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

Q 11 :
The number of functions $f : {1, 2, 3, 4} \to {a \in : Z | a | \leq 8}$ satisfying $f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$ is [2023]

2
1
4
3

1

2

3

4

(1)

$f : {1, 2, 3, 4} \to {a \in Z : | a | \leq 8}$

$f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$

$f (n + 1)$ must be divisible by $n$

$f (4) \Rightarrow$ -6, -3, 0, 3, 6

$f (3) \Rightarrow$ -8, -6, -4, -2, 0, 2, 4, 6, 8

$f (2) \Rightarrow$ -8, ................, 8

$f (1) \Rightarrow$ -8, .............. , 8

$\frac{f (4)}{3} must be odd since f (3) is even.$

$∴$ Only two solutions possible.

$\begin{matrix} f (4) & f (3) & f (2) & f (1) \\ - 3 & 2 & 0 & 1 \\ 3 & 0 & 1 & 0 \end{matrix}$

Q 12 :
Let $f (x) = 2 x^{n} + λ, λ \in R, n \in N, a n d f (4) = 133, f (5) = 255 .$ Then the sum of all the positive integer divisors of $(f (3) - f (2))$ is [2023]

59
60
61
58

1

2

3

4

(2)

$f (x) = 2 x^{n} + λ, λ \in R, n \in N$

$f (4) = 2 {(4)}^{n} + λ = 133$ ... $(i)$

$f (5) = 2 {(5)}^{n} + λ = 255$ ... $(ii)$

(ii) - (i), we get

$\Rightarrow f (5) - f (4) = 2 {(5)}^{n} - 2 {(4)}^{n} = 122; 5^{n} - 4^{n} = 61$

which gives $n = 3$ .

Then $f (3) = 2 {(3)}^{3} + λ, f (2) = 2 {(2)}^{3} + λ$

$f (3) = 54 + λ,$ $f (2) = 16 + λ .$ So, $f (3) - f (2) = 38$

Positive divisions of 38 are 1, 2, 19, 38 and whose sum = 1 + 2 + 19 + 38 = 60

Q 13 :
Let $f : R \to R$ be a function defined by $f (x) = \log_{\sqrt{m}} {\sqrt{2} (\sin x - \cos x) + m - 2},$ for some $m$ , such that the range of $f$ is [0, 2]. Then the value of $m$ is [2023]

3
5
4
2

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(2)

Since, $- \sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$

$∴$ $- 2 \leq \sqrt{2} (\sin x - \cos x) \leq 2$

Let $\sqrt{2} (\sin x - \cos x) = t$

$- 2 \leq t \leq 2$ ...(i)

$f (x) = \log_{\sqrt{m}} (t + m - 2)$

We have, $0 \leq f (x) \leq 2$

$\Rightarrow 0 \leq \log_{\sqrt{m}} (t + m - 2) \leq 2$

$1 \leq t + m - 2 \leq m$

$- m + 3 \leq t \leq 2$ ...(ii)

From (i) and (ii), we get $- m + 3 = - 2 \Rightarrow m = 5$

Q 14 :
Let $f : R \to R$ be a function such that $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 1} .$ Then [2023]

$f (x)$ is many-one in $(- \infty, - 1)$
$f (x)$ is one-one in $(- \infty, \infty)$
$f (x)$ is many-one in $(1, \infty)$
$f (x)$ is one-one in $[1, \infty)$ but not in $(- \infty, \infty)$

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(4)

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

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

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

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

The graph of the function is, By horizontal line test, we can say that

[IMAGE 4]

$f (x) is one-one in [1, \infty) but not in (- \infty, \infty) .$

Q 15 :
The domain of $f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)}, x \in ℝ$ is [2023]

$(2, \infty) - {3}$
$ℝ - {3}$
$ℝ - {- 1, 3}$
$(- 1, \infty) - {3}$

1

2

3

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(1)

Given, $f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)} = \frac{\log_{(x + 1)} (x - 2)}{x^{2} - 2 x - 3}$

