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

Q 41 :

Let for some $α \in R, R \to R$ be a function satisfying $f (x + y) = f (x) + 2 y^{2} + y + α x y$ for all $x, y \in ℝ$ .

If f(0) = –1 and f(1) = 2, then the value of $\sum_{n = 1}^{5} (α + f (n))$ is: [2026]

110
140
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170

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(2)

$f (x + y) = f (x) + 2 y^{2} + y + α x y$ ; $x, y \in R$

Put x = 0, $f (y) = f (0) + 2 y^{2} + y + α (0) (y)$

$\Rightarrow f (y) = - 1 + 2 y^{2} + y$ [ $∵ f (0) = - 1$ ]

$\Rightarrow f (y) = 2 y^{2} + 1$ $∴ f (n) = 2 n^{2} + n - 1$

$\Rightarrow f (2) = 8 + 2 - 1 = 9$

Now, f(1) = 2

Put x = 1, y = 1; f(2) = f(1) + 2 + 1 + $α$

$\Rightarrow 9 = 5 + α$

$\Rightarrow α = 4$

Now, $\sum_{n = 1}^{5} (α + f (n)) = \sum_{n = 1}^{5} (4 + (2 n^{2} + n - 1))$

$= \sum_{n = 1}^{5} (2 n^{2} + n + 3)$

$= 2 \sum_{n = 1}^{5} n^{2} + \sum_{n = 1}^{5} n + 3 \sum_{n = 1}^{5}$

$= 2 \times [\frac{5 (6) \times 11}{6}] + \frac{5 \times 6}{2} + 3 \times 5$

$= 2 \times 55 + 15 + 15 = 140$

Q 42 :

Let $f : R \to R$ be defined as $f (x) = \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3}$ . Then f is : [2026]

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

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(4)

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

$\Rightarrow \lim_{x \to \infty} f (x) = \lim_{x \to \infty} \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3} = \frac{2}{3}$

$\lim_{x \to - \infty} f (x) = \lim_{x \to - \infty} \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3} = \frac{2}{3}$

Since, $\lim_{x \to \infty} f (x) = \lim_{x \to - \infty} f (x) = \frac{2}{3}$ , the function must change direction to return to that value i.e., $\frac{2}{3}$ . This means, it fails the horizontal line test.

Thus, the given function is not one-one.

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

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

Since, $D \geq 0, {(y + 3)}^{2} - 4 {(3 y - 2)}^{2} \geq 0$

$\Rightarrow (7 y - 1) (5 y - 7) \leq 0 \Rightarrow y \in [\frac{1}{7}, \frac{7}{5}]$

Since, range of f(x) is a finite interval, f(x) is not onto.

Hence, f(x) is neither one-one nor onto.

Q 43 :

If the domain of the function $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$ is $(- \infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of a + b + c + d + e is ________. [2026]

(4)

We have, $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$

$\log_{0.6} | \frac{2 x - 5}{x^{2} - 4} | \geq 0$

$\Rightarrow | \frac{2 x - 5}{x^{2} - 4} | \leq 1$

$\Rightarrow | 2 x - 5 | \leq | x^{2} - 4 |; x \neq \frac{5}{2}, 2, - 2$

Also, $| 2 x - 5 | \leq | x^{2} - 4 |$

$\Rightarrow {(2 x - 5)}^{2} \leq {(x^{2} - 4)}^{2}$

$\Rightarrow (x^{2} - 4 + 2 x - 5) (x^{2} - 4 - 2 x + 5) \geq 0$

$\Rightarrow (x^{2} + 2 x - 9) (x^{2} - 2 x + 1) \geq 0$

$\Rightarrow (x^{2} + 2 x - 9) {(x - 1)}^{2} \geq 0$

Since, ${(x - 1)}^{2} \geq 0$

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

$\Rightarrow x \in (- \infty, - 1 - \sqrt{10}] \cup [1} \cup (- 1 + \sqrt{10}, \infty)$

exclude x = 2, $\frac{5}{2}$ ( $∵$ –2 is already excluded)

Hence, $x \in (- \infty, - 1 - \sqrt{10}] \cup [1} \cup (- 1 + \sqrt{10}, 5 / 2) \cup (5 / 2, \infty)$

$x \in (- \infty, a] \cup [b} \cup [c, d) \cup (e, \infty)$ (Given)

$∴ a = - 1 - \sqrt{10}, b = 1, c = - 1 + \sqrt{10}, d = 5 / 2, e = 5 / 2$

$∴$ a + b + c + d + c = 4.

