GATE EC 2020 Set 1

The correct prepositional usage for accidents is 'killed in accidents'. For specific diseases, the standard idiom is 'die of a disease'. Therefore, the correct pair is 'in, of'.

The sentence uses the correlative conjunction 'not only', which must be paired with 'but also' to maintain grammatical parallelism and complete the meaning.

The relationship in the first pair is antonymous: 'Explicit' (clearly stated) is the opposite of 'Implicit' (implied). Similarly, 'Express' (to put into words or show feelings) is antonymous to 'Repress' (to restrain or prevent the expression of feelings). While 'Suppress' is similar, 'Repress' is the more direct psychological antonym in this context.

The passage states that the French-speaking couple filed complaints about "formal in-flight announcements in English lasting 15 seconds, as opposed to informal 5 second messages in French." This indicates that the English announcements were longer (15 seconds) compared to the French messages (5 seconds), which was the cause of their upset.Therefore, the correct option is (C).

A function $f(x)$ is superadditive if $f(x_1+x_2) \geq f(x_1) + f(x_2)$ for all $x_1, x_2$ in its domain.
Let's test each option for $x > 1$:(A) $f(x) = e^x$
We need to check if $e^{x_1+x_2} \geq e^{x_1} + e^{x_2}$ for $x_1, x_2 > 1$.
We can rewrite the inequality as $e^{x_1}e^{x_2} - e^{x_1} - e^{x_2} \geq 0$.
Adding 1 to both sides, we get $e^{x_1}e^{x_2} - e^{x_1} - e^{x_2} + 1 \geq 1$.
This can be factored as $(e^{x_1}-1)(e^{x_2}-1) \geq 1$.
Since $x_1 > 1$, $e^{x_1} > e \approx 2.718$, so $e^{x_1}-1 > 1.718$.
Similarly, since $x_2 > 1$, $e^{x_2}-1 > 1.718$.
Therefore, $(e^{x_1}-1)(e^{x_2}-1) > 1.718 \times 1.718 \approx 2.95 > 1$.
The inequality holds true. So, $f(x) = e^x$ is a superadditive function for $x > 1$.(B) $f(x) = \sqrt{x}$
We need to check if $\sqrt{x_1+x_2} \geq \sqrt{x_1} + \sqrt{x_2}$.
Squaring both sides (since both sides are positive for $x_1, x_2 > 1$):
$x_1+x_2 \geq (\sqrt{x_1} + \sqrt{x_2})^2$
$x_1+x_2 \geq x_1 + x_2 + 2\sqrt{x_1x_2}$
$0 \geq 2\sqrt{x_1x_2}$
This is false for $x_1, x_2 > 1$ (as $2\sqrt{x_1x_2}$ would be positive). Thus, $f(x) = \sqrt{x}$ is not superadditive; it is subadditive.(C) $f(x) = 1/x$
We need to check if $1/(x_1+x_2) \geq 1/x_1 + 1/x_2$.
$1/(x_1+x_2) \geq (x_2+x_1)/(x_1x_2)$
Since $x_1, x_2 > 1$, $x_1x_2$ and $x_1+x_2$ are positive. We can cross-multiply:
$x_1x_2 \geq (x_1+x_2)^2$
$x_1x_2 \geq x_1^2 + 2x_1x_2 + x_2^2$
$0 \geq x_1^2 + x_1x_2 + x_2^2$
This is false for $x_1, x_2 > 1$ (as $x_1^2 + x_1x_2 + x_2^2$ would be positive). Thus, $f(x) = 1/x$ is not superadditive; it is subadditive.(D) $f(x) = e^{-x}$
We need to check if $e^{-(x_1+x_2)} \geq e^{-x_1} + e^{-x_2}$.
This is equivalent to $1/e^{x_1+x_2} \geq 1/e^{x_1} + 1/e^{x_2}$.
Let $y_1 = e^{x_1}$ and $y_2 = e^{x_2}$. Since $x_1, x_2 > 1$, $y_1, y_2 > e > 1$.
The inequality becomes $1/(y_1y_2) \geq 1/y_1 + 1/y_2$.
This is the same form as $1/x$ and will also be false for $y_1, y_2 > 1$. Thus, $f(x) = e^{-x}$ is not superadditive; it is subadditive.The final answer is $\boxed{A}$

