Q1GATE 2026NAT

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as follows:

f(x)=\left\{\begin{array}{lr} c_{1} e^{x}-c_{2} \log _{{e}}\left(\frac{1}{x}\right), & \text { if } x>0 \\ 3 & \text { otherwise } \end{array}\right.

where $c_{1}, c_{2} \in \mathbb{R}$ . If $f$ is continuous at $x=0$ , then $c_{1}+c_{2}=$ ________. (answer in integer)

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📖 Explanation

For a function to be continuous at a point, in this case $x=0$ , its Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function's value at that point must all be equal.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$
The function is given as:
$

f(x) = \begin{cases} c_1 e^x - c_2 \log_e \left(\frac{1}{x}\right), & \text{if } x > 0 \\ 3, & \text{otherwise} \end{cases}

$First, let's simplify the expression for$ x > 0 $using the logarithm property$ \log_e(1/x) = -\log_e(x)$:
$f(x) = c_1 e^x + c_2 \log_e(x), \quad \text{for } x > 0$
Now, let's evaluate the limits and the function value.

Left-Hand Limit (LHL) and Function Value:
For $x \le 0$ , the function is defined as $f(x)=3$ .
Therefore, $\lim_{x \to 0^-} f(x) = 3$ and $f(0) = 3$ .
Right-Hand Limit (RHL):
For $x > 0$ , we have $f(x) = c_1 e^x + c_2 \log_e(x)$ .
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (c_1 e^x + c_2 \log_e(x))$
As $x \to 0^+$ , we know that $e^x \to 1$ and $\log_e(x) \to -\infty$ . The limit becomes $c_1(1) + c_2(-\infty )$ . For this limit to be finite and equal to 3 (to ensure continuity), the term that goes to infinity must be eliminated. This is only possible if its coefficient, $c_2$ , is zero.
So, we must have $c_2 = 0$ .
Solving for the Constants:
With $c_2 = 0$ , the RHL simplifies to:
$\lim_{x \to 0^+} (c_1 e^x) = c_1 e^0 = c_1$
For continuity, we equate the LHL, RHL, and $f(0)$ :
$3 = c_1 = 3$
This gives us $c_1 = 3$ .
Final Calculation:
We found that $c_1 = 3$ and $c_2 = 0$ . The required sum is:
$c_1 + c_2 = 3 + 0 = 3$

Q2GATE 2026MSQ

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as follows: $f(x)=\left(\frac{|x|}{2}-x\right)\left(x-\frac{|x|}{2}\right)$ Which of the following statements is/are true?

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📖 Explanation

To analyze the function and its properties, we first need to understand its definition, which depends on the absolute value of $x$ . We'll break it down into two cases.

Step 1: Express $f(x)$ as a piecewise function.

The function is $f(x) = \left(\frac{|x|}{2} - x\right)\left(x - \frac{|x|}{2}\right)$ .

Case 1: For $x \geq 0$ , we have $|x| = x$ .
The function becomes:
$f(x) = \left(\frac{x}{2} - x\right)\left(x - \frac{x}{2}\right) = \left(-\frac{x}{2}\right)\left(\frac{x}{2}\right) = -\frac{x^2}{4}$
Case 2: For $x < 0$ , we have $|x| = -x$ .
The function becomes:
$f(x) = \left(\frac{-x}{2} - x\right)\left(x - \frac{-x}{2}\right) = \left(-\frac{3x}{2}\right)\left(\frac{x}{2}\right) = -\frac{3x^2}{4}$

So, we have the piecewise function:
$

f(x) = \begin{cases} -\frac{x^2}{4} & \text{if } x \geq 0 \\ -\frac{3x^2}{4} & \text{if } x < 0 \end{cases}

Step 2: Analyze for Local Maximum/Minimum (Options A and B).

For $x \ge 0$ , $f(x) = -x^2/4$ is a downward-opening parabola.
For $x < 0$ , $f(x) = -3x^2/4$ is also a downward-opening parabola.
At $x=0$ , $f(0) = 0$ . For any other $x$ (whether positive or negative), $x^2 > 0$ , so both $-x^2/4$ and $-3x^2/4$ are negative.
This means $f(x) \le f(0)$ for all $x$ . Therefore, the function has a local (and global) maximum at $x=0$ .
Since there is a local maximum and no other critical points, there is no local minimum.

