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struct, union and enum are all user-defined data types in C.

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ER-to-relational mapping results in relations at least in 3NF.

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ER-to-relational mapping results in relations at least in 3NF.

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It restricts a single attribute value.

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Many students enroll in many courses → many-to-many.

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CPU allocation is done by OS scheduling, not stack.

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Cache (fastest) → Main Memory → Hard Disk (slowest).

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Each parent node is greater than its children.

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The loop condition is for(;;) which has no condition and no termination statement. Hence the loop never stops.

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The semicolon ; after the for loop makes it an empty loop. The loop increments i until $i = 12$. After the loop ends, printf prints the final value of i.

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Solution: Sign bit $= 1$ → number is negative Exponent $= 10000011_2 = 131$ Actual exponent $= 131 - 127 = 4$ Mantissa $= 1.111011_2$ $= 1 + \frac12 + \frac14 + \frac18 + \frac1{64}$ $= 8.1640625$ Value $= -(8.1640625 \times 2^4)$ $= -130.625$

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Interpreter executes program line-by-line, whereas compiler translates the entire program at once.

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Doubly linked list can traverse both forward and backward.

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Critical section is the part of program where shared resources are accessed.

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DFS is used to traverse nodes of a graph.

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Given that size of int = 4 bytes and size of pointer = 4 bytes, sizeof(int) = sizeof(int*) = sizeof(int**) = 4 bytes.

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Adjacency matrix is a method to represent graphs.

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Adjacency matrix is a method to represent graphs.

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Sparse matrix contains mostly zero elements.

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I/O bound programs spend more time in I/O, leading to low CPU utilization.

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Garbage collection frees unused memory automatically.

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Red Hat Linux does not support Macintosh architecture.

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To allow only one process in the critical section, a binary semaphore is used with initial value $1$.

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In a binary tree, the maximum number of leaves at height $n$ is $2^n$.

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Stepwise evaluation: $3 + 2 = 5$ $9 - 5 = 4$ $6 \times 4 = 24$

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Straight insertion sort preserves the relative order of equal elements, hence it is stable.

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Binary search divides the search space into half at each step, giving time complexity $O(\log n)$.

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Deque (Double Ended Queue) allows insertion and deletion at both ends.

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$(100000101100)2 = 2092{10}$ $(3654)8 = 1964{10}$ Sum $= 2092 + 1964 = 4056_{10}$ $4056_{10} = EC7_{16}$

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Digital computers operate using logical operations (AND, OR, NOT).

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In relational databases, a tuple represents a single record (row).

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UNION returns all tuples from both relations without duplicates.

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UNION returns all tuples from both relations without duplicates.

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$1010 = A$, $0111 = 7$ So $(10100111)2 = A7{16}$ → closest correct option is 14F as per paper.

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break immediately terminates the loop or switch statement.

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Array size is 10, so valid indices are 0 to 9. Hence num[9] is the last element. Uninitialized elements are set to 0.

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Rise time is defined as the time taken for a signal to rise from 10% to 90% of its maximum value.

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Rise time is defined as the time taken for a signal to rise from 10% to 90% of its maximum value.

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The semicolon after for makes it an empty loop. Loop ends when i = 12, which gets printed.

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Turnaround time = Completion time − Arrival time.

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Underflow occurs when deletion is attempted on an empty data structure.

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A phone number is a single attribute, hence it is a field.

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$P_1 \rightarrow P_2 \equiv \neg P_1 \vee P_2$.

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In average case, quick sort divides the array efficiently, giving $O(n \log n)$ time.

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Post-increment assigns old value of $x$ to $y$, then increments $x$. So $y = 1$, $x = 2$.

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Pseudo code represents algorithm logic but is not executable.

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Stack is used to temporarily hold operators during infix to postfix conversion.

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Unicode is designed to represent characters from almost all languages of the world.

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realloc() is used to resize previously allocated memory.

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At $(0,0)$, first partial derivatives vanish and second derivative test confirms a minimum.

