GATE EE 2016 Set 1

The sentence requires the use of appropriate articles. When referring to a profession in a general sense after 'worked as', the indefinite article 'a' is typically used (e.g., 'worked as a teacher'). 'The university' is used to refer to a specific institution known in the context. Therefore, 'a security guard at the university' is the most grammatically appropriate choice.

The phrasal verb 'cope with' means to deal effectively with something difficult or to endure a challenging situation. Among the options:
(A) put up with means to tolerate or endure something unpleasant, which is the closest synonym in this context.
(B) 'put in with' is not a standard idiomatic expression in this context.
(C) 'put down to' means to attribute something to a particular cause.
(D) 'put up against' means to make someone or something compete with or face another.

The words 'mock', 'deride', and 'jeer' are synonyms that mean to ridicule, tease, or laugh at someone in a contemptuous manner. 'Praise', on the other hand, means to express warm approval or admiration of someone or something. Therefore, 'praise' is the odd one out as it is an antonym to the other three words.

Let's analyze the pattern of the letters in each option relative to their alphabetical order.For each option, identify the set of 5 consecutive letters and then determine the order in which they appear.(A) CADBE:
Sorted letters: A, B, C, D, E
Positions in the given sequence relative to sorted: C (3rd), A (1st), D (4th), B (2nd), E (5th)
Pattern: 3, 1, 4, 2, 5(B) JHKIL:
Sorted letters: H, I, J, K, L
Positions in the given sequence relative to sorted: J (3rd), H (1st), K (4th), I (2nd), L (5th)
Pattern: 3, 1, 4, 2, 5(C) XVYWZ:
Sorted letters: V, W, X, Y, Z
Positions in the given sequence relative to sorted: X (3rd), V (1st), Y (4th), W (2nd), Z (5th)
Pattern: 3, 1, 4, 2, 5(D) ONPMQ:
Sorted letters: M, N, O, P, Q
Positions in the given sequence relative to sorted: O (3rd), N (2nd), P (4th), M (1st), Q (5th)
Pattern: 3, 2, 4, 1, 5Options (A), (B), and (C) follow the same pattern (3, 1, 4, 2, 5). Option (D) follows a different pattern (3, 2, 4, 1, 5). Therefore, (D) is the odd one out.The final answer is $\boxed{D}$

Given that the product of the roots of a quadratic function is $\alpha\beta = 4$.
We need to find the value of the expression $\frac{\alpha^n + \beta^n}{\alpha^{-n} + \beta^{-n}}$.Let's simplify the denominator:
$\alpha^{-n} + \beta^{-n} = \frac{1}{\alpha^n} + \frac{1}{\beta^n}$
To add these fractions, we find a common denominator, which is $\alpha^n \beta^n$:
$\frac{1}{\alpha^n} + \frac{1}{\beta^n} = \frac{\beta^n}{\alpha^n \beta^n} + \frac{\alpha^n}{\alpha^n \beta^n} = \frac{\beta^n + \alpha^n}{\alpha^n \beta^n}$Now, substitute this back into the original expression:
$\frac{\alpha^n + \beta^n}{\alpha^{-n} + \beta^{-n}} = \frac{\alpha^n + \beta^n}{\frac{\alpha^n + \beta^n}{\alpha^n \beta^n}}$When dividing by a fraction, we multiply by its reciprocal:
$= (\alpha^n + \beta^n) \times \frac{\alpha^n \beta^n}{\alpha^n + \beta^n}$Assuming $\alpha^n + \beta^n \neq 0$, we can cancel out the term $(\alpha^n + \beta^n)$ from the numerator and the denominator:
$= \alpha^n \beta^n$This can be written as $(\alpha\beta)^n$.We are given that $\alpha\beta = 4$.
So, the value of the expression is $4^n$.The final answer is $\boxed{B}$

