Solution

For real values:

$1 - x^2 - y^2 \ge 0$

$x^2 + y^2 \le 1$

Minimum value of $\sqrt{1 - x^2 - y^2}$ is $0$ (when $x^2+y^2=1$)

Maximum value is $1$ (when $x=y=0$)

So exponent varies from $0$ to $1$

$e^0 = 1$

$e^1 = e$

Hence range is $[1, e]$

Solution

Given:

$a^4 = e$

$b^2 = e$

Also $a^3 b = ba$

Multiply by $a$:

$a^4 b = baa$

$b = baa$

$ab = a(baa) = (ab)^{-1}$

Hence $(ab)^2 = e$

Therefore order of $ab = 2$

Solution

$\iint_R y\sin(ny)\,dA=\int_{1}^{2}\!\!dx\int_{0}^{\pi} y\sin(ny)\,dy$

$=\int_{0}^{\pi} y\sin(ny)\,dy$

Let $u=y,\; dv=\sin(ny)\,dy \Rightarrow v=-\frac{\cos(ny)}{n}$

$\int_{0}^{\pi} y\sin(ny)\,dy=\left[-\frac{y\cos(ny)}{n}\right]_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos(ny)\,dy$

$=-\frac{\pi\cos(n\pi)}{n}+\frac{1}{n^2}\left[\sin(ny)\right]_{0}^{\pi}=-\frac{\pi(-1)^n}{n}$

Solution

First octant: $0\le x\le 2$, $0\le y\le 3$, $0\le z\le 9-y^2$

$V=\int_{0}^{2}\!\!dx\int_{0}^{3}(9-y^2)\,dy=2\left[9y-\frac{y^3}{3}\right]_{0}^{3}=2(27-9)=36$

Solution

Use polar: $x=r\cos\theta,\; y=r\sin\theta$

Cylinder: $r^2=2r\cos\theta \Rightarrow r=2\cos\theta,\; -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$

$V=\iint_D (x^2+y^2)\,dA=\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} r^2\cdot r\,dr\,d\theta$

$=\int_{-\pi/2}^{\pi/2}\left[\frac{r^4}{4}\right]_{0}^{2\cos\theta}d\theta=\int_{-\pi/2}^{\pi/2}4\cos^4\theta\,d\theta$

$=8\int_{0}^{\pi/2}\cos^4\theta\,d\theta=8\cdot\frac{3\pi}{16}=\frac{3\pi}{2}$

Solution

$x_{n+1}=\frac{1}{2}\left(x_n+\frac{9}{4x_n}\right)$ is Newton iteration for $\sqrt{\frac{9}{4}}$

$\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5$

Solution

Hence $\tan^{-1}(\cdot)\to\frac{\pi}{2}$

Solution

Using Green’s theorem:

$\oint_C (M\,dx+N\,dy)=\iint_D\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$

Here $M=y^2,\; N=x^2$ so $\frac{\partial N}{\partial x}=2x,\; \frac{\partial M}{\partial y}=2y$

Integral $=\iint_D 2(x-y)\,dA$ over triangle with vertices $(0,0),(1,0),(0,1)$

By symmetry $\iint_D x\,dA=\iint_D y\,dA$ so $\iint_D (x-y)\,dA=0$

Hence value $=0$

Solution

$y_{k+1}=ay_k \Rightarrow y_k=ca^k$

Solution

Characteristic equation:

$(m^2-a^2)^3=0$

Roots: $m=\pm a$ each of multiplicity 3.

