4

(3)

Given function is $f (x) = \frac{2^{x} - 2^{- x}}{2^{x} + 2^{- x}}$

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

$\Rightarrow f' (x) = \frac{2}{{(2^{2 x} + 1)}^{2}} \times 2 \times 2^{2 x} \times \log_{e} (2) > 0$

$\Rightarrow$ f(x) is always increasing.

Hence it is one-one.

Since $f (- \infty) = - 1$ and $f (\infty) = 1 \Rightarrow f (x) \in (- 1, 1) \neq (- \infty, 1)$

Thus, the function f(x) is one-one but not onto.

Q 40 :

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ . If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5}$ $| f (i) |$ is [2025]

675
545
735
715

1

2

3

4

(1)

Given, $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ ... (i)

Also, $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ ... (ii)

Put x = y = 0 in (ii), we get

f(0) = f(0) + f(0) + 1 – 0 $\Rightarrow$ f(0) = –1 ... (iii)

Using (i), f(0) = b = –1

Put y = – x in (ii), we get

$f (x - x) = f (x) + f (- x) + 1 + \frac{2}{7} x^{2}$

$\Rightarrow - 1 = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x - 1 + (2 + 3 a) x^{2} - (\frac{a + 2}{a - 1}) x - 1 + 1 + \frac{2}{7} x^{2}$

$\Rightarrow (2 (2 + 3 a) + \frac{2}{7}) x^{2} = 0 \Rightarrow a = \frac{- 5}{7}$

$∴ f (x) = \frac{- 1}{7} x^{2} - \frac{3}{4} x - 1 = \frac{- 1}{28} (4 x^{2} + 21 x + 28)$

Now, $28 \sum_{i = 1}^{5} | f (i) | = 28 [| f (1) | + | f (2) | + . . . + | f (5) |]$

$= 28 \times \frac{1}{28} [4 (1^{2} + 2^{2} + . . . + 5^{2}) + 21 (1 + 2 + . . . + 5) + 28 (5)]$

= 675

Topic Question Set

Q 11 :

Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps $f : A \to B$ , such that $f (1) + f (3) = 14$ , is: [2024]

Q 12 :

The function $f (x) = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}, x \in R$ is [2024]

Q 13 :

Let $f (x) = \frac{1}{7 - \sin 5 x}$ be a function defined on R. Then the range of the function $f (x)$ is equal to : [2024]

Q 14 :

Let [ $t$ ] be the greatest integer less than or equal to $t$ . Let A be the set of all prime factors of 2310 and $f : A \to Z be the function f (x) = [\log_{2} (x^{2} + [\frac{x^{3}}{5}])] .$ The number of one-to-one functions from A to the range of $f$ is [2024]

Q 15 :

Let $f (x) = {\begin{matrix} - a if - a \leq x \leq 0 \\ x + a if 0 < x \leq a \end{matrix}$ where $a > 0$ and $g (x) = \frac{(f (| x |) - | f (x) |)}{2}$ . Then the function $g :$ $[- a, a] \to [- a, a]$ is [2024]

Q 16 :

If the domain of the function $f (x) = \frac{\sqrt{x^{2} - 25}}{(4 - x^{2})} + \log_{10} (x^{2} + 2 x - 15)$ is $(- \infty, α) \cup [β, \infty),$ then $α^{2} + β^{3}$ is equal to: [2024]

Q 17 :

Let $f (x) = {\begin{matrix} x - 1, & x is even, \\ 2 x, & x is odd, \end{matrix} x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $\lim_{x \to a^{-}} {\frac{| x |^{3}}{a} - [\frac{x}{a}]},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is equal to: [2024]

Q 18 :

The function $f : N - {1} \to N$ ; defined by $f (n)$ = the highest prime factor of $n$ , is: [2024]

Q 19 :

Let $f : R - {\frac{- 1}{2}} \to R$ and $g : R - {\frac{- 5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{| x | + 1}{2 x + 5}$ . Then, the domain of the function $f o g$ is [2024]

(4)

Q 20 :

If $f (x) = {\begin{matrix} 2 + 2 x, & - 1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{matrix}; g (x) = {\begin{matrix} - x, & - 3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{matrix},$ then range of $(f o g) (x)$ is [2024]

Q 21 :

