JEE Advanced Maths PYP Relations and Functions

Q 1 :

If the function $f : ℝ \to ℝ$ is defined by $f (x) =$ $| x |$ $(x - \sin x),$ then which of the following statements is TRUE ? [2020]

$f$ is one-one, but NOT onto
$f$ is onto, but NOT one-one
$f$ is BOTH one-one and onto
$f$ is NEITHER one-one NOR onto

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(3)

$f (x) is a non-periodic, continuous and odd function$

$f (x) = {\begin{matrix} - x^{2} + x \sin x, & x < 0 \\ x^{2} - x \sin x, & x \geq 0 \end{matrix}$

$f' (x) =$ ${\begin{matrix} - 2 x + \sin x + x \cos x, & x < 0 \\ 2 x - \sin x - x \cos x, & x \geq 0 \end{matrix}$

$f'' (x) =$ ${\begin{matrix} - (x - \sin x) - x (1 - \cos x), & x < 0 \\ (x - \sin x) + x (1 - \cos x), & x > 0 \end{matrix}$

$∵$ $x - \sin x < 0$ $if x < 0$ and

$1 - \cos x > 0,$ $\forall x \in ℝ$

$∴$ $- (x - \sin x) - x (1 - \cos x) > 0$ $if x < 0$

and $(x - \sin x) + x (1 - \cos x) > 0$ $if x > 0$

$\Rightarrow f' (x) > 0,$ $\forall x \in ℝ$ $\Rightarrow f (x) is increasing in ℝ$

$\Rightarrow f (x) is one-one$

$∴$ $\lim_{x \to - \infty} (- x^{2}) (1 - \frac{\sin x}{x}) = - \infty$

$∴$ $\lim_{x \to \infty} x^{2} (1 - \frac{\sin x}{x}) = \infty$

$\Rightarrow Range of f (x) = ℝ$

$\Rightarrow f (x) is an onto function.$

Q 2 :

The function $f : [0, 3] \to [1, 29]$ , defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1,$ is [2012]

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

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(2)

$Given: f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1$

$\Rightarrow f' (x) = 6 x^{2} - 30 x + 36$

$= 6 (x^{2} - 5 x + 6) = 6 (x - 2) (x - 3)$

$∵$ $f' (x) > 0,$ $\forall x \in [0, 2)$ and $f' (x) < 0,$ $\forall x \in (2, 3)$

$∴$ $f (x) is increasing on [0, 2) and decreasing on (2, 3)$

$∴$ $f (x) is many-one on [0, 3]$

Also, $f (0) = 1,$ $f (2) = 29,$ $f (3) = 28$

$∴$ $Absolute minimum = 1 and Absolute maximum = 29$

$∴$ $Range of f = [1, 29] = codomain$

$∴$ $f is onto$

Q 3 :

Let $f$ , $g$ and $h$ be real-valued functions defined on the interval [0, 1] by $f (x) = e^{x^{2}} + e^{- x^{2}},$ $g (x) = x e^{x^{2}} + e^{- x^{2}}$ and $h (x) = x^{2} e^{x^{2}} + e^{- x^{2}} .$ If $a$ , $b$ and $c$ denote, respectively, the absolute maximum of $f$ , $g$ and $h$ on [0, 1], then [2010]

$a = b$ and $c \neq b$
$a = c$ and $a \neq b$
$a \neq b$ and $c \neq b$
$a = b = c$

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(4)

$f (x) = e^{x^{2}} + e^{- x^{2}}$ $\Rightarrow f' (x) = 2 x (e^{x^{2}} - e^{- x^{2}}) \geq 0,$ $\forall x \in [0, 1]$

$∴$ $f (x) is an increasing function on [0, 1]$

$∴$ $f_{\max} = f (1) = e + \frac{1}{e} = a;$ $g (x) = x e^{x^{2}} + e^{- x^{2}}$

$\Rightarrow g' (x) = (2 x^{2} + 1) e^{x^{2}} - 2 x e^{- x^{2}} \geq 0,$ $\forall x \in [0, 1]$

$∴$ $g (x) is an increasing function on [0, 1]$

$∴$ $g_{\max} = g (1) = e + \frac{1}{e} = b$

$h (x) = x^{2} e^{x^{2}} + e^{- x^{2}}$

$\Rightarrow h' (x) = 2 x [e^{x^{2}} (1 + x^{2}) - e^{- x^{2}}] \geq 0,$ $\forall x \in [0, 1]$

$∴$ $h (x) is an increasing function on [0, 1]$

$∴$ $h_{\max} = h (1) = e + \frac{1}{e} = c$

$∴$ $a = b = c$

Q 4 :

