PGDBA 2022 Solutions

These solutions are in interactive format. If you wish to attempt them as timed mock, then please visit this link.

Important Links

Important Links before we start

Interactive Quiz - Question 142

Read the passage and answer the following questions
One of the problems we face in working out where we stand on surveillance is that none of us know exactly how we are being surveilled, and what the coming years might bring. Surveillance technology is developing at breakneck speed, and what seemed science-fiction 10 years ago is today old news. As a thought experiment, consider a hypothetical government that demands that every citizen wears a biometric bracelet that monitors body temperature and heart-rate 24 hours a day. The resulting data is hoarded and analysed by government algorithms. The algorithms will know that you are sick even before you know it, and they will also know where you have been, and who you have met. The chains of infection could be drastically shortened, and even cut altogether. Such a system could arguably stop the epidemic in its tracks within days. Sounds wonderful, right?

The downside is, of course, that this would give legitimacy to a terrifying new surveillance system. If you know, for example, that I clicked on a Fox News link rather than a CNN link, that can teach you something about my political views and perhaps even my personality. But if you can monitor what happens to my body temperature, blood pressure and heart-rate as I watch the video clip, you can learn what makes me laugh, what makes me cry, and what makes me really, really angry.
It is crucial to remember that anger, joy, boredom and love are biological phenomena just like fever and a cough. The same technology that identifies coughs could also identify laughs. If corporations and governments start harvesting our biometric data en masse, they can get to know us far better than we know ourselves, and they can then not just predict our feelings but also manipulate our feelings and sell us anything they want - be it a product or a politician. Biometric monitoring would make Cambridge Analytica's data hacking tactics look like something from the Stone Age. Imagine North Korea in 2030, when every citizen has to wear a biometric bracelet 24 hours a day. If you listen to a speech by the Great Leader and the bracelet picks up the tell-tale signs of anger, you are done for.

You could, of course, make the case for biometric surveillance as a temporary measure taken during a state of emergency. It would go away once the emergency is over. But temporary measures have a nasty habit of outlasting emergencies, especially as there is always a new emergency lurking on the horizon. My home country of Israel, for example, declared a state of emergency during its 1948 War of Independence, which justified a range of temporary measures from press censorship and land confiscation to special regulations for making pudding (I kid you not). The War of Independence has long been won, but Israel never declared the emergency over, and has failed to abolish many of the "temporary" measures of 1948 (the emergency pudding decree was mercifully abolished in 2011).

Q1. In the context of the passage, which of the following statements about biometric surveillance are TRUE?
1. Biometric surveillance could arguably stop an epidemic very fast.
2. Biometric surveillance turns emotions into data.
3. Biometric surveillance can predict our actions.
4. Biometric surveillance can manipulate our feelings.

Interactive Quiz - Question 143

Q2. How does the author define data surveillance?

Interactive Quiz - Question 144

Q3. How might authoritarian governments use biometric surveillance to remain in power?

Interactive Quiz - Question 145

Q4. Many individuals wear smart watches that monitor heart rate, oxygen level and other physical activity. In the context of the passage, does this imply that governments are already practicing biometric surveillance?

Interactive Quiz - Question 146

Q5. What does the author want to convey through the anecdote of the Emergency Pudding Decree?

Interactive Quiz - Question 147

Read the passage and answer the following questions
Doomsayers of the past two centuries have blamed, among other things, novels, the radio, jazz, rock 'n roll, television, horror films, Dungeons & Dragons, video games, the internet, smartphones and social media for the sad decline of the young. John Protzko, a psychologist... wondered whether things might not be quite so gloomy as they seemed. To try to bring some rigour to the question, he went hunting for examples of a cognitive experiment called the marshmallow test. This test, first performed at Stanford University in the 1960s, measures how good young children are at self-control - specifically, whether or not they can defer a small but immediate reward, such as a marshmallow, in favour of a bigger one later. It was one of the first examples of a standardised psychological test, so it gave him plenty of historical data to work with.
The set-up is simple. A child is taken into a room and presented with a choice of sugary snacks. A researcher explains that the child can choose his favourite treat and eat it whenever he likes - but, if he waits 15 minutes, he can have two instead. The researcher then leaves the room. Age is the strongest predictor of successfully resisting the temptation [to take the treat immediately]. Among children of the same age, however, doing well on the test is associated with plenty of good things later in life, from healthy weight to longer school attendance and better exam results.

