GATE EC 2018 Set 1

The sentence requires two homophones: 'piece' and 'peace'. 'Piece' refers to a portion or part of something (e.g., a piece of cake), while 'peace' refers to a state of tranquility or harmony. The correct combination is 'piece' for the first blank and 'peace' for the second blank. Thus, option (B) is correct.

The phrase "Even though" indicates a contrast between the two blanks. "Vast scope for improvement" suggests that the tourism sector has significant potential to get better, while "neglected area" indicates that it is currently not receiving the attention it deserves. This pair provides the most logical contrast. Other options do not fit the context or the required contrast.

A number is divisible by $3$ if the sum of its digits is divisible by $3$.
For the number $715\blacksquare423$, the sum of the digits is:
$7 + 1 + 5 + \blacksquare + 4 + 2 + 3 = 22 + \blacksquare$
For $22 + \blacksquare$ to be divisible by $3$, the possible values for $\blacksquare$ (where $0 \le \blacksquare \le 9$) are:

• If $22 + \blacksquare = 24$, then $\blacksquare = 2$
• If $22 + \blacksquare = 27$, then $\blacksquare = 5$
• If $22 + \blacksquare = 30$, then $\blacksquare = 8$

The smallest whole number among these is $2$. Note: The term "thousandths place" in the question is a typo for "thousands place" as the number is an integer.

The given expression is an infinite geometric series: $1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \dots$
Here, the first term $a = 1$ and the common ratio $r = \frac{1/4}{1} = \frac{1}{4}$.
Since $|r| < 1$, the sum of an infinite geometric series is given by:$$S = \frac{a}{1 - r}$$$$S = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$Therefore, the correct option is (D).

Let $H$ be the height of the lamp post and $h$ be the height of the person.
Given:
Height of person, $h = 1.5$ m
Distance of person from lamp post, $d = 3$ m
Length of shadow, $s = 2 \times h = 2 \times 1.5 = 3$ mUsing the property of similar triangles formed by the lamp post and the person with their respective shadows:$$\frac{H}{d + s} = \frac{h}{s}$$$$\frac{H}{3 + 3} = \frac{1.5}{3}$$$$\frac{H}{6} = 0.5$$$$H = 6 \times 0.5 = 3 \text{ m}$$Thus, the height of the lamp post is 3 meters. The correct option is (B).

The formula for compound interest is $A = P(1 + \frac{r}{100})^n$, where:
$A$ = Future value (Rs. 10,00,000)
$P$ = Principal amount (amount to be deposited now)
$r$ = Annual interest rate (10%)
$n$ = Number of years (5)Substituting the values:$$10,00,000 = P(1 + \frac{10}{100})^5$$$$10,00,000 = P(1.1)^5$$$$10,00,000 = P(1.61051)$$$$P = \frac{10,00,000}{1.61051} \approx 6,20,921.32$$Among the given options, the closest minimum amount is Rs. 6,21,000. Therefore, the correct option is (B).

Let the equal mass of alloys A and B taken be $m$ kg.Alloy A:
Ratio of Gold : Copper = 2:3
Mass of Gold in alloy A = $\frac{2}{2+3} \times m = \frac{2}{5}m$
Mass of Copper in alloy A = $\frac{3}{2+3} \times m = \frac{3}{5}m$Alloy B:
Ratio of Gold : Copper = 3:7
Mass of Gold in alloy B = $\frac{3}{3+7} \times m = \frac{3}{10}m$
Mass of Copper in alloy B = $\frac{7}{3+7} \times m = \frac{7}{10}m$Alloy C:
Alloy C is formed by melting equal masses of A and B. The total mass of alloy C is $m+m=2m$.
Total mass of Gold in alloy C = Mass of Gold in A + Mass of Gold in B
$= \frac{2}{5}m + \frac{3}{10}m = \frac{4m + 3m}{10} = \frac{7m}{10}$Total mass of Copper in alloy C = Mass of Copper in A + Mass of Copper in B
$= \frac{3}{5}m + \frac{7}{10}m = \frac{6m + 7m}{10} = \frac{13m}{10}$The ratio of gold to copper in alloy C is:
$\frac{\text{Total mass of Gold}}{\text{Total mass of Copper}} = \frac{7m/10}{13m/10} = \frac{7}{13}$So, the ratio is 7:13.

