GATE EC 2016 Set 1

Mount Everest is a specific, unique entity, so the superlative adjective 'highest' must be preceded by the definite article 'the'. Option (D) is incorrect because the phrase 'one of the' must be followed by a plural noun (e.g., 'one of the highest peaks'). Therefore, option (A) is the only grammatically correct choice.

The sentence requires a verb to complete the question. 'Lose' is a verb meaning to be deprived of or cease to have or retain something. 'Loose' is primarily an adjective meaning not tight. 'Loss' is a noun. 'Louse' is a singular noun for a type of parasite. Thus, 'lose' is the correct choice.

The word 'Despite' indicates a logical contrast. If a medicine is effective, one would expect it to be widely prescribed. The phrase 'it is not ______________ widely' provides the necessary contrast to the positive attribute in the first blank. Therefore, 'effectiveness' and 'prescribed' form the most logical pair.

• Total fruits = $5,692,000$
• Unripe fruits = $15\%$ of $5,692,000 = 0.15 \times 5,692,000 = 853,800$
• Ripe fruits = $85\%$ of $5,692,000 = 0.85 \times 5,692,000 = 4,838,200$

2. Calculate the number of apples in each category:

• Apples in unripe fruits = $45\%$ of $853,800 = 0.45 \times 853,800 = 384,210$
• Oranges in ripe fruits = $66\%$, so Apples in ripe fruits = $(100\% - 66\%) = 34\%$
• Apples in ripe fruits = $34\%$ of $4,838,200 = 0.34 \times 4,838,200 = 1,644,988$

3. Calculate total apples:

• Total apples = Apples in unripe + Apples in ripe
• Total apples = $384,210 + 1,644,988 = 2,029,198$

Therefore, the correct option is (A).

Let the distance from 'I' to Arun be $d$.
From the problem statement:

• Distance to Ahmed = $5$ km
• Distance to Susan = $7$ km
• Arun is farther than Ahmed: $d > 5$
• Arun is closer than Susan: $d < 7$

So, the possible distance $d$ must satisfy $5 < d < 7$.
Checking the options:
(A) $3.00$ (False)
(B) $4.99$ (False)
(C) $6.02$ (True, as it lies between $5$ and $7$)
(D) $7.01$ (False)Thus, the correct option is (C).

• $P(\text{Infected and No Symptoms}) = P(\text{Infected}) \times P(\text{No Symptoms} | \text{Infected})$
• $P(\text{Infected and No Symptoms}) = 0.5 \times 0.7 = 0.35 = 35\%$

Therefore, the correct option is (C).

Let's analyze each inference:
(i) "His friends were always asking him to help them." The passage states "He had always assisted them in times of need", which implies they asked for help, but "always asking" is an overstatement and not directly inferable.
(ii) "He felt that when in need of help, his friends would let him down." The passage states "he was confident that they would reciprocate. However, the events of the last week proved him wrong." This implies his confidence was misplaced, and they did let him down. So, this feeling is a valid consequence of the events.
(iii) "He was sure that his friends would help him when in need." This is explicitly stated in the passage: "he was confident that they would reciprocate." This is a valid inference about his prior belief.
(iv) "His friends did not help him last week." This is directly implied by "the events of the last week proved him wrong" regarding his confidence that they would reciprocate. This is a valid inference about what happened.Both (iii) and (iv) are directly inferable from the passage. (iii) describes his initial state of mind, and (iv) describes the event that contradicted it.Therefore, the logically valid inferences are (iii) and (iv).The final answer is $\boxed{B}$

Let's analyze the given statements:Combining these two, we get the age order: Shiva > Leela > Pavithra.Now let's evaluate the options:
(A) "When Shiva plays chess with Leela and Pavithra, he often loses." We have no information about Shiva's chess playing ability, so this statement cannot be confirmed as true.
(B) "Leela is the oldest of the three." This is false, as Shiva is older than Leela.
(C) "Shiva is a better chess player than Pavithra." We have no information about Shiva's chess playing ability relative to Pavithra, so this statement cannot be confirmed as true.
(D) "Pavithra is the youngest of the three." Based on the age order Shiva > Leela > Pavithra, Pavithra is indeed the youngest. This statement must be TRUE.The final answer is $\boxed{D}$

