GATE 2021 CS (Set 2) question paper PDF and answer key

Use the identity $a^2-b^2=(a-b)(a+b)$.

Here,

$$a=x-\frac12,\qquad b=x-\frac32.$$

So,

$$a-b=1$$

and

$$a+b=2x-2.$$

Therefore,

$$\left(x-\frac12\right)^2-\left(x-\frac32\right)^2 = 2x-2.$$

Given

$$2x-2=x+2,$$

we get

$$x=4.$$

Therefore, the correct answer is Option B.

Take a cube of side length $a$. An edge vector can be represented as

$$(a,0,0)$$

and the body diagonal vector as

$$(a,a,a).$$

Then

$$\cos\theta = \frac{(a,0,0)\cdot(a,a,a)} {\|(a,0,0)\|\|(a,a,a)\|} = \frac{a^2}{a\cdot a\sqrt{3}} = \frac{1}{\sqrt{3}}.$$

Therefore, the correct answer is Option B.

Use $y,z$ as the select lines of the 4×1 multiplexer.

For each $yz$ combination:

• $yz=00$: minterms $m_0,m_4$ are both included, so $I_0=1$.
• $yz=01$: minterms $m_1,m_5$ are both included, so $I_1=1$.
• $yz=10$: minterm $m_6$ is included but $m_2$ is not, so $I_2=x$.
• $yz=11$: minterm $m_3$ is included but $m_7$ is not, so $I_3=\overline{x}$.

The circuit matching this assignment is Option A.

Therefore, the correct answer is Option A.

Since $L_1$ is regular, $\overline{L_1}$ is also regular.

For option B:

$$\overline{\overline{L_1} \cup \overline{L_2}} = L_1 \cap L_2$$

by De Morgan's law. The intersection of a regular language and a context-free language is context-free. So B is context-free.

For option C:

$$L_2 \cup \overline{L_2} = \Sigma^*$$

so the expression becomes

$$L_1 \cup \Sigma^* = \Sigma^*,$$

which is regular, hence context-free. So C is context-free.

For option D:

$$(L_1 \cap L_2) \cup (\overline{L_1} \cap L_2) = (L_1 \cup \overline{L_1}) \cap L_2 = \Sigma^* \cap L_2 = L_2.$$

So D is context-free.

Option A is not necessarily context-free because context-free languages are not closed under complement in general.

Therefore, the correct answers are B, C, and D.

The processor frequency is

$$2\text{ MHz}=2\times 10^6$$

cycles per second.

DMA uses 0.5% of processor cycles:

$$0.005 \times 2\times 10^6 = 10000$$

cycles per second.

Each DMA cycle transfers one 8-bit character, so the data transfer rate is

$$10000 \times 8 = 80000$$

bits per second.

Therefore, the answer is 80000.