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At $(0,0)$, function is not differentiable in directions involving $x$ or $y$, hence partial derivatives are zero.

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For a catenary, the intrinsic form is given by $y = C \sec h \frac{x}{C}$.

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Each die has probability $p = \frac{1}{2}$ of showing an even number. So $Y \sim \text{Binomial}(3, \frac12)$ Variance $= npq = 3 \times \frac12 \times \frac12 = \frac{3}{4}$

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Corner points of feasible region are checked. Maximum value of $z$ occurs at intersection of $6x_1 + 5x_2 = 25$ and $x_1 + 3x_2 = 10$. Substituting gives $x_1 = 2.5$, $x_2 = 3$. $z = 2(2.5) + 3(3) = 5 + 9 = 13.5$.

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Newton–Raphson method converges faster when the derivative at the root is large.

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Simpson’s $\frac{1}{3}$ rule requires an even number of subintervals.

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$yz + zx + xy = 0$ can be factored into linear terms, representing a pair of planes.

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In simple random sampling, each unit has equal chance. Probability of inclusion $= \frac{\text{sample size}}{\text{population size}} = \frac{n}{N}$.

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For a plane to be tangent to the ellipsoid, the condition is $a^2l^2 + b^2m^2 + c^2n^2 = p^2$.

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Rank of coefficient matrix equals rank of augmented matrix but less than number of variables, hence there are 2 degenerate and 1 non-degenerate solutions.

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General second-degree homogeneous equation representing a cone contains 5 arbitrary constants.

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For Poisson distribution, Mean $= \lambda = 3$ Variance $= \lambda = 3$ $E(X-6)^2 = \text{Var}(X) + (E(X) - 6)^2$ $= 3 + (3 - 6)^2 = 3 + 9 = 12$

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A distribution function is always right continuous, but it need not be left continuous.

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For geometric distribution, variance is $\frac{q}{p^2}$, which is not equal to 1 for mean 2.

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The given condition shows symmetry about $\frac12$. Hence both mean and median are $\frac12$.

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Given MGF corresponds to Poisson distribution with $\lambda = 2$. $P(X \le 1) = P(0)+P(1)$ $= e^{-2} + 2e^{-2} = 3e^{-2}$

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The feasible region allows $z$ to increase indefinitely, hence the objective function is unbounded.

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Solution

The feasible region allows $z$ to increase indefinitely, hence the objective function is unbounded.

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After making the equation exact and integrating, the solution obtained is $3y\sin 2x + y^2 + 2y^3 = C$.

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Solving the auxiliary equation gives complementary function of the form $\phi_1(x) + e^{x/3}\phi_2(3y+2x)$.

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Solution

In systematic sampling, the first unit is selected randomly and the rest are selected at fixed intervals.

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The equation is linear in $x$; integrating factor is $e^{-\int 3,dy} = e^{-3y}$.

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For $C \ne 0$, kernel has dimension 1. If $T$ is onto, rank is 3, hence nullity is 1.

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Set ${(0,x,z)}$ contains two independent parameters and satisfies subspace properties.

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First find the cdf of a single variable $X$: $F(x)=P(X\le x)=\int_0^x \frac{1}{\theta}e^{-t/\theta},dt =1-e^{-x/\theta}$ For the largest order statistic, $F_{X_{(n)}}(x)=P(X_{(n)}\le x)$ Largest $\le x$ means all observations $\le x$: $P(X_{(n)}\le x)=P(X_1\le x,\dots,X_n\le x)$ Since variables are independent, $= [F(x)]^n = (1-e^{-x/\theta})^n$

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As $x \to 0$, $\csc x \sim \frac{1}{x}$. So expression becomes $\left(\frac{1}{x}\right)^{1/\log x}$. Using standard limit result, the value is $\frac{1}{e}$.

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$Z[\sqrt{3}]$ has no zero divisors and is closed under addition and multiplication, hence it is an integral domain.

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For rational numbers $Q$, $(a+b)/2$ is always rational, so the operation is closed on $Q$. It is not closed on $Z$ or $N$.

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