Hence complementary function:

$(c_1+c_2x+c_3x^2)e^{ax}+(c_4+c_5x+c_6x^2)e^{-ax}$

Since RHS is $e^{ax}$ and $m=a$ has multiplicity 3,

Particular integral = $x^3\frac{e^{ax}}{3!\,(2a)^3} =\frac{x^2e^{ax}}{8a^2}$

Solution

Distance formula:

$d=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}$

Plane: $x+2y+z-4=0$

$d=\frac{|1(1)+2(0)+1(-2)-4|}{\sqrt{1^2+2^2+1^2}}$

$=\frac{|1-2-4|}{\sqrt6}=\frac{5}{\sqrt6} =\frac{5\sqrt6}{6}$

Solution

Lagrange basis for node $x_3=1$:

$L_3(0)=\frac{(0+2)(0+1)(0-2)}{(1+2)(1+1)(1-2)}$

$=\frac{(2)(1)(-2)}{(3)(2)(-1)} =\frac{-4}{-6}=\frac{2}{3}$

Solution

For Randomized Block Design:

Error d.f. = $(t-1)(b-1)$

Here $t=4$, $b=5$

Error d.f. = $(4-1)(5-1)=3 \times 4 = 12$

Solution

From $-3x_1 + x_2 \ge 3$

$x_2 \ge 3 + 3x_1$

But from $x_1 + x_2 \le 1$

$x_2 \le 1 - x_1$

So we need:

$3 + 3x_1 \le 1 - x_1$

$4x_1 \le -2$

$x_1 \le -\frac{1}{2}$

This contradicts $x_1 \ge 0$

Hence problem is infeasible.

Solution

Level of significance:

$\alpha = P(X \ge 1 \mid \theta=2)$

$= \int_{1}^{\infty} 2e^{-2x} dx$

$= \left[-e^{-2x}\right]_{1}^{\infty}$

$= e^{-2}$

$= \frac{1}{e^2}$

Solution

Solution

For exponential distribution:

Mean $= \frac{1}{\lambda}$

Variance $= \frac{1}{\lambda^2}$

Given mean $=3 \Rightarrow \lambda=\frac{1}{3}$

Variance $=\frac{1}{(1/3)^2}=9$

Solution

Total items = 100

Probability both defective:

$=\frac{20}{100}\times\frac{19}{99}$

$=\frac{380}{9900}=\frac{19}{495}$

Solution

Even values: $2,4,6,\dots$

$P(\text{even})=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots$

$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots$

This is geometric series:

$=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}$

Solution

Correlation coefficient $r=\pm\sqrt{b_{xy}b_{yx}}$

Hence it is geometric mean of two regression coefficients.

Solution

Mean of uniform integers $1$ to $8$:

$=\frac{1+8}{2}=4.5$

Solution

Let $p=P(X=1)$

Then $P(X=0)=1-p$

$E(X)=p$

$V(X)=p(1-p)$

Given $p=3p(1-p)$

$1=3(1-p)$

$1=3-3p$

$3p=2$

$p=\frac{2}{3}$

Therefore $P(X=0)=1-p=\frac{1}{3}$

Solution

Solution

$Z=\frac{X-\mu}{\sigma} \sim N(0,1)$

Square of standard normal variable follows $\chi^2$ distribution with 1 d.f.

Solution

$t=\frac{\bar{d}}{s_d/\sqrt{n}}$

$=\frac{15}{5/\sqrt{9}}=\frac{15}{5/3}=\frac{15\times3}{5}=9$

Solution

By property of arithmetic mean:

$\sum (x_i-\bar{x})=0$

Solution

If each observation is multiplied by a constant $k$, then new S.D. = $|k| \times$ old S.D.

Here $k=2$

New S.D. = $2 \times 8 = 16$

Solution

For numbers $1,2,\dots,n$:

Mean $=\frac{n+1}{2}$

Variance $=\frac{n^2-1}{12}$

S.D. $=\sqrt{\frac{n^2-1}{12}}$

Solution

For normal curve, points of inflexion are at

$\mu \pm \sigma$

Solution

Standard error of sample proportion:

$SE=\sqrt{\frac{PQ}{n}}$

Solution

Thus $\mu=\mu_0$ with $\sigma^2$ known is simple.