Dr Protzko...polled 260 experts in child cognitive development, inviting them to predict what he might find. Just over half thought that children would have become worse at delaying gratification - perhaps thinking about a plethora of recent studies into the supposedly deleterious effects of modern technology. Another third predicted no change.

Only $16 \%$ of the experts made the correct prediction. This is, that children have become steadily and significantly better at the test over the past half century. In 1967 , the average waiting time before succumbing to temptation was around three minutes. By 2017, that had risen to eight minutes - an increase of about a minute a decade. And that increase seems to be happening at all levels of ability. The most impulsive children are improving at the same rate as the most prudent.
The rate of increase caught Dr Protzko's eye as well. That rate, a fifth of a standard deviation every decade, is about the same improvement as has been seen in IQ tests over the past 80 years.... The cause of this increase in IQ, which is dubbed the Flynn effect after the psychologist who brought it to the world's attention, remains mysterious - as does whether $\mathrm{Dr}$ Protzko's results are related to it. $I Q$ is associated with the ability to delay gratification, but the correlation is far from perfect.

Q6. Which of the following best reflects the main argument of the passage?

Interactive Quiz - Question 148

Q7. From the passage, what can one infer about the impact of the 'Flynn Effect' on children's performance in the Marshmallow Test?

Interactive Quiz - Question 149

Q8. In the context of the passage, what could the 'plethora of recent studies into the supposedly deleterious effects of modern technology' have concluded?

Interactive Quiz - Question 150

Q9. From the passage it can be inferred that those doing well in the 'Marshmallow Test' are more likely to do well later in adult life because,

Interactive Quiz - Question 151

Q10. In the context of the passage, what do doomsayers intend to convey when they talk of 'the sad decline of the young'?

Interactive Quiz - Question 152

Q11. Choose the combination of words that appropriately completes this sentence: 'You tried to ______ money from me, and now that I've refused, you ______ and _______ me.'

Interactive Quiz - Question 153

Q12. Fill in the blank:
The adjective for metal is metallic. But not for iron, which is _________.

Interactive Quiz - Para Jumbles Q154

Q13. Arrange the sentences in the correct order:
1. Both at the same level of credibility, both at the same level of fakery.
2. And so, at Disneyland, along with Mickey Mouse and the kindly Bears, there must also be, in tactile evidence, Metaphysical Evil (The Haunted Mansion) and Historical Evil (The Pirates), and in the waxwork museums, alongside the Venuses de Milo, we must find grave robbers, Dracula, Jack the Ripper and the Phantom of the Opera.
3. The ideology of this America wants to establish reassurance through imitation.
4. But profit defeats ideology, because the consumers want to be thrilled not only by the guarantee of the Good but also by the shudder of the Bad.

Interactive Quiz - Para Jumbles Q155

Q14. Arrange the sentences in the correct order:
1. Mr. Hernandez is not the first Latin American head of state to be accused of drug trafficking.
2. On 15^th February, 2022, Mr. Juan Orlando Hernandez, the outgoing President of Honduras, was arrested and taken away in handcuffs.
3. But the rot goes particularly deep in Honduras.
4. The arrest was in response to an extradition request from the US relating to a drug trafficking case.

Interactive Quiz - Para Jumbles Q156

Q15. Arrange the sentences in the correct order:
1. He pioneered a new method for making such shapes understandable: instead of drawing them as solids, he made them see-through skeletons, as if constructed of wooden beams.
2. Most of Leonardo's drawings for Pacioli's book, which was finished in 1498, are variations of five shapes known as Platonic solids.
3. He also illustrated more complex shapes such as rhombicuboctahedron, which has twenty-six facets, eight of them equilateral triangles that are bordered by squares.
4. These are polyhedrons that have the same number of faces meeting at each vertex: pyramids, cubes, octahedrons (eight faces), dodecahedrons (twelve) and icosahedrons (twenty).

Interactive Math Quiz - Question 167

Q26. In how many ways can the 26 letters of the English alphabet be arranged such that no two vowels are next to each other in the arrangement?

Solution:
We have 21 consonants and 5 vowels. To ensure no two vowels are adjacent:

Arrange the 21 consonants. This can be done in $21!$ ways.
Place the 5 vowels in the gaps created between the consonants (including before the first consonant and after the last consonant). There are $22$ gaps available.
Choose 5 gaps from the 22 for placing the vowels. This can be done in $^{22} C_{5}$ ways.
Arrange the 5 vowels in these 5 selected gaps. This can be done in $5!$ ways.