If $f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3}$ and $(f o f) (x) = g (x)$ , where $g : R - {\frac{2}{3}} \to R - {\frac{2}{3}},$ then $(g o g o g) (4)$ is equal to [2024]

Q 22 :

Consider the function $f : R \to R$ defined by $f (x) = \frac{2 x}{\sqrt{1 + 9 x^{2}}} .$ If the composition of $\underset{10 times}{f, \underset{⏟}{(f o f o f o . . . o f)}} (x)$ $= \frac{2^{10} x}{\sqrt{1 + 9 α x^{2}}},$ then the value of $\sqrt{3 α + 1}$ is equal to _________ . [2024]

Q 23 :

If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m,$ $n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq {(2022)}^{2}$ is equal to ___________ . [2024]

Q 24 :

Let $A = {(x, y) : 2 x + 3 y = 23, x, y \in N}$ and $B = {x : (x, y) \in A}$ . Then the number of one-one functions from A to B is equal to ________ . [2024]

Q 25 :

Let A = {1, 2, 3, ..., 7} and let P(A) denote the power set of A. If the number of functions $f : A \to P (A)$ such that $a \in f (a)$ , $\forall$ $a \in A$ is $m^{n}$ , $m$ and $n \in N$ and $m$ is least, then $m + n$ is equal to _____. [2024]

Q 26 :

Let $f, g : R \to R$ be defined as: [2024]

$f (x) = | x - 1 |$ and $g (x) =$ ${\begin{matrix} e^{x}, & x \geq 0 \\ x + 1, & x \leq 0 \end{matrix} .$

Then the function $f (g (x))$ is

Q 27 :

Let $f : R \to R$ and $g : R \to R$ be defined as:

$f (x)$ $= {\begin{matrix} \log_{e} x, & x > 0 \\ e^{- x}, & x \leq 0 \end{matrix}$ and $g (x)$ $= {\begin{matrix} x, & x \geq 0 \\ e^{x}, & x < 0 \end{matrix} .$

Then, $g o f : R \to R$ is [2024]

Q 28 :

If the function $f : (- \infty, - 1] \to (a, b]$ defined by $f (x) = e^{x^{3} - 3 x + 1}$ is one-one and onto, then the distance of the point $P (2 b + 4, a + 2)$ from the line $x + e^{- 3} y = 4$ is : [2024]

Q 29 :

If the domain of the function

$f (x) = \frac{1}{\sqrt{10 + 3 x - x^{2}}} + \frac{1}{\sqrt{x + | x |}}$ is (a, b), then

${(1 + a)}^{2} + b^{2}$ is equal to: [2025]

Q 30 :

If the domain of the function

$f (x) = \log_{e} (\frac{2 x - 3}{5 + 4 x}) + \sin^{- 1} (\frac{4 + 3 x}{2 - x}) is [α, β), then α^{2} + 4 β$

is equal to [2025]

Q 31 :

If the domain of the function $f (x) = \log_{7} (1 - \log_{4} (x^{2} - 9 x + 18))$ is $(α, β) \cup (γ, δ)$ , then $α + β + γ + δ$ is equal to [2025]

Q 32 :

Let f be a function such that $f (x) + 3 f (\frac{24}{x}) = 4 x, x \neq 0$ . Then f(3) + f(8) is equal to [2025]

Q 33 :

Let $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f o g : [2, 4] \to R$ is $[α, β]$ then $\frac{1}{β - α}$ is equal to [2025]

Q 34 :

Consider the sets $A = {(x, y) \in R \times R : x^{2} + y^{2} = 25}$ , $B = {(x, y) \in R \times R : x^{2} + 9 y^{2} = 144}$ , $C = {(x, y) \in Z \times Z : x^{2} + y^{2} \leq 4}$ and $D = A \cap B$ . The total number of one-one functions from the set D to the set C is: [2025]

Q 35 :

If the range of the function $f (x) = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$ , is $(- \infty, α] \cup [β, \infty)$ , then $α^{2} + β^{2}$ is equal to : [2025]

Q 36 :

Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16}. Then the number of many-one functions f : A $\to$ B such that $I \in f (A)$ is equal to : [2025]

Q 37 :

Let $f (x) = \log_{e} x$ and $g (x) = \frac{x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2}{2 x^{2} - 2 x + 1}$ . Then the domain of fog is [2025]