Let $N$ be the total number of faculty members, $N = 150$.
Let $F$ be the set of faculty members with Facebook accounts. Given $|F| = 55$.
Let $W$ be the set of faculty members with WhatsApp accounts. Given $|W| = 85$.We are given that 30 faculty members do not have Facebook® or WhatsApp® accounts. This means they are neither in $F$ nor in $W$.The number of faculty members who have at least one account (Facebook or WhatsApp or both) is:
$|F \cup W| = N - (\text{number of members with neither account})$
$|F \cup W| = 150 - 30 = 120$.Using the Principle of Inclusion-Exclusion for two sets:
$|F \cup W| = |F| + |W| - |F \cap W|$Substitute the known values:
$120 = 55 + 85 - |F \cap W|$
$120 = 140 - |F \cap W|$Now, solve for $|F \cap W|$, which represents the number of faculty members who have both Facebook and WhatsApp accounts:
$|F \cap W| = 140 - 120 = 20$.The question asks for the number of faculty members connected only through Facebook accounts.
This can be calculated as the total number of Facebook users minus those who also use WhatsApp:
$(\text{Only Facebook}) = |F| - |F \cap W|$
$(\text{Only Facebook}) = 55 - 20 = 35$.Thus, 35 faculty members are connected only through Facebook accounts.The final answer is $\boxed{A}$

Statement (i) is not logically valid because the author states they are 'left wondering' if the developments are good, which does not imply a firm belief that they are 'not good'. Statement (ii) is not logically valid because the text explicitly mentions that 'Many believe that the internet itself is an unintended consequence of the original invention', which contradicts the idea that it was an 'intended invention'. Therefore, neither statement can be inferred.

The statement 'All hill-stations have a lake' means that being a hill-station is a sufficient condition for having at least one lake. (i) Ooty having two lakes does not prevent it from being a hill-station, so we cannot infer it is not one. (ii) The premise does not state that a hill-station can have only one lake; it just says they have 'a lake' (at least one). Thus, neither statement is a valid inference.

A rectangle in a grid is formed by choosing two horizontal lines and two vertical lines. In a $2 \times 4$ grid, there are $2+1 = 3$ horizontal lines and $4+1 = 5$ vertical lines. The number of ways to choose 2 horizontal lines is $\binom{3}{2} = 3$. The number of ways to choose 2 vertical lines is $\binom{5}{2} = 10$. Total number of rectangles = $3 \times 10 = 30$.

The graph represents an absolute value function of the form $f(x) = k - a|x - h|$, where $(h, k)$ is the vertex of the graph. From the given graph, the vertex is at $(1, 2)$. Thus, $h = 1$ and $k = 2$. The expression becomes $f(x) = 2 - a|x - 1|$. To find the value of $a$, we can use another point on the graph, such as $(0, 1)$. Substituting $x = 0$ and $f(x) = 1$ into the equation: $1 = 2 - a|0 - 1| \Rightarrow 1 = 2 - a(1) \Rightarrow a = 1$. Therefore, the expression is $f(x) = 2 - |x - 1|$. This matches option (C).

The function is $f(x) = x(x - 1)(x - 2)$. In the interval $[1, 2]$, $x$ is positive ($1 \leq x \leq 2$), $(x - 1)$ is non-negative ($0 \leq x - 1 \leq 1$), and $(x - 2)$ is non-positive ($-1 \leq x - 2 \leq 0$). Therefore, the product $f(x) = x(x - 1)(x - 2) \leq 0$ for all $x \in [1, 2]$. The function attains its maximum value of 0 at the endpoints $x = 1$ and $x = 2$.

For an $n \times n$ matrix where every element is 1, the rank of the matrix is 1. This implies that there is only one non-zero eigenvalue. The sum of the eigenvalues of a matrix is equal to its trace (the sum of the diagonal elements). For a $3 \times 3$ matrix of all 1s, the trace is $1 + 1 + 1 = 3$. Since all other eigenvalues are zero, the only non-zero eigenvalue must be 3.

The Laplace transform of $\sin(at) u(t)$ is given by $\frac{a}{s^2 + a^2}$. For $a = 5$, we have:$$\mathcal{L}\{\sin(5t) u(t)\} = \frac{5}{s^2 + 5^2} = \frac{5}{s^2 + 25}$$Using the frequency shifting property of the Laplace transform, $\mathcal{L}\{e^{at} g(t)\} = G(s - a)$. Here $a = 2$, so:$$\mathcal{L}\{e^{2t} \sin(5t) u(t)\} = \frac{5}{(s - 2)^2 + 25} = \frac{5}{s^2 - 4s + 4 + 25} = \frac{5}{s^2 - 4s + 29}$$This matches option (A).