Step 3: Analyze the Derivative $f'(x)$ (Options C and D).

First, let's find the derivative for each piece.

For $x \geq 0$ :
$f'(x) = \frac{d}{dx}\left(-\frac{x^2}{4}\right) = -\frac{2x}{4} = -\frac{x}{2}$
For $x < 0$ :
$f'(x) = \frac{d}{dx}\left(-\frac{3x^2}{4}\right) = -\frac{6x}{4} = -\frac{3x}{2}$

Now, let's check the continuity and differentiability of $f'(x)$ at $x=0$ .

Continuity of $f'$ at $x=0$ (Option C):
We check the limits of $f'(x)$ as $x$ approaches 0 from both sides.
- Right-hand limit: $\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} \left(-\frac{x}{2}\right) = 0$
- Left-hand limit: $\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \left(-\frac{3x}{2}\right) = 0$
  Since the left-hand and right-hand limits are equal, $f'(x)$ is continuous at $x=0$ . As it's also continuous for all other $x$ , $f'$ is continuous over $\mathbb{R}$ .
Differentiability of $f'$ (Option D):
The function for the derivative is

f'(x) = \begin{cases} -x/2 & \text{if } x \geq 0 \\ -3x/2 & \text{if } x < 0 \end{cases}

. The "derivative of the derivative," $f''(x)$ , would be:
* $f''(x) = -1/2$ for $x > 0$
* $f''(x) = -3/2$ for $x < 0$
Since the second derivative has different values on the left and right of $x=0$ , the first derivative $f'(x)$ is not differentiable at $x=0$ .

Based on the analysis:

$f$ has a local maximum (at $x=0$ ). (Statement A is true)
$f'$ is continuous over $\mathbb{R}$ . (Statement C is true)
$f'$ is not differentiable over $\mathbb{R}$ (specifically at $x=0$ ). (Statement D is true)

Q3GATE 2026NAT

Consider a function $f : (0,1) \rightarrow \{0,1\}$ defined as follows. For a real number $r \in (0,1)$ , $f(r) = 1$ if the second digit after the decimal point in $r$ is one of the four digits $2, 3, 6,$ and $7$ . Otherwise, $f(r)$ is equal to $0$ . The number of points in $(0,1)$ at which $f$ is discontinuous is ______. (Answer in integer)

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Q4GATE 2025MCQ

$g(. )$ is a function from $A$ to $B$ , $f(. )$ is a function from $B$ to $C$ , and their composition defined as $f(g(. ))$ is a mapping from $A$ to $C$ . If $f(. )$ and $f(g(. ))$ are onto (surjective) functions, which ONE of the following is TRUE about the function $g(. )$ ?

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📖 Explanation

Let's analyze the properties of surjective functions in composition.
Given $f: B \to C$ and $f \circ g: A \to C$ are both surjective functions. This means that for every element in $C$ , there exists at least one element in $B$ (for $f$ ) and at least one element in $A$ (for $f \circ g$ ) that maps to it.

Consider a counterexample where $g$ is neither one-to-one nor onto, but $f$ and $f \circ g$ are onto.
Let $A = \{1, 2, 3\}$ , $B = \{4, 5, 6\}$ , and $C = \{8, 9\}$ .
Define $g: A \to B$ as $g = \{(1, 4), (2, 4), (3, 5)\}$ . Here, $g$ is not one-to-one (since $g(1)=g(2)=4$ ) and not onto (since $6 \in B$ has no preimage in $A$ ).
Define $f: B \to C$ as $f = \{(4, 8), (5, 9), (6, 8)\}$ . Here, $f$ is onto because $Range(f) = \{8, 9\} = C$ .
Now, let's find the composition $f \circ g: A \to C$ :
$f(g(1)) = f(4) = 8$
$f(g(2)) = f(4) = 8$
$f(g(3)) = f(5) = 9$
So, $f \circ g = \{(1, 8), (2, 8), (3, 9)\}$ . The $Range(f \circ g) = \{8, 9\} = C$ , which means $f \circ g$ is also surjective.