Total number of ways:

^{22} C_{5} \times 21! \times 5!

Therefore, the correct answer is:

D .^{22} C_{5} \times 21! \times 5!

Interactive Math Quiz - Question 168

Q27. If a circle passes through the points of intersection of the coordinate axes with the lines

x - λ y + 1 = 0

and

x - 2 y + 3 = 0

, then

λ

equals:

Solution:
The equations of the lines are in intercept form:

\frac{x}{- 1} + \frac{y}{1 / λ} = 1, \frac{x}{- 3} + \frac{y}{3 / 2} = 1.

Using the concept of the power of a point:

O A \cdot O B = O C \cdot O D,

where

O A = - 3

O B = - 1

O C = \frac{1}{λ}

, and

O D = \frac{3}{2}

- 3 \cdot - 1 = \frac{1}{λ} \cdot \frac{3}{2} .

Simplifying:

λ = \frac{1}{2} .

Another possibility arises when

\frac{1}{λ}

and

\frac{3}{2}

coincide:

λ = \frac{2}{3} .

Hence, the two possible values for

λ

are:

\frac{2}{3} or \frac{1}{2} .

Therefore, the correct answer is:

B .

Interactive Math Quiz - Question 169

Q28. For any real number

x

, let

[x]

denote the greatest integer

m

such that

m \leq x

. Then the value of

\int_{0}^{π} [2 \sin x] d x

is:

Solution:
We need to evaluate:

\int_{0}^{π} [2 \sin x] d x,

where

[2 \sin x]

represents the greatest integer less than or equal to

2 \sin x

.

The function

2 \sin x

varies as follows over

[0, π]

: -

2 \sin x = 0

for

x = 0

and

x = π

, -

2 \sin x = 1

for

x \in [\frac{π}{6}, \frac{π}{2}] \cup [\frac{π}{2}, \frac{5 π}{6}]

.

Breaking the integral into intervals:

\int_{0}^{π} [2 \sin x] d x = \int_{0}^{π / 6} 0 d x + \int_{π / 6}^{π / 2} 1 d x + \int_{π / 2}^{5 π / 6} 1 d x + \int_{5 π / 6}^{π} 0 d x .

Evaluating each term:

\int_{0}^{π / 6} 0 d x = 0, \int_{π / 6}^{π / 2} 1 d x = \frac{π}{2} - \frac{π}{6}, \int_{π / 2}^{5 π / 6} 1 d x = \frac{5 π}{6} - \frac{π}{2}, \int_{5 π / 6}^{π} 0 d x = 0.

Summing up:

\int_{0}^{π} [2 \sin x] d x = (\frac{π}{2} - \frac{π}{6}) + (\frac{5 π}{6} - \frac{π}{2}) = \frac{2 π}{3} .

Therefore, the correct answer is:

\frac{2 π}{3} .

Interactive Math Quiz - Question 170

Q29. The number of terms in the expansion of

(1 - x)^{51} {(1 + x + x^{2} + x^{3} + x^{4})}^{50}

is:

Solution:
Consider the given expression:

(1 - x)^{51} {(1 + x + x^{2} + x^{3} + x^{4})}^{50} .

Simplify:

(1 - x)^{51} {(1 + x + x^{2} + x^{3} + x^{4})}^{50} = (1 - x) \cdot (1 - x)^{50} \cdot {(1 + x + x^{2} + x^{3} + x^{4})}^{50} .

Using the formula for the sum of a geometric series:

1 + x + x^{2} + x^{3} + x^{4} = \frac{1 - x^{5}}{1 - x} .

Thus:

(1 - x)^{51} {(1 + x + x^{2} + x^{3} + x^{4})}^{50} = (1 - x) \cdot (1 - x^{5})^{50} .

Expand

(1 - x^{5})^{50}

, which has

51

terms:

(1 - x^{5})^{50} = 1 + terms with powers of x^{5} .

Multiply

(1 - x)

with these

51

terms. Since there is no overlap in powers of

x

, the total number of terms becomes:

51 \cdot 2 = 102.

Therefore, the correct answer is:

102 .