Let's break down the passage to identify the chronological order of events: This indicates that the East Asian crisis happened first and led to the excess liquidity that contributed to the sub-prime lending crisis. The sub-prime lending crisis followed the East Asian crisis. The banking crisis followed the sub-prime lending crisis. The global financial crisis is the overarching event that started with the sub-prime lending crisis and was closely linked to the banking crisis.Combining these points, the correct sequence of events is:
East Asian crisis $\rightarrow$ Subprime lending crisis $\rightarrow$ Banking crisis $\rightarrow$ Global financial crisis.Comparing this with the given options:
(A) East Asian crisis$\rightarrow$ subprime lending crisis $\rightarrow$ banking crisis$\rightarrow$ global financial crisis. - This matches our derived sequence.
(B) Subprime lending crisis$\rightarrow$ global financial crisis$\rightarrow$ banking crisis$\rightarrow$ East Asian crisis. - Incorrect order.
(C) Banking crisis$\rightarrow$ subprime lending crisis$\rightarrow$ global financial crisis $\rightarrow$ East Asian crisis. - Incorrect order.
(D) Global financial crisis $\rightarrow$ East Asian crisis $\rightarrow$ banking crisis $\rightarrow$ subprime lending crisis. - Incorrect order.The final answer is $\boxed{A}$

At quarter past three (3:15), the minute hand is exactly at 3 (15 minutes mark).
The hour hand moves $360^{\circ}$ in 12 hours, so it moves $30^{\circ}$ in 1 hour, or $0.5^{\circ}$ per minute.
At 3:00, the hour hand is at 3. In 15 minutes, it moves $15 \times 0.5^{\circ} = 7.5^{\circ}$ past 3.
So, at 3:15, the minute hand is at $3 \times 30^{\circ} = 90^{\circ}$ (from 12 o'clock position).
The hour hand is at $90^{\circ} + 7.5^{\circ} = 97.5^{\circ}$ (from 12 o'clock position).
The angle between the hour hand and the minute hand is $|97.5^{\circ} - 90^{\circ}| = 7.5^{\circ}$.Alternatively, using the formula for the angle between hands at H hours and M minutes:
Angle $= |30H - \frac{11}{2}M|$
For 3:15, H=3, M=15
Angle $= |30(3) - \frac{11}{2}(15)| = |90 - \frac{165}{2}| = |90 - 82.5| = 7.5^{\circ}$.Thus, the angle between the hour hand and the minute hand is $7.5^{\circ}$.

For a rectangle inscribed in a circle, the maximum possible area occurs when the rectangle is a square.
Let the side length of the square be $s$. The diagonal of the square is equal to the diameter of the circle.
Diameter $= 2a$.
By Pythagorean theorem for the square: $s^2 + s^2 = (2a)^2$
$2s^2 = 4a^2$
$s^2 = 2a^2$
So, the area of the square (rectangle of maximum area) is $A_{square} = s^2 = 2a^2$.The area of the circle is $A_{circle} = \pi r^2 = \pi a^2$.The shaded portion is the area of the circle minus the area of the inscribed rectangle (square).
Area of shaded portion $= A_{circle} - A_{square} = \pi a^2 - 2a^2$.Therefore, the correct option is (C).