Let's analyze each statement based on the provided paragraph:(i) Even as a junior cricketer, John was considered a good captain. The paragraph states the Board recognized his "potential as a leader" but also that his "on-field temper has always been a matter of concern... since his junior days." This concern suggests he was not considered a complete or good captain at that time. So, this statement cannot be inferred.(ii) Finding a complete captain is a challenge. The last sentence, "After many years, it appears that the team has finally found a complete captain," strongly implies that the search was long and difficult. This makes the inference that finding one is a challenge a logical conclusion. So, this statement can be inferred.(iii) Fans and the Cricket Board have differing views on what they want in a captain. The paragraph says John's aggression "has filled stadia with die-hard fans," while for the Board, his temper was a "matter of concern." This contrast indicates differing perspectives on his aggressive behavior. However, the core desire for a successful captain might be the same. The inference is plausible but might be considered less direct than (ii) and (iv). The question is about what they want in a captain, and the text only provides evidence on the single trait of aggression.(iv) Over the past three seasons John has accumulated big scores. The paragraph states that "Until recently, it appeared that he found it difficult to convert his aggression into big scores." It then contrasts this with the last three seasons, where he has become a "successful batsman-captain." A successful batsman, by definition, scores well and likely accumulates big scores. This is a strong inference. So, this statement can be inferred.Based on the analysis, statements (ii) and (iv) are the most strongly and directly supported inferences from the text. Statement (i) is contradicted, and statement (iii) is plausible but less certain than (ii) and (iv). Therefore, the best option is the one that includes (ii) and (iv).

This is a problem of conditional probability that can be solved using Bayes' theorem.
Let's define the events:

• $B$: The event that the cab involved in the accident is Blue.
• $G$: The event that the cab involved in the accident is Green.
• $W_B$: The event that the witness identifies the cab as Blue.

From the problem statement, we can extract the following probabilities:

• $P(G) = 0.85$ (85% of cabs are green)
• $P(B) = 1 - P(G) = 1 - 0.85 = 0.15$ (15% of cabs are blue)

2. Conditional probabilities of the witness's testimony:

• The witness is correct 80% of the time. This means:
• $P(W_B | B) = 0.80$ (Probability the witness says Blue, given the cab was Blue)
• $P(W_G | G) = 0.80$ (Probability the witness says Green, given the cab was Green)
• The witness is incorrect 20% of the time. This means:
• $P(W_B | G) = 1 - P(W_G | G) = 1 - 0.80 = 0.20$ (Probability the witness says Blue, given the cab was Green)

We want to find the probability that the cab was actually Blue, given that the witness identified it as Blue. This is $P(B | W_B)$.Using Bayes' theorem:$$P(B | W_B) = \frac{P(W_B | B) \times P(B)}{P(W_B)}$$First, we need to calculate the total probability of the witness identifying the cab as Blue, $P(W_B)$, using the law of total probability:$$P(W_B) = P(W_B | B) \times P(B) + P(W_B | G) \times P(G)$$Plugging in the values:$$P(W_B) = (0.80 \times 0.15) + (0.20 \times 0.85)$$$$P(W_B) = 0.12 + 0.17$$$$P(W_B) = 0.29$$Now, we can calculate $P(B | W_B)$:$$P(B | W_B) = \frac{P(W_B | B) \times P(B)}{P(W_B)} = \frac{0.12}{0.29}$$$$P(B | W_B) \approx 0.41379$$Converting this to a percentage, we get approximately 41.38%.
The closest option is 41%.

The passage states that the region has remained inaccessible to tourists primarily due to 'poor connectivity'. The company's plan to introduce 'helicopter service' is a direct measure to improve connectivity. Therefore, it can be logically inferred that this improved connectivity will lead to an increase in tourists.

• Option (A) is not mentioned or implied in the text.
• Option (B) is a concern or prediction expressed by environmentalists, not a factual inference from the premises.
• Option (C) is not supported by any information in the text regarding the current state of the reefs.
• Option (D) is the most logical inference as it addresses the causal link between the barrier (poor connectivity) and the proposed solution (helicopter service).