Interactive Math Quiz - Question 171

Q30. A straight line segment

A B

of length

p

moves with its ends on the axes. Let

C

be a point on

A B

such that

A C : B C = 1 : 3

. Then the equation of the locus of

C

is:

Solution:
Let the endpoints of

A B

on the axes be

A (a, 0)

and

B (0, b)

.
The coordinates of

C

dividing

A B

in the ratio

1 : 3

are:

h = \frac{1 \cdot b + 3 \cdot 0}{1 + 3} = \frac{b}{4}, k = \frac{1 \cdot 0 + 3 \cdot a}{1 + 3} = \frac{3 a}{4} .

Thus,

a = \frac{4 k}{3}

and

b = 4 h

.
Since

A B

is a straight line of length

p

, the relation between

a

and

b

is:

a^{2} + b^{2} = p^{2} .

Substituting

a

and

b

{(\frac{4 k}{3})}^{2} + (4 h)^{2} = p^{2} .

Simplify:

\frac{16 k^{2}}{9} + 16 h^{2} = p^{2} \Rightarrow 16 (9 h^{2} + k^{2}) = 9 p^{2} .

Replace

h

and

k

with

x

and

y

16 (9 x^{2} + y^{2}) = 9 p^{2} .

Therefore, the correct answer is:

B .

Interactive Math Quiz - Question 172

Q31. If

f (x) = \sin (\log_{10} x)

and

h (x) = \cos (\log_{10} x)

, then

\frac{1}{2} (h (\frac{x}{y}) + h (x y)) - f (x) f (y)

equals:

Solution:
Given:

f (x) = \sin (\log_{10} x), h (x) = \cos (\log_{10} x) .

The expression becomes:

\frac{1}{2} [h (\frac{x}{y}) + h (x y)] - f (x) f (y) .

Substitute:

h (\frac{x}{y}) = \cos (\log_{10} \frac{x}{y}), h (x y) = \cos (\log_{10} (x y)) .

Thus:

\frac{1}{2} [\cos (\log_{10} \frac{x}{y}) + \cos (\log_{10} (x y))] - \sin (\log_{10} x) \sin (\log_{10} y) .

Using trigonometric identities:

\cos (A) + \cos (B) = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}),

and

\cos (A) \cos (B) - \sin (A) \sin (B) = \cos (A + B) .

Let

\log_{10} x = a

and

\log_{10} y = b

. Then:

\frac{1}{2} [2 \cos (a) \cos (b)] - \sin (a) \sin (b) = \cos (a) \cos (b) - \sin (a) \sin (b) .

Simplify:

\cos (a + b) = \cos (\log_{10} (x) + \log_{10} (y)) = \cos (\log_{10} (x y)) .

Therefore, the correct answer is:

A .

Interactive Math Quiz - Question 173

Q32. Let

a_{1}, a_{2} \in R

be such that

| a_{1} - a_{2} | = 10

. Consider:

f (x) = | \begin{array}{ccc} 1 & a_{2} & a_{1} \\ 1 & a_{2} & 2 a_{1} - x \\ 1 & 2 a_{2} - x & a_{1} \end{array} |

Then, the largest value of

f (x)

is:

Solution:
Given:

f (x) = | \begin{array}{ccc} 1 & a_{2} & a_{1} \\ 1 & a_{2} & 2 a_{1} - x \\ 1 & 2 a_{2} - x & a_{1} \end{array} |

Expanding the determinant:

f (x) = - (a_{1} - x) (a_{2} - x) .

Simplifying:

f (x) = - a_{1} a_{2} + (a_{1} + a_{2}) x - x^{2} .

To find the maximum value, calculate

f^{'} (x)

f^{'} (x) = - 2 x + (a_{1} + a_{2}) .

Setting

f^{'} (x) = 0

x = \frac{a_{1} + a_{2}}{2} .

x = \frac{a_{1} + a_{2}}{2}

, calculate

f (x)

f (\frac{a_{1} + a_{2}}{2}) = - (\frac{a_{1} - a_{2}}{2}) (\frac{a_{2} - a_{1}}{2}) = \frac{{(a_{2} - a_{1})}^{2}}{4} .

Given

| a_{1} - a_{2} | = 10

f (\frac{a_{1} + a_{2}}{2}) = \frac{100}{4} = 25.

Therefore, the largest value of

f (x)

is:

C .

Interactive Math Quiz - Question 174

Q33. Consider a function

f (x) = x^{2} + p x + q

such that the roots of

f (x) = 0

are positive and distinct. Let the arithmetic mean, geometric mean, and harmonic mean of the two roots be

a

b

, and

c

, respectively. Then, which of the following statements is TRUE?

