GATE EC 2023 Set 1

The sentence requires a noun that refers to a person's moral sense of right and wrong.

• Conscience (noun) means an inner feeling or voice viewed as acting as a guide to the rightness or wrongness of one's behavior.
• Conscious (adjective) means being aware of and responding to one's surroundings.
• Consensus (noun) means a general agreement.
• Consent (noun/verb) means permission for something to happen or agreement to do something.

Therefore, "conscience" is the most appropriate fit.

This is an analogy based on the branches of government and the institutions that represent them.

• Parliament is the primary institution of the Legislature (the law-making branch).
• Similarly, Courts are the primary institutions of the Judiciary (the branch that interprets laws).

Thus, the correct pair is Courts : Judiciary.

To find the smallest number with distinct digits whose digits add up to 45, we first determine the minimum number of digits required. The sum of the largest distinct digits $9+8+7+6+5+4+3+2+1 = 45$. This uses 9 digits. Any number with fewer than 9 digits cannot have a sum of 45 without repeating digits (e.g., $9 \times 5 = 45$ but digits are not distinct). Thus, the smallest number must have 9 digits. To minimize a 9-digit number, we place the smallest digits in the highest place values. Using the digits $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, the smallest number is $123,456,789$.Checking the options:
(A) $123555789$: Digits are not distinct ($5$ is repeated).
(B) $123457869$: Digits are distinct and sum to 45, but it is larger than $123456789$.
(C) $123456789$: Digits are distinct, sum to 45, and it is the smallest possible 9-digit number.
(D) $99999$: Digits are not distinct.Therefore, the correct option is (C).

Let $R$ be the set of students who like romantic movies and $C$ be the set of students who like comedy movies.
Total students $N = 100$.From (i), the number of students who like neither romantic nor comedy movies is 30. Therefore, the number of students who like at least one of the two is:$$|R \cup C| = 100 - 30 = 70$$From (ii), the number of students who like romantic movies is twice the number of students who like comedy movies:$$|R| = 2|C|$$From (iii), the number of students who like both is 20:$$|R \cap C| = 20$$Using the principle of inclusion-exclusion:$$|R \cup C| = |R| + |C| - |R \cap C|$$$$70 = 2|C| + |C| - 20$$$$70 + 20 = 3|C|$$$$90 = 3|C| \implies |C| = 30$$Now, find the number of students who like romantic movies:$$|R| = 2|C| = 2 \times 30 = 60$$Thus, 60 students like romantic movies.

A rectangle is a quadrilateral with four right angles. Note that every square is also a rectangle.Let's count the rectangles in the figure:Total number of rectangles = $1 + 4 + 2 + 2 + 1 = 10$.The diagonals divide the squares into triangles, but they do not form any additional rectangles.Therefore, the correct option is (C).

Let $A$ be the set of animals, $F$ be the set of ferocious creatures, and $C$ be the set of creatures with claws.
From the passage:Evaluating the options:
(A) $C \subseteq A$: Not necessarily true. Both $C$ and $A$ are subsets of $F$, but we don't know if $C$ is a subset of $A$.
(B) Some $C$ are not $F$: This contradicts the statement $C \subseteq F$.
(C) Some not $F$ are $C$: This also contradicts $C \subseteq F$.
(D) Some $F$ are $C$: Since $C \subseteq F$ and $C$ is non-empty, there must be elements in $F$ that are also in $C$. This is a valid logical inference.

The given graph is symmetric about the y-axis, which implies that the function must be an even function, i.e., $f(x) = f(-x)$.Since only option (A) represents an even function that matches the symmetric nature of the graph, (A) is the correct choice.

The passage begins with the phrase "When I was a kid...", which indicates that the narrator is currently an adult reflecting on their past. Therefore, it is an adult's memory of their childhood preferences. Option (A) is incorrect because the narrator is no longer a child. Option (C) is a misinterpretation of the phrase "partial to", which means "having a liking for", not "reading in parts". Option (D) is an external value judgment not supported by the text alone.