If the functions $f (x)$ and $g (x)$ are defined on $ℝ \to ℝ$ such that

$f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix};$ $g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

then $(f - g) (x)$ is [2005]

one-one & onto
neither one-one nor onto
one-one but not onto
onto but not one-one

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(1)

$Given f (x) and g (x) defined on ℝ \to ℝ$

and $f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix}$

$g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

$∴$ $(f - g) : ℝ \to ℝ such that$

$(f - g) (x) =$ ${\begin{matrix} - x, & x \in rational \\ x, & x \in irrational \end{matrix}$

Since $(f - g) : ℝ \to ℝ$ , for any $x$ , there is only one value of $(f (x) - g (x))$ whether $x$ is rational or irrational. Moreover, as $x \in ℝ$ , $f (x) - g (x)$ also belongs to $R$ . Therefore $(f - g)$ is one-one onto.

Q 5 :

If $f : [0, \infty) \to [0, \infty)$ and $f (x) = \frac{x}{1 + x},$ then $f$ is [2003]

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

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(2)

$Given: f : [0, \infty) \to [0, \infty)$ and $f (x) = \frac{x}{x + 1}$

$∴$ $f' (x) = \frac{(1 + x) - x}{{(1 + x)}^{2}} = \frac{1}{{(1 + x)}^{2}} > 0,$ $\forall x$

$∴$ $f is an increasing function \Rightarrow f is one-one.$

Now, $D_{f} = [0, \infty)$

For range, let $\frac{x}{1 + x} = y$ $\Rightarrow x = \frac{y}{1 - y}$

Now, $x \geq 0 \Rightarrow 0 \leq y < 1$

$∴$ $R_{f} = [0, 1) \neq Co-domain,$

$∴$ $f is not onto.$

Q 6 :

Let function $f : R \to R$ be defined by $f (x) = 2 x + \sin x$ for $x \in R,$ then $f$ is [2002]

one-to-one and onto
one-to-one but NOT onto
onto but NOT one-to-one
neither one-to-one nor onto

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(1)

$Given: f (x) = 2 x + \sin x,$ $x \in R$

$\Rightarrow f' (x) = 2 + \cos x .$

Now, $- 1 \leq \cos x \leq 1$

$\Rightarrow 1 \leq 2 + \cos x \leq 3$

$∴$ $f' (x) > 0,$ $\forall x \in R$

$\Rightarrow f (x) is strictly increasing and therefore one-one.$

$Also,$ $\lim_{x \to \infty} f (x) = \infty$ and $\lim_{x \to - \infty} f (x) = - \infty$

$∴$ $Range of f (x) = R = domain of f (x) \Rightarrow f (x) is onto.$

$Hence,$ $f (x) is one-one and onto.$

Q 7 :

Suppose $f (x) = {(x + 1)}^{2}$ for $x \geq - 1 .$ If $g (x)$ is the function whose graph is the reflection of the graph of $f (x)$ with respect to the line $y = x$ , then $g (x)$ equals [2002]

$- \sqrt{x} - 1, x \geq 0$
$\frac{1}{{(x + 1)}^{2}}, x > - 1$
$\sqrt{x + 1}, x \geq - 1$
$\sqrt{x} - 1, x \geq 0$

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(4)

$Given: f (x) = {(x + 1)}^{2}, x \geq - 1$

If $g (x)$ is the reflection of $f (x)$ in the line $y = x$ , then it can be obtained by interchanging $x$ and $y$ in $f (x)$

i.e., $y = {(x + 1)}^{2}$ changes to $x = {(y + 1)}^{2}$

$\Rightarrow y + 1 = \sqrt{x}$ $[y + 1 \neq - \sqrt{x}, since y \geq - 1]$

$\Rightarrow y = \sqrt{x} - 1,$ $defined ∀ x \geq 0$

[IMAGE 1199]

$∴$ $g (x) = \sqrt{x} - 1,$ $\forall x \geq 0$

Q 8 :

Let $E$ = {1, 2, 3, 4} and $F$ = {1, 2}. Then the number of onto functions from $E$ to $F$ is [2001]

14
16
12
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(1)

E = {1, 2, 3, 4} and F = {1, 2}

From E to F we can define, in all, 2 × 2 × 2 × 2 = 16 functions (2 options for each element of E), out of which 2 are into, when all the elements of E either map to 1 or to 2.