Solution:
Given:

f (x) = x^{2} + p x + q

, and the roots are positive and distinct.
The arithmetic mean (AM), geometric mean (GM), and harmonic mean (HM) of the roots satisfy:

AM \geq GM \geq HM .

Since the roots are distinct, we have

a > b > c

, where

a, b,

and

c

are the arithmetic, geometric, and harmonic means, respectively.
The parabola

f (x)

is decreasing in the interval before

x = - \frac{p}{2}

, which includes

a, b,

and

c

.
Therefore:

f (c) > f (b) > f (a) .

Hence, the correct option is:

C .

Interactive Math Quiz - Question 175

Q34. Suppose

E_{1}

and

E_{2}

are two independent events, each having probability

p

. If

P (E_{1} \cup E_{2}) = \frac{5}{9}

, what is the value of

p

Solution:
Given:

P (E_{1}) = P (E_{2}) = p

, and

E_{1}

and

E_{2}

are independent events.
By the formula:

P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2})

Since

E_{1}

and

E_{2}

are independent:

P (E_{1} \cap E_{2}) = p^{2} .

Substituting the values:

p + p - p^{2} = \frac{5}{9} .

Simplify:

2 p - p^{2} = \frac{5}{9} .

Rearrange to form a quadratic equation:

9 p^{2} - 18 p + 5 = 0.

Solve using the quadratic formula:

p = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{18 \pm \sqrt{324 - 180}}{18} .

Simplify:

p = \frac{18 \pm 12}{18} .

Discard

p = \frac{30}{18}

as it exceeds 1:

p = \frac{1}{3} .

Thus, the correct answer is:

C .

Interactive Math Quiz - Question 176

Q35. Let

P

be a

19 \times 19

matrix whose entries in both the diagonals are all equal to 1 and all other entries are equal to 0. Then,

rank (P)

is equal to:

Solution:
The given matrix

P

is a

19 \times 19

matrix where the entries in both diagonals are all equal to 1 and all other entries are 0.
The matrix has identical bottom 9 rows to the top 9 rows, meaning that these rows do not contribute to the rank.
Thus, the rank of the matrix is the number of linearly independent rows, which is 10.
Therefore, the rank of

P

10

, so the correct answer is Option 'B'.

Interactive Math Quiz - Question 177

Q36. The limit

lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x}}

is:

Solution:
We are asked to evaluate the following limit:

lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x}} = y (say)

Expanding using the binomial approximation for small

x

{(1 + \frac{\tan x}{x} - 1)}^{\frac{1}{x}} \approx (1 + \frac{1}{x} (\frac{\tan x - x}{x}))

Applying L'Hôpital's Rule twice, we can deduce:

lim_{x \to 0} \frac{\tan x - x}{x^{2}} = 0

Hence, the limit approaches

1

, so the correct answer is Option 'A'.

Interactive Math Quiz - Question 178

Q37. If the line

\frac{x}{a} + \frac{y}{b} = 1

is a tangent to the curve

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1,

then which of the following is TRUE?

Solution:
Given curve is

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1

The slope of the curve is:

\frac{2}{3} x^{- \frac{1}{3}} + \frac{2}{3} y^{- \frac{1}{3}} \frac{d y}{d x} = 0

\Rightarrow \frac{d y}{d x} = - {(\frac{y}{x})}^{\frac{1}{3}}

The slope of the given line is:

m = - \frac{b}{a}

At the point of tangency

(x_{1}, y_{1})

, we have:

- {(\frac{y_{1}}{x_{1}})}^{\frac{1}{3}} = \frac{- b}{a}

This implies:

a^{3} y_{1} - b^{3} x_{1} = 0

Also, the point

(x_{1}, y_{1})

satisfies the equation of the line:

b x_{1} + a y_{1} = a b

Solving the system of equations, we find:

x_{1} = \frac{a^{3}}{a^{2} + b^{2}}, y_{1} = \frac{b^{3}}{a^{2} + b^{2}}

Substituting these values into the equation of the curve:

{(\frac{a^{3}}{a^{2} + b^{2}})}^{\frac{2}{3}} + {(\frac{b^{3}}{a^{2} + b^{2}})}^{\frac{2}{3}} = 1

Simplifying, we get:

(a^{2} + b^{2})^{\frac{1}{3}} = 1

Hence:

a^{2} + b^{2} = 1

Therefore, the correct answer is Option 'A'.