Thus, no more than 700 testing kits will be required.

When a sheet of paper is folded $n$ times, the number of creases formed is $2^n - 1$.
In this problem, the rectangular sheet is folded 5 times.
Therefore, the total number of creases visible when the sheet is unfolded will be $2^5 - 1 = 32 - 1 = 31$.Thus, option (C) is the correct answer.

To minimize the length of the error vector $\mathbf{e} = \mathbf{v}_1 - \alpha \mathbf{v}_2$, the error vector must be orthogonal to the vector $\mathbf{v}_2$ (the projection theorem).The condition for orthogonality is:$$\mathbf{e} \cdot \mathbf{v}_2 = 0$$$$(\mathbf{v}_1 - \alpha \mathbf{v}_2) \cdot \mathbf{v}_2 = 0$$$$\mathbf{v}_1 \cdot \mathbf{v}_2 - \alpha (\mathbf{v}_2 \cdot \mathbf{v}_2) = 0$$$$\alpha = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2}$$Given $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$:$$\mathbf{v}_1 \cdot \mathbf{v}_2 = (1)(2) + (2)(1) + (0)(3) = 2 + 2 + 0 = 4$$$$\mathbf{v}_2 \cdot \mathbf{v}_2 = (2)^2 + (1)^2 + (3)^2 = 4 + 1 + 9 = 14$$Substituting these values into the expression for $\alpha$:$$\alpha = \frac{4}{14} = \frac{2}{7}$$Therefore, the value of $\alpha$ that minimizes the length of the error vector is $\frac{2}{7}$.

The rate of increase of a scalar field $f(x, y, z)$ in the direction of a vector $\mathbf{v}$ is given by the directional derivative $D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}$, where $\mathbf{u}$ is the unit vector in the direction of $\mathbf{v}$.$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (yz, xz, xy)$$$$\nabla f|_{(0,2,1)} = (2 \cdot 1, 0 \cdot 1, 0 \cdot 2) = (2, 0, 0)$$$$|\mathbf{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$$$\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left( \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \right)$$$$D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = (2, 0, 0) \cdot \left( \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \right) = 2 \cdot \frac{2}{3} + 0 \cdot \frac{1}{3} + 0 \cdot \frac{2}{3} = \frac{4}{3}$$Thus, the rate of increase is $\frac{4}{3}$.

We are given $w^4 = 16j$.
First, express $16j$ in polar form:
Magnitude: $|16j| = 16$.
Angle: The complex number $j$ lies on the positive imaginary axis, so its angle is $\pi/2$ radians.
Thus, $16j = 16e^{j(\pi/2 + 2k\pi)}$ for integer $k$.
Now, $w^4 = 16e^{j(\pi/2 + 2k\pi)}$.
To find $w$, we take the fourth root:
$w = (16e^{j(\pi/2 + 2k\pi)})^{1/4}$
$w = 16^{1/4} e^{j(\pi/2 + 2k\pi)/4}$
$w = 2 e^{j(\pi/8 + k\pi/2)}$Let's find the four distinct roots for $k = 0, 1, 2, 3$:
For $k=0$: $w_0 = 2 e^{j(\pi/8)}$
For $k=1$: $w_1 = 2 e^{j(\pi/8 + \pi/2)} = 2 e^{j(5\pi/8)}$
For $k=2$: $w_2 = 2 e^{j(\pi/8 + \pi)} = 2 e^{j(9\pi/8)}$
For $k=3$: $w_3 = 2 e^{j(\pi/8 + 3\pi/2)} = 2 e^{j(\pi/8 + 12\pi/8)} = 2 e^{j(13\pi/8)}$Now, let's check the given options:
(A) $2e^{j2\pi/8} = 2e^{j\pi/4}$. This is not among the calculated roots.
(B) $2e^{j\pi/8}$. This is one of the roots ($w_0$).
(C) $2e^{j5\pi/8}$. This is one of the roots ($w_1$).
(D) $2e^{j9\pi/8}$. This is one of the roots ($w_2$).Therefore, option (A) cannot be a value of $w$.