Q 38 :

Let $f (x) = \frac{2^{x + 2} + 16}{2^{2 x + 1} + 2^{x + 4} + 32}$ . Then the value of $8 (f (\frac{1}{15}) + f (\frac{2}{15}) + . . . + f (\frac{59}{15}))$ is equal to [2025]

Q 39 :

The function $f : (- \infty, \infty) \to (- \infty, 1)$ , defined by $f (x) = \frac{2^{x} - 2^{- x}}{2^{x} + 2^{- x}}$ is : [2025]

Q 40 :

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ . If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5}$ $| f (i) |$ is [2025]

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Topic Question Set

Q 11 : Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps f:A→B, such that f(1)+f(3)=14, is: [2024]

Q 12 : The function f(x)=x2+2x-15x2-4x+9,x∈R is [2024]

Q 13 : Let f(x)=17-sin5x be a function defined on R. Then the range of the function f(x) is equal to : [2024]

Q 14 : Let [t] be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and f:A→Z be the function f(x)=[log2(x2+[x35])].The number of one-to-one functions from A to the range of f is [2024]

Q 15 : Let f(x)={-a if -a≤x≤0x+a if 0<x≤a where a>0 and g(x)=(f(|x|)-|f(x)|)2. Then the function g: [-a,a]→[-a,a] is [2024]

Q 16 : If the domain of the function f(x)=x2-25(4-x2)+log10(x2+2x-15) is (-∞,α)∪[β,∞), then α2+β3 is equal to: [2024]

Q 17 : Let f(x)={x-1,x is even,2x,x is odd,x∈N. If for some a∈N,f(f(f(a)))=21, then limx→a-{|x|3a-[xa]}, where [t] denotes the greatest integer less than or equal to t, is equal to: [2024]

Q 18 : The function f:N-{1}→N; defined by f(n) = the highest prime factor of n, is: [2024]

Q 19 : Let f:R-{-12}→R and g:R-{-52}→R be defined as f(x)=2x+32x+1 and g(x)=|x|+12x+5. Then, the domain of the function fog is [2024]

(4)

Q 20 : If f(x)={2+2x,-1≤x<01-x3,0≤x≤3;g(x)={-x,-3≤x≤0x,0<x≤1, then range of (fog)(x) is [2024]

Q 21 : If f(x)=4x+36x-4, x≠23 and (fof)(x)=g(x), where g:R-{23}→R-{23}, then (gogog)(4) is equal to [2024]

Q 22 : Consider the function f:R→R defined by f(x)=2x1+9x2. If the composition of f,(fofofo...of)⏟10 times (x) =210x1+9αx2, then the value of 3α+1 is equal to _________ . [2024]

Q 23 : If a function f satisfies f(m+n)=f(m)+f(n) for all m, n∈N and f(1)=1, then the largest natural number λ such that ∑k=12022f(λ+k)≤(2022)2 is equal to ___________ . [2024]

Q 24 : Let A={(x,y):2x+3y=23,x,y∈N} and B={x:(x,y)∈A}. Then the number of one-one functions from A to B is equal to ________ . [2024]

Q 25 : Let A = {1, 2, 3, ..., 7} and let P(A) denote the power set of A. If the number of functions f:A→P(A) such that a∈f(a), ∀ a∈A is mn, m and n∈N and m is least, then m+n is equal to _____. [2024]

Q 26 : Let f,g:R→R be defined as: [2024] f(x)=|x-1| and g(x)={ex,x≥0x+1,x≤0. Then the function f(g(x)) is

Q 27 : Let f:R→R and g:R→R be defined as: f(x)={logex,x>0e-x,x≤0 and g(x)={x,x≥0ex,x<0. Then, gof:R→R is [2024]

Q 28 : If the function f:(-∞,-1]→(a,b] defined by f(x)=ex3-3x+1 is one-one and onto, then the distance of the point P(2b+4, a+2) from the line x+e-3y=4 is : [2024]

Q 29 : If the domain of the function f(x)=110+3x–x2+1x+|x| is (a, b), then (1+a)2+b2 is equal to: [2025]

Q 30 : If the domain of the function f(x)=loge(2x–35+4x)+sin–1(4+3x2–x) is [α,β), then α2+4β is equal to [2025]