In this example, $f$ and $f \circ g$ are surjective, but $g$ is neither one-to-one nor onto. Therefore, $g(.)$ is not required to be a one-to-one or onto function for the given conditions.

Q5GATE 2024MCQ

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x)=\max \left\{x, x^3\right\}, x \in \mathbb{R}$ , where $\mathbb{R}$ is the set of all real numbers. The set of all points where $f(x)$ is NOT differentiable is

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📖 Explanation

The function $f(x) = \max\{x, x^3\}$ means $f(x)$ takes the larger of $x$ and $x^3$ .
To find where $f(x)$ is not differentiable, we need to identify points where $x = x^3$ or where there is a sharp turn in the graph.
Set $x = x^3$ , which can be rewritten as $x^3 - x = 0$ .
Factor out $x$ : $x(x^2 - 1) = 0$ .
Further factor $x^2 - 1$ as $(x-1)(x+1)$ : $x(x-1)(x+1) = 0$ .
This gives three solutions: $x=0$ , $x=1$ , and $x=-1$ . These are the points where the two functions $y=x$ and $y=x^3$ intersect.
At these intersection points, the function $f(x)$ might have sharp corners if the slopes of $y=x$ and $y=x^3$ are different.
The derivative of $y=x$ is $1$ . The derivative of $y=x^3$ is $3x^2$ .
At $x=-1$ : $1 \neq 3(-1)^2 = 3$ . Not differentiable.
At $x=0$ : $1 \neq 3(0)^2 = 0$ . Not differentiable.
At $x=1$ : $1 \neq 3(1)^2 = 3$ . Not differentiable.
Therefore, the function $f(x)$ is not differentiable at $x=-1, 0, 1$ .

Q6GATE 2024NAT

Let $A$ and $B$ be non-empty finite sets such that there exist one-to-one and onto functions (i) from $A$ to $B$ and (ii) from $A \times A$ to $A \cup B$ . The number of possible values of $|A|$ is ___

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📖 Explanation

Let $|A| = n$ and $|B| = m$ .
Given there's a one-to-one and onto function from $A$ to $B$ , it implies that $|A| = |B|$ , so $n=m$ .
Given there's a one-to-one and onto function from $A \times A$ to $A \cup B$ .
The cardinality of $A \times A$ is $|A \times A| = |A| \times |A| = n \times n = n^2$ .
The cardinality of $A \cup B$ is $|A \cup B| = |A| + |B| - |A \cap B|$ .
Since $n=m$ , if $A$ and $B$ are distinct sets, $|A \cup B| = n+n = 2n$ (assuming $A \cap B = \emptyset$ ).
If $A=B$ , then $|A \cup B| = n$ .
Case 1: $A = B$ . Then $|A \cup B| = n$ . So, $n^2 = n$ . This gives $n(n-1)=0$ , so $n=0$ or $n=1$ . Since $A$ and $B$ are non-empty, $n=1$ .
Case 2: $A \neq B$ and $A \cap B = \emptyset$ . Then $|A \cup B| = n+n = 2n$ . So, $n^2 = 2n$ . This gives $n^2 - 2n = 0$ , so $n(n-2)=0$ , thus $n=0$ or $n=2$ . Since $A$ and $B$ are non-empty, $n=2$ .
The number of possible values for $|A|$ (which is $n$ ) are 1 and 2.

Q7GATE 2023MSQ

Let $f:A\rightarrow B$ be an onto (or surjective) function, where $A$ and $B$ are nonempty sets. Define an equivalence relation $\sim$ on the set $A$ as $a_1\sim a_2 \text{ if } f(a_1)=f(a_2),$ where $a_1, a_2 \in A$ . Let $\varepsilon =\{[x]:x \in A\}$ be the set of all the equivalence classes under $\sim$ . Define a new mapping $F: \varepsilon \rightarrow B$ as $F([x]) = f(x),$ for all the equivalence classes $[x]$ in $\varepsilon$ Which of the following statements is/are TRUE?