$∴$ Number of onto functions = 16 − 2 = 14

Q 9 :

The domain of definition of the function $f (x)$ is given by the equation $2^{x} + 2^{y} = 2$ is [2000]

$0 < x \leq 1$
$0 \leq x \leq 1$
$- \infty < x \leq 0$
$- \infty < x < 1$

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(4)

$Given: 2^{x} + 2^{y} = 2,$ $\forall x, y \in R$

$but$ $2^{x}, 2^{y} > 0,$ $\forall x, y \in R$

$∴$ $2^{x} = 2 - 2^{y} < 2$ $\Rightarrow 0 < 2^{x} < 2$ $\Rightarrow x < 1$

$∴$ $Domain = (- \infty, 1)$

Q 10 :

Let $f : R \to R$ be any function. Define $g : R \to R$ by $g (x) =$ $| f (x) |$ for all $x$ . Then $g$ is [2000]

onto if $f$ is onto
one-one if $f$ is one-one
continuous if $f$ is continuous
differentiable if $f$ is differentiable

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(3)

$Let h (x) =$ $| x |$

$∴$ $g (x) =$ $| f (x) |$ $= h (f (x))$

Since the composition of two continuous functions is continuous, therefore $g$ is continuous if $f$ is continuous.

Q 11 :

Let the set of all relations $R$ on the set ${a, b, c, d, e, f},$ such that $R$ is reflexive and symmetric, and $R$ contains exactly 10 elements, be denoted by $S$ . Then the number of elements in $S$ is ______. [2025]

(105)

[IMAGE 1200]

For relation to be reflexive, all the diagonal elements must be taken.

Out of the remaining 30 elements, there are 15 pairs, and we need 2 pairs such that R contains exactly 10 elements and is both reflexive and symmetric.

Hence, Number of ways $= {}^{15}C_{2} = 105$

Q 12 :

Let $X$ be a set with exactly 5 elements and $Y$ be a set with exactly 7 elements. If $α$ is the number of one-one functions from $X$ to $Y$ and $β$ is the number of onto functions from $Y$ to $X$ , then the value of $\frac{1}{5!} (β - α)$ is ________. [2018]

(119)

$Here n (X) = 5 and n (Y) = 7$

$Number of one-one functions = α = {}^{7}C_{5} \times 5!$

$and$ $Number of onto functions Y \to X = β$

[IMAGE 1201]

$= \frac{7!}{3! 4!} \times 5! + \frac{7!}{{(2!)}^{3} 3!} \times 5! = ({}^{7}C_{3} + 3 \times {}^{7}C_{3}) 5!$

$= 4 \times {}^{7}C_{3} \times 5!$

$\Rightarrow \frac{β - α}{5!} = 4 \times {}^{7}C_{3} - {}^{7}C_{5} = 4 \times 35 - 21 = 119$

Q 13 :

Let $ℕ$ denote the set of all natural numbers, and $ℤ$ denote the set of all integers. Consider the functions $f : ℕ \to ℤ$ and $g : ℤ \to ℕ$ defined by

$f (n) =$ ${\begin{matrix} \frac{(n + 1)}{2} & if n is odd, \\ \frac{(4 - n)}{2} & if n is even, \end{matrix}$ and

$g (n) =$ ${\begin{matrix} 3 + 2 n & if n \geq 0, \\ - 2 n & if n < 0 . \end{matrix}$

Define $(g \circ f) (n) = g (f (n))$ for all $n \in ℕ$ , and $(f \circ g) (n) = f (g (n))$ for all $n \in ℤ$ .

Then which of the following statement(s) is (are) TRUE? [2025]

$g \circ f$ is NOT one-one and $g \circ f$ is NOT onto
$f \circ g$ is NOT one-one but $f \circ g$ is onto
$g$ is one-one and $g$ is onto
$f$ is NOT one-one but $f$ is onto

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Select one or more options

(1, 4)

$Given,$ $f (n) =$ ${\begin{matrix} \frac{(n + 1)}{2} & if n is odd \\ \frac{(4 - n)}{2} & if n is even \end{matrix}$

$f (n) = {(1, 1), (2, 1), (3, 2), (4, 0), (5, 3), (6, - 1), \dots}$

$∴$ $f (n) is many-one and onto function.$

$g (n) =$ ${\begin{matrix} 3 + 2 n & if n \geq 0 \\ - 2 n & if n < 0 \end{matrix}$

$g (n) = {(- 3, 6), (- 2, 4), (- 1, 2), (0, 3), (1, 5), (2, 7), (3, 9), (4, 15), \dots}$

$∴$ $g (n) is one-one and into function.$

$f (g (n)) = 2 + n,$ $n \in ℕ$ $\Rightarrow f \circ g is one-one and into$

$g (f (n)) =$ ${\begin{matrix} 4 + n & if n is odd natural number \\ 7 - n & if n = 2, 4 \\ n - 4 & if n is even natural number and n \geq 6 \end{matrix}$