Let $f(z) = \frac{z+2}{z^2+2z+2}$.
First, find the poles of $f(z)$ by setting the denominator to zero:
$z^2+2z+2 = 0$
Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$z = \frac{-2 \pm \sqrt{2^2-4(1)(2)}}{2(1)}$
$z = \frac{-2 \pm \sqrt{4-8}}{2}$
$z = \frac{-2 \pm \sqrt{-4}}{2}$
$z = \frac{-2 \pm 2j}{2}$
So, the poles are $z_1 = -1 + j$ and $z_2 = -1 - j$.Next, determine which poles lie inside the contour $C$.
The contour $C$ is a circle defined by $|z + 1 - j| = 1$. This is a circle centered at $z_c = -1 + j$ with radius $R = 1$.Let's check $z_1 = -1 + j$:
$|(-1 + j) + 1 - j| = |0| = 0$. Since $0 < 1$, the pole $z_1 = -1 + j$ is inside the contour $C$.Let's check $z_2 = -1 - j$:
$|(-1 - j) + 1 - j| = |-2j| = 2$. Since $2 > 1$, the pole $z_2 = -1 - j$ is outside the contour $C$.So, only $z_1 = -1 + j$ is inside the contour.
We can rewrite the denominator as $(z - (-1+j))(z - (-1-j)) = (z - z_1)(z - z_2)$.
The integral is $\oint_C \frac{z+2}{(z - z_1)(z - z_2)} dz$.
By Cauchy's Residue Theorem, the integral is $2\pi j \times \text{Res}(f, z_1)$.
The residue at $z_1$ is:
$\text{Res}(f, z_1) = \lim_{z \to z_1} (z - z_1) \frac{z+2}{(z - z_1)(z - z_2)}$
$\text{Res}(f, z_1) = \lim_{z \to z_1} \frac{z+2}{z - z_2}$
Substitute $z_1 = -1 + j$ and $z_2 = -1 - j$:
$\text{Res}(f, -1 + j) = \frac{(-1 + j) + 2}{(-1 + j) - (-1 - j)}$
$= \frac{1 + j}{-1 + j + 1 + j}$
$= \frac{1 + j}{2j}$
$= \frac{1}{2j} + \frac{j}{2j}$
$= \frac{-j}{2} + \frac{1}{2}$ (since $1/j = -j$)
$= \frac{1 - j}{2}$Now, calculate the integral:
Integral $= 2\pi j \times \frac{1 - j}{2}$
$= \pi j (1 - j)$
$= \pi j - \pi j^2$
$= \pi j - \pi (-1)$
$= \pi j + \pi$
$= \pi (1 + j)$The final answer is $\pi(1 + j)$.

We are given that $B = P^{-1}AP$. This means that matrices $A$ and $B$ are similar matrices.Eigenvalues of similar matrices:
Similar matrices have the same eigenvalues. Therefore, the eigenvalues of $A$ are the same as the eigenvalues of $B$, which are $\{\lambda_k \mid 1 \leq k \leq n\}$.Eigenvectors of similar matrices:
Let $\lambda_k$ be an eigenvalue of $B$ with corresponding eigenvector $v_k$. So, $Bv_k = \lambda_k v_k$.
We have the relationship $B = P^{-1}AP$. We want to find the eigenvector $x_k$ of $A$ corresponding to the eigenvalue $\lambda_k$.
So, $Ax_k = \lambda_k x_k$.
Substitute $A = PBP^{-1}$ into the equation:
$(PBP^{-1})x_k = \lambda_k x_k$
Multiply both sides by $P^{-1}$ from the left:
$P^{-1}(PBP^{-1})x_k = P^{-1}(\lambda_k x_k)$
$(P^{-1}P)BP^{-1}x_k = \lambda_k P^{-1}x_k$
$IBP^{-1}x_k = \lambda_k P^{-1}x_k$
$B(P^{-1}x_k) = \lambda_k (P^{-1}x_k)$Comparing this equation with $Bv_k = \lambda_k v_k$, we can see that $P^{-1}x_k$ must be an eigenvector of $B$ corresponding to $\lambda_k$. Thus, $P^{-1}x_k = v_k$.
To find $x_k$, multiply by $P$ from the left:
$P(P^{-1}x_k) = Pv_k$
$x_k = Pv_k$So, the eigenvectors of $A$ are $\{Pv_k \mid 1 \leq k \leq n\}$.Combining both, the sets of eigenvalues and eigenvectors of $A$ are $\{\lambda_k \mid 1 \leq k \leq n\}$ and $\{Pv_k \mid 1 \leq k \leq n\}$, respectively.

