UGC NET Computer Science 2021 Previous Year Questions (PYQs) – Page 1 of 10
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Which of the statements given below is/are correct?
It is always important and useful to include an $alt$ attribute on $img$ tag in HTML because
A. users who cannot see the image due to vision impairment can have a textual description of the image which can be spoken aloud by a screen reader.
B. If the image fails to load due to slow connection, broken path, etc. then alt text is displayed instead.
C. SEO Search Engine Optimization benefits.
Choose the correct answer from the options given below:
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Solution
The $alt$ attribute is used inside the $img$ tag in HTML.
It provides alternative text for an image.
Now check each statement.
Statement A:
If a user has vision impairment, then screen readers can read the $alt$ text aloud.
So, the user can understand what the image represents.
Therefore, statement A is correct.
Statement B:
If the image does not load because of slow internet, wrong image path or broken link, then the browser can display the $alt$ text.
Therefore, statement B is correct.
Statement C:
Search engines cannot understand images like humans.
So, $alt$ text helps search engines understand the image content.
It also gives SEO benefits.
Therefore, statement C is correct.
Hence, A, B and C are correct.
What does the following function $f()$ in C return?
int f(unsigned int N)
{
unsigned int counter = 0;
while (N > 0)
{
counter += N & 1;
N = N >> 1;
}
return counter == 1;
}
int f(unsigned int N)
{
unsigned int counter = 0;
while (N > 0)
{
counter += N & 1;
N = N >> 1;
}
return counter == 1;
}
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Solution
Given function takes an unsigned integer $N$.
Inside the function:
$counter=0$
Now the loop runs while:
$N>0$
In each iteration:
$counter += N\ \&\ 1$
Here, $N\ \&\ 1$ checks the last bit of $N$.
If the last bit is $1$, then $counter$ increases by $1$.
If the last bit is $0$, then $counter$ does not increase.
After that:
$N=N>>1$
This shifts the bits of $N$ one position to the right.
So, the loop checks all bits of $N$ one by one.
Therefore, $counter$ stores the number of $1$ bits in the binary representation of $N$.
Finally, the function returns:
$counter==1$
This condition is true only when the binary representation of $N$ contains exactly one $1$.
A number which has exactly one $1$ in binary form is a power of $2$.
For example:
$2=(10)_2$
$4=(100)_2$
$8=(1000)_2$
So, the function returns $1$ if $N$ is a power of $2$, otherwise it returns $0$.
Inside the function:
$counter=0$
Now the loop runs while:
$N>0$
In each iteration:
$counter += N\ \&\ 1$
Here, $N\ \&\ 1$ checks the last bit of $N$.
If the last bit is $1$, then $counter$ increases by $1$.
If the last bit is $0$, then $counter$ does not increase.
After that:
$N=N>>1$
This shifts the bits of $N$ one position to the right.
So, the loop checks all bits of $N$ one by one.
Therefore, $counter$ stores the number of $1$ bits in the binary representation of $N$.
Finally, the function returns:
$counter==1$
This condition is true only when the binary representation of $N$ contains exactly one $1$.
A number which has exactly one $1$ in binary form is a power of $2$.
For example:
$2=(10)_2$
$4=(100)_2$
$8=(1000)_2$
So, the function returns $1$ if $N$ is a power of $2$, otherwise it returns $0$.
Consider the following recursive function $F()$ in Java that takes an integer value and returns a string value:
public static String F(int N){
if (N <= 0) return "-";
return F(N - 3) + N + F(N - 2) + N;
}
The value of $F(5)$ is:
public static String F(int N){
if (N <= 0) return "-";
return F(N - 3) + N + F(N - 2) + N;
}
The value of $F(5)$ is:
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Solution
Given function is $F(N)=F(N-3)+N+F(N-2)+N$ and if $N\le 0$, then $F(N)=-$.
Now,
$F(1)=F(-2)+1+F(-1)+1=-1-1$
$F(2)=F(-1)+2+F(0)+2=-2-2$
$F(3)=F(0)+3+F(1)+3=-3-1-13$
Now find $F(5)$:
$F(5)=F(2)+5+F(3)+5$
$F(5)=(-2-2)+5+(-3-1-13)+5$
$F(5)=-2-25-3-1-135$
Now,
$F(1)=F(-2)+1+F(-1)+1=-1-1$
$F(2)=F(-1)+2+F(0)+2=-2-2$
$F(3)=F(0)+3+F(1)+3=-3-1-13$
Now find $F(5)$:
$F(5)=F(2)+5+F(3)+5$
$F(5)=(-2-2)+5+(-3-1-13)+5$
$F(5)=-2-25-3-1-135$
Match List I with List II:
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. JNDI | I. Runtime support for running Java program |
| B. RMI | II. The API in support of naming and directory services. |
| C. JDK | III. The methods provided by the Java development kit and runtime support for calling remote methods. |
| D. JRE | IV. The compiler and class libraries to develop Java applications. |
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Solution
JNDI means Java Naming and Directory Interface. It provides API support for naming and directory services. So, $A-II$.