Q 31 : If the domain of the function f(x)=log7(1–log4(x2–9x+18)) is (α,β)∪(γ,δ), then α+β+γ+δ is equal to [2025]

Q 32 : Let f be a function such that f(x)+3f(24x)=4x, x≠0. Then f(3) + f(8) is equal to [2025]

Q 33 : Let f,g : (1,∞)→R be defined as f(x)=2x+35x+2 and g(x)=2–3x1–x. If the range of the function fog : [2,4]→R is [α,β] then 1β–α is equal to [2025]

Q 34 : Consider the sets A={(x,y)∈R×R : x2+y2=25}, B={(x,y)∈R×R : x2+9y2=144}, C={(x,y)∈Z×Z : x2+y2≤4} and D=A∩B. The total number of one-one functions from the set D to the set C is: [2025]

Q 35 : If the range of the function f(x)=5–xx2–3x+2, x≠1,2, is (–∞,α]∪[β,∞), then α2+β2 is equal to : [2025]

Q 36 : Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16}. Then the number of many-one functions f : A → B such that I∈f(A) is equal to : [2025]

Q 37 : Let f(x) = logex and g(x)=x4–2x3+3x2–2x+22x2–2x+1. Then the domain of fog is [2025]

Q 38 : Let f(x)=2x+2+1622x+1+2x+4+32. Then the value of 8(f(115)+f(215)+...+f(5915)) is equal to [2025]

Q 39 : The function f : (–∞,∞)→(–∞,1), defined by f(x)=2x–2–x2x+2–x is : [2025]

Q 40 : Let f : R→R be a function defined by f(x)=(2+3a)x2+(a+2a–1)x+b, a≠1. If f(x+y)=f(x)+f(y)+1–27xy, then the value of 28∑i=15|f(i)| is [2025]

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Q 11 :

Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps $f : A \to B$ , such that $f (1) + f (3) = 14$ , is: [2024]

Q 12 :

The function $f (x) = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}, x \in R$ is [2024]

Q 13 :

Let $f (x) = \frac{1}{7 - \sin 5 x}$ be a function defined on R. Then the range of the function $f (x)$ is equal to : [2024]

Q 14 :

Let [ $t$ ] be the greatest integer less than or equal to $t$ . Let A be the set of all prime factors of 2310 and $f : A \to Z be the function f (x) = [\log_{2} (x^{2} + [\frac{x^{3}}{5}])] .$ The number of one-to-one functions from A to the range of $f$ is [2024]

Q 15 :

Let $f (x) = {\begin{matrix} - a if - a \leq x \leq 0 \\ x + a if 0 < x \leq a \end{matrix}$ where $a > 0$ and $g (x) = \frac{(f (| x |) - | f (x) |)}{2}$ . Then the function $g :$ $[- a, a] \to [- a, a]$ is [2024]

Q 16 :

If the domain of the function $f (x) = \frac{\sqrt{x^{2} - 25}}{(4 - x^{2})} + \log_{10} (x^{2} + 2 x - 15)$ is $(- \infty, α) \cup [β, \infty),$ then $α^{2} + β^{3}$ is equal to: [2024]

Q 17 :

Let $f (x) = {\begin{matrix} x - 1, & x is even, \\ 2 x, & x is odd, \end{matrix} x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $\lim_{x \to a^{-}} {\frac{| x |^{3}}{a} - [\frac{x}{a}]},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is equal to: [2024]

Q 18 :

The function $f : N - {1} \to N$ ; defined by $f (n)$ = the highest prime factor of $n$ , is: [2024]

Q 19 :

Let $f : R - {\frac{- 1}{2}} \to R$ and $g : R - {\frac{- 5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{| x | + 1}{2 x + 5}$ . Then, the domain of the function $f o g$ is [2024]

Q 20 :

If $f (x) = {\begin{matrix} 2 + 2 x, & - 1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{matrix}; g (x) = {\begin{matrix} - x, & - 3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{matrix},$ then range of $(f o g) (x)$ is [2024]

Q 21 :

If $f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3}$ and $(f o f) (x) = g (x)$ , where $g : R - {\frac{2}{3}} \to R - {\frac{2}{3}},$ then $(g o g o g) (4)$ is equal to [2024]

Q 23 :

If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m,$ $n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq {(2022)}^{2}$ is equal to ___________ . [2024]