For a quadratic equation of the form $Ax^2 + Bx + C = 0$, the roots are equal if the discriminant $D = B^2 - 4AC = 0$. In the given equation $ax^2 - bx + c = 0$:Now, let's evaluate the options:

• (A) $\beta = b/a$: Incorrect, as $\beta = b/2a$.
• (B) $\beta^2 = ac$: $\beta^2 = (b/2a)^2 = b^2/4a^2$. Substituting $b^2 = 4ac$, we get $\beta^2 = 4ac/4a^2 = c/a$. Thus, $\beta^2 = ac$ is incorrect.
• (C) $\beta^3 = bc/(2a^2)$:
• LHS: $\beta^3 = (b/2a)^3 = \frac{b^3}{8a^3}$.
• RHS: $\frac{bc}{2a^2} = \frac{b \cdot (b^2/4a)}{2a^2} = \frac{b^3/4a}{2a^2} = \frac{b^3}{8a^3}$.
• Since LHS = RHS, this option is correct.
• (D) $b^2 \neq 4ac$: Incorrect, as $b^2 = 4ac$ is the condition for equal roots.

Step 1: Calculate the total number of students enrolled in school $P$ over the 5-year period from the bar chart.
Sum $P = 3 + 5 + 5 + 6 + 4 = 23$ (in thousands).
Average $P = \frac{23}{5}$.Step 2: Calculate the absolute difference in enrollment between school $P$ and school $Q$ for each year.

• 2014: $|3 - 4| = 1$
• 2015: $|5 - 7| = 2$
• 2016: $|5 - 8| = 3$
• 2017: $|6 - 7| = 1$
• 2018: $|4 - 5| = 1$

Step 3: Calculate the average of these differences.
Sum of differences = $1 + 2 + 3 + 1 + 1 = 8$.
Average of difference = $\frac{8}{5}$.Step 4: Find the ratio of the average number of students in school $P$ to the average of the difference.
Ratio = $\frac{23/5}{8/5} = \frac{23}{8} = 23 : 8$.Therefore, the correct option is (B).

In $\mathbb{R}^4$, the dimension of the space is 4.
(A) True: A set of vectors spans a space only if it contains at least as many linearly independent vectors as the dimension of the space. Six vectors might be linearly dependent and fail to span $\mathbb{R}^4$.
(B) True: In any $n$-dimensional space, any set of more than $n$ vectors is linearly dependent. Since $6 > 4$, these vectors must be linearly dependent.
(C) False: For a set of 4 vectors to form a basis for $\mathbb{R}^4$, they must be linearly independent. Not any arbitrary subset of four vectors from the given six will necessarily be linearly independent.
(D) True: In an $n$-dimensional space, any set of $n$ vectors that spans the space is automatically a basis.

(A) True: By definition, a vector field is solenoidal if its divergence is zero.
(B) True: The curl of a vector field is itself a vector field.
(C) False: A vector field is irrotational if its curl is zero ($\nabla \times \vec{A} = 0$). The condition $\nabla^2 \vec{A} = 0$ is the vector Laplace equation.
(D) True: This is a standard vector identity relating the curl of a curl to the gradient of the divergence and the Laplacian.

$$\frac{\partial f}{\partial x} = -\cos y\, e^{1-x\cos y} + z\,e^{-1/(1+y^2)}.$$At $(1,0,e)$: $\cos 0 = 1$, so the first term is $-e^{0} = -1$, and the second is $e\cdot e^{-1} = 1$.$$\left.\frac{\partial f}{\partial x}\right|_{(1,0,e)} = -1+1 = 0.$$Answer: (B) $0$

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients:$$\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0$$The characteristic equation is found by replacing $\frac{d^2y}{dx^2}$ with $m^2$, $\frac{dy}{dx}$ with $m$, and $y$ with $1$:$$m^2 - 6m + 9 = 0$$This is a perfect square trinomial:$$(m - 3)^2 = 0$$So, we have repeated roots: $m_1 = m_2 = 3$.
For repeated real roots $m$, the general solution is of the form:$$y(x) = (C_1 + C_2x)e^{mx}$$Substituting $m=3$:$$y(x) = (C_1 + C_2x)e^{3x}$$Comparing this with the given options, option (C) matches.Therefore, the correct answer is (C).

Set $x[n]=\delta[n]$. Then $h[n] = \max_{k\le n}|\delta[k]|$.