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📖 Explanation

The function $F: \varepsilon \rightarrow B$ is defined as $F([x]) = f(x)$ .

F is well-defined: For $F$ to be well-defined, if $[x_1] = [x_2]$ , then $F([x_1]) = F([x_2])$ . By definition of $\sim$ , $[x_1] = [x_2]$ implies $x_1 \sim x_2$ , which means $f(x_1) = f(x_2)$ . Thus, $F([x_1]) = f(x_1) = f(x_2) = F([x_2])$ . So, $F$ is well-defined. Statement (A) is FALSE.
F is onto (surjective): Since $f: A \rightarrow B$ is onto, for every $y \in B$ , there exists an $x \in A$ such that $f(x) = y$ . Then, for this $y$ , there exists an equivalence class $[x] \in \varepsilon$ such that $F([x]) = f(x) = y$ . So, $F$ is onto. Statement (B) is TRUE.
F is one-to-one (injective): Assume $F([x_1]) = F([x_2])$ . This implies $f(x_1) = f(x_2)$ . By the definition of the equivalence relation $\sim$ , $f(x_1) = f(x_2)$ means $x_1 \sim x_2$ , which implies $[x_1] = [x_2]$ . So, $F$ is one-to-one. Statement (C) is TRUE.
F is bijective: Since $F$ is both one-to-one and onto, it is a bijective function. Statement (D) is TRUE.

Q8GATE 2022MCQ

Which one of the following is the closed form for the generating function of the sequence $\{a_n \}_{n \geq 0}$ defined below?

a_n= \left \{ \begin{matrix} n+1, &n \text{ is odd} \\ 1,& \text{otherwise} \end{matrix} \right.

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📖 Explanation

The sequence is $a_n = n+1$ if $n$ is odd, and $a_n = 1$ if $n$ is otherwise (i.e., $n$ is even).
Let's list the first few terms of the sequence starting from $n=0$ :
$a_0 = 1$ (n=0, even)
$a_1 = 1+1 = 2$ (n=1, odd)
$a_2 = 1$ (n=2, even)
$a_3 = 3+1 = 4$ (n=3, odd)
$a_4 = 1$ (n=4, even)
$a_5 = 5+1 = 6$ (n=5, odd)
So the sequence is $\{1, 2, 1, 4, 1, 6, 1, 8, \dots\}$ .

The generating function is $G(x) = \sum_{n=0}^{\infty } a_n x^n = a_0 x^0 + a_1 x^1 + a_2 x^2 + a_3 x^3 + \dots$
$G(x) = 1 + 2x + 1x^2 + 4x^3 + 1x^4 + 6x^5 + 1x^6 + \dots$
We can separate this into even and odd terms:
Terms for even $n$ : $1x^0 + 1x^2 + 1x^4 + 1x^6 + \dots = 1 + x^2 + x^4 + x^6 + \dots = \sum_{k=0}^{\infty } (x^2)^k = \frac{1}{1-x^2}$ .
Terms for odd $n$ : $2x^1 + 4x^3 + 6x^5 + \dots = \sum_{k=0}^{\infty } (2(2k+1)x^{2k+1}) = 2x + 4x^3 + 6x^5 + \dots$
Let $S_{odd}(x) = 2x + 4x^3 + 6x^5 + \dots = 2x(1 + 2x^2 + 3x^4 + \dots)$ .
We know that $\sum_{k=0}^{\infty } (k+1)y^k = \frac{1}{(1-y)^2}$ .
Let $y = x^2$ . Then $1 + 2x^2 + 3x^4 + \dots = \sum_{k=0}^{\infty } (k+1)(x^2)^k = \frac{1}{(1-x^2)^2}$ .
So, $S_{odd}(x) = 2x \cdot \frac{1}{(1-x^2)^2} = \frac{2x}{(1-x^2)^2}$ .
Combining the even and odd terms:
$G(x) = \frac{1}{1-x^2} + \frac{2x}{(1-x^2)^2}$ .
To combine these, find a common denominator $(1-x^2)^2$ :
$G(x) = \frac{1-x^2}{(1-x^2)^2} + \frac{2x}{(1-x^2)^2} = \frac{1-x^{2+}2x}{(1-x^2)^2}$ .
This result does not directly match option (A) $\frac{x(1+x^2)}{(1-x^2)^2}+\frac{1}{1-x}$ .
Let's check option (A) by simplifying it:
$\frac{x(1+x^2)}{(1-x^2)^2}+\frac{1}{1-x} = \frac{x+x^3}{(1-x^2)^2} + \frac{1-x^2}{(1-x^2)^2} = \frac{x+x^{3+}1-x^2}{(1-x^2)^2} = \frac{1+x-x^{2+}x^3}{(1-x^2)^2}$ .
This is not matching.