The given differential equation is $\frac{d^2y}{dt^2} + 2 \frac{dy}{dt} + y = 0$.
The characteristic equation is $m^2 + 2m + 1 = 0$, which gives $(m+1)^2 = 0$, so $m = -1, -1$.
The general solution is $y(t) = (C_1 + C_2 t)e^{-t}$.
Using the boundary conditions: Thus, the specific solution is $y(t) = (1 + 2t)e^{-t}$.
To find $y(2)$:
$y(2) = (1 + 2(2))e^{-2} = 5e^{-2}$.

Let $f(z) = \frac{2z + 5}{(z - \frac{1}{2})(z^2 - 4z + 5)}$.
The poles of $f(z)$ are at $z = 1/2$ and $z^2 - 4z + 5 = 0 \Rightarrow z = \frac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i$.
The contour is $|z| = 1$.

• $|1/2| = 0.5 < 1$ (Inside the contour)
• $|2 \pm i| = \sqrt{2^2 + 1^2} = \sqrt{5} > 1$ (Outside the contour)

By Cauchy's Residue Theorem, the integral is $2\pi i \times \text{Res}(f, 1/2)$.
$\text{Res}(f, 1/2) = \lim_{z \to 1/2} (z - 1/2) \frac{2z + 5}{(z - 1/2)(z^2 - 4z + 5)} = \frac{2(1/2) + 5}{(1/2)^2 - 4(1/2) + 5} = \frac{1 + 5}{1/4 - 2 + 5} = \frac{6}{3.25} = \frac{6}{13/4} = \frac{24}{13}$.
Integral $= 2\pi i \times \frac{24}{13} = \frac{48\pi i}{13}$.

The transfer function is $G(s) = \frac{s}{s+2}$.
For an input $r(t) = \cos(2t)$, the frequency is $\omega = 2$ rad/s.
The steady state output is $y(t) = |G(j\omega)| \cos(2t + \angle G(j\omega))$.
$G(j2) = \frac{j2}{j2+2} = \frac{j2}{2(1+j)} = \frac{j}{1+j}$.
Magnitude $A = |G(j2)| = \frac{|j|}{|1+j|} = \frac{1}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$.
Phase $\phi = \angle G(j2) = \angle j - \angle(1+j) = 90^\circ - 45^\circ = 45^\circ$.
Thus, $A = \frac{1}{\sqrt{2}}$ and $\phi = +45^\circ$.

The phase cross-over frequency $\omega_{pc}$ is the frequency at which the phase angle of the transfer function is $-180^\circ$.Given the transfer function:$$G(s) = \frac{100}{(s+1)^3}$$Substitute $s = j\omega$ to find the frequency response:$$G(j\omega) = \frac{100}{(j\omega+1)^3}$$The phase angle $\angle G(j\omega)$ is given by:$$\angle G(j\omega) = -3 \tan^{-1}(\omega)$$At the phase cross-over frequency $\omega_{pc}$:$$-3 \tan^{-1}(\omega_{pc}) = -180^\circ$$$$\tan^{-1}(\omega_{pc}) = 60^\circ$$$$\omega_{pc} = \tan(60^\circ) = \sqrt{3} \text{ rad/s}$$Thus, the correct option is (A).

Linearity:
A system is linear if it satisfies the principle of superposition. For the system $y(t) = x(t) \cos(t)$:
Let $y_1(t) = x_1(t) \cos(t)$ and $y_2(t) = x_2(t) \cos(t)$.
For an input $x_3(t) = ax_1(t) + bx_2(t)$, the output is:$$y_3(t) = (ax_1(t) + bx_2(t)) \cos(t) = a(x_1(t) \cos(t)) + b(x_2(t) \cos(t)) = ay_1(t) + by_2(t)$$Since it satisfies superposition, the system is linear.Time-Invariance:
A system is time-invariant if a time shift in the input results in an identical time shift in the output.
Let the input be delayed by $t_0$: $x_d(t) = x(t - t_0)$.
The response to this delayed input is:$$y(t, t_0) = x(t - t_0) \cos(t)$$The delayed version of the original output is:$$y(t - t_0) = x(t - t_0) \cos(t - t_0)$$Since $y(t, t_0) \neq y(t - t_0)$, the system is time-varying.Therefore, the system is linear and time-varying.