In a cascode configuration, the overall transconductance $g_m$ is determined by the input transistor $M_1$, as the same current flows through both transistors: $g_m = \frac{\partial i_D}{\partial v_{GS1}} = g_{m1}$. The output resistance $r_o$ looking into the drain of $M_2$ is boosted by the intrinsic gain of $M_2$: $r_o = r_{o2} + r_{o1} + g_{m2}r_{o2}r_{o1} \approx g_{m2}r_{o2}r_{o1}$. Thus, option (C) is correct.

A transimpedance amplifier (TIA) is a current-to-voltage converter. To accurately sense the input current without loading the source, it must have a very low input impedance ($R_{in} \to 0$). To provide the output voltage to a subsequent stage without signal loss, it must have a very low output impedance ($R_{out} \to 0$).

A system is linear if it satisfies the principles of superposition and homogeneity. Thus, option (C) represents a non-linear (specifically, an affine) system.

Let's evaluate the statements:
(A) True: Both Routh-Hurwitz and Nyquist criteria can be used to find the range of gain $K$ for which a system remains stable.
(B) True: For minimum-phase systems, there is a unique relationship between magnitude and phase. Thus, the Nyquist plot (polar plot) can be inferred from the Bode magnitude plot.
(C) True: The Routh criterion is designed for characteristic equations that are polynomials. Transport lag ($e^{-sT}$) makes the equation transcendental, which Routh cannot handle directly. Nyquist, being a frequency-domain graphical method, handles it easily.
(D) False: The closed-loop frequency response (magnitude and phase) can be obtained from the open-loop frequency response (Nyquist plot) using constant M-circles and constant N-circles.
Therefore, statement (D) is false.

Given $p(s) = s^3 + a_2 s^2 + a_1 s + a_0$.
The derivative is $p'(s) = 3s^2 + 2a_2 s + a_1$.
$p'(s)$ is a quadratic polynomial. If it has no real roots, its discriminant must be negative ($D < 0$). Since the leading coefficient (3) is positive, $p'(s)$ must be strictly positive for all real values of $s$ ($p'(s) > 0, \forall s \in \mathbb{R}$).
If the derivative of a function is always positive, the function is strictly increasing.
A strictly increasing cubic polynomial with real coefficients must cross the x-axis exactly once.
Therefore, $p(s)$ has exactly one real root.

In a p-n junction diode at equilibrium, the net current for each carrier type (electrons and holes) must be zero. This means that for each carrier, the drift current and diffusion current must be equal in magnitude and opposite in direction.For electrons: $J_{n,total} = J_{n,drift} + J_{n,diff} = 0 \implies J_{n,drift} = -J_{n,diff}$.
Thus, on average, electrons drift and diffuse in opposite directions. Statement (D) claims they are in the same direction, which is NOT TRUE.Regarding other statements:
(A) Hole diffusion is P → N, electron diffusion is N → P (electron current is P → N). Both are in the same direction. (TRUE)
(B) Hole drift is N → P, electron drift is P → N (electron current is N → P). Both are in the same direction. (TRUE)
(C) Holes drift N → P and electrons drift P → N, which are opposite directions. (TRUE)

The given circuit is a static CMOS gate. Let's analyze the pull-up (PMOS) and pull-down (NMOS) networks.Pull-up Network (PMOS):
It consists of two parallel branches:

• Branch 1: PMOS with gate $\bar{X}$ and PMOS with gate $Y$ in series. It conducts when $\bar{X}=0$ and $Y=0$, i.e., $X=1$ and $Y=0$.
• Branch 2: PMOS with gate $X$ and PMOS with gate $\bar{Y}$ in series. It conducts when $X=0$ and $\bar{Y}=0$, i.e., $X=0$ and $Y=1$.