Let's re-examine the odd part sum:
$S_{odd}(x) = \sum_{k=0}^{\infty } (2k+2)x^{2k+1}$ .
No, it's $a_n = n+1$ for odd $n$ . So $a_{2k+1} = (2k+1)+1 = 2k+2$ .
$S_{odd}(x) = \sum_{k=0}^{\infty } (2k+2)x^{2k+1}$ .
$S_{odd}(x) = 2x + 4x^3 + 6x^5 + \dots$
We know that $\sum_{k=0}^{\infty } (k+1)y^k = \frac{1}{(1-y)^2}$ .
Let $S_1 = 1 + 2y + 3y^2 + \dots = \frac{1}{(1-y)^2}$ .
Then $y S_1 = y + 2y^2 + 3y^3 + \dots$ .
We need $2x + 4x^3 + 6x^5 + \dots = 2x(1 + 2x^2 + 3x^4 + \dots)$ .
Let $y = x^2$ .
So, $S_{odd}(x) = 2x \sum_{k=0}^{\infty } (k+1)(x^2)^k = 2x \frac{1}{(1-x^2)^2} = \frac{2x}{(1-x^2)^2}$ .
This is correct.
So $G(x) = \frac{1}{1-x^2} + \frac{2x}{(1-x^2)^2} = \frac{(1-x^2)+2x}{(1-x^2)^2} = \frac{1+2x-x^2}{(1-x^2)^2}$ .

Let's check the options again.
Option (A): $\frac{x(1+x^2)}{(1-x^2)^2}+\frac{1}{1-x} = \frac{x+x^3 + (1-x^2)}{(1-x^2)^2} = \frac{1+x-x^{2+}x^3}{(1-x^2)^2}$ . Still not a match.