$g (f (2)) = g (f (1)) = 5$

$∴$ $g \circ f is many-one and into$

Q 14 :

Let $a \in R$ and let $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a .$ Then [2014]

$f (x)$ has three real roots if $a > 4$
$f (x)$ has only one real root if $a > 4$
$f (x)$ has three real roots if $a < - 4$
$f (x)$ has three real roots if $- 4 < a < 4$

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Select one or more options

(2, 4)

$f (x) = x^{5} - 5 x + a$

$f (x) = 0$ $\Rightarrow x^{5} - 5 x + a = 0$ $\Rightarrow a = 5 x - x^{5} = g (x)$

$\Rightarrow g (x) = 0$ $when$ $x = 0, 5^{1 / 4}, - 5^{1 / 4}$ and $g' (x) = 0 \Rightarrow x = 1, - 1$

Also, $g (- 1) = - 4$ $and$ $g (1) = 4$

Thus, the graph of $g (x)$ will be as shown below.

[IMAGE 1202]

From the graph, it is clear that if $a \in (- 4, 4),$

then $g (x) = a$ or $f (x) = 0$ has 3 real roots.

If $a > 4$ or $a < - 4$

then $f (x) = 0$ has only one real root.

$∴$ Options (2) and (4) are the correct options.

Q 15 :

The function $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$ has a local minimum or a local maximum at $x =$ [2013]

$- 2$
$\frac{- 2}{3}$
$2$
$\frac{2}{3}$

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Select one or more options

(1, 2)

Given: $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$

Critical points of $f (x)$ can be obtained by solving

$| x | = 0,$ $| x + 2 | = 0$ and $| | x + 2 | - 2 | x | |$ $= 0,$

which give $x = 0, - 2, 2, - \frac{2}{3}$

$∴$ $f (x) =$ ${\begin{matrix} - 2 x - 4, & x \leq - 2 \\ 2 x + 4, & - 2 < x \leq - \frac{2}{3} \\ - 4 x & - \frac{2}{3} < x \leq 0 \\ 4 x & 0 < x \leq 2 \\ 2 x + 4 & x > 2 \end{matrix}$

Graph of $y = f (x)$ is as follows:

[IMAGE 1203]

From the graph, $f (x)$ has local minimum at $x = - 2$ and $x = 0$ and $f (x)$ has local maximum at $x = - \frac{2}{3}$

Q 16 :

Let $f : (- 1, 1) \to ℝ$ be such that $f (\cos 4 θ) = \frac{2}{2 - \sec^{2} θ}$ for $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2}) .$ Then the value(s) of $f (\frac{1}{3})$ is (are) [2012]

$1 - \sqrt{\frac{3}{2}}$
$1 + \sqrt{\frac{3}{2}}$
$1 - \sqrt{\frac{2}{3}}$
$1 + \sqrt{\frac{2}{3}}$

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Select one or more options

(1, 2)

$Given: f (\cos 4 θ) = \frac{2}{2 - \sec^{2} θ} = \frac{2 \cos^{2} θ}{2 \cos^{2} θ - 1}$

$= \frac{1 + \cos 2 θ}{\cos 2 θ} = 1 + \frac{1}{\cos 2 θ}$

$Let$ $\cos 4 θ = \frac{1}{3}$ $\Rightarrow 2 \cos^{2} 2 θ - 1 = \frac{1}{3}$ $\Rightarrow \cos 2 θ = \pm \sqrt{\frac{2}{3}}$

$∴$ $f (\cos 4 θ) = 1 + \frac{1}{\cos 2 θ} = 1 \pm \sqrt{\frac{3}{2}}$ or $f (\frac{1}{3}) = 1 \pm \sqrt{\frac{3}{2}}$

Q 17 :

Match the statements given in Column-I with the intervals/union of intervals given in Column-II. [2011]

Column-I Column-II

(A) The set { $Re (\frac{2 i z}{1 - z^{2}}) :$ $z$ is a complex number,
$| z | = 1,$ $z \neq \pm 1$ } is (p) $(- \infty, - 1) \cup (1, \infty)$

(B) The domain of the function
$f (x) = \sin^{- 1} (\frac{8 {(3)}^{x - 2}}{1 - 3^{2 (x - 1)}})$ is (q) $(- \infty, - 0) \cup (0, \infty)$

