PGDBA 2023 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 112

Read the passage and answer the following questions
In the animal kingdom, mimics are a dime a dozen. Stick insects pretend to be twigs. Hawk-moth caterpillars resemble venomous snakes. Edible heliconid butterflies disguise themselves with the wing patterns of noxious ones, and noxious ones copy each other to make it easier for predators to learn what not to eat. All these examples, though, are visual. Auditory mimicry is rarer. But, as he describes in Current Biology, Danilo Russo of the University of Naples Federico II thinks he has found a novel case of it. Some bats, he believes, mimic angry bees, wasps and hornets in order to scare away owls that might otherwise eat them. Dr Russo first noticed the propensity of greater mouse-eared bats to buzz a few years ago, when he was collecting them...to study their ecology. The noise struck him as similar to the sound of hornets that inhabited the area of southern Italy he was working in. That led him to wonder whether bat buzzing was a form of mimicry which helped its practitioners to scare off would-be predators. To test this idea, he... and a colleague...first recorded the buzzing that captured bats made when handled. Then, having donned suitable protective clothing, they embarked on the more dangerous task of recording the buzzing made, en masse, by four different species of Hymenoptera: European paper wasps; buff-tailed bumblebees; European hornets; and domestic honeybees.... For the next part of their experiment Dr Russo and Dr Ancillotto recruited the services of 16 captive owls-eight barn and eight tawny. Both of these species are known to hunt bats. The researchers put the owls, one at a time, in an enclosure equipped with branches for them to perch on, and also two boxes with holes in them. The boxes resembled the sorts of cavities in trees that owls would explore in the wild for food. They placed a loudspeaker alongside one of the boxes and, after the birds had settled in, broadcast through it five seconds of uninterrupted bat buzzing and a similar amount of insect buzzing three times in a row for each noise. As a control, they broadcast in like manner several non-buzzing sounds made by bats. During the broadcasts (which occurred in random order) and for five minutes thereafter, they videoed the owls. The videos were then analysed, by an independent observer, without benefit of their soundtracks. The results were unequivocal. When they heard both the bat buzzings and the hornet buzzings the owls moved as far from the speakers as they could manage. In contrast, when the non-buzzing bat sounds were played, they crept closer. Dr Russo and Dr Ancillotto believe this is the first reported case of a mammal using acoustic mimicry to scare away a predator. They strongly suspect, however, that it is not unique. Anecdotes suggest several birds and also small mammals, such as dormice-particularly species that dwell in trees and, like dormice, in rock cavities-make buzzing noises when their hidey-holes are disturbed. This has not yet been documented formally as acoustic mimicry. But, given the propensity for venomous buzzing insects to dwell in those sorts of places too, and also the fear that these insects generate in other species, human beings included, Dr Russo thinks this may well be what is going on. He therefore predicts that when these other buzzes are recorded and analysed the results will show that acoustic mimicry by vertebrates of stinging insects is far more widespread than currently realised.

Q1. In the examples of defensive mechanism exhibited by animals given below, which is the odd one out?

Interactive Quiz - Question 113

Q2. From the passage it can be inferred that the species Hymenoptera use the following mechanism for defence:

Interactive Quiz - Question 114

Q3. Which of the following outcomes of his experiments would have disproved Mr. Danilo Russo's hypothesis that mimicry was a defensive mechanism?

Interactive Quiz - Question 115

Q4. From the passage it can be inferred that the following animal/bird would not be able to use acoustic mimicry as a defensive mechanism:

Interactive Quiz - Question 116

Q5. Which of the following mechanisms to prevent robbery would be the best example of human beings using mimicry as a defence mechanism?