There might be a slight difference in the formula, or a common algebraic simplification that makes one of the options equivalent.
Let's look at the solution PDF. It says $G(x) = \frac{1}{(1-x)} + x \frac{d}{dx}(\frac{x}{(1-x^2)})$ . This is a different approach.
Let's use the method shown in the PDF:
$a_n = 1$ for even $n$ , $a_n = n+1$ for odd $n$ .
Sum of even terms: $1+x^{2+}x^{4+}\dots = \frac{1}{1-x^2}$ .
Sum of odd terms: $2x+4x^{3+}6x^{5+}\dots$
The PDF states this is $x(1+2x^{2+}3x^{4+}\dots) + (x^{2+}x^{4+}x^{6+}\dots)$ . No.
The solution PDF has a different formulation for the generating function's odd part:
$G(x) = (1+x^{2+}x^{4+}\dots) + (2x+4x^{3+}6x^{5+}\dots)$
The first part is $\frac{1}{1-x^2}$ .
The second part is $2x(1+2x^{2+}3x^{4+}\dots) = 2x \cdot \frac{1}{(1-x^2)^2}$ . This is what I derived.
So $G(x) = \frac{1}{1-x^2} + \frac{2x}{(1-x^2)^2} = \frac{1-x^{2+}2x}{(1-x^2)^2}$ .
Now let's check the algebra in option A:
$\frac{x(1+x^2)}{(1-x^2)^2}+\frac{1}{1-x} = \frac{x+x^3}{(1-x^2)^2} + \frac{1-x^2}{(1-x)(1+x)} \cdot \frac{1+x}{1+x} = \frac{x+x^3}{(1-x^2)^2} + \frac{(1-x^2)(1+x)}{(1-x^2)^2}$
This is $\frac{x+x^{3+}1+x-x^{2-}x^3}{(1-x^2)^2} = \frac{1+2x-x^2}{(1-x^2)^2}$ .
So, my derived form matches option (A) after simplification.
The PDF solution shows a very confusing derivation.
It says $x \frac{d}{dx}(\frac{x}{(1-x^2)})$ .
$\frac{d}{dx}(\frac{x}{1-x^2}) = \frac{1(1-x^2) - x(-2x)}{(1-x^2)^2} = \frac{1-x^{2+}2x^2}{(1-x^2)^2} = \frac{1+x^2}{(1-x^2)^2}$ .
So $x \frac{d}{dx}(\frac{x}{(1-x^2)}) = \frac{x(1+x^2)}{(1-x^2)^2}$ .
This is the first term of option (A).
Then it adds $\frac{1}{1-x}$ . This term is $\frac{1+x}{1-x^2}$ .
The solution attempts to sum $G(x) = \sum a_n x^n = \sum_{n \text{ even}} x^n + \sum_{n \text{ odd}} (n+1)x^n$ .
The even sum is $\frac{1}{1-x^2}$ .
The odd sum is $2x+4x^{3+}6x^{5+}\dots = 2x(1+2x^{2+}3x^{4+}\dots) = 2x \frac{1}{(1-x^2)^2}$ .
So, $G(x) = \frac{1}{1-x^2} + \frac{2x}{(1-x^2)^2} = \frac{1-x^{2+}2x}{(1-x^2)^2}$ .
Now, check option A: $\frac{x(1+x^2)}{(1-x^2)^2}+\frac{1}{1-x}$ .
The $\frac{1}{1-x}$ term needs to be converted to the denominator $(1-x^2)^2$ .
$\frac{1}{1-x} = \frac{1+x}{1-x^2} = \frac{(1+x)(1-x^2)}{(1-x^2)^2} = \frac{1-x^{2+}x-x^3}{(1-x^2)^2}$ .
Then option A is $\frac{x+x^{3+}1-x^{2+}x-x^3}{(1-x^2)^2} = \frac{1+2x-x^2}{(1-x^2)^2}$ .
This matches my derived $G(x)$ . is indeed correct.
The PDF solution shows $G(x) = \frac{1}{(1-x)} + x \frac{d}{dx}(\frac{x}{(1-x^2)})$ . This is incorrect.
The first term should be $\frac{1}{1-x^2}$ not $\frac{1}{1-x}$ .
The PDF formula for the odd terms $a_n = n+1$ as $x \frac{d}{dx}(\frac{x}{(1-x^2)})$ is actually for the sum of $n x^n$ , not $(n+1)x^n$ .
$x \frac{d}{dx}(\frac{1}{1-x^2}) = x \frac{2x}{(1-x^2)^2} = \frac{2x^2}{(1-x^2)^2}$ .
The PDF's sum of odd terms seems to be for $a_n = n x^n$ . No, it's $a_n = (n+1)x^n$ .
My derivation of $G(x)$ is correct, and it matches option A after simplifying option A.

Q9GATE 2021MSQ

Consider the following sets, where $n\geq 2$ : S1: Set of all nxn matrices with entries from the set $\{a, b, c\}$ S2: Set of all functions from the set $\{0,1,2,...,n^{2-}1\}$ to the set $\{0, 1, 2 \}$ Which of the following choice(s) is/are correct?