(C) If
$f (θ) =$ $| \begin{matrix} 1 & \tan θ & 1 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} |$ $,$
then the set
${f (θ) : 0 \leq θ < \frac{π}{2}}$ is (r) $[2, \infty)$

(D) If $f (x) =$ $x^{3 / 2}$
$(3 x - 10),$ $x \geq 0$ then
$f (x)$ is increasing in (s) $(- \infty, - 1] \cup [1, \infty)$

(t) $(- \infty, 0] \cup [2, \infty)$

(A) → (s), (B) → (t), (C) → (r), (D) → (r)
(A) → (r), (B) → (t), (C) → (r), (D) → (s)
(A) → (r), (B) → (r), (C) → (t), (D) → (s)
(A) → (t), (B) → (r), (C) → (r), (D) → (s)

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(1)

(A) → (s), (B) → (t), (C) → (r), (D) → (r)

$(A) Let z = x + i y . Given that | z | = 1, i.e. x^{2} + y^{2} = 1, x \neq \pm 1$

Then $Re (\frac{2 i z}{1 - z^{2}})$ $= Re (\frac{2 i z}{z . \bar{z} - z^{2}})$

$= Re (\frac{2 i}{\bar{z} - z})$ $= Re (\frac{2 i}{- 2 i y})$ $= Re (\frac{- 1}{y})$ $= - \frac{1}{y},$

where, $x = \sqrt{1 - y^{2}}$

$- 1 \leq y \leq 1 \Rightarrow - \frac{1}{y} \geq 1 or - \frac{1}{y} \leq - 1$

$∴$ $Re (\frac{2 i z}{1 - z^{2}}) \in (- \infty, - 1] \cup [1, \infty)$

$∴$ $A \to s$

$(B) For the domain of f (x) = \sin^{- 1} (\frac{8 \cdot 3^{x - 2}}{1 - 3^{2 (x - 1)}})$

We should have

$- 1 \leq (\frac{8 \cdot 3^{x - 2}}{1 - 3^{2 (x - 1)}}) \leq 1$ $\Rightarrow - 1 \leq \frac{8 \cdot 3^{x}}{9 - 3^{2 x}} \leq 1$

$\Rightarrow \frac{8 \cdot 3^{x}}{9 - 3^{2 x}} \geq - 1$ $\Rightarrow \frac{8 \cdot 3^{x} + 9 - 3^{2 x}}{9 - 3^{2 x}} \geq 0$

$\Rightarrow \frac{(3^{x} - 9) (3^{x} + 1)}{(3^{x} - 3) (3^{x} + 3)} \geq 0$

We know that $3^{x} > 0$

[IMAGE 1204]

Hence, $x \in (- \infty, 0) \cup (2, \infty)$ $\dots (i)$

Also, $\frac{8 \cdot 3^{x}}{9 - 3^{2 x}} \leq 1$ $\Rightarrow \frac{8 \cdot 3^{x} - 9 + 3^{2 x}}{9 - 3^{2 x}} \leq 0$

$\Rightarrow \frac{(3^{x} + 9) (3^{x} - 1)}{(3^{x} - 3) (3^{x} + 3)} \geq 0$

[IMAGE 1205]

Hence, $x \in (- \infty, 0] \cup (1, \infty)$ $\dots (i i)$

From (i) and (ii), we get $x \in (- \infty, 0] \cup [2, \infty)$

$∴$ $B \to t$

$(C)$ $f (θ) =$ $| \begin{matrix} 1 & \tan θ & 1 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} |$

Applying $R_{1} \to R_{1} + R_{3}$ $=$ $| \begin{matrix} 0 & 0 & 2 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} |$

$= 2 (1 + \tan^{2} θ) = 2 \sec^{2} θ \geq 2,$ $0 \leq θ < \frac{π}{2}$

$∴$ $C \to r$

$(D) f (x) = x^{3 / 2} (3 x - 10),$ $x \geq 0$

$∴$ $f' (x) = \frac{3}{2} x^{1 / 2} (3 x - 10) + x^{3 / 2} (3)$

For $f (x)$ to be increasing, $f' (x) \geq 0$

$\Rightarrow 3 x^{1 / 2} [\frac{3 x - 10}{2} + x] \geq 0$ $\Rightarrow 3 x^{1 / 2} (5 x - 10) \geq 0$

[IMAGE 1206]

Hence, $f (x) is increasing on [2, \infty)$

$∴$ $D \to r$

Q 18 :

Let $f (x) = \frac{x^{2} - 6 x + 5}{x^{2} - 5 x + 6}$

Match the expressions/statements in Column I with the expressions/statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [2007]