We are given three equations:From equation (1):
$q^{-a} = \frac{1}{r} \Rightarrow \frac{1}{q^a} = \frac{1}{r} \Rightarrow q^a = r$From equation (2):
$r^{-b} = \frac{1}{s} \Rightarrow \frac{1}{r^b} = \frac{1}{s} \Rightarrow r^b = s$From equation (3):
$s^{-c} = \frac{1}{q} \Rightarrow \frac{1}{s^c} = \frac{1}{q} \Rightarrow s^c = q$Now, substitute the expression for $r$ from the modified equation (1) into the modified equation (2):
$(q^a)^b = s \Rightarrow q^{ab} = s$Next, substitute this expression for $s$ into the modified equation (3):
$(q^{ab})^c = q \Rightarrow q^{abc} = q^1$Since the bases are the same ($q$), the exponents must be equal:
$abc = 1$The final answer is $\boxed{C}$

To find the ratio of work done by Q and R after 7 days, we first determine their total work capacities and individual rates. $25 \text{ days} \times 12 \text{ hours/day} = 300 \text{ hours}$. $50 \text{ days} \times 12 \text{ hours/day} = 600 \text{ hours}$. Let the total work be $W = 600$ units.
Rate of Q ($r_Q$) = $\frac{600}{300} = 2$ units/hour.
Rate of R ($r_R$) = $\frac{600}{600} = 1$ unit/hour.

• Q took 2 days of sick leave, so Q worked for $7 - 2 = 5$ days.
• Total hours worked by Q = $5 \text{ days} \times 12 \text{ hours/day} = 60$ hours.
• Work done by Q ($W_Q$) = $60 \text{ hours} \times 2 \text{ units/hour} = 120$ units.
• R worked for all 7 days at 18 hours/day.
• Total hours worked by R = $7 \text{ days} \times 18 \text{ hours/day} = 126$ hours.
• Work done by R ($W_R$) = $126 \text{ hours} \times 1 \text{ unit/hour} = 126$ units.

5. Ratio of work done:
Ratio = $W_Q : W_R = 120 : 126$.
Dividing both sides by 6: $\frac{120}{6} : \frac{126}{6} = 20 : 21$.Thus, the correct option is (C).

To find the expression for the inverse matrix $M^{-1}$, we start with the given matrix equation:$$M^4 = I$$Since $M^4 = I$, the matrix $M$ is invertible. Multiplying both sides of the equation by $M^{-1}$ from the left (or right), we get:$$M^{-1} M^4 = M^{-1} I$$$$M^3 = M^{-1}$$Thus, the inverse of $M$ is given by:$$M^{-1} = M^3$$We are also given that $M^4 = I$. This implies that for any integer $k \ge 1$:$$M^{4k} = (M^4)^k = I^k = I$$Now, we can express $M^{-1}$ in terms of $k$ by multiplying it by $M^{4k}$ (since multiplying by $I$ does not change the matrix):$$M^{-1} = M^{-1} \cdot I$$$$M^{-1} = M^{-1} \cdot M^{4k}$$Using the laws of exponents for matrices:$$M^{-1} = M^{4k - 1}$$To match the given options, we can rewrite $M^{-1}$ by utilizing the relation $M^4 = I$ again:$$M^{-1} = M^3 \cdot I = M^3 \cdot M^{4k} = M^{4k + 3}$$Let us verify this for any natural number $k$:$$M^{4k+3} = M^{4k} \cdot M^3 = I \cdot M^3 = M^3 = M^{-1}$$This matches Option C.