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📖 Explanation

Let $|S_1|$ be the cardinality of $S_1$ and $|S_2|$ be the cardinality of $S_2$ .
$S_1$ : Set of all $n \times n$ matrices with entries from $\{a, b, c\}$ . There are $n^2$ entries, and each can be one of 3 values. So, $|S_1| = 3^{n^2}$ .
$S_2$ : Set of all functions from $\{0, 1, \dots, n^{2-}1\}$ to $\{0, 1, 2\}$ . The domain has $n^2$ elements, and the codomain has 3 elements. So, $|S_2| = 3^{n^2}$ .
Since $|S_1| = |S_2|$ , there exists a bijection between $S_1$ and $S_2$ . If a bijection exists, a surjection also exists.
Therefore, "There exists a surjection from S1 to S2" is correct, and "There exists a bijection from S1 to S2" is correct.

Q10GATE 2017NAT

If the ordinary generating function of a sequence $\{a_{n}\}_{n=0}^{\infty } \; is \; \frac{1+z}{(1-z)^{3}}$ then $a_{3}-a_{0}$ is equal to ______.

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📖 Explanation

The given ordinary generating function is $G(z) = \sum_{n=0}^{\infty } a_n z^n = \frac{1+z}{(1-z)^3}$ .
This can be written as $G(z) = (1+z)(1-z)^{-3}$ .
Using the binomial expansion for $(1-z)^{-k-1} = \sum_{n=0}^{\infty } \binom{n+k}{k} z^n$ , we get:
$(1-z)^{-3} = 1 + 3z + 6z^2 + 10z^3 + \dots$
Now, multiply by $(1+z)$ :
$G(z) = (1+z)(1 + 3z + 6z^2 + 10z^3 + \dots)$
$G(z) = (1 + 3z + 6z^2 + 10z^3 + \dots) + (z + 3z^2 + 6z^3 + 10z^4 + \dots)$
$G(z) = 1 + 4z + 9z^2 + 16z^3 + \dots$
From this series, the coefficient of $z^0$ is $a_0 = 1$ , and the coefficient of $z^3$ is $a_3 = 16$ .
The required value is $a_3 - a_0 = 16 - 1 = 15$ .

Q11GATE 2015MCQ

If $g(x)=1-x$ and $h(x)=\frac{x}{x-1}$ , then $\frac{g(h(x))}{h(g(x))}$ is:

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📖 Explanation

First, we evaluate the composite functions.
To find $g(h(x))$ , we substitute $h(x)$ into $g(x)$ :
$g(h(x)) = 1 - h(x) = 1 - \frac{x}{x-1} = \frac{(x-1) - x}{x-1} = \frac{-1}{x-1}$

To find $h(g(x))$ , we substitute $g(x)$ into $h(x)$ :
$h(g(x)) = \frac{g(x)}{g(x)-1} = \frac{1-x}{(1-x)-1} = \frac{1-x}{-x}$

Now, we compute the required ratio:
$\frac{g(h(x))}{h(g(x))} = \frac{-1/(x-1)}{(1-x)/(-x)} = \frac{-1}{x-1} \cdot \frac{-x}{1-x} = \frac{x}{(x-1)(1-x)}$

This expression is equivalent to $\frac{h(x)}{g(x)} = \frac{x/(x-1)}{1-x} = \frac{x}{(x-1)(1-x)}$ .

Q12GATE 2014NAT

Let S denote the set of all functions f: ${\{0,1\}}^{4} \rightarrow \{0,1\}$ . Denote by N the number of functions from S to the set {0,1}. The value of $log_{2} log_{2}N$ is______.

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📖 Explanation

Let the domain of functions in set $S$ be $A = {\{0,1\}}^{4}$ . The size of $A$ is $|A| = 2^4 = 16$ .
Each function $f \in S$ maps elements from $A$ to $\{0,1\}$ . So, for each of the $|A|$ inputs, there are 2 possible outputs.
The total number of functions in $S$ is $|S| = 2^{|A|} = 2^{16}$ .
The set $N$ represents the number of functions from $S$ to $\{0,1\}$ . Similar to before, the size of the domain is $|S|$ , and the codomain is $\{0,1\}$ .
Therefore, $N = 2^{|S|} = 2^{2^{16}}$ .
We need to calculate $\log_{2} \log_{2} N$ .
Substitute the value of $N$ :
$\log_{2} \log_{2} (2^{2^{16}}) = \log_{2} (2^{16}) = 16$

The value of $\log_{2} \log_{2}N$ is 16.
[CORRECT_OPTION: 16]

Q13GATE 2014MCQ

Consider the set of all functions f:{0,1,...,2014} $\rightarrow$ {0,1...,2014} such that f(f(i))=i, for 0 $\leq$ i $\leq$ 2014 . Consider the following statements. P. For each such function it must be the case that for every i, f(i) = i, Q. For each such function it must be the case that for some i,f(i) = i, R. Each such function must be onto. Which one of the following is CORRECT?