Column I Column II

(A) If $- 1 < x < 1$ , then $f (x)$ satisfies (p) $0 < f (x) < 1$

(B) If $1 < x < 2$ , then $f (x)$ satisfies (q) $f (x) < 0$

(C) If $3 < x < 5$ , then $f (x)$ satisfies (r) $f (x) > 0$

(D) If $x > 5$ , then $f (x)$ satisfies (s) $f (x) < 1$

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

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(2)

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

$f (x) = \frac{x^{2} - 6 x + 5}{x^{2} - 5 x + 6} = \frac{(x - 5) (x - 1)}{(x - 2) (x - 3)}$

(A) If $- 1 < x < 1$ then $f (x) = \frac{(- ve) (- ve)}{(- ve) (- ve)} = + ve$

$∴$ $f (x) > 0 (r)$

Also, $f (x) - 1 = \frac{- x - 1}{x^{2} - 5 x + 6} = \frac{- (x + 1)}{(x - 2) (x - 3)}$

For $- 1 < x < 1,$ $f (x) - 1 = \frac{- (+ ve)}{(- ve) (- ve)} = - ve$

$\Rightarrow f (x) - 1 < 0 \Rightarrow f (x) < 1 (s)$

$∴$ $0 < f (x) < 1 (p)$

(B) If $1 < x < 2,$ then $f (x) = \frac{(- ve) (+ ve)}{(- ve) (- ve)} = - ve$

$∴$ $f (x) < 0 (q)$ and so, $f (x) < 1 (s)$

(C) If $3 < x < 5,$ then $f (x) = \frac{(- ve) (+ ve)}{(+ ve) (+ ve)} = - ve$

$∴$ $f (x) < 0 (q)$ and so, $f (x) < 1 (s)$

(D) For $x > 5,$ $f (x) > 0 (r)$

Also, $f (x) - 1 = \frac{- (x + 1)}{(x - 2) (x - 3)} < 0$

For $x > 5$ , $f (x) < 1 (s)$

$∴$ $0 < f (x) < 1 (p)$

Q 19 :

Let $S = {1, 2, 3, 4, 5, 6}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:

(i) $R$ has exactly 6 elements.

(ii) For each $(a, b) \in R$ , we have $| a - b |$ $\geq 2$ .

Let $Y = {R \in X : The range of R has exactly one element}$ and

$Z = {R \in X : R is a function from S to S} .$

Let $n (A)$ denote the number of elements in a set A. [2024]

Q. If $n (X) = {}^{n}C_{6}$ , then the value of $m$ is _______.

(20)

Given $S = {1, 2, 3, 4, 5, 6}$

Let $R : S \to S$ be a relation such that

Number of elements in $R = 6$

and for each $(a, b) \in R :$ $| a - b |$ $\geq 2$

$X \to$ set of all such relations $R : S \to S$

[IMAGE 1207]

Total number of ordered pairs $(a, b)$ such that $| a - b |$ $\geq 2 = 20$

$∴$ $n (X)$ = number of elements in $X = {}^{20}C_{6}$

$∴$ $m = 20$

Q 20 :

Let $S = {1, 2, 3, 4, 5, 6}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:

(i) $R$ has exactly 6 elements.

(ii) For each $(a, b) \in R$ , we have $| a - b |$ $\geq 2$

Let $Y = {R \in X : The range of R has exactly one element}$ and

$Z = {R \in X : R is a function from S to S}$

Let $n (A)$ denote the number of elements in a set $A$ . [2024]

Q. If the value of $n (Y) + n (Z)$ is $k^{2}$ , then $| k |$ is _____.

(36)

Given $S = {1, 2, 3, 4, 5, 6}$ $;$ $R : S \to S$

Number of elements in $R = 6$

and for each $(a, b) \in R;$ $| a - b |$ $\geq 2$

$X \to$ set of all such relations $R : S \to S$

If [IMAGE 1208]

Total number of ordered pairs $(a, b)$ such that $| a - b |$ $\geq 2 = 20$

$∴$ $n (X)$ = number of elements in $X = {}^{20}C_{6}$

$∴$ $m = 20$

$Y = {R \in X : The range of R has exactly one element} .$

From above, if range of $R$ has exactly one element, then maximum number of elements in $R$ will be $4$

$∴$ $n (Y) = 0$

$Z = {R \in X : R is a function from S to S}$

$n (Z) = {}^{4}C_{1} \times {}^{3}C_{1} \times {}^{3}C_{1} \times {}^{3}C_{1} \times {}^{3}C_{1} \times {}^{4}C_{1} = {(36)}^{2}$