Interactive Quiz - Question 117

Read the passage and answer the following questions
Familiar though his name may be to us, the storyteller in his living immediacy is by no means a present force. He has already become something remote from us and something that is getting even more distant. To present someone like Leskov as a storyteller does not mean bringing him closer to us but, rather, increasing our distance from him. Viewed from a certain distance, the great, simple outlines which define the storyteller stand out in him, or rather, they become visible in him, just as in a rock a human head or an animal's body may appear to an observer at the proper distance and angle of vision. This distance and this angle of vision are prescribed for us by an experience which we may have almost every day. It teaches us that the art of storytelling is coming to an end. Less and less frequently do we encounter people with the ability to tell a tale properly. More and more often there is embarrassment all around when the wish to hear a story is expressed. It is as if something that seemed inalienable to us, the securest among our possessions, were taken from us: the ability to exchange experiences. The earliest symptom of a process whose end is the decline of storytelling is the rise of the novel at the beginning of modern times. What distinguishes the novel from the story (and from the epic in the narrower sense) is its essential dependence on the book. The dissemination of the novel became possible only with the invention of printing. What can be handed on orally, the wealth of the epic, is of a different kind from what constitutes the stock in trade of the novel. What differentiates the novel from all other forms of prose literature -the fairy tale, the legend, even the novella-is that it neither comes from oral tradition nor goes into it. This distinguishes it from storytelling in particular. The storyteller takes what he tells from experience-his own or that reported by others. And he in turn makes it the experience of those who are listening to his tale. The novelist has isolated himself. The birthplace of the novel is the solitary individual, who is no longer able to express himself by giving examples of his most important concerns, is himself uncounseled, and cannot counsel others. To write a novel means to carry the incommensurable to extremes in the representation of human life. In the midst of life's fullness, and through the representation of this fullness, the novel gives evidence of the profound perplexity of the living. Even the first great book of the genre, Don Quixote, teaches how the spiritual greatness, the boldness, the helpfulness of one of the noblest of men, Don Quixote, are completely devoid of counsel and do not contain the slightest scintilla of wisdom. If now and then, in the course of the centuries, efforts have been made-most effectively, perhaps, in Wilhelm Meisters Wanderjahre-to implant instruction in the novel, these attempts have always amounted to a modification of the novel form. The Bildungsroman, on the other hand, does not deviate in any way from the basic structure of the novel. By integrating the social process with the development of a person, it bestows the most frangible justification on the order determining it. The legitimacy it provides stands in direct opposition to reality. Particularly in the Bildungsroman, it is this inadequacy that is actualized.

Q6. Why is the 'art of storytelling' coming to an end?

Interactive Quiz - Question 118

Q7. What is the difference between the novel and the story?

Interactive Quiz - Question 119

Q8. It can be inferred from the passage that:

Interactive Quiz - Question 120

Q9. What does the author mean by 'frangible justification' in the passage?

Interactive Quiz - Question 121

Q10. What does the author mean by, "Even the first great book of the genre, Don Quixote, teaches how the spiritual greatness, the boldness, the helpfulness of one of the noblest of men, Don Quixote, are completely devoid of counsel and do not contain the slightest scintilla of wisdom."

Interactive Quiz - Para Jumbles

Q11. Para Jumbles - Arrange the sentences in the correct order:

1. Then, when he spoke, he did not explicitly reject the result, though he did not concede, either.
2. For months, he had insinuated that Luiz Inacio Lula da Silva, his rival, could only win if it was rigged.
3. When Lula won by a mere 1.8 per cent points, many people expected Mr. Bolsonaro to contest the result.
4. When Jair Bolsonaro lost his bid for re-election on October 30 he said nothing for 44 hours.

Interactive Quiz - Para Jumbles

Q12. Para Jumbles - Arrange the sentences in the correct order:

1. The technocracy of professional sport has managed to impose a soccer of lightning speed and brute strength, a soccer that negates joy, kills fantasy, and outlaws daring.
2. Luckily, on the field you can still see, even if only once in a long while, some insolent rascal who sets aside the script and commits the blunder of dribbling past the entire opposing side, the referee, and the crowd in the stands, all for the carnal delight of embracing the forbidden adventure of freedom.
3. Play has become spectacle, with few protagonists and many spectators, soccer for watching.
4. And that spectacle has become one of the most profitable businesses in the world, organized not to facilitate play but to impede it.

Interactive Quiz - Para Jumbles

Q13. Para Jumbles - Arrange the sentences in the correct order:

1. For example, the $2008-2009$ U.S. recession triggered by the bursting of the subprime mortgage bubble was extremely severe.
2. However, while the economic downturn that technically began in December 2007 and ended in June 2009 was the longest in the U.S. since World War II, it only lasted about 18 months.
3. The major difference between a recession and a depression is that a depression is much more severe and long-lasting.
4. The U.S. unemployment rate nearly doubled from about 5 % to 9.5 % and the S&P 500 dropped by more than 50 % from peak to trough.