The output is pulled to $V_{DD}$ (logic 1) when $(X=1, Y=0)$ or $(X=0, Y=1)$, which is $X \oplus Y$.Pull-down Network (NMOS):
It consists of two parallel branches:

• Branch 1: NMOS with gate $\bar{X}$ and NMOS with gate $\bar{Y}$ in series. It conducts when $\bar{X}=1$ and $\bar{Y}=1$, i.e., $X=0$ and $Y=0$.
• Branch 2: NMOS with gate $X$ and NMOS with gate $Y$ in series. It conducts when $X=1$ and $Y=1$.

The output is pulled to ground (logic 0) when $(X=0, Y=0)$ or $(X=1, Y=1)$, which is $X \odot Y$.Since the output is 1 for XOR inputs and 0 for XNOR inputs, the logic function is $f(X, Y) = X \oplus Y$ (XOR).

The given sum of products (SOP) expression is:
$F = \bar{A}\bar{B}\bar{C} + \bar{A}B\bar{C} + A\bar{B}\bar{C}$In minterm notation, this corresponds to:
$F = m_0 + m_2 + m_4 = \sum m(0, 2, 4)$The product of sums (POS) form consists of the maxterms whose indices are not present in the SOP form. For a 3-variable function, the indices are $0, 1, \dots, 7$. The missing indices are $1, 3, 5, 6, 7$.Thus, $F = \prod M(1, 3, 5, 6, 7)$.Converting these indices to maxterm expressions:

• $M_1 (001) = (A + B + \bar{C})$
• $M_3 (011) = (A + \bar{B} + \bar{C})$
• $M_5 (101) = (\bar{A} + B + \bar{C})$
• $M_6 (110) = (\bar{A} + \bar{B} + C)$
• $M_7 (111) = (\bar{A} + \bar{B} + \bar{C})$

Therefore, $F = (A + B + \bar{C}) \cdot (A + \bar{B} + \bar{C}) \cdot (\bar{A} + B + \bar{C}) \cdot (\bar{A} + \bar{B} + C) \cdot (\bar{A} + \bar{B} + \bar{C})$. This matches option (C).

On a standard Smith chart (normalized impedance chart):Therefore, P: Short Circuit, Q: Matched Load, R: Open Circuit. This corresponds to option (C).

Let's evaluate the implications:

• S2 implies S1: A fundamental theorem in linear algebra states that if an $n \times n$ matrix has $n$ distinct eigenvalues, it is guaranteed to have $n$ linearly independent eigenvectors (and is thus diagonalizable). For a $4 \times 4$ matrix, 4 distinct eigenvalues $\implies$ 4 LI eigenvectors. So, S2 $\implies$ S1 is TRUE.
• S1 implies S2: A matrix can have 4 LI eigenvectors even if eigenvalues are repeated (e.g., the Identity matrix $I_{4 \times 4}$ has 4 LI eigenvectors but only one distinct eigenvalue, 1). So, S1 $\implies$ S2 is FALSE.
• S1 implies S3: A matrix can be diagonalizable (have 4 LI eigenvectors) and still be singular (e.g., a zero matrix has 4 LI eigenvectors but is singular). So, S1 $\implies$ S3 is FALSE.
• S3 implies S2: An invertible matrix can have repeated eigenvalues (e.g., $I_{4 \times 4}$ is invertible but has repeated eigenvalues). So, S3 $\implies$ S2 is FALSE.

Thus, the only true statement is (C).

Given $f(x, y) = \frac{ax^2 + by^2}{xy} = \frac{ax}{y} + \frac{by}{x}$.First, find the partial derivatives:
$\frac{\partial f}{\partial x} = \frac{a}{y} - \frac{by}{x^2}$
$\frac{\partial f}{\partial y} = -\frac{ax}{y^2} + \frac{b}{x}$Now, evaluate them at $x = 1$ and $y = 2$:
$\left. \frac{\partial f}{\partial x} \right|_{(1,2)} = \frac{a}{2} - \frac{b(2)}{1^2} = \frac{a}{2} - 2b$
$\left. \frac{\partial f}{\partial y} \right|_{(1,2)} = -\frac{a(1)}{2^2} + \frac{b}{1} = -\frac{a}{4} + b$Given $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$ at $(1,2)$:
$\frac{a}{2} - 2b = -\frac{a}{4} + b$
Multiply by 4 to clear denominators:
$2a - 8b = -a + 4b$
$3a = 12b$
$a = 4b$Thus, the correct relation is $a = 4b$, which is option (D).