Let $X$ be a Poisson-distributed random variable with parameter (mean) $\lambda > 0$. The probability mass function of $X$ is given by:$$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \dots$$For a Poisson distribution, the mean and variance are both equal to $\lambda$:$$E[X] = \lambda$$$$\text{Var}(X) = \lambda$$The second moment of $X$, denoted as $E[X^2]$, is related to the mean and variance by the formula:$$\text{Var}(X) = E[X^2] - (E[X])^2$$Substituting the known values into this relation:$$\lambda = E[X^2] - \lambda^2$$We are given that the second moment $E[X^2] = 2$. Substituting this into the equation yields a quadratic equation in terms of $\lambda$:$$\lambda = 2 - \lambda^2$$$$\lambda^2 + \lambda - 2 = 0$$Factoring the quadratic equation:$$(\lambda + 2)(\lambda - 1) = 0$$This gives two possible roots:$$\lambda = -2 \quad \text{or} \quad \lambda = 1$$Since the parameter $\lambda$ of a Poisson distribution must be strictly positive ($\lambda > 0$), we discard the negative root. Thus:$$\lambda = 1$$The mean of the random variable is $\lambda = 1$.Correct Answer: 1 (which lies within the allowed range of 0.9 to 1.1).

To determine the correct option, let us analyze each statement individually:

• Analysis: This statement is false. Continuity at a point does not guarantee differentiability.
• Counterexample: Consider the absolute value function $f(x) = |x|$ at $x_0 = 0$.
• The function is continuous at $x = 0$ because:

$$\lim_{x \to 0} |x| = f(0) = 0$$

• However, the left-hand derivative at $x = 0$ is $-1$ and the right-hand derivative is $+1$. Since the one-sided derivatives are not equal, $f(x)$ is not differentiable at $x = 0$.
• Thus, P is false.

2. Statement Q: "If $f(x)$ is continuous at $x = x_0$, then it may not be differentiable at $x = x_0$."

• Analysis: This statement is true. As demonstrated by the counterexample $f(x) = |x|$ at $x_0 = 0$, a function can be continuous at a point without being differentiable there.
• Thus, Q is true.

3. Statement R: "If $f(x)$ is differentiable at $x = x_0$, then it is also continuous at $x = x_0$."

• Analysis: This is a fundamental theorem in calculus. If a function $f(x)$ is differentiable at $x = x_0$, then the limit:

$$f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$$ exists and is finite.

• We can write:

$$f(x) - f(x_0) = \frac{f(x) - f(x_0)}{x - x_0} \cdot (x - x_0)$$

• Taking the limit as $x \to x_0$ on both sides:

$$\lim_{x \to x_0} [f(x) - f(x_0)] = \lim_{x \to x_0} \left[ \frac{f(x) - f(x_0)}{x - x_0} \right] \cdot \lim_{x \to x_0} (x - x_0)$$ $$\lim_{x \to x_0} [f(x) - f(x_0)] = f'(x_0) \cdot 0 = 0$$ $$\lim_{x \to x_0} f(x) = f(x_0)$$

• This satisfies the definition of continuity at $x = x_0$. Therefore, differentiability implies continuity.
• Thus, R is true.

Conclusion

• P is false.
• Q is true.
• R is true.

This corresponds to Option B.

Solutions to the Laplace equation $\nabla^2 f = 0$ are known as harmonic functions. A fundamental property of harmonic functions is the Maximum Principle, which states that a non-constant harmonic function cannot attain its maximum or minimum in the interior of its domain; these extrema must occur on the boundary. This is a direct consequence of the mean value property of harmonic functions.

The function $F(x)$ is defined as the integral of $f(x)$ from $-5$ to $x$: $F(x) = \int_{-5}^{x} f(y) dy$. By the Fundamental Theorem of Calculus, $F'(x) = f(x)$.Comparing these characteristics with the given options, only graph (C) correctly represents $F(x)$ as it starts at 0, decreases to a minimum at $x=0$, and returns to 0 at $x=5$.