RMI means Remote Method Invocation. It supports calling methods on remote objects. So, $B-III$.
JDK means Java Development Kit. It contains compiler and class libraries to develop Java applications. So, $C-IV$.
JRE means Java Runtime Environment. It provides runtime support for running Java programs. So, $D-I$.
Final matching is $A-II,\ B-III,\ C-IV,\ D-I$.
RMI means Remote Method Invocation. It supports calling methods on remote objects. So, $B-III$.
JDK means Java Development Kit. It contains compiler and class libraries to develop Java applications. So, $C-IV$.
JRE means Java Runtime Environment. It provides runtime support for running Java programs. So, $D-I$.
Final matching is $A-II,\ B-III,\ C-IV,\ D-I$.
Match List I with List II
| List I | List II |
|---|---|
| A. Imperative | I. Declarative, clausal representation, theorem proving |
| B. Object-oriented | II. Side-effect-free, declarative, expression evaluation |
| C. Logic | III. Imperative, abstract data type |
| D. Functional | IV. Command-based, procedural |
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Solution
Imperative programming is command-based and procedural, so $A-IV$.
Object-oriented programming is based on objects, classes and abstract data types, so $B-III$.
Logic programming is declarative and uses clausal representation and theorem proving, so $C-I$.
Functional programming is side-effect-free and based on expression evaluation, so $D-II$.
Final matching is $A-IV,\ B-III,\ C-I,\ D-II$.
Object-oriented programming is based on objects, classes and abstract data types, so $B-III$.
Logic programming is declarative and uses clausal representation and theorem proving, so $C-I$.
Functional programming is side-effect-free and based on expression evaluation, so $D-II$.
Final matching is $A-IV,\ B-III,\ C-I,\ D-II$.
Suppose you have eight black and white images taken with a $1$-megapixel camera and one $8$-color image taken by an $8$-megapixel camera. How much hard disk space in total do you need to store these images on your computer?
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Solution
A black and white image needs $1$ bit per pixel.
Size of one $1$-megapixel black and white image $=1\times 10^6$ bits.
Size of $8$ such images $=8\times 10^6$ bits $=1\ MB$.
An $8$-color image needs $\log_2 8=3$ bits per pixel.
Size of one $8$-megapixel $8$-color image $=8\times 10^6\times 3=24\times 10^6$ bits $=3\ MB$.
Total space $=1\ MB+3\ MB=4\ MB$.
Size of one $1$-megapixel black and white image $=1\times 10^6$ bits.
Size of $8$ such images $=8\times 10^6$ bits $=1\ MB$.
An $8$-color image needs $\log_2 8=3$ bits per pixel.
Size of one $8$-megapixel $8$-color image $=8\times 10^6\times 3=24\times 10^6$ bits $=3\ MB$.
Total space $=1\ MB+3\ MB=4\ MB$.
Which of the statements given below are correct?
The midpoint or Bresenham algorithm for rasterizing lines is optimized relative to DDA algorithm in that
A. it avoids round-off operations.
B. it is incremental.
C. it uses only integer arithmetic.
D. all straight lines can be displayed as straight exact.
Choose the correct answer from the options given below:
The midpoint or Bresenham algorithm for rasterizing lines is optimized relative to DDA algorithm in that
A. it avoids round-off operations.
B. it is incremental.
C. it uses only integer arithmetic.
D. all straight lines can be displayed as straight exact.
Choose the correct answer from the options given below:
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Solution
Bresenham line drawing algorithm is optimized compared to DDA algorithm.
DDA uses floating point calculations and round-off operations.
Bresenham avoids round-off operations, so statement A is correct.
Bresenham is incremental because the next pixel is calculated from the previous pixel, so statement B is correct.
Bresenham uses only integer arithmetic, so statement C is correct.
But on raster display, every straight line cannot be displayed exactly as a mathematical straight line because pixels are discrete, so statement D is false.
Therefore, A, B and C only are correct.
DDA uses floating point calculations and round-off operations.
Bresenham avoids round-off operations, so statement A is correct.
Bresenham is incremental because the next pixel is calculated from the previous pixel, so statement B is correct.
Bresenham uses only integer arithmetic, so statement C is correct.
But on raster display, every straight line cannot be displayed exactly as a mathematical straight line because pixels are discrete, so statement D is false.
Therefore, A, B and C only are correct.
What is the transformation matrix $M$ that transforms a square in the $xy$-plane defined by $(1,1)^T,(-1,1)^T,(-1,-1)^T$ and $(1,-1)^T$ to a parallelogram whose corresponding vertices are $(2,1)^T,(0,1)^T,(-2,-1)^T$ and $(0,-1)^T$?