Logarithmic function will be defined

$x - 2 > 0 \Rightarrow x > 2 and x + 1 > 0 \Rightarrow x > - 1$

$\Rightarrow x + 1 \neq 1 \Rightarrow x \neq 0 and x > 0$

In denominator, $x^{2} - 2 x - 3 \neq 0$

$\Rightarrow (x - 3) (x + 1) \neq 0 \Rightarrow x \neq - 1, 3 .$

So, $x \in (2, \infty) - {3}$

Q 16 :
Consider a function $f : ℕ \to ℝ,$ satisfying $f (1) + 2 f (2) + 3 f (3) + . . . + x f (x) = x (x + 1) f (x); x \geq 2$ with $f (1) = 1 .$ Then $\frac{1}{f (2022)} + \frac{1}{f (2028)}$ is equal to [2023]

8400
8200
8100
8000

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2

3

4

(3)

Let us consider a function $f : N \to R$ satisfying

$f (1) + 2 f (2) + 3 f (3) \dots + x f (x) = x (x + 1) f (x)$

where $x \geq 2$ with $f (1) = 1$ . We have for $x \geq 2$

$f (1) + 2 f (2) + 3 f (3) + \dots + x f (x) = x (x + 1) f (x)$

Now we will replace $x$ by $(x + 1)$ , we get

$x (x + 1) f (x) + (x + 1) f (x + 1) = (x + 1) (x + 2) f (x + 1)$

$\Rightarrow \frac{x}{f (x + 1)} + \frac{1}{f (x)} = \frac{x + 2}{f (x)} \Rightarrow x f (x) = (x + 1) f (x + 1) x \geq 2$

$f (2) = \frac{1}{4}, f (3) = \frac{1}{6}$

Now, $f (2022) = \frac{1}{4044}$

Similarly, $f (2028) = \frac{1}{4056}$

So, $\frac{1}{f (2022)} + \frac{1}{f (2028)} = 4044 + 4056 = 8100$

Q 17 :
The range of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is [2023]

$[2 \sqrt{2}, \sqrt{11}]$
$[\sqrt{5}, \sqrt{10}]$
$[\sqrt{5}, \sqrt{13}]$
$[\sqrt{2}, \sqrt{7}]$

1

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(2)

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

Let $y = \sqrt{3 - x} + \sqrt{2 + x}$

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

$= 5 + 2 \sqrt{6 + x - x^{2}}$

$= 5 + 2 \sqrt{\frac{25}{4} - {(x - \frac{1}{2})}^{2}}$

$y_{m a x}^{2}$ when ${(x - \frac{1}{2})}^{2} = 0$

$∴$ $y_{\max}^{2} = 5 + 2 \times \frac{5}{2} = 10 \Rightarrow y_{\max} = \sqrt{10}$

$y_{m i n}^{2}$ when $\sqrt{\frac{25}{4} - {(x - \frac{1}{2})}^{2}} = 0 \Rightarrow y_{\min} = \sqrt{5}$

$Range of f (x) = [\sqrt{5}, \sqrt{10}]$

Q 18 :
If the domain of the function $f (x) = \frac{[x]}{1 + x^{2}},$ where [x] is greatest integer $\leq x,$ is [2,6), then its range is [2023]

$(\frac{5}{37}, \frac{2}{5}]$
$(\frac{5}{26}, \frac{2}{5}] - {\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}}$
$(\frac{5}{26}, \frac{2}{5}]$
$(\frac{5}{37}, \frac{2}{5}] - {\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}}$

1

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(1)

We have, $f (x) = \frac{[x]}{1 + x^{2}} and domain = [2, 6)$

$∴$ $f (x) =$ ${\begin{matrix} \frac{2}{1 + x^{2}}; & [2, 3) \\ \frac{3}{1 + x^{2}}; & [3, 4) \\ \frac{4}{1 + x^{2}}; & [4, 5) \\ \frac{5}{1 + x^{2}}; & [5, 6) \end{matrix}$

For $x \in [2, 6)$ , $f (x) > 0$ and it is a decreasing function.