• For $n<0$: every $|\delta[k]|$ with $k\le n<0$ is $0$, so $h[n]=0$.
• For $n\ge 0$: the range $k\le n$ includes $k=0$, where $|\delta[0]|=1$; all other terms are $0$, so $h[n]=1$.

Thus $h[n]=u[n]$, the unit step. (It is not 0 or 1 for all $n$, and it is not $\delta[n]$ because the maximum 'remembers' the past value 1.)Answer: (C) unit step signal $u[n]$

For an intrinsic semiconductor, the intrinsic Fermi level $E_i$ is given by:$$E_i = \frac{E_c + E_v}{2} + \frac{kT}{2} \ln\left(\frac{N_v}{N_c}\right)$$Where:
$E_c$ is the conduction band edge.
$E_v$ is the valence band edge.
$kT$ is the thermal energy.
$N_v$ is the effective density of states in the valence band.
$N_c$ is the effective density of states in the conduction band.The mid-bandgap energy level is $E_{mid} = \frac{E_c + E_v}{2}$.The shift of the intrinsic Fermi level from the mid-bandgap energy level is:$$\Delta E_i = E_i - E_{mid} = \frac{kT}{2} \ln\left(\frac{N_v}{N_c}\right)$$Given:
Temperature $T = 300$ K.
Thermal voltage $V_T = \frac{kT}{q} = 26$ mV $= 0.026$ V.
So, $kT = qV_T = 1.6 \times 10^{-19} \text{ C} \times 0.026 \text{ V} = 0.026$ eV.
Effective density of states for holes ($N_v$) is twice that of electrons ($N_c$), so $N_v = 2N_c$.Substitute these values into the equation for $\Delta E_i$:$$\Delta E_i = \frac{kT}{2} \ln\left(\frac{2N_c}{N_c}\right)$$$$\Delta E_i = \frac{kT}{2} \ln(2)$$Using $kT = 0.026$ eV:$$\Delta E_i = \frac{0.026 \text{ eV}}{2} \ln(2)$$$$\Delta E_i = 0.013 \text{ eV} \times 0.6931$$ (since $\ln(2) \approx 0.6931$)$$\Delta E_i \approx 0.0090103 \text{ eV}$$Converting to meV:$$\Delta E_i \approx 9.01 \text{ meV}$$Therefore, the intrinsic Fermi level is shifted from the mid-bandgap energy level by approximately 9.01 meV.The correct answer is (B).

In a forward-biased $pn$ junction, the Shockley-Read-Hall (SRH) recombination rate is given by $U = \frac{np - n_i^2}{\tau_p(n + n_1) + \tau_n(p + p_1)}$. For equal capture cross-sections and assuming $n_1, p_1 \approx n_i$, the maximum recombination rate $U_{max}$ occurs where $n \approx p \approx n_i e^{V_a/2V_T}$. This condition is typically satisfied near the center of the depletion region, not at the edges. Therefore, statement (B) is false.

When an op-amp is operated in a positive feedback configuration, it acts as a comparator with hysteresis or simply saturates to its supply rails. Because the feedback is positive, any small difference at the input is amplified until the output reaches one of the saturation limits ($+V_{sat}$ or $-V_{sat}$). Given the supply voltages are $\pm 5$ V, the output $V_o$ will be a constant value of either $+5$ V or $-5$ V depending on the initial state and input conditions.

The output is open, so no current flows in the $4\,\Omega$ resistor and $V_{TH}$ equals the voltage $V_2$ at the right-hand node (bottom wire = reference).Let $V_1$ be the left node (1 A source, $1\,\Omega$). The 2 V source has its $+$ terminal on the right, so the node between the source and the $2\,\Omega$ series resistor is at $V_1+2$.KCL at $V_1$: $1 = \dfrac{V_1}{1} + \dfrac{(V_1+2)-V_2}{2} \Rightarrow 3V_1 - V_2 = 0$.KCL at $V_2$: $\dfrac{(V_1+2)-V_2}{2} + 2 = \dfrac{V_2}{2} \Rightarrow V_1 - 2V_2 = -6$.From the first, $V_1 = V_2/3$; substituting, $V_2/3 - 2V_2 = -6 \Rightarrow V_2 = 3.6$ V.Answer: (C) $3.6$ V