Interactive Math Quiz - Question 179

Q38. Let

P

Q

R

be any three sets. Consider the following two statements:

(I) $P - (Q - R) = (P - Q) \cup R$
(II) $P - (Q \cup R) = (P - Q) - R$

Which of the following is TRUE?

Solution:
Statement (I):

P - (Q - R) = P - (Q \cap R^{'}) = P \cap (Q^{'} \cup R)

And

(P - Q) \cup R = (P \cap Q^{'}) \cup R = (P \cup R) \cap (Q^{'} \cup R)

R \subset P

, then:

(P \cup R) \cap (Q^{'} \cup R) = P \cap (Q^{'} \cup R)

Hence, Statement (I) is true if

R \subset P

. Statement (II):

P - (Q \cup R) = P \cap (Q \cup R)^{'} = P \cap (Q^{'} \cap R^{'}) = P \cap Q^{'} \cap R^{'}

And

(P - Q) - R = (P \cap Q^{'}) \cap R^{'} = P \cap Q^{'} \cap R^{'}

Hence, Statement (II) is always correct. Therefore, the correct answer is Option 'D' (II is always correct and I is correct if and only if

R \subset P

Interactive Math Quiz - Question 180

Q39. The domain of the function

f (x) = \sqrt{\frac{1 - | x |}{2 - | x |}}

is:

Solution:
The valid domain of

f (x)

is where

\frac{1 - | x |}{2 - | x |} ⩾ 0

and

2 - | x | \neq 0

(i.e.,

x \neq 2

and

x \neq - 2

).
Solving the inequality:

\frac{1 - | x |}{2 - | x |} ⩾ 0 and x \neq 2, - 2

The domain is:

(- \infty, - 2) \cup [- 1, 1] \cup (2, \infty)

Therefore, the correct answer is Option 'B'.

Interactive Math Quiz - Question 181

Q40. For which values of

λ \in R

will the system of linear equations

\begin{aligned} x + y + z & = 2 \\ x + 2 y + z & = - 2 \\ x + y + (λ - 5) z & = λ \end{aligned}

have a unique solution?

Solution:
The system of equations can be written in matrix form as:

A x = B

where

A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & (λ - 5) \end{matrix}], B = [\begin{matrix} 2 \\ - 2 \\ λ \end{matrix}]

The augmented matrix is:

[\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 1 & 2 & 1 & - 2 \\ 1 & 1 & (λ - 5) & λ \end{array}]

After row operations, we get:

[\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & - 4 \\ 0 & 0 & (λ - 6) & (λ - 2) \end{array}]

For the system to have a unique solution, the rank of the coefficient matrix must be equal to the rank of the augmented matrix and both should be equal to the number of variables (3). The rank of the matrix will be 3 if

λ \neq 6

, because the third row will have a non-zero pivot for all

λ \neq 6

. Therefore, the solution is valid when

λ \neq 6

. Thus, the correct answer is Option 'B'.

Interactive Math Quiz - Question 182

Q41. If

α

and

β

are the two roots of the quadratic equation

x^{2} + x + 1 = 0

then the equation whose roots are

α^{2022}

and

β^{2022}

is:

Solution:
The roots

α

and

β

satisfy the equation:

x^{2} + x + 1 = 0

which implies:

α + β = - 1 and α β = 1

Also,

α^{2} + α + 1 = 0

and

β^{2} + β + 1 = 0

imply:

α^{3} = 1 and β^{3} = 1

Therefore:

α^{2022} = {(α^{3})}^{674} = 1 and similarly β^{2022} = 1

Thus:

α^{2022} + β^{2022} = 2

The quadratic equation whose roots are

α^{2022}

and

β^{2022}

has the form:

x^{2} - 2 x + 1 = 0

Therefore, the correct answer is Option 'B'.

Interactive Math Quiz - Question 183

Q42. The area of the region bounded by the curves

y^{2} = 4 x

and

y = 2 x

is:

Solution:
The intersection points of the curves are found by setting the equations equal:

(2 x)^{2} = 4 x \Rightarrow x^{2} = x

\Rightarrow x (x - 1) = 0 \Rightarrow x = 0 or x = 1

The intersection points are

(0, 0)

and

(1, 2)

. To calculate the area, integrate with respect to

y

, where

x

ranges from

\frac{y^{2}}{4}

\frac{y}{2}

, and

y

ranges from 0 to 2:

Area = \int_{0}^{2} (\frac{y}{2} - \frac{y^{2}}{4}) d y

= {[\frac{y^{2}}{4} - \frac{y^{3}}{12}]}_{0}^{2}

= 1 - \frac{8}{12} = \frac{1}{3}

Therefore, the correct answer is Option 'C'.