At $x = 2, f (x) = \frac{2}{5}$ and at $x = 6, f (x) = \frac{5}{37}$

$Hence, Range = (\frac{5}{37}, \frac{2}{5}]$

Q 19 :
Let $f : R - {2, 6} \to R$ be real valued function defined as $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12} .$ Then range of $f$ is [2023]

$(- \infty, - \frac{21}{4}] \cup [\frac{21}{4}, \infty)$
$(- \infty, - \frac{21}{4}] \cup [0, \infty)$
$(- \infty, - \frac{21}{4}) \cup (0, \infty)$
$(- \infty, - \frac{21}{4}] \cup [1, \infty)$

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(2)

[IMAGE 5]

Let $y = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12}$

By cross multiplying, we get

$y x^{2} - 8 x y + 12 y - x^{2} - 2 x - 1 = 0$

$\Rightarrow x^{2} (y - 1) - x (8 y + 2) + (12 y - 1) = 0$

Case I: When $y \neq 1$ ; $D \geq 0$

$\Rightarrow {(8 y + 2)}^{2} - 4 (y - 1) (12 y - 1) \geq 0$

$\Rightarrow y (4 y + 21) > 0$

$\Rightarrow y \in (- \infty, - \frac{21}{4}] \cup [0, \infty) - {1}$

Case II: When $y = 1$

$x^{2} + 2 x + 1 = x^{2} - 8 x + 12$

$\Rightarrow 10 x = 11$

$\Rightarrow x = \frac{11}{10}$

So, $y$ can be 1. Hence, $y \in (- \infty, - \frac{21}{4}] \cup [0, \infty)$

Q 20 :
If domain of the function

$\log_{e} (\frac{6 x^{2} + 5 x + 1}{2 x - 1}) + \cos^{- 1} (\frac{2 x^{2} - 3 x + 4}{3 x - 5})$ is $(α, β) \cup (γ, δ],$ then 18 $(α^{2} + β^{2} + γ^{2} + δ^{2})$ is equal to ____________. [2023]

0%

(20)

For domain, $\frac{6 x^{2} + 5 x + 1}{2 x - 1} > 0$

$\Rightarrow \frac{(3 x + 1) (2 x + 1)}{2 x - 1} > 0 \Rightarrow x \in (\frac{- 1}{2}, \frac{- 1}{3}) \cup (\frac{1}{2}, \infty)$ $...(A)$

and [IMAGE 6]

$\Rightarrow \frac{2 x^{2} - 1}{3 x - 5} \geq 0 and \frac{2 x^{2} - 6 x + 9}{3 x - 5} \leq 0 \Rightarrow 3 x - 5 < 0$

$\Rightarrow x \in [\frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}}] \cup (\frac{5}{3}, \infty) ...(B)$ [IMAGE 7]

$x < \frac{5}{3}$ $...(C)$

$∴$ $A \cap B \cap C \equiv (\frac{- 1}{2}, \frac{- 1}{3}) \cup (\frac{1}{2}, \frac{1}{\sqrt{2}}]$

$So, 18 (α^{2} + β^{2} + γ^{2} + δ^{2}) = 18 (\frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2})$

$= 18 + 2 = 20$

Topic Question Set

Q 11 :
The number of functions $f : {1, 2, 3, 4} \to {a \in : Z | a | \leq 8}$ satisfying $f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$ is [2023]

Q 12 :
Let $f (x) = 2 x^{n} + λ, λ \in R, n \in N, a n d f (4) = 133, f (5) = 255 .$ Then the sum of all the positive integer divisors of $(f (3) - f (2))$ is [2023]

Q 13 :
Let $f : R \to R$ be a function defined by $f (x) = \log_{\sqrt{m}} {\sqrt{2} (\sin x - \cos x) + m - 2},$ for some $m$ , such that the range of $f$ is [0, 2]. Then the value of $m$ is [2023]

Q 14 :
Let $f : R \to R$ be a function such that $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 1} .$ Then [2023]

Q 15 :
The domain of $f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)}, x \in ℝ$ is [2023]

Q 16 :
Consider a function $f : ℕ \to ℝ,$ satisfying $f (1) + 2 f (2) + 3 f (3) + . . . + x f (x) = x (x + 1) f (x); x \geq 2$ with $f (1) = 1 .$ Then $\frac{1}{f (2022)} + \frac{1}{f (2028)}$ is equal to [2023]

Q 17 :
The range of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is [2023]

Q 18 :
If the domain of the function $f (x) = \frac{[x]}{1 + x^{2}},$ where [x] is greatest integer $\leq x,$ is [2,6), then its range is [2023]