From the figure: EN is active-low and tied to 0, so the MUX is enabled. The data inputs are $I_0 = R$, $I_1 = 0$, $I_2 = R$, $I_3 = 1$, with $S_1 = P$, $S_0 = Q$.$$F = \overline{P}\,\overline{Q}\,I_0 + \overline{P}Q\,I_1 + P\overline{Q}\,I_2 + PQ\,I_3 = \overline{P}\,\overline{Q}R + P\overline{Q}R + PQ.$$Combining the first two terms: $\overline{Q}R(\overline{P}+P) = \overline{Q}R$, so$$F = PQ + \overline{Q}R.$$Check the others: (B) gives $F=1$ for $P=1,Q=0,R=0$, but the MUX outputs $I_2=R=0$; (C) gives $F=1$ for $P=0,Q=1$, but the MUX outputs $I_1=0$; (D) gives $F=1$ for $P=Q=R=0$, but the MUX outputs $I_0=R=0$.Answer: (A) $PQ + \overline{Q}R$

By the argument principle, as $s$ traverses a closed contour $\Gamma$, the image $G(s)$ encircles the origin $N = Z - P$ times in the same sense as $\Gamma$, where $Z$ and $P$ are the zeros and poles of $G(s)$ inside $\Gamma$. (Encirclements of $-1+j0$ would concern $1+G(s)$, whose zeros are not shown, so options (C) and (D) cannot be decided from this map.)From the figure:

• Inside $\Gamma$: three zeros (one on the negative real axis just inside the contour and a complex-conjugate pair) and two poles (on the real axis on either side of the origin).
• Outside $\Gamma$: one zero on the negative real axis, a real pole on the positive axis and a complex pole pair.

So $Z = 3$, $P = 2$ and $N = Z - P = 1$.The arrow on $\Gamma$ points upward on its left side, i.e. $\Gamma$ is traversed clockwise. Hence $G(\Gamma)$ encircles the origin once in the clockwise direction.Answer: (B)

Let $P_e$ be the probability of error in decoding a single bit. Given $P_e = \alpha$.
The probability of a bit being decoded correctly is $P_c = 1 - P_e = 1 - \alpha$.The system transmits a block of $N$ bits. The error events for each bit are independent.The received block is declared erroneous if at least one of its bits is decoded wrongly.
It is easier to calculate the probability of the complementary event: the block is not erroneous.The block is not erroneous if and only if all $N$ bits are decoded correctly.
Since the error events are independent, the probability that all $N$ bits are decoded correctly is the product of the probabilities of each bit being correct:
$P(\text{all bits correct}) = (P_c)^N = (1 - \alpha)^N$The probability that the received block is erroneous is $1$ minus the probability that all bits are correct:
$P(\text{block erroneous}) = 1 - P(\text{all bits correct})$
$P(\text{block erroneous}) = 1 - (1 - \alpha)^N$This matches option (D).The final answer is \boxed{\text{1 - (1-\alpha)^N}}

The reflection coefficient on a Smith chart is given by $\Gamma = \frac{Z - Z_0}{Z + Z_0}$, where $Z_0$ is the characteristic impedance. For normalized impedance $z = \frac{Z}{Z_0} = jx$ (where $x = X/Z_0$), the reflection coefficient is $\Gamma = \frac{jx - 1}{jx + 1}$. The magnitude of the reflection coefficient is $|\Gamma| = \left| \frac{jx - 1}{jx + 1} \right| = \frac{\sqrt{x^2 + (-1)^2}}{\sqrt{x^2 + 1^2}} = 1$. On the Smith chart, $|\Gamma| = 1$ represents a circle of radius 1 centered at the origin (0, 0).