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📖 Explanation

A function $f: A \to A$ such that $f(f(i)) = i$ for all $i \in A$ is called an involution.
Statement P claims that for every such function, $f(i) = i$ for all $i$ . This is false. Consider the function $f(0)=1, f(1)=0$ , and $f(i)=i$ for $i \geq 2$ . Here $f(f(0))=f(1)=0$ , $f(f(1))=f(0)=1$ , and $f(f(i))=i$ for $i \geq 2$ . This function satisfies the given condition but $f(0) \ne 0$ and $f(1) \ne 1$ .
Statement Q claims that for each such function, there must be some $i$ such that $f(i)=i$ . This is true. If $f(i) \ne i$ for all $i$ , then $f$ would be a fixed-point free involution. However, if the domain is finite and has an odd number of elements (like $\{0, 1, ..., 2014\}$ , which has 2015 elements), a fixed-point free involution is impossible. Thus, at least one fixed point must exist.
Statement R claims that each such function must be onto. This is true because if $f(f(i))=i$ , then $f$ is its own inverse, and any function with an inverse is a bijection (both injective and surjective/onto).

Q14GATE 2014MCQ

Let A be a finite set having $x$ elements and let B be a finite set having $y$ elements. What is the number of distinct functions mapping B into A.

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Q15GATE 2012MCQ

How many onto (or surjective) functions are there from an n-element ( $n \geq 2$ ) set to a 2-element set?

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📖 Explanation

To find the number of onto (surjective) functions from an $n$ -element set to a 2-element set, we first consider the total number of functions.

Let the $n$ -element set be $A$ and the 2-element set be $B = \{y_1, y_2\}$ .
For each of the $n$ elements in set $A$ , there are 2 choices of elements in set $B$ to map to.
Thus, the total number of functions from $A$ to $B$ is $2^n$ .

An onto function requires that every element in the codomain (set $B$ ) must be mapped to by at least one element from the domain (set $A$ ).
In this case, since the codomain has only two elements ( $y_1$ and $y_2$ ), the functions that are not onto are those where either:

All elements from $A$ map to $y_1$ . (This is 1 function)
All elements from $A$ map to $y_2$ . (This is 1 function)

These two cases are the only functions that are not onto.
Therefore, the number of onto functions is the total number of functions minus the number of functions that are not onto:
Number of onto functions = (Total functions) - (Functions where all map to $y_1$ ) - (Functions where all map to $y_2$ )
Number of onto functions = $2^n - 1 - 1 = 2^n - 2$ .

Q16GATE 2005MCQ

Let $f$ be a function from a set A to a set B, $g$ a function from B to C, and $h$ a function from A to C, such that $h(a) = g(f(a))$ for all $a \in A$ . Which of the following statements is always true for all such functions $f$ and $g$ ?

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Q17GATE 2005MCQ

Let f: B $\rightarrow$ C and g: A $\rightarrow$ B be two functions and let h = f $\cdot$ g. Given that h is an onto function which one of the following is TRUE?

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Q18GATE 2003MCQ

Let : A $\rightarrow$ B be injective (one-to-one) function. Define $g:2^{A}\rightarrow 2^{B}$ as: g(C) = {f (x)| x $\in$ C), for all subsets C of A. Define $g:2^{B}\rightarrow 2^{A}$ as : h(D) = {x | x $\in$ A, f (x) $\in$ D}, for all subsets D of B. Which of the following statements is always true?

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Q19MCQ

The number of functions from an m element set to an n element set is

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Q20MCQ

Consider the function y=|x| in the interval [-1, 1]. In this interval, the function is

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