Q 19 :
Let $f : R - {2, 6} \to R$ be real valued function defined as $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12} .$ Then range of $f$ is [2023]

Q 20 :
If domain of the function

$\log_{e} (\frac{6 x^{2} + 5 x + 1}{2 x - 1}) + \cos^{- 1} (\frac{2 x^{2} - 3 x + 4}{3 x - 5})$ is $(α, β) \cup (γ, δ],$ then 18 $(α^{2} + β^{2} + γ^{2} + δ^{2})$ is equal to ____________. [2023]

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Topic Question Set

Q 11 : The number of functions f:{1,2,3,4}→{a∈:Z|a|≤8} satisfying f(n)+1nf(n+1)=1, ∀n∈{1,2,3} is [2023]

Q 12 : Let f(x)=2xn+λ,λ∈R,n∈N, and f(4)=133,f(5)=255. Then the sum of all the positive integer divisors of (f(3)-f(2)) is [2023]

Q 13 : Let f:R→R be a function defined by f(x)=logm{2(sinx-cosx)+m-2}, for some m, such that the range of f is [0, 2]. Then the value of m is [2023]

Q 14 : Let f:R→R be a function such that f(x)=x2+2x+1x2+1. Then [2023]

Q 15 : The domain of f(x)=log(x+1)(x-2)e2logex-(2x+3),x∈ℝ is [2023]

Q 16 : Consider a function f:ℕ→ℝ, satisfying f(1)+2f(2)+3f(3)+...+xf(x)=x(x+1)f(x);x≥2 with f(1)=1. Then 1f(2022)+1f(2028) is equal to [2023]

Q 17 : The range of the function f(x)=3-x+2+x is [2023]

Q 18 : If the domain of the function f(x)=[x]1+x2, where [x] is greatest integer ≤x, is [2,6), then its range is [2023]

Q 19 : Let f:R-{2,6}→R be real valued function defined as f(x)=x2+2x+1x2-8x+12. Then range of f is [2023]

Q 20 : If domain of the function loge(6x2+5x+12x-1)+cos-1(2x2-3x+43x-5) is (α,β)∪(γ,δ], then 18(α2+β2+γ2+δ2) is equal to ____________. [2023]

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Q 11 :
The number of functions $f : {1, 2, 3, 4} \to {a \in : Z | a | \leq 8}$ satisfying $f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$ is [2023]

Q 12 :
Let $f (x) = 2 x^{n} + λ, λ \in R, n \in N, a n d f (4) = 133, f (5) = 255 .$ Then the sum of all the positive integer divisors of $(f (3) - f (2))$ is [2023]

Q 13 :
Let $f : R \to R$ be a function defined by $f (x) = \log_{\sqrt{m}} {\sqrt{2} (\sin x - \cos x) + m - 2},$ for some $m$ , such that the range of $f$ is [0, 2]. Then the value of $m$ is [2023]

Q 14 :
Let $f : R \to R$ be a function such that $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 1} .$ Then [2023]

Q 15 :
The domain of $f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)}, x \in ℝ$ is [2023]

Q 16 :
Consider a function $f : ℕ \to ℝ,$ satisfying $f (1) + 2 f (2) + 3 f (3) + . . . + x f (x) = x (x + 1) f (x); x \geq 2$ with $f (1) = 1 .$ Then $\frac{1}{f (2022)} + \frac{1}{f (2028)}$ is equal to [2023]

Q 17 :
The range of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is [2023]

Q 18 :
If the domain of the function $f (x) = \frac{[x]}{1 + x^{2}},$ where [x] is greatest integer $\leq x,$ is [2,6), then its range is [2023]

Q 19 :
Let $f : R - {2, 6} \to R$ be real valued function defined as $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12} .$ Then range of $f$ is [2023]

Q 20 :
If domain of the function

$\log_{e} (\frac{6 x^{2} + 5 x + 1}{2 x - 1}) + \cos^{- 1} (\frac{2 x^{2} - 3 x + 4}{3 x - 5})$ is $(α, β) \cup (γ, δ],$ then 18 $(α^{2} + β^{2} + γ^{2} + δ^{2})$ is equal to ____________. [2023]