Complex exponentials of the form $e^{st}$ are eigenfunctions of linear time-invariant (LTI) systems. For a continuous-time LTI system with impulse response $h(t)$, if the input is $x(t) = e^{j\omega_0 t}$, the output $y(t)$ is given by:$$y(t) = h(t) * e^{j\omega_0 t} = \int_{-\infty}^{\infty} h(\tau) e^{j\omega_0 (t-\tau)} d\tau = e^{j\omega_0 t} \int_{-\infty}^{\infty} h(\tau) e^{-j\omega_0 \tau} d\tau = H(j\omega_0) e^{j\omega_0 t}$$where $H(j\omega_0)$ is the frequency response of the system at $\omega_0$. Since the output is a scalar multiple of the input, $e^{j\omega_0 t}$ is an eigenfunction. Sinusoids like $\cos$ and $\sin$ are not strictly eigenfunctions because they can undergo a phase shift, which is not a simple scalar multiplication. $e^{j\omega_0 t} u(t)$ is not an eigenfunction due to the windowing effect of the step function.

A sampled signal $x[n] = x(nT_s)$ is periodic if there exists an integer $N > 0$ such that $x[n+N] = x[n]$ for all $n$. This implies $x((n+N)T_s) = x(nT_s)$. For a continuous-time signal with period $T$, this condition is satisfied if $NT_s$ is an integer multiple of $T$, i.e., $NT_s = kT$ for some integer $k$. This simplifies to $\frac{T}{T_s} = \frac{N}{k}$. Thus, the ratio of the signal period to the sampling period must be a rational number.
(A) $\frac{T}{T_s} = \sqrt{2}$ is irrational.
(B) $\frac{T}{T_s} = 1.2 = \frac{6}{5}$ is rational. This satisfies the condition.
(C) and (D) are incorrect as it depends on the rationality of the ratio.

The z-transform of $a^n u[n]$ is $\frac{z}{z-a}$ with ROC $|z| > |a|$.
The z-transform of $b^n u[n]$ is $\frac{z}{z-b}$ with ROC $|z| > |b|$.
For the sum $x[n] = a^n u[n] + b^n u[n]$, the ROC is the intersection of the individual ROCs: $|z| > |a| \cap |z| > |b|$.
Given $0 < |a| < |b| < 1$, the intersection is $|z| > \max(|a|, |b|)$, which is $|z| > |b|$.

For a two-port network defined by its transmission (ABCD) parameters, the condition for reciprocity is $AD - BC = 1$. The transmission matrix is given as $T = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$. The determinant of this matrix is $\det(T) = AD - BC$. Therefore, for a reciprocal network, $\det(T) = 1$.

Let the continuous-time sinusoid be $x(t) = \cos(2\pi f_s t)$ where $f_s = 33$ Hz.
Let the periodic Dirac impulse train be $p(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT_p)$ where $T_p = 1/f_p$ and $f_p = 46$ Hz.The product signal is $y(t) = x(t)p(t)$. When a continuous-time signal is multiplied by an impulse train, the spectrum of the resulting signal consists of shifted versions of the original signal's spectrum, centered at multiples of the impulse train's frequency.The spectrum of $x(t)$ has impulses at $\pm f_s$.
The spectrum of $y(t)$ will have frequency components at $k f_p \pm f_s$ for integer values of $k$.Given $f_s = 33$ Hz and $f_p = 46$ Hz.
Let's list some of the absolute frequencies present in $y(t)$:
For $k=0$: $|0 \cdot f_p \pm f_s| = f_s = 33$ Hz
For $k=1$:
$|1 \cdot f_p - f_s| = |46 - 33| = 13$ Hz
$|1 \cdot f_p + f_s| = |46 + 33| = 79$ Hz
For $k=2$:
$|2 \cdot f_p - f_s| = |2(46) - 33| = |92 - 33| = 59$ Hz
$|2 \cdot f_p + f_s| = |2(46) + 33| = |92 + 33| = 125$ HzThe signal $y(t)$ is passed through an ideal analog low-pass filter with a cutoff frequency $f_c = 23$ Hz.
This filter will only allow frequency components less than or equal to $23$ Hz to pass.From the list of frequencies:

• $13$ Hz: This component will pass because $13 \le 23$ Hz.
• $33$ Hz: This component will not pass because $33 > 23$ Hz.
• $79$ Hz, $59$ Hz, $125$ Hz, etc.: These components will also not pass as they are all greater than $23$ Hz.