An all-pass system is defined by having a constant magnitude response $|H(e^{j\omega})| = 1$ for all frequencies $\omega$. The given poles are $p_1 = 0.25 \angle 0^\circ$ (inside the unit circle, $|p_1| = 0.25$) and $p_2 = 2 \angle 30^\circ$ (outside the unit circle, $|p_2| = 2$).
For the ROC to include $|z| = 1$, it must be the ring $0.25 < |z| < 2$.
An ROC that is a ring (bounded by poles on both sides) corresponds to a non-causal, two-sided impulse response.
Therefore, the system is stable only if the impulse response is two-sided.

Given $x(t)$ is a periodic function with period $T = 10$. Its Fourier series representation is:$$x(t) = \sum_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t}, \text{ where } \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{10}$$When $x(t)$ is considered as a periodic function with period $T' = 40$, its Fourier series representation is:$$x(t) = \sum_{n=-\infty}^{\infty} b_n e^{j n \omega_0' t}, \text{ where } \omega_0' = \frac{2\pi}{T'} = \frac{2\pi}{40} = \frac{\omega_0}{4}$$Comparing the two series for the same function $x(t)$:$$\sum_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t} = \sum_{n=-\infty}^{\infty} b_n e^{j n \frac{\omega_0}{4} t}$$For the terms to match, the frequencies must be identical. This occurs when $k \omega_0 = n \frac{\omega_0}{4}$, which implies $n = 4k$. Thus, the coefficients are related as follows:

• $b_n = a_k$ when $n = 4k$
• $b_n = 0$ when $n$ is not a multiple of 4

Now, we evaluate the sum of the absolute values of the coefficients for the second series:$$\sum_{n=-\infty}^{\infty} |b_n| = \sum_{k=-\infty}^{\infty} |b_{4k}| = \sum_{k=-\infty}^{\infty} |a_k|$$Given that $\sum_{k=-\infty}^{\infty} |a_k| = 16$, it follows that:$$\sum_{n=-\infty}^{\infty} |b_n| = 16$$Therefore, the correct option is (C).

The given amplitude modulated signal is:$$s(t) = \cos(2000 \pi t) + 4 \cos(2400 \pi t) + \cos(2800 \pi t)$$Rearranging the terms to identify the carrier component:$$s(t) = 4 \cos(2400 \pi t) + [\cos(2800 \pi t) + \cos(2000 \pi t)]$$Using the trigonometric identity $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$:$$\cos(2800 \pi t) + \cos(2000 \pi t) = 2 \cos(2400 \pi t) \cos(400 \pi t)$$Substituting this back into the expression for $s(t)$:$$s(t) = 4 \cos(2400 \pi t) + 2 \cos(400 \pi t) \cos(2400 \pi t)$$$$s(t) = 4 [1 + 0.5 \cos(400 \pi t)] \cos(2400 \pi t)$$This is a standard AM signal of the form $s(t) = A_c [1 + \mu \cos(\omega_m t)] \cos(\omega_c t)$, where:

• Carrier amplitude $A_c = 4$
• Modulation index $\mu = 0.5$

The power of the carrier signal is:$$P_c = \frac{A_c^2}{2} = \frac{4^2}{2} = 8 \text{ W}$$The total power in the sidebands (which represents the power contributed by the message signal in the modulated wave) is:$$P_{sb} = P_c \frac{\mu^2}{2} = 8 \times \frac{0.5^2}{2} = 8 \times 0.125 = 1 \text{ W}$$Alternatively, the sideband power can be calculated as the sum of the powers of the individual sideband components:$$P_{sb} = \text{Power of } \cos(2000 \pi t) + \text{Power of } \cos(2800 \pi t) = \frac{1^2}{2} + \frac{1^2}{2} = 0.5 + 0.5 = 1 \text{ W}$$The ratio of the power of the message signal (sideband power) to the power of the carrier signal is:$$\text{Ratio} = \frac{P_{sb}}{P_c} = \frac{1}{8} = 0.125$$