Solution

Velocity of centre of mass:

$\vec{V}=\frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}$

$=\frac{100(2\hat{i}-7\hat{j}+3\hat{k})+20(-10\hat{i}+35\hat{j}-3\hat{k})}{120}$

$=\frac{(200-200)\hat{i}+(-700+700)\hat{j}+(300-60)\hat{k}}{120}$

$=\frac{240\hat{k}}{120}=2\hat{k}$

Solution

Given: $K = 2m_0c^2$

Total energy $E = K + m_0c^2 = 3m_0c^2$

$E=\gamma m_0 c^2$

$\gamma=3$

Relativistic mass $=\gamma m_0=3m_0$

Ratio $=\frac{m}{m_0}=3$

Solution

Laplace equation: $\nabla^2\phi=0$

(a) $\nabla^2=2+10-12=0$ ✔

(b) $\nabla^2=0$ ✔

(c) $\nabla^2=0-2+2=0$ ✔

(d) $\nabla^2=4-8+2=-2 \ne 0$ ✘

Solution

E-R modelling starts from general view → top-down approach.

Solution

Recursive call without stopping condition → infinite printing until stack overflow.

Solution

Floyd’s algorithm finds shortest path between all pairs.

Solution

System CPU time = kernel execution time.

Solution

Array name arr represents base address.

Size of int = 32 bit = 4 bytes.

Base address = 2000.

After arr++, pointer moves by one integer = 4 bytes.

New address = 2000 + 4 = 2004.

Solution

Start: []

push(1) → [1]

push(2) → [1,2]

pop → 2 → [1]

push(1) → [1,1]

push(2) → [1,1,2]

pop → 2 → [1,1]

pop → 1 → [1]

pop → 1 → []

push(2) → [2]

pop → 2 → []

Popped sequence: 2,2,1,1,2

Solution

Solution:

EDGE → GSM
UMTS → GSM evolution
EHSPA → GSM evolution
FHSS → not GSM

Solution

$(1111111)_2 = 127$

$(4543)_8 = 4×512 + 5×64 + 4×8 + 3 = 2403$

Total = 127 + 2403 = 2530

2530 in hexadecimal = 9B2

Solution

FDDI uses dual ring topology.

Solution

Von-Neumann architecture uses same memory for data and instructions.

Solution

New process is created using fork().

Solution

Minimum nodes follow Fibonacci relation:

$N(h)=N(h-1)+N(h-2)+1$

$N(6)=20$

Solution

With self-loops allowed, each vertex can connect to n vertices.

Total edges = n × n = n²

Solution

Banker’s Algorithm is used for deadlock avoidance in operating systems.

Solution

Degree of a vertex in an undirected graph is the number of edges incident on it.

Solution

Solution

Malicious software is commonly called malware.

Solution

For mutual exclusion, only one process can enter critical section at a time.

Solution

Cassette tape uses sequential access method.

Solution

The statement suggests foreign movies cause harm (moral decline), otherwise there is no reason to ban them.

Thus Assumption I is implicit.

Assumption II is not required.

Solution

She moves 3 km West and 4 km North.

Distance from house:

$\sqrt{3^2+4^2}=5$ km

Direction = North-West

Solution

Placing positions according to directions and tracing movement, M lies directly above (North of) Y.

Solution

Pattern: subtract 8 each time

98 − 8 = 90

90 − 8 = 82

82 − 8 = 74

74 − 8 = 66

66 − 8 = 58

Solution

Solution:

2 = 1³ + 1

9 = 2³ + 1

28 = 3³ + 1

65 = 4³ + 1

Next = 5³ + 1 = 125 + 1 = 126

Solution

Solution:

Standardized regression coefficient:

$\beta = b \left(\frac{S_x}{S_y}\right)$

Solution

Vehicle I → B (Engineer) and G (Teacher)

Vehicle II → A (Teacher), F (Teacher)

C cannot travel with A and F (given).

Two same profession cannot travel together.

Since D is doctor and C is doctor, they cannot be in same vehicle.