Q 24 :

Let $A = {(x, y) : 2 x + 3 y = 23, x, y \in N}$ and $B = {x : (x, y) \in A}$ . Then the number of one-one functions from A to B is equal to ________ . [2024]

Q 25 :

Let A = {1, 2, 3, ..., 7} and let P(A) denote the power set of A. If the number of functions $f : A \to P (A)$ such that $a \in f (a)$ , $\forall$ $a \in A$ is $m^{n}$ , $m$ and $n \in N$ and $m$ is least, then $m + n$ is equal to _____. [2024]

Q 26 :

Let $f, g : R \to R$ be defined as: [2024]

$f (x) = | x - 1 |$ and $g (x) =$ ${\begin{matrix} e^{x}, & x \geq 0 \\ x + 1, & x \leq 0 \end{matrix} .$

Then the function $f (g (x))$ is

Q 27 :

Let $f : R \to R$ and $g : R \to R$ be defined as:

$f (x)$ $= {\begin{matrix} \log_{e} x, & x > 0 \\ e^{- x}, & x \leq 0 \end{matrix}$ and $g (x)$ $= {\begin{matrix} x, & x \geq 0 \\ e^{x}, & x < 0 \end{matrix} .$

Then, $g o f : R \to R$ is [2024]

Q 28 :

If the function $f : (- \infty, - 1] \to (a, b]$ defined by $f (x) = e^{x^{3} - 3 x + 1}$ is one-one and onto, then the distance of the point $P (2 b + 4, a + 2)$ from the line $x + e^{- 3} y = 4$ is : [2024]

Q 29 :

If the domain of the function

$f (x) = \frac{1}{\sqrt{10 + 3 x - x^{2}}} + \frac{1}{\sqrt{x + | x |}}$ is (a, b), then

${(1 + a)}^{2} + b^{2}$ is equal to: [2025]

Q 30 :

If the domain of the function

$f (x) = \log_{e} (\frac{2 x - 3}{5 + 4 x}) + \sin^{- 1} (\frac{4 + 3 x}{2 - x}) is [α, β), then α^{2} + 4 β$

is equal to [2025]

Q 31 :

If the domain of the function $f (x) = \log_{7} (1 - \log_{4} (x^{2} - 9 x + 18))$ is $(α, β) \cup (γ, δ)$ , then $α + β + γ + δ$ is equal to [2025]

Q 32 :

Let f be a function such that $f (x) + 3 f (\frac{24}{x}) = 4 x, x \neq 0$ . Then f(3) + f(8) is equal to [2025]

Q 33 :

Let $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f o g : [2, 4] \to R$ is $[α, β]$ then $\frac{1}{β - α}$ is equal to [2025]

Q 34 :

Consider the sets $A = {(x, y) \in R \times R : x^{2} + y^{2} = 25}$ , $B = {(x, y) \in R \times R : x^{2} + 9 y^{2} = 144}$ , $C = {(x, y) \in Z \times Z : x^{2} + y^{2} \leq 4}$ and $D = A \cap B$ . The total number of one-one functions from the set D to the set C is: [2025]

Q 35 :

If the range of the function $f (x) = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$ , is $(- \infty, α] \cup [β, \infty)$ , then $α^{2} + β^{2}$ is equal to : [2025]

Q 36 :

Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16}. Then the number of many-one functions f : A $\to$ B such that $I \in f (A)$ is equal to : [2025]

Q 37 :

Let $f (x) = \log_{e} x$ and $g (x) = \frac{x^{4} - 2 x^{3} + 3 x^{2} - 2 x + 2}{2 x^{2} - 2 x + 1}$ . Then the domain of fog is [2025]

Q 38 :

Let $f (x) = \frac{2^{x + 2} + 16}{2^{2 x + 1} + 2^{x + 4} + 32}$ . Then the value of $8 (f (\frac{1}{15}) + f (\frac{2}{15}) + . . . + f (\frac{59}{15}))$ is equal to [2025]

Q 39 :

The function $f : (- \infty, \infty) \to (- \infty, 1)$ , defined by $f (x) = \frac{2^{x} - 2^{- x}}{2^{x} + 2^{- x}}$ is : [2025]

Q 40 :

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ . If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5}$ $| f (i) |$ is [2025]