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Solution
We need a transformation that maps each original point to the corresponding new point.
Check the coordinates:
$(1,1)\to(2,1)$
$(-1,1)\to(0,1)$
$(-1,-1)\to(-2,-1)$
$(1,-1)\to(0,-1)$
From these mappings, we can observe that $y'=y$.
Also, $x'=x+y$.
So, in homogeneous coordinate form:
$\left[\begin{matrix}x'\\y'\\1\end{matrix}\right]=\left[\begin{matrix}1&1&0\\0&1&0\\0&0&1\end{matrix}\right]\left[\begin{matrix}x\\y\\1\end{matrix}\right]$
Now verify using $(1,1)$:
$x'=1+1=2,\ y'=1$
So, $(1,1)\to(2,1)$.
Therefore, the required transformation matrix is:
$M=\left[\begin{matrix}1&1&0\\0&1&0\\0&0&1\end{matrix}\right]$
Check the coordinates:
$(1,1)\to(2,1)$
$(-1,1)\to(0,1)$
$(-1,-1)\to(-2,-1)$
$(1,-1)\to(0,-1)$
From these mappings, we can observe that $y'=y$.
Also, $x'=x+y$.
So, in homogeneous coordinate form:
$\left[\begin{matrix}x'\\y'\\1\end{matrix}\right]=\left[\begin{matrix}1&1&0\\0&1&0\\0&0&1\end{matrix}\right]\left[\begin{matrix}x\\y\\1\end{matrix}\right]$
Now verify using $(1,1)$:
$x'=1+1=2,\ y'=1$
So, $(1,1)\to(2,1)$.
Therefore, the required transformation matrix is:
$M=\left[\begin{matrix}1&1&0\\0&1&0\\0&0&1\end{matrix}\right]$
Suppose a Bezier curve $P(t)$ is defined by the following four control points in the $xy$-plane:
$P_0=(-2,0),\ P_1=(-2,4),\ P_2=(2,4),\ P_3=(2,0)$
Then which of the following statements are correct?
A. Bezier curve $P(t)$ has degree $3$.
B. $P\left(\frac{1}{2}\right)=(0,3)$.
C. Bezier curve $P(t)$ may extend outside the convex hull of its control points.
$P_0=(-2,0),\ P_1=(-2,4),\ P_2=(2,4),\ P_3=(2,0)$
Then which of the following statements are correct?
A. Bezier curve $P(t)$ has degree $3$.
B. $P\left(\frac{1}{2}\right)=(0,3)$.
C. Bezier curve $P(t)$ may extend outside the convex hull of its control points.
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Solution
Since there are $4$ control points, the Bezier curve is cubic.
So, degree $=4-1=3$. Hence, statement A is true.
For cubic Bezier curve,
$P(t)=(1-t)^3P_0+3t(1-t)^2P_1+3t^2(1-t)P_2+t^3P_3$
At $t=\frac{1}{2}$,
$P\left(\frac{1}{2}\right)=\frac{1}{8}P_0+\frac{3}{8}P_1+\frac{3}{8}P_2+\frac{1}{8}P_3$
$x=\frac{1}{8}(-2)+\frac{3}{8}(-2)+\frac{3}{8}(2)+\frac{1}{8}(2)=0$
$y=\frac{1}{8}(0)+\frac{3}{8}(4)+\frac{3}{8}(4)+\frac{1}{8}(0)=3$
So, $P\left(\frac{1}{2}\right)=(0,3)$. Hence, statement B is true.
Bezier curve always lies inside the convex hull of its control points.
So, statement C is false.
Given below are two statements:
Statement I: The maximum number of sides that a triangle might have when clipped to a rectangular viewport is $6$.
Statement II: In $3D$ graphics, the perspective transformation is nonlinear in $z$.
In light of the above statements, choose the correct answer from the options given below.
Statement I: The maximum number of sides that a triangle might have when clipped to a rectangular viewport is $6$.
Statement II: In $3D$ graphics, the perspective transformation is nonlinear in $z$.
In light of the above statements, choose the correct answer from the options given below.
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Solution
When a triangle is clipped against a rectangular viewport, new vertices may be created at clipping boundaries.
A triangle has $3$ sides initially.
After clipping with a rectangle, the resulting polygon can have at most $6$ sides.
So, Statement I is true.
In $3D$ graphics, perspective transformation involves division by depth value $z$.
Because of this division by $z$, the transformation is not linear in $z$.
So, Statement II is also true.
Hence, both Statement I and Statement II are true.
A triangle has $3$ sides initially.
After clipping with a rectangle, the resulting polygon can have at most $6$ sides.
So, Statement I is true.
In $3D$ graphics, perspective transformation involves division by depth value $z$.
Because of this division by $z$, the transformation is not linear in $z$.
So, Statement II is also true.
Hence, both Statement I and Statement II are true.
UGC NET Computer Science
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