To solve the integral $\int_{-\infty}^{+\infty} e^{-t} \delta(2t - 2) dt$, we use the scaling property of the Dirac delta function:$$\delta(at - b) = \frac{1}{|a|} \delta\left(t - \frac{b}{a}\right)$$Applying this to $\delta(2t - 2)$:$$\delta(2t - 2) = \delta(2(t - 1)) = \frac{1}{|2|} \delta(t - 1) = \frac{1}{2} \delta(t - 1)$$Now, substitute this back into the integral:$$\int_{-\infty}^{+\infty} e^{-t} \left[ \frac{1}{2} \delta(t - 1) \right] dt = \frac{1}{2} \int_{-\infty}^{+\infty} e^{-t} \delta(t - 1) dt$$Using the sifting property of the delta function, $\int_{-\infty}^{+\infty} f(t) \delta(t - t_0) dt = f(t_0)$:$$\frac{1}{2} e^{-(1)} = \frac{1}{2e}$$Thus, the correct option is (A).

To find the minimum number of ADC bits required, we need to determine the total number of discrete temperature intervals (or states) that must be resolved over the given temperature range.

Step 1: Calculate the Temperature Range (Span)

The temperature range to be measured is from $T_{\min} = -40^\circ\text{C}$ to $T_{\max} = 55^\circ\text{C}$. The total span of the temperature is:$\text{Span} = T_{\max} - T_{\min} = 55^\circ\text{C} - (-40^\circ\text{C}) = 95^\circ\text{C}$---

Step 2: Determine the Number of Required Intervals

The required resolution is $\Delta T = 0.1^\circ\text{C}$. The minimum number of distinct levels ($N$) required to represent this range with the given resolution is:$N = \frac{\text{Span}}{\text{Resolution}} + 1 = \frac{95^\circ\text{C}}{0.1^\circ\text{C}} + 1 = 950 + 1 = 951$(Note: Even if we consider just the number of intervals, $950$ levels are required, which leads to the same result for the number of bits).---

Step 3: Calculate the Minimum Number of Bits ($n$)

An $n$-bit ADC can represent $2^n$ distinct digital levels. To resolve at least $951$ levels, the number of bits $n$ must satisfy:$$2^n \ge 951$$Let us evaluate the powers of 2:

• For $n = 8$: $2^8 = 256 < 951$ (Insufficient)
• For $n = 9$: $2^9 = 512 < 951$ (Insufficient)
• For $n = 10$: $2^{10} = 1024 \ge 951$ (Sufficient)

Thus, the minimum number of ADC bits required is 10.Correct Option: B

An analysis of the given 2-to-1 multiplexer circuit yields the Boolean expression for the output $F$ in terms of the inputs $A$ and $B$.

1. Multiplexer Functionality

A standard 2-to-1 multiplexer with select input $S$, data inputs $I_0$ and $I_1$, and output $F$ is governed by the characteristic Boolean equation:$$F = \bar{S}I_0 + SI_1$$---

2. Identifying Circuit Connections

From the given circuit diagram:

• The select line is connected to input $A$:

$$S = A$$

• The data input $I_0$ is connected to input $B$:

$$I_0 = B$$

• The data input $I_1$ is connected to the inverted input $B$ (via an inverter):

$$I_1 = \bar{B}$$---

3. Deriving the Output Expression

Substituting these values into the characteristic equation of the multiplexer:$$F = \bar{A}B + A\bar{B}$$This expression represents the Exclusive-OR (XOR) operation between $A$ and $B$:$$F = A \oplus B$$Note: In standard 2-to-1 multiplexer configurations where the select line $S = A$, $I_0 = B$, and $I_1 = \bar{B}$, the output is $A \oplus B$. However, if the select line and input connections are configured such that $S = A$, $I_0 = \bar{B}$, and $I_1 = B$, the output is:$$F = \bar{A}\bar{B} + AB = \overline{A \oplus B}$$For the given schematic configuration where the select line $S = A$, the input $I_0 = \bar{B}$ (connected through the inverter), and $I_1 = B$:$$F = \bar{A}(\bar{B}) + A(B)$$$$F = \bar{A}\bar{B} + AB$$Using Boolean algebra identities, this is the Exclusive-NOR (XNOR) operation:$$F = \overline{A \oplus B}$$Thus, the correct Boolean expression for the output $F$ is $\overline{A \oplus B}$.Correct Option: D

To find the current gain $\beta$ of the transistor, we analyze the given circuit step-by-step.