Therefore, the output signal from the low-pass filter will only contain the frequency component at $13$ Hz.The fundamental frequency of the output signal is the lowest non-zero frequency component present in the output. In this case, it is $13$ Hz.The final answer is $\boxed{13}$.

The energy band diagram shows a 'New Energy Level' at $0.01$ eV below the conduction band ($E_c$). This energy level is very close to the conduction band.Therefore, the diagram represents an intrinsic semiconductor doped with pentavalent atoms to form an n-type semiconductor.Let's evaluate the options:
(A) Intrinsic semiconductor doped with pentavalent atoms to form n-type semiconductor: This matches our analysis.
(B) Intrinsic semiconductor doped with trivalent atoms to form n-type semiconductor: Trivalent atoms form p-type semiconductors, not n-type.
(C) Intrinsic semiconductor doped with pentavalent atoms to form p-type semiconductor: Pentavalent atoms form n-type semiconductors, not p-type.
(D) Intrinsic semiconductor doped with trivalent atoms to form p-type semiconductor: Trivalent atoms form p-type semiconductors, but the energy level shown is for an n-type material.The final answer is $\boxed{A}$.

Let's analyze each statement regarding the effects of reducing channel length ($L$) in a MOSFET (short-channel effects):P: As channel length reduces, OFF-state current increases.

• Explanation: When the channel length is reduced, short-channel effects become prominent. These include Drain-Induced Barrier Lowering (DIBL) and subthreshold leakage. DIBL causes the potential barrier for electrons at the source end to be lowered by the drain voltage, leading to an increase in subthreshold current (OFF-state current) even when the gate-source voltage is below the threshold. Therefore, this statement is CORRECT.

Q: As channel length reduces, output resistance increases.

• Explanation: The output resistance ($r_o$) of a MOSFET in saturation is given by $r_o \approx 1/(\lambda I_D)$, where $\lambda$ is the channel length modulation parameter and $I_D$ is the drain current. As the channel length $L$ reduces, the channel length modulation effect becomes more significant, meaning $\lambda$ increases. An increase in $\lambda$ leads to a decrease in output resistance. Therefore, this statement is INCORRECT.

R: As channel length reduces, threshold voltage remains constant.

• Explanation: Due to short-channel effects like DIBL and velocity saturation, the threshold voltage ($V_{TH}$) of a MOSFET typically decreases as the channel length reduces. The gate loses some control over the channel, and the drain voltage influences the channel more, effectively lowering the threshold voltage. Therefore, this statement is INCORRECT.

S: As channel length reduces, ON current increases.

• Explanation: The ON current ($I_{ON}$) in a MOSFET is generally proportional to $W/L$ (width-to-length ratio) and $(V_{GS} - V_{TH})^2$ (in saturation) or $(V_{GS} - V_{TH})V_{DS}$ (in linear region). For a given $W$ and $V_{GS}$, reducing $L$ directly increases the $W/L$ ratio, which tends to increase the ON current. Also, the decrease in $V_{TH}$ (as per short-channel effects) further contributes to an increase in $(V_{GS} - V_{TH})$, thus increasing the ON current. Therefore, this statement is CORRECT.

We are asked to identify the INCORRECT statements. Based on our analysis, statements Q and R are incorrect.Comparing with the options:
(A) P and Q
(B) P and S
(C) Q and R
(D) R and SThe correct option is (C).The final answer is $\boxed{C}$.

In the given circuit, the op-amp is in a negative feedback configuration. Due to the virtual short property, the voltage at the inverting terminal ($V_-$) is equal to the voltage at the non-inverting terminal ($V_+$). $$I_0 = \left( \frac{\beta}{\beta+1} \right) \frac{V_{ref}}{R}$$This matches option (B).

$$V_- = 10 \times \frac{2\text{ k}\Omega}{8\text{ k}\Omega + 2\text{ k}\Omega} = 10 \times \frac{2}{10} = 2\text{ V}$$