GATE 2026 EC question paper PDF and answer key

$\log_{p^{1/n}} y = \frac{\log y}{\frac{1}{n}\log p} = n\log_p y, \qquad \log_{y^{1/n}} p = n\log_y p$

So the product is $n^2 (\log_p y)(\log_y p) = n^2 = 16$, giving $n = 4$ (since $n > 1$).

The statement has the form "for every student $s$, there exists a student $t$ with $|s - t| \le 10$".

Its negation is "there exists a student $s$ such that for every other student $t$, $|s - t| > 10$", which is option (B).

Under an applied forward bias $V_a$, the quasi-Fermi levels separate by $E_{fn} - E_{fp} = qV_a = 2$ eV. Option (C) is in volts, which is not a unit of energy.

$R$ is the square with vertices $(0,0), (1,1), (2,0), (1,-1)$, i.e. $|x - 1| + |y| \le 1$, with area 2. Put $u = x - 1$: $\iint_R (x^2 + y^2 - 1)\,dA = \iint (u^2 + 2u + y^2)\,dA$ By symmetry $\iint 2u\,dA = 0$, and $\iint u^2\,dA = \int_{-1}^{1} u^2 \cdot 2(1 - |u|)\,du = \frac{1}{3}$. Likewise $\iint y^2\,dA = \frac{1}{3}$.

$\text{Result} = \frac{2}{3} \approx 0.67$

For an inverting amplifier with a T-network in the feedback path: $\frac{v_o}{v_i} = -\frac{R_2 + R_3 + \frac{R_2 R_3}{R_4}}{R_1} = -\frac{50 + 50 + 50}{50} = -3$ The magnitude is 3.00.