Interactive Math Quiz - Question 184

Q43. The highest power of 7 that divides

2022!

is:

Solution:
The highest power of

a

that divides

b!

is calculated as:

⌊ \frac{b}{a} ⌋ + ⌊ \frac{b}{a^{2}} ⌋ + ⌊ \frac{b}{a^{3}} ⌋ + \dots

Applying this for

a = 7

and

b = 2022

⌊ \frac{2022}{7} ⌋ + ⌊ \frac{2022}{49} ⌋ + ⌊ \frac{2022}{343} ⌋ + ⌊ \frac{2022}{2401} ⌋

= 288 + 41 + 5 + 0 + \dots = 334

Therefore, the correct answer is Option 'B'.

Interactive Math Quiz - Question 185

Q44. Consider the function

f : R \to R

defined by

f (x) = max {x, x^{2}} - min {x, x^{2}} .

Then

f

is differentiable:

Solution:
The function

f (x)

can be split into piecewise segments:

f (x) = {\begin{cases} x^{2} - x & if x < - 1, \\ x^{2} - x & if - 1 \leq x < 0, \\ x - x^{2} & if 0 \leq x < 1, \\ x^{2} - x & if x \geq 1. \end{cases}

The derivative of

f (x)

in each region is:

f^{'} (x) = {\begin{cases} 2 x - 1 & if x < - 1, \\ 2 x - 1 & if - 1 \leq x < 0, \\ 1 - 2 x & if 0 \leq x < 1, \\ 2 x - 1 & if x \geq 1. \end{cases}

x = 0

, the right-hand limit (RHL) is

1 - 2 (0) = 1

and the left-hand limit (LHL) is

2 (0) - 1 = - 1

. Hence,

f (x)

is not differentiable at

x = 0

. At

x = 1

, the RHL is

2 (1) - 1 = 1

and the LHL is

1 - 2 (1) = - 1

. Hence,

f (x)

is not differentiable at

x = 1

. Therefore, the function

f (x)

is differentiable everywhere except at

x = 0

and

x = 1

. Thus, the correct answer is Option 'C'.

Interactive Math Quiz - Question 186

Q45. Let

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

be the distinct fifth roots of 1. Then the value of

a_{1}^{2022} + a_{2}^{2022} + a_{3}^{2022} + a_{4}^{2022} + a_{5}^{2022}

is:

Solution:
The

a_{i}

's are the distinct fifth roots of unity. That means:

a_{i}^{5} = 1 \forall i \in {1, 2, 3, 4, 5} .

One of these roots is

a_{1} = 1

, and the other roots are powers of a primitive fifth root of unity, denoted

ω

. So the roots are:

a_{1} = 1, a_{2} = ω, a_{3} = ω^{2}, a_{4} = ω^{3}, a_{5} = ω^{4} .

We want to compute:

a_{1}^{2022} + a_{2}^{2022} + a_{3}^{2022} + a_{4}^{2022} + a_{5}^{2022} .

Since

a_{i}^{5} = 1

for all

i

, we can reduce the exponents modulo 5:

a_{1}^{2022} = a_{1}^{2} = 1^{2} = 1, a_{2}^{2022} = ω^{2022} = ω^{2}, a_{3}^{2022} = ω^{4044} = ω^{4},

a_{4}^{2022} = ω^{8088} = ω^{3}, a_{5}^{2022} = ω^{10110} = ω .

Now, sum these values:

a_{1}^{2022} + a_{2}^{2022} + a_{3}^{2022} + a_{4}^{2022} + a_{5}^{2022} = 1 + ω + ω^{2} + ω^{3} + ω^{4} .

The sum of all the distinct fifth roots of unity is 0, so:

1 + ω + ω^{2} + ω^{3} + ω^{4} = 0.

Thus, the value of the sum is 0, and the correct answer is Option 'A'.