$n (Y) + n (Z) = 0 + {(36)}^{2} = k^{2}$

$\Rightarrow$ $| k |$ $= 36$

Topic Question Set

Q 1 :

If the function $f : ℝ \to ℝ$ is defined by $f (x) =$ $| x |$ $(x - \sin x),$ then which of the following statements is TRUE ? [2020]

Q 2 :

The function $f : [0, 3] \to [1, 29]$ , defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1,$ is [2012]

Q 4 :

If the functions $f (x)$ and $g (x)$ are defined on $ℝ \to ℝ$ such that

$f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix};$ $g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

then $(f - g) (x)$ is [2005]

Q 5 :

If $f : [0, \infty) \to [0, \infty)$ and $f (x) = \frac{x}{1 + x},$ then $f$ is [2003]

Q 6 :

Let function $f : R \to R$ be defined by $f (x) = 2 x + \sin x$ for $x \in R,$ then $f$ is [2002]

Q 7 :

Suppose $f (x) = {(x + 1)}^{2}$ for $x \geq - 1 .$ If $g (x)$ is the function whose graph is the reflection of the graph of $f (x)$ with respect to the line $y = x$ , then $g (x)$ equals [2002]

Q 8 :

Let $E$ = {1, 2, 3, 4} and $F$ = {1, 2}. Then the number of onto functions from $E$ to $F$ is [2001]

Q 9 :

The domain of definition of the function $f (x)$ is given by the equation $2^{x} + 2^{y} = 2$ is [2000]

Q 10 :

Let $f : R \to R$ be any function. Define $g : R \to R$ by $g (x) =$ $| f (x) |$ for all $x$ . Then $g$ is [2000]

Q 11 :

Let the set of all relations $R$ on the set ${a, b, c, d, e, f},$ such that $R$ is reflexive and symmetric, and $R$ contains exactly 10 elements, be denoted by $S$ . Then the number of elements in $S$ is ______. [2025]

Q 12 :

Let $X$ be a set with exactly 5 elements and $Y$ be a set with exactly 7 elements. If $α$ is the number of one-one functions from $X$ to $Y$ and $β$ is the number of onto functions from $Y$ to $X$ , then the value of $\frac{1}{5!} (β - α)$ is ________. [2018]

Q 14 :

Let $a \in R$ and let $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a .$ Then [2014]

Q 15 :

The function $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$ has a local minimum or a local maximum at $x =$ [2013]

Q 16 :

Let $f : (- 1, 1) \to ℝ$ be such that $f (\cos 4 θ) = \frac{2}{2 - \sec^{2} θ}$ for $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2}) .$ Then the value(s) of $f (\frac{1}{3})$ is (are) [2012]

Want Us to Email you About Special Offers & Updates?

	Column-I		Column-II
(A)	The set { $Re (\frac{2 i z}{1 - z^{2}}) :$ $z$ is a complex number, $\| z \| = 1,$ $z \neq \pm 1$ } is	(p)	$(- \infty, - 1) \cup (1, \infty)$
(B)	The domain of the function $f (x) = \sin^{- 1} (\frac{8 {(3)}^{x - 2}}{1 - 3^{2 (x - 1)}})$ is	(q)	$(- \infty, - 0) \cup (0, \infty)$
(C)	If $f (θ) =$ $\| \begin{matrix} 1 & \tan θ & 1 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} \|$ $,$ then the set ${f (θ) : 0 \leq θ < \frac{π}{2}}$ is	(r)	$[2, \infty)$
(D)	If $f (x) =$ $x^{3 / 2}$ $(3 x - 10),$ $x \geq 0$ then $f (x)$ is increasing in	(s)	$(- \infty, - 1] \cup [1, \infty)$
		(t)	$(- \infty, 0] \cup [2, \infty)$

	Column I		Column II
(A)	If $- 1 < x < 1$ , then $f (x)$ satisfies	(p)	$0 < f (x) < 1$
(B)	If $1 < x < 2$ , then $f (x)$ satisfies	(q)	$f (x) < 0$
(C)	If $3 < x < 5$ , then $f (x)$ satisfies	(r)	$f (x) > 0$
(D)	If $x > 5$ , then $f (x)$ satisfies	(s)	$f (x) < 1$

Topic Question Set

Q 1 : If the function f:ℝ→ℝ is defined by f(x)=|x|(x-sinx), then which of the following statements is TRUE ? [2020]

Q 2 : The function f:[0,3]→[1,29], defined by f(x)=2x3-15x2+36x+1, is [2012]

Q 3 : Let f, g and h be real-valued functions defined on the interval [0, 1] by f(x)=ex2+e-x2, g(x)=xex2+e-x2 and h(x)=x2ex2+e-x2. If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then [2010]

