GATE DA 2025 Set 1

The relationship is of synonyms. Courage is synonymous with Bravery. Similarly, Yearning is synonymous with Longing.

The sentence describes a past action that was in progress before another past action occurred. The past perfect continuous tense ('had been playing') is used to show that the action (playing tennis) started in the past and continued up until another time in the past (when it started to rain).

We need to count the number of pixels where the intensity $U \le 4$.From the given grid:
Row 1: 0, 1, 0, 2 (All 4 are $\le 4$)
Row 2: 4, 7, 3, 3 (Values 4, 3, 3 are $\le 4$ $\rightarrow$ 3 pixels)
Row 3: 5, 5, 4, 4 (Values 4, 4 are $\le 4$ $\rightarrow$ 2 pixels)
Row 4: 6, 7, 3, 2 (Values 3, 2 are $\le 4$ $\rightarrow$ 2 pixels)Total count = $4 + 3 + 2 + 2 = 11$.

The number associated with a region is the sum of the numbers of the shapes that cover it.

• Rectangle = 1
• Triangle = 2
• Ellipse = 3

Region P: Intersection of Rectangle, Triangle, and Ellipse.
$P = 1 + 2 + 3 = 6$Region Q: Intersection of Triangle and Ellipse (outside Rectangle).
$Q = 2 + 3 = 5$Region R: Intersection of Rectangle and Triangle (outside Ellipse).
$R = 1 + 2 = 3$Thus, $P=6, Q=5, R=3$.

Let the original length be $L$ and width be $W$.
Original Area $A = L \times W$.New width $W' = W + 0.10W = 1.1W$.
Length remains $L$.New Area $A' = L \times W' = L \times (1.1W) = 1.1(LW) = 1.1A$.The new area is $110\%$ of the original area, which means the area increases by $10\%$.

Matching the statements with appropriate responses:

• P. This house is in a mess. $\rightarrow$ 3. No problem, let me clear it up for you. ('Clear it up' implies cleaning or tidying).
• Q. I am not happy with the marks given to me. $\rightarrow$ 4. Don't worry, I will take it up with your teacher. ('Take it up with' implies discussing a grievance).
• R. Politics is a subject I avoid talking about. $\rightarrow$ 1. Alright, I won't bring it up during our conversations. ('Bring it up' implies mentioning a topic).
• S. I don't know what this word means. $\rightarrow$ 2. Well, you can easily look it up. ('Look it up' implies searching for a definition).

Correct match: P-3, Q-4, R-1, S-2.

Let $W(a)$ be the weight function. It is given that $W(a)$ is strictly increasing for $a \le 50$ and strictly decreasing for $a \ge 50$.
We are given $W(a_1) = W(a_2)$ with $a_1 < a_2$.
Since the function is monotonic on either side of 50, the only way for the weights to be equal at two different ages is if one age is in the increasing phase and the other is in the decreasing phase.
Therefore, $a_1$ must be less than 50 and $a_2$ must be greater than 50.
Thus, $a_1 < 50 < a_2$.

A regular dodecagon consists of 12 congruent isosceles triangles meeting at the center. The central angle of each triangle is $360^\circ / 12 = 30^\circ$. The two equal sides are the radius $r$.
The area of one such triangle is given by:$$Area_{\Delta} = \frac{1}{2} r^2 \sin(30^\circ) = \frac{1}{2} r^2 \left(\frac{1}{2}\right) = \frac{r^2}{4}$$The total area of the 12 triangles is:$$Total Area = 12 \times \frac{r^2}{4} = 3r^2$$We want to form squares of side $r$. The area of one such square is $r^2$.
The number of squares that can be formed is:$$N_{squares} = \frac{Total Area}{Area_{square}} = \frac{3r^2}{r^2} = 3$$The number of triangles required to form one square is:$$N_{triangles/square} = \frac{Area_{square}}{Area_{\Delta}} = \frac{r^2}{r^2/4} = 4$$Thus, 3 squares can be formed, and 4 triangles are required for each square.

Given equation:$$3^{x^2} = 27 \times 9^x$$Rewrite terms with base 3:$$27 = 3^3, \quad 9^x = (3^2)^x = 3^{2x}$$Substitute back:$$3^{x^2} = 3^3 \times 3^{2x} = 3^{2x+3}$$Equating the exponents:$$x^2 = 2x + 3 \implies x^2 - 2x = 3$$We need to find the value of:$$\frac{2^{x^2}}{(2^x)^2} = \frac{2^{x^2}}{2^{2x}} = 2^{x^2 - 2x}$$Substitute $x^2 - 2x = 3$:$$2^{x^2 - 2x} = 2^3$$Thus, the value is $2^3$.