Interactive Math Quiz - Question 187

Q46. Suppose

2 \sin θ - 5 \cos θ = \sqrt{13}

. Then the expression

(2 \cos θ + 5 \sin θ)

equals:

Solution:
Let

2 \cos θ + 5 \sin θ = x

. The given equation is:

2 \sin θ - 5 \cos θ = \sqrt{13} .

Squaring both the equations and adding them:

4 \cos^{2} θ + 25 \sin^{2} θ + 20 \sin θ \cos θ = x^{2}

and

4 \sin^{2} θ + 25 \cos^{2} θ - 20 \sin θ \cos θ = 13.

Adding these equations:

4 (\sin^{2} θ + \cos^{2} θ) + 25 (\sin^{2} θ + \cos^{2} θ) = x^{2} + 13,

which simplifies to:

29 = x^{2} + 13,

so:

x^{2} = 16 \Rightarrow x = \pm 4.

Therefore, the correct answer is option 'D'.

Interactive Math Quiz - Question 188

Q47. If

P

and

Q

are two matrices such that

P Q = Q

and

Q P = P

, then

P^{3} + Q^{3}

is equal to:

Solution:
Given that

P Q = Q

and

Q P = P

, we have:

P Q P = Q P \Rightarrow P^{2} = Q P,

P^{3} = P Q P = Q P = P .

Also, from

Q P = P

, we get:

Q P Q = P Q \Rightarrow Q^{2} = P Q,

Q^{3} = Q P Q = P Q = Q .

Therefore:

P^{3} + Q^{3} = P + Q .

Thus, the correct answer is option 'C'.

Interactive Math Quiz - Question 189

Q48. For what value of

m

will the equation

\frac{x^{2} - b x}{a x - c} = \frac{m - 1}{m + 1}

have roots equal in magnitude but opposite in sign?

Solution:
Let the equation have roots equal in magnitude but opposite in sign, say

α

and

- α

(x - α) (x + α) = 0 \Rightarrow x^{2} - α^{2} = 0.

The given equation is:

(m + 1) x^{2} - (m + 1) b x = (m - 1) a x - (m - 1) c .

Simplifying:

x^{2} - b x - \frac{(m - 1)}{(m + 1)} a x + \frac{(m - 1)}{(m + 1)} c = 0.

x^{2} - (b + \frac{(m - 1) a}{m + 1}) x + \frac{(m - 1)}{m + 1} c = 0.

For there to be no term in

x

, we have:

b + \frac{(m - 1) a}{m + 1} = 0.

Solving:

\frac{(1 - m) a}{1 + m} = b \Rightarrow a - a m = b + b m \Rightarrow \frac{a - b}{a + b} = m .

Thus, the correct value of

m

\frac{a - b}{a + b}

, which corresponds to option 'B'.

Interactive Math Quiz - Question 190

Q49. An airplane is observed to be approaching a point that is at a distance of 20 km from the point of observation such that the angle of elevation is

60^{\circ}

. Then the height of the airplane above the ground is:

Solution:
Using basic trigonometry, we know that the height

h

of the airplane is given by:

h = 20 \sin 60^{\circ} = 20 \times \frac{\sqrt{3}}{2} = 10 \sqrt{3} .

Therefore, the height of the airplane above the ground is

10 \sqrt{3} km

, which corresponds to option 'C'.

Interactive Math Quiz - Question 191

Q50. Which of the following equations will have positive integer solutions?

Solution:
Let's first analyze each option: **Option A**:

x^{2} + y^{2} = 2025

. We can find integer solutions using the Pythagorean theorem:

45^{2} = 2025 \Rightarrow 45^{2} = 9^{2} \cdot 5^{2} = 9^{2} (3^{2} + 4^{2})

= 27^{2} + 36^{2} = 2025

So, the correct solution is

x = 27, y = 36

, which are positive integers. Therefore, **Option A** is correct. **Option B**:

x^{2} + y^{2} = 2023

. Since 2023 is odd, one of

x

y

must be odd, and the other must be even. However, the equation does not allow for integer solutions, as shown in the detailed working. **Option C**:

x^{2} - y^{2} = 2022

. The equation can be factored as:

(x - y) (x + y) = 2022

This equation cannot have integer solutions where both

x

and

y

are positive integers, as explained in the working. **Option D**:

x y (x - y) = 2021

. Since 2021 is an odd number, and none of the cases where

x

and

y

are both even or odd allow for integer solutions, there are no integer solutions for this equation. Therefore, the correct answer is **Option A**.

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PGDBA 2022 Solutions

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