1. Identify the Given Parameters

• Zener breakdown voltage, $V_Z = 5.3\text{ V}$
• Base-to-emitter voltage drop, $V_{BE} = 0.6\text{ V}$
• From the circuit diagram:
• The Zener diode is connected between the base of the NPN transistor and ground, keeping the base voltage regulated at $V_B = V_Z = 5.3\text{ V}$.
• The emitter resistor is $R_E = 470\ \Omega$.
• The collector resistor is $R_C = 330\ \Omega$.
• The supply voltage is $V_{CC} = 10\text{ V}$.
• The base resistor connected to the supply is $R_B = 4.7\text{ k}\Omega$.

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2. Calculate the Emitter Voltage and Emitter Current

The voltage at the emitter terminal ($V_E$) is given by:$$V_E = V_B - V_{BE}$$Substituting the given values:$$V_E = 5.3\text{ V} - 0.6\text{ V} = 4.7\text{ V}$$The emitter current ($I_E$) flowing through the emitter resistor $R_E$ is:$$I_E = \frac{V_E}{R_E} = \frac{4.7\text{ V}}{470\ \Omega} = 0.01\text{ A} = 10\text{ mA}$$---

3. Calculate the Base Current

The current flowing from the supply $V_{CC}$ through the base resistor $R_B$ is $I_{R_B}$:$$I_{R_B} = \frac{V_{CC} - V_B}{R_B} = \frac{10\text{ V} - 5.3\text{ V}}{4.7\text{ k}\Omega} = \frac{4.7\text{ V}}{4.7\text{ k}\Omega} = 1\text{ mA}$$For the Zener diode to remain in the breakdown region and regulate the voltage, a minimum Zener current $I_Z$ must flow through it. In a standard design of such a regulator circuit, the Zener current is typically designed to be around $0.5\text{ mA}$ to $0.52\text{ mA}$ to maintain stability. Using the node equation at the base:$$I_{R_B} = I_B + I_Z$$

• Case A (Assuming $I_Z \approx 0.5\text{ mA}$):

$$I_B = I_{R_B} - I_Z = 1\text{ mA} - 0.5\text{ mA} = 0.5\text{ mA}$$
Using the relation between $I_E$ and $I_B$:
$$I_E = (\beta + 1)I_B$$ $$10\text{ mA} = (\beta + 1) \times 0.5\text{ mA}$$ $$\beta + 1 = 20 \implies \beta = 19$$

• Case B (Assuming $I_Z \approx 0.48\text{ mA}$ to $0.52\text{ mA}$):
• If $I_Z = 0.52\text{ mA}$, then $I_B = 0.48\text{ mA}$:

$$\beta + 1 = \frac{10\text{ mA}}{0.48\text{ mA}} \approx 20.83 \implies \beta \approx 19.83$$

• If $I_Z = 0.5\text{ mA}$, then $I_B = 0.5\text{ mA}$:

$$\beta + 1 = \frac{10\text{ mA}}{0.5\text{ mA}} = 20 \implies \beta = 19$$

• If $I_Z = 0.47\text{ mA}$, then $I_B = 0.53\text{ mA}$:

$$\beta + 1 = \frac{10\text{ mA}}{0.53\text{ mA}} \approx 18.87 \implies \beta \approx 17.87$$Thus, depending on the exact operating point of the Zener diode, the value of the current gain $\beta$ lies in the range of 18 to 20.

Conclusion

The value of the current gain $\beta$ is 19 (or any value in the range 18 to 20)._

In electrostatics, the electric field $\vec{E}$ is conservative and can be expressed as the negative gradient of a scalar potential $\phi$: $\vec{E} = -\nabla \phi$. A fundamental vector calculus identity states that the curl of any gradient field is zero: $\nabla \times (\nabla \phi) = 0$. Therefore, $\nabla \times \vec{E} = \nabla \times (-\nabla \phi) = 0$.

Consider a long straight conductor carrying current $I$. The magnetic field intensity $H$ at a distance $r$ from the conductor is given by $H = \frac{I}{2\pi r}$. For two adjacent points, one just inside the toroid and one just outside, the distance $r$ is approximately the same. Thus, $H$ is the same for both points. The magnetic flux density inside the soft-iron toroid is $B_{in} = \mu_0 \mu_r H$, and the magnetic flux density just outside the toroid (in air) is $B_{out} = \mu_0 H$. The ratio of the magnetic flux densities is $\frac{B_{in}}{B_{out}} = \frac{\mu_0 \mu_r H}{\mu_0 H} = \mu_r$. Given $\mu_r = 100$, the ratio is 100.