From the bar chart, we can determine the frequency distribution of the number of patients per shift ($X$):

• $X = 5$: 20 shifts
• $X = 6$: 40 shifts
• $X = 7$: 30 shifts
• $X = 8$: 10 shifts

Total number of shifts $N = 20 + 40 + 30 + 10 = 100$.The average number of patients per shift, $E[X]$, is calculated as:$E[X] = \frac{\sum f_i x_i}{\sum f_i} = \frac{20(5) + 40(6) + 30(7) + 10(8)}{100}$$E[X] = \frac{100 + 240 + 210 + 80}{100} = \frac{630}{100} = 6.3$The earnings function is given by $E = 1000(X - 0.2)$.
The average earning is the expected value of the earnings function:$E[\text{Earnings}] = E[1000(X - 0.2)] = 1000(E[X] - 0.2)$Substituting $E[X] = 6.3$:$E[\text{Earnings}] = 1000(6.3 - 0.2) = 1000(6.1) = 6100$Thus, the average amount earned per shift is ₹ 6,100.

This is a direct application of the Law of Iterated Expectations (also known as the Tower Property). The law states that for any random variables $X$ and $Y$ (assuming expectations exist):$$E[E[X|Y]] = E[X]$$Intuitively, the expected value of the conditional expectation of $X$ given $Y$ is simply the overall expected value of $X$.

An upper triangular matrix is already in row echelon form. If the question refers to the standard Gaussian elimination process (forward elimination) to transform a matrix into upper triangular form, applying it to a matrix that is already upper triangular involves checking entries below the diagonal (which are zero) and performing no arithmetic updates. However, the algorithmic complexity to traverse and verify the structure is $O(n^2)$.If the question implies solving a system of linear equations $Ax=b$ where $A$ is upper triangular, the method used is Backward Substitution. The number of operations for backward substitution is:

• For $x_n$: 1 division.
• For $x_{n-1}$: 1 multiplication, 1 subtraction, 1 division.
• ...
• Total operations $\approx \sum_{i=1}^n i = \frac{n(n+1)}{2} \approx O(n^2)$.

If the question implies Gauss-Jordan elimination to convert the upper triangular matrix to a diagonal (or identity) matrix (Reduced Row Echelon Form), we must eliminate the entries above the diagonal. This process is symmetric to forward elimination and also requires $O(n^2)$ operations.In all relevant interpretations involving non-trivial work on an upper triangular matrix in the context of Gaussian elimination/solvers, the complexity is $O(n^2)$.

Given that the sum of the elements in each row of $A \in \mathbb{R}^{n \times n}$ is 1. This implies that the vector $\mathbf{1} = [1, 1, \dots, 1]^T$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda = 1$, i.e., $A\mathbf{1} = 1\mathbf{1} = \mathbf{1}$.The matrix $B$ is defined as $B = A^3 - 2A^2 + A$. Let's evaluate the action of $B$ on the vector $\mathbf{1}$:$$B\mathbf{1} = (A^3 - 2A^2 + A)\mathbf{1} = A^3\mathbf{1} - 2A^2\mathbf{1} + A\mathbf{1}$$Since $A\mathbf{1} = \mathbf{1}$, it follows that $A^2\mathbf{1} = A(A\mathbf{1}) = A\mathbf{1} = \mathbf{1}$ and $A^3\mathbf{1} = A(A^2\mathbf{1}) = A\mathbf{1} = \mathbf{1}$.
Substituting these into the expression for $B\mathbf{1}$:$$B\mathbf{1} = \mathbf{1} - 2(\mathbf{1}) + \mathbf{1} = \mathbf{0}$$This shows that $x = \mathbf{1}$ is a non-trivial solution to the homogeneous equation $Bx = 0$. In linear algebra, if a homogeneous system of linear equations $Bx = 0$ has at least one non-trivial solution, it must have infinitely many solutions (specifically, any scalar multiple $c\mathbf{1}$ is also a solution).Therefore, the equation $Bx = 0$ has infinitely many solutions.

The given function is $f(x) = \frac{e^x - e^{-x}}{2}$, which is the definition of the hyperbolic sine function, $\sinh(x)$.The derivatives of $\sinh(x)$ cycle as follows:
$f(x) = \sinh(x)$
$f'(x) = \cosh(x)$
$f''(x) = \sinh(x)$
$f'''(x) = \cosh(x)$In general:
$f^{(k)}(x) = \sinh(x)$ if $k$ is even.
$f^{(k)}(x) = \cosh(x)$ if $k$ is odd.We need to find the $10^{th}$ derivative at $x=0$, i.e., $f^{(10)}(0)$.
Since $k=10$ is even, $f^{(10)}(x) = \sinh(x)$.Evaluating at $x=0$:
$f^{(10)}(0) = \sinh(0) = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0$.Alternatively, using the Maclaurin series expansion for $e^x$ and $e^{-x}$:
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$$f(x) = \frac{1}{2} [(1 + x + \frac{x^2}{2!} + \dots) - (1 - x + \frac{x^2}{2!} - \dots)]$
$f(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots$This series contains only odd powers of $x$. The coefficient of $x^{10}$ is 0. Since the coefficient of $x^k$ in the Maclaurin series is $\frac{f^{(k)}(0)}{k!}$, we have:
$\frac{f^{(10)}(0)}{10!} = 0 \implies f^{(10)}(0) = 0$.