Q 44 :

Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f : A $\to$ A such that $f (1) \geq 3, f (3) \leq 4$ , and f(2) + f(3) = 5, is ________. [2026]

(72)

A = {1, 2, 3, 4, 5, 6}

f : A $\to$ A

Case I : (f(2), f(3)) = (4, 1) or (1, 4), then Choices of values of f(1) $\in$ {3, 5, 6} = 3

Number of ways 2 x 3 x 6 = 36

Case II : (f(2), f(3) = (2, 3) or (3, 2), then Choices of values f(1) $\in$ {4, 5, 6} = 3

Number of ways = 2 x 3 x 6 = 36

Total number of functions = 36 + 36 = 72.

Q 45 :

$f (x)$ is a differentiable function satisfy the relationship $f^{2} (x) + f^{2} (y) + 2 (x y - 1) = f^{2} (x + y)$ $\forall x, y \in R .$ Also $f (x) > 0$ $\forall x \in R$ , and $f (\sqrt{2}) = 2$ . then $f (\sqrt{7}) =$

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(1)

$Put x = 0 and y = 0 \Rightarrow f^{2} (0) = 2$

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{f^{2} (x + h) - f^{2} (x)}{[f (x + h) + f (x)] h}$

$= \frac{1}{2 f (x)} \lim_{h \to 0} \frac{f^{2} (h) + 2 (x h - 1)}{h}$ $= \frac{1}{2 f (x)} \lim_{h \to 0} [\frac{2 x h}{h} + \frac{f^{2} (h) - 2}{h}]$

$= \frac{1}{2 f (x)} [2 x + \lim_{h \to 0} \frac{f^{2} (h) - f^{2} (0)}{h}]$ $= \frac{1}{2 f (x)} [2 x + \lim_{h \to 0} \frac{f (h) - f (0)}{h} (f (h) + f (0))]$

$f' (x) = \frac{1}{2 f (x)} [2 x + f' (0) \cdot 2 f (0)]$

$∴$ $f (x) \cdot f' (x) = x + f (0) \cdot f' (0)$

$\Rightarrow f (x) \cdot f' (x) = x + λ,$ $where λ = f (0) \cdot f' (0)$

$Integrating both sides$

$∴$ $\frac{f^{2} (x)}{2} = \frac{x^{2}}{2} + λ x + C$

$f^{2} (x) = x^{2} + 2 λ x + C$ $At x = 0, f^{2} (0) = 2 \Rightarrow C = 2$

$x = \sqrt{f^{2} (\sqrt{2})} \Rightarrow λ = 0$

$∴$ $f^{2} (x) = x^{2} + 2$

$\Rightarrow f (x) = \sqrt{x^{2} + 2}, (f (x) > 0)$

Q 46 :

$Number of correct statements among the following.$

$Statement I: If f, g : R \to R are defined as$ $f (x) =$ ${\begin{matrix} x, & if x is rational \\ 0, & if x is irrational \end{matrix}$ $and$ $g (x) =$ ${\begin{matrix} 0, & if x is rational \\ x, & if x is irrational \end{matrix},$

$then f - g is one-one and onto.$

$Statement II: f (x) = \sin x + \cos a x is a periodic function, then a must be rational.$

$Statement III: If f : A \to B and g : B \to C are functions such that g \circ f : A \to C is one-one, then f must be one-one.$

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(4)

$f - g =$ $x$ $x is rational$

$= - x$ $x is irrational$

Q 47 :

$Graph of y = f (x) is given below$

$Then graph of y = \frac{- 1}{| f (x) |} is best represented by$

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(3)

Draw graphs

Q 48 :

Match the range of functions given in Column I with Column II.

Column I Column II

A $f (x) =$ $| \sin x |$ $, x \in R$ p [0, 2]

B $f (x) =$ $| 3 - x |$ $+$ $| 2 + x |$ $, x \in [0, 4]$ q [5, 7]

C $f (x) = x^{4} + 2 x^{2} + 5, x \in [- 1, 1]$ r [0, 1]

D $f (x) = \cos^{2} x, x \in R$ s [5, 8]

(A) → (p), (B) → (s), (C) → (q), (D) → (r)
(A) → (q), (B) → (p), (C) → (r), (D) → (s)
(A) → (p), (B) → (q), (C) → (s), (D) → (r)
(A) → (r), (B) → (q), (C) → (s), (D) → (r)

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(4)

$A) f' (x) = (1 - x) (2 x + 1) e^{x (1 - x)} \geq 0, range is [0, 1]$

$B) f (x) =$ ${\begin{matrix} 1 - 2 x, & x < - 2 \\ 5, & - 2 \leq x < 3 \\ 2 x - 1, & x \geq 3 \end{matrix}$ $Min. value of f (x) = 5$

$Max. value of f (x) = 2 (4) - 1 = 7$

$C) f (x) = {(x^{2} + 1)}^{2} + 4, Minimum at x = 0$

$D) f' (x) = 2 x^{3} (2 - x^{2}) e^{- x^{2}} \Rightarrow Decreasing in [- 1, 0]$

Q 49 :

The Domain of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is

[-2, 3]
[-3, 2]
[-3, -2]
[-2, 2]

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(1)

$3 - x \geq 0$ $2 + x \geq 0,$ $x - 3 \leq 0$ $x \geq - 2,$ $x \leq 3$

Topic Question Set

Q 41 :

Let for some $α \in R, R \to R$ be a function satisfying $f (x + y) = f (x) + 2 y^{2} + y + α x y$ for all $x, y \in ℝ$ .