Q 4 : If the functions f(x) and g(x) are defined on ℝ→ℝ such that f(x)={0,x∈rationalx,x∈irrational; g(x)={0,x∈irrationalx,x∈rational then (f-g)(x) is [2005]

Q 5 : If f:[0,∞)→[0,∞) and f(x)=x1+x, then f is [2003]

Q 6 : Let function f:R→R be defined by f(x)=2x+sinx for x∈R, then f is [2002]

Q 7 : Suppose f(x)=(x+1)2 for x≥-1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y=x, then g(x) equals [2002]

Q 8 : Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is [2001]

Q 9 : The domain of definition of the function f(x) is given by the equation 2x+2y=2 is [2000]

Q 10 : Let f:R→R be any function. Define g:R→R by g(x)=|f(x)| for all x. Then g is [2000]

Q 11 : Let the set of all relations R on the set {a,b,c,d,e,f}, such that R is reflexive and symmetric, and R contains exactly 10 elements, be denoted by S. Then the number of elements in S is ______. [2025]

Q 12 : Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If α is the number of one-one functions from X to Y and β is the number of onto functions from Y to X, then the value of 15!(β-α) is ________. [2018]

Q 14 : Let a∈R and let f:R→R be given by f(x)=x5-5x+a. Then [2014]

Q 15 : The function f(x)=2|x|+|x+2|-||x+2|-2|x|| has a local minimum or a local maximum at x= [2013]

Q 16 : Let f:(-1,1)→ℝ be such that f(cos4θ)=2 2-sec2θ for θ∈(0,π4)∪(π4,π2). Then the value(s) of f(13) is (are) [2012]

Want Us to Email you About Special Offers & Updates?

Q 1 :

If the function $f : ℝ \to ℝ$ is defined by $f (x) =$ $| x |$ $(x - \sin x),$ then which of the following statements is TRUE ? [2020]

Q 2 :

The function $f : [0, 3] \to [1, 29]$ , defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1,$ is [2012]

Q 4 :

If the functions $f (x)$ and $g (x)$ are defined on $ℝ \to ℝ$ such that

$f (x) =$ ${\begin{matrix} 0, & x \in rational \\ x, & x \in irrational \end{matrix};$ $g (x) =$ ${\begin{matrix} 0, & x \in irrational \\ x, & x \in rational \end{matrix}$

then $(f - g) (x)$ is [2005]

Q 5 :

If $f : [0, \infty) \to [0, \infty)$ and $f (x) = \frac{x}{1 + x},$ then $f$ is [2003]

Q 6 :

Let function $f : R \to R$ be defined by $f (x) = 2 x + \sin x$ for $x \in R,$ then $f$ is [2002]

Q 7 :

Suppose $f (x) = {(x + 1)}^{2}$ for $x \geq - 1 .$ If $g (x)$ is the function whose graph is the reflection of the graph of $f (x)$ with respect to the line $y = x$ , then $g (x)$ equals [2002]

Q 8 :

Let $E$ = {1, 2, 3, 4} and $F$ = {1, 2}. Then the number of onto functions from $E$ to $F$ is [2001]

Q 9 :

The domain of definition of the function $f (x)$ is given by the equation $2^{x} + 2^{y} = 2$ is [2000]

Q 10 :

Let $f : R \to R$ be any function. Define $g : R \to R$ by $g (x) =$ $| f (x) |$ for all $x$ . Then $g$ is [2000]

Q 11 :

Let the set of all relations $R$ on the set ${a, b, c, d, e, f},$ such that $R$ is reflexive and symmetric, and $R$ contains exactly 10 elements, be denoted by $S$ . Then the number of elements in $S$ is ______. [2025]

Q 12 :

Let $X$ be a set with exactly 5 elements and $Y$ be a set with exactly 7 elements. If $α$ is the number of one-one functions from $X$ to $Y$ and $β$ is the number of onto functions from $Y$ to $X$ , then the value of $\frac{1}{5!} (β - α)$ is ________. [2018]

Q 14 :

Let $a \in R$ and let $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a .$ Then [2014]

Q 15 :

The function $f (x) = 2$ $| x |$ $+$ $| x + 2 |$ $-$ $| | x + 2 | - 2 | x | |$ has a local minimum or a local maximum at $x =$ [2013]

Q 16 :

Let $f : (- 1, 1) \to ℝ$ be such that $f (\cos 4 θ) = \frac{2}{2 - \sec^{2} θ}$ for $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2}) .$ Then the value(s) of $f (\frac{1}{3})$ is (are) [2012]