To determine which statements are logically equivalent, we analyze each one:Comparing the statements:

• $S_1 \equiv p \rightarrow q \equiv \neg p \vee q$
• $S_3 \equiv \neg p \vee q$

Since both $S_1$ and $S_3$ are equivalent to $\neg p \vee q$, we have $S_1 \equiv S_3$.Checking other options:

• $S_2 (\neg p \wedge q) \not\equiv S_3 (\neg p \vee q)$
• $S_2 (\neg p \wedge q) \not\equiv S_4 (\neg p \vee \neg q)$
• $S_1 (p \rightarrow q) \not\equiv S_4 (\neg p \vee \neg q)$

Therefore, the correct option is (A).

If a decomposition is not dependency-preserving, checking whether an update satisfies the functional dependencies requires joining the decomposed relations to reconstruct the original relation (or a view of it) where the dependencies can be verified. Therefore, the Join operation will be executed more frequently.

The expression is evaluated as follows:Thus, the query returns the names of all owners who own a red car made by ABC.

We insert the keys one by one using $h(x) = 3x \pmod{10}$ and linear probing for collisions. Table: [_, _, _, 1, _, _, _, _, _, _] Table: [_, _, 4, 1, _, _, _, _, _, _] Table: [_, _, 4, 1, _, 5, _, _, _, _] Table: [_, _, 4, 1, _, 5, _, _, 6, _]

• Probe 3: Occupied (by 1).
• Probe 4: Empty. Placed at 4.

Table: [_, _, 4, 1, 14, 5, _, _, 6, _]

• Probe 6: Empty. Placed at 6.

Table: [_, _, 4, 1, 14, 5, 15, _, 6, _]Key 14 is at index 4. Key 15 is at index 6.

The median $m$ of a random variable $X$ is the value such that $F_X(m) = 0.5$. Given that the median is 3, we have:$$F_X(3) = 0.5$$Since $F_X(x)$ is non-zero and non-one in the interval $[t, 4]$, and the median is 3, we assume $t \le 3 \le 4$. Using the given CDF formula for this interval:$$\frac{3 - t}{4 - t} = 0.5$$$$3 - t = 0.5(4 - t)$$$$3 - t = 2 - 0.5t$$$$1 = 0.5t$$$$t = 2$$Checking the validity: If $t=2$, the interval is $[2, 4]$. The value 3 lies within this interval. Thus, $t=2$ is the correct value.

Given $X = aZ + b$ where $Z \sim N(0, 1)$. Thus $E[Z] = 0$ and $Var(Z) = E[Z^2] - (E[Z])^2 = 1$.$$E[X] = E[aZ + b] = aE[Z] + b = a(0) + b = b$$Given $E[X] = 1$, we have $b = 1$.$$X - E[X] = (aZ + b) - b = aZ$$$$E[(X - E[X])Z] = E[(aZ)Z] = aE[Z^2] = a(1) = a$$Given $E[(X - E[X])Z] = -2$, we have $a = -2$.$$E[(X - E[X])^2] = E[(aZ)^2] = a^2 E[Z^2] = a^2$$Given $E[(X - E[X])^2] = 4$, we have $a^2 = 4$. Since $a = -2$, $(-2)^2 = 4$, which is consistent.Thus, $a = -2$ and $b = 1$.

For an exponential distribution with rate parameter $\lambda$, the PDF is $f(x) = \lambda e^{-\lambda x}$ for $x \ge 0$, and the CDF is $F(x) = 1 - e^{-\lambda x}$.
The survival function is $P(X \ge x) = 1 - F(x) = e^{-\lambda x}$.
Given:$$P(X \ge 2) = 0.25$$$$e^{-2\lambda} = 0.25 = \frac{1}{4}$$Taking natural logarithm on both sides:$$-2\lambda = \ln(1/4) = \ln(2^{-2}) = -2\ln 2$$$$\lambda = \ln 2$$Thus, the value of $\lambda$ is $\ln 2$.

Given the update rule for a misclassified point $(x_n, y_n)$:$$w_{new} = w_{old} + y_n x_n$$$$b_{new} = b_{old} + y_n$$We are considering the specific case where $y_n = +1$ and the current prediction is incorrect, i.e., $f(x_n; w_{old}, b_{old}) < 0$.
The new function value at $x_n$ is:$$f(x_n; w_{new}, b_{new}) = w_{new}^\top x_n + b_{new}$$Substituting the update rules:$$= (w_{old} + y_n x_n)^\top x_n + (b_{old} + y_n)$$$$= w_{old}^\top x_n + y_n x_n^\top x_n + b_{old} + y_n$$$$= (w_{old}^\top x_n + b_{old}) + y_n \|x_n\|^2 + y_n$$$$= f(x_n; w_{old}, b_{old}) + y_n (\|x_n\|^2 + 1)$$Since $y_n = +1$ and $\|x_n\|^2 \ge 0$, the term $y_n (\|x_n\|^2 + 1) = \|x_n\|^2 + 1 \ge 1 > 0$.
Therefore:$$f(x_n; w_{new}, b_{new}) > f(x_n; w_{old}, b_{old})$$