After proper arrangement, C must travel in Vehicle III.

Solution

Replace symbols:

8 × 2 + 12 ÷ 3 − 9

Now apply BODMAS:

8 × 2 = 16

12 ÷ 3 = 4

16 + 4 − 9 = 11

Solution

Pattern: ×2 + increasing numbers

7 × 2 + 1 = 15

15 × 2 + 2 = 32

32 × 2 + 3 = 67

67 × 2 + 4 = 138

138 × 2 + 5 = 281

Solution

Pattern: ×3, ÷2 alternately

40 × 3 = 120

120 ÷ 2 = 60

60 × 3 = 180

180 ÷ 2 = 90

90 × 3 = 270

270 ÷ 2 = 135

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Given:

$T(x,y,z) = (2x,\; x-y,\; 5x+4y+z)$

Write system:

$u = 2x$

$v = x - y$

$w = 5x + 4y + z$

From first:

$x = \frac{u}{2}$

From second:

$y = x - v = \frac{u}{2} - v$

From third:

$z = w - 5x - 4y$

$z = w - \frac{5u}{2} - 4\left(\frac{u}{2} - v\right)$

$z = w - \frac{5u}{2} - 2u + 4v$

$z = w - \frac{9u}{2} + 4v$

Thus inverse:

$T^{-1}(x,y,z) = \left(\frac{x}{2}, \frac{x - 2y}{2}, \frac{-9x + 8y + 2z}{2}\right)$

Solution

Trace of $A = 3 + (-2) = 1$

Determinant of $A = (3)(-2) - (-8) = -6 + 8 = 2$

Characteristic equation of $A$ is:

$\lambda^2 - (\text{trace})\lambda + \det = 0$

$\lambda^2 - \lambda + 2 = 0$

By Cayley–Hamilton theorem:

$A^2 - A + 2I = 0$

Comparing with $A^2 - kA + 2I = 0$

$k = 1$

Solution

Put $p = \frac{dy}{dx}$

$y = 2xp + yp^2$

$y(1 - p^2) = 2xp$

$y = \frac{2xp}{1 - p^2}$

This is Clairaut’s form. Hence general solution is

$2xc - y + c^2 = 0$

Solution

Reciprocal cone coefficients are reciprocals of given coefficients.

So equation becomes

$\frac{x^2}{1} + \frac{y^2}{1/2} + \frac{z^2}{1/3} = 0$

Multiplying gives

$6x^2 + 3y^2 + 2z^2 = 0$

Solution

If equally inclined then $l = m = n$

$l^2 + m^2 + n^2 = 1$

$3l^2 = 1$

$l = \pm\frac{1}{\sqrt{3}}$

Solution

Matrix form:

$\begin{bmatrix} -1 & -1 \\ 3 & 8 \\ 9 & -11 \end{bmatrix}$

Columns are independent ⇒ Rank = 2

Nullity = 2 - 2 = 0

Solution

For perpendicular focal chords of parabola, required sum = $\frac{1}{4}$

Solution

Solution

Actual integral:

$\int_{0}^{1} (x^3 - cx^2)\,dx = \left[\frac{x^4}{4} - \frac{cx^3}{3}\right]_0^1 = \frac{1}{4} - \frac{c}{3}$

Trapezoidal rule with one interval:

$T = \frac{1}{2}[f(0) + f(1)]$

$f(0) = 0$

$f(1) = 1 - c$

$T = \frac{1}{2}(1 - c)$

Since exact:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2}(1 - c)$

Solving:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2} - \frac{c}{2}$

Multiply by 12:

$3 - 4c = 6 - 6c$

$2c = 3$

$c = 1.5$

Solution

First compute:

$2 * 3 = 2 + 3 + (2)(3) = 11$

Inverse element $x$ satisfies:

$11 * x = 0$ (identity element is $0$)

$11 + x + 11x = 0$

$x(12) = -11$

$x = -\frac{11}{12}$