If f(0) = –1 and f(1) = 2, then the value of $\sum_{n = 1}^{5} (α + f (n))$ is: [2026]

Q 42 :

Let $f : R \to R$ be defined as $f (x) = \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3}$ . Then f is : [2026]

Q 43 :

If the domain of the function $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$ is $(- \infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of a + b + c + d + e is ________. [2026]

Q 44 :

Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f : A $\to$ A such that $f (1) \geq 3, f (3) \leq 4$ , and f(2) + f(3) = 5, is ________. [2026]

Q 45 :

$f (x)$ is a differentiable function satisfy the relationship $f^{2} (x) + f^{2} (y) + 2 (x y - 1) = f^{2} (x + y)$ $\forall x, y \in R .$ Also $f (x) > 0$ $\forall x \in R$ , and $f (\sqrt{2}) = 2$ . then $f (\sqrt{7}) =$

Q 47 :

$Graph of y = f (x) is given below$

$Then graph of y = \frac{- 1}{| f (x) |} is best represented by$

Q 49 :

The Domain of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is

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	Column I		Column II
A	$f (x) =$ $\| \sin x \|$ $, x \in R$	p	[0, 2]
B	$f (x) =$ $\| 3 - x \|$ $+$ $\| 2 + x \|$ $, x \in [0, 4]$	q	[5, 7]
C	$f (x) = x^{4} + 2 x^{2} + 5, x \in [- 1, 1]$	r	[0, 1]
D	$f (x) = \cos^{2} x, x \in R$	s	[5, 8]

Topic Question Set

Q 41 : Let for some α∈R, R→R be a function satisfying f(x+y)=f(x)+2y2+y+αxy for all x,y∈ℝ. If f(0) = –1 and f(1) = 2, then the value of ∑n=15(α+f(n)) is: [2026]

Q 42 : Let f:R→R be defined as f(x)=2x2–3x+23x2+x+3. Then f is : [2026]

Q 43 : If the domain of the function f(x)=log(0.6)(|2x–5x2–4|) is (–∞,a]∪{b}∪[c,d)∪(e,∞), then the value of a + b + c + d + e is ________. [2026]

Q 44 : Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f : A → A such that f(1)≥3, f(3)≤4, and f(2) + f(3) = 5, is ________. [2026]

Q 45 : f(x) is a differentiable function satisfy the relationship f2(x)+f2(y)+2(xy-1)=f2(x+y) ∀x,y∈R. Also f(x)>0 ∀ x∈R, and f(2)=2. then f(7)=

Q 47 : Graph of y=f(x) is given below Then graph of y=-1|f(x)| is best represented by

Q 48 : Match the range of functions given in Column I with Column II. Column I Column II A f(x)=|sinx|, x∈R p [0, 2] B f(x)=|3-x|+|2+x|, x∈[0,4] q [5, 7] C f(x)=x4+2x2+5, x∈[-1,1] r [0, 1] D f(x)=cos2x, x∈R s [5, 8]

Q 49 : The Domain of the function f(x)=3-x+2+x is

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Q 41 :

Let for some $α \in R, R \to R$ be a function satisfying $f (x + y) = f (x) + 2 y^{2} + y + α x y$ for all $x, y \in ℝ$ .

If f(0) = –1 and f(1) = 2, then the value of $\sum_{n = 1}^{5} (α + f (n))$ is: [2026]

Q 42 :

Let $f : R \to R$ be defined as $f (x) = \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3}$ . Then f is : [2026]

Q 43 :

If the domain of the function $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$ is $(- \infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of a + b + c + d + e is ________. [2026]

Q 44 :

Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f : A $\to$ A such that $f (1) \geq 3, f (3) \leq 4$ , and f(2) + f(3) = 5, is ________. [2026]

Q 45 :

$f (x)$ is a differentiable function satisfy the relationship $f^{2} (x) + f^{2} (y) + 2 (x y - 1) = f^{2} (x + y)$ $\forall x, y \in R .$ Also $f (x) > 0$ $\forall x \in R$ , and $f (\sqrt{2}) = 2$ . then $f (\sqrt{7}) =$

Q 47 :

$Graph of y = f (x) is given below$

$Then graph of y = \frac{- 1}{| f (x) |} is best represented by$

Q 49 :

The Domain of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is