In a semiconductor, if the Fermi energy level lies within the conduction band, it indicates that the concentration of electrons in the conduction band is very high, exceeding the effective density of states in the conduction band. This condition is known as a degenerate n-type semiconductor. In such a material, the conduction band is heavily populated with electrons, making the material highly conductive due to electron flow.Therefore, the correct option is (A).

At absolute zero temperature ($T = 0 K$), an intrinsic semiconductor behaves like a perfect insulator. In this state, all electrons are tightly bound in the valence band, completely filling all available energy states in the valence band. Consequently, there are no free electrons to move into the conduction band, meaning all energy states in the conduction band are empty of electrons.Therefore, the correct statement is that all energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.Option (A) accurately describes this condition.

For a series RLC circuit, the resonant frequency $\omega_0$ and the quality factor $Q$ are given by:$$\omega_0 = \frac{1}{\sqrt{LC}}$$$$Q = \frac{\omega_0 L}{R} = \frac{1}{R} \sqrt{\frac{L}{C}}$$Given:
$Q = 1000$
$\omega_0 = 10^6$ rad/sLet's check each option:Option (A): $R = 1 \Omega, L = 1 \mu H = 10^{-6} H, C = 1 \mu F = 10^{-6} F$
$\omega_0 = \frac{1}{\sqrt{(10^{-6})(10^{-6})}} = \frac{1}{\sqrt{10^{-12}}} = \frac{1}{10^{-6}} = 10^6$ rad/s (Matches)
$Q = \frac{\omega_0 L}{R} = \frac{(10^6)(10^{-6})}{1} = \frac{1}{1} = 1$ (Does not match $Q=1000$)Option (B): $R = 0.1 \Omega, L = 1 \mu H = 10^{-6} H, C = 1 \mu F = 10^{-6} F$
$\omega_0 = 10^6$ rad/s (Matches, as L and C are the same as in A)
$Q = \frac{\omega_0 L}{R} = \frac{(10^6)(10^{-6})}{0.1} = \frac{1}{0.1} = 10$ (Does not match $Q=1000$)Option (C): $R = 0.01 \Omega, L = 1 \mu H = 10^{-6} H, C = 1 \mu F = 10^{-6} F$
$\omega_0 = 10^6$ rad/s (Matches)
$Q = \frac{\omega_0 L}{R} = \frac{(10^6)(10^{-6})}{0.01} = \frac{1}{0.01} = 100$ (Does not match $Q=1000$)Option (D): $R = 0.001 \Omega, L = 1 \mu H = 10^{-6} H, C = 1 \mu F = 10^{-6} F$
$\omega_0 = 10^6$ rad/s (Matches)
$Q = \frac{\omega_0 L}{R} = \frac{(10^6)(10^{-6})}{0.001} = \frac{1}{0.001} = 1000$ (Matches)Thus, option (D) provides the correct values for $R$, $L$, and $C$ that satisfy the given quality factor and center frequency.The final answer is $\boxed{D}$

In a MOS capacitor, the depletion width ($W_{\text{dep}}$) varies with the gate voltage ($V_{\text{g}}$) as follows:Therefore, the depletion width starts small (near zero) in accumulation, increases as $V_{\text{g}}$ goes from $V_{\text{fb}}$ to $V_{\text{t}}$, and then saturates at its maximum value in inversion.Graph (B) correctly depicts this behavior: $W_{\text{dep}}$ is low in accumulation, increases from $V_{\text{fb}}$ to $V_{\text{t}}$, and then flattens out for $V_{\text{g}} > V_{\text{t}}$.The final answer is $\boxed{B}$.