Interactive Quiz - Fill in the Blanks

Q14. Fill in the blank:

Arriving late for the interview __________ my chances of getting the job.

Interactive Quiz - Fill in the Blanks

Q15. Fill in the blank:

In case of any discrepancy, please __________ to us by email.

Interactive Math Quiz - Question 127

Logical reasoning Set

A museum has five halls. Each hall has a separate entry ticket which must be purchased right before entering that hall. The ticket price for each hall is non-zero and unique. Moreover, the ticket price for each hall is a multiple of 30. Hall-3 and Hall-1 have the maximum and minimum ticket prices, respectively. Ticket price of Hall-5 is twice that of Hall-4. A visitor must visit Hall-1 first before deciding the sequence (clockwise, i.e. 1->5->4->3->2 OR anti-clockwise, i.e. 1->2->3->4->5) of visiting the remaining halls. Three photographers - Amardeep, Barnali, and Chanchal, visited all the halls by purchasing tickets worth Rs 450 each. Moving from one hall to another took five minutes for each of them. Table 1 below provides the time spent by the three photographers in each hall. Visiting a hall is considered complete only after a photographer spends the specified time (as given in Table 1) in that hall. Consider the time taken to purchase the ticket as negligible. The floor plan of the museum is given in Fig. 1 (each circle represents a hall, e.g. circle numbered 1 represents Hall-1).

The following additional information is given:
1. Barnali's sequence of visiting the halls was different from Amardeep's and Chanchal's.
2. Chanchal entered Hall-1 at 10:08 Hrs. She spent Rs 120 by 10:40 Hrs.
3. Amardeep left Hall-2 two minutes after Chanchal entered that hall. Amardeep entered Hall-5 five minutes after Barnali entered that hall.

Q16. Who was the last photographer to complete visiting Hall-3, and at what time did the last photographer complete visiting Hall-3?

Solution:
Given:

Total ticket expenditure = ₹450
Ticket prices are multiples of ₹30
Hall-5 ticket price = $2 \times$ Hall-4
Hall-3 has the highest price, and Hall-1 has the lowest

Let’s assume ticket prices:

Hall-1: ₹30 (minimum)
Hall-2: ₹60
Hall-4: ₹90
Hall-5: ₹180 (twice Hall-4)
Hall-3: ₹90

By 11:35 Hrs, Barnali visited three halls: Hall-1, Hall-2, and Hall-4.

Total spent = ₹30 + ₹60 + ₹90 = ₹210

Correct Answer: D (₹210)

Interactive Math Quiz - Question 128

Q17. At what time did Chanchal complete visiting all five halls?

Interactive Math Quiz - Question 129

Q18. In which of the following halls were Barnali and Chanchal present at the same time?

Interactive Math Quiz - Question 130

Q19. Choose the correct statement.

Interactive Math Quiz - Question 131

Q20. Who was the last photographer to complete visiting Hall-3, and at what time did the last photographer complete visiting Hall-3?

Interactive Math Quiz - Question 26

Q26. Let $M = [\begin{array}{ccc} 0 & 1 & - α \\ - 1 & 0 & 5 \\ α & - 5 & 0 \end{array}]$ and $y = (\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \end{array})$ , where $α$ is non-zero and $y_{1}, y_{2}, y_{3}$ are real numbers. Let $M x = y$ for some vector $x$ . Then the value of $x^{T} y$ is:

Solution:
Let

M = [\begin{array}{ccc} 0 & 1 & - α \\ - 1 & 0 & 5 \\ α & - 5 & 0 \end{array}]

, and

x = [\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}]

.

From

M x = y

, we get:

y = [\begin{matrix} x_{2} - α x_{3} \\ - x_{1} + 5 x_{3} \\ α x_{1} - 5 x_{2} \end{matrix}]

To compute

x^{T} y

x^{T} y = [\begin{array}{ccc} x_{1} & x_{2} & x_{3} \end{array}] \cdot [\begin{matrix} x_{2} - α x_{3} \\ - x_{1} + 5 x_{3} \\ α x_{1} - 5 x_{