For a lossless dielectric medium, the phase velocity $v_p$ is given by:$$v_p = \frac{c}{\sqrt{\epsilon_r \mu_r}}$$Given $\epsilon_r = 4$ and $\mu_r = 1$.
Substituting these values:$$v_p = \frac{c}{\sqrt{4 \times 1}} = \frac{c}{\sqrt{4}} = \frac{c}{2}$$Thus, $v_p < c$.For a lossless and non-dispersive dielectric medium (where $\epsilon_r$ and $\mu_r$ are constant with frequency), the group velocity $v_g$ is equal to the phase velocity $v_p$.
Therefore, $v_g = v_p = \frac{c}{2}$.
Thus, $v_g < c$.Combining these results, both $v_p < c$ and $v_g < c$.
This matches option (D).The final answer is $\boxed{D}$

Let's analyze the input and output stages of the given circuit:Input Stage:
The input signal $V_{IN}$ is applied to the gate of the first MOSFET. The source of the first MOSFET is connected to a current source (implied by $V_1$ and $R_1$ if it were a common source, but here it's a common gate configuration for the first transistor if we consider $V_1$ as a bias for the source). However, looking at the overall structure, $V_{IN}$ is applied to the source of the first transistor, and its gate is biased by $V_1$. This configuration is a common-gate (CG) amplifier stage. A common-gate amplifier has a low input impedance.Output Stage:
The output $V_{OUT}$ is taken from the drain of the second MOSFET. The gate of the second MOSFET is biased by $V_2$. The drain is connected to $V_{DD}$ through a resistor (or current source) and the source is connected to the output node. This configuration is a common-drain (CD) amplifier stage, also known as a source follower. However, the diagram shows $V_{OUT}$ at the drain of the second transistor, which is a common-source (CS) configuration if $V_2$ biases the gate and the source is grounded or connected to a current source. Let's re-examine the connections.Let's trace the signal path:Let's re-evaluate the input and output impedances based on the overall circuit behavior.Input Impedance ($Z_{in}$):
The input is applied to the source of the first MOSFET, which is configured as a common-gate stage (gate is biased at $V_1$). A common-gate stage has a very low input impedance, typically $Z_{in} \approx 1/g_m$, where $g_m$ is the transconductance of the MOSFET. Thus, the circuit has a low input impedance.Output Impedance ($Z_{out}$):
The output is taken from the drain of the second MOSFET, which is configured as a common-source stage (gate is driven by the first stage, source is biased at $V_2$). The output impedance of a common-source stage with a resistive load $R_L$ (here, $R_1$ in the output part) is approximately $R_L$ in parallel with the MOSFET's output resistance $r_o$. If the load is a current source, the output impedance would be high. Given the resistor $R_1$ at the output, the output impedance is high. More precisely, for a common-source stage with an active load or current source load, the output impedance is very high. The diagram shows $R_1$ connected to $V_{DD}$ at the output, which acts as a load. The output impedance of a CS stage with a current source load (or high resistance load) is high.Let's consider the overall behavior based on input and output impedances:

• Low Input Impedance implies it is a current-controlled device (it 'accepts' current easily).
• High Output Impedance implies it is a current source (it 'delivers' current).

Therefore, the circuit behaves as a current-controlled current source.Let's verify the options:
(A) Voltage controlled voltage source: High input impedance, low output impedance.
(B) Voltage controlled current source: High input impedance, high output impedance.
(C) Current controlled voltage source: Low input impedance, low output impedance.
(D) Current controlled current source: Low input impedance, high output impedance.Based on the analysis, the circuit has low input impedance (common-gate input) and high output impedance (common-source output with a load that makes it behave like a current source). This matches the characteristics of a current-controlled current source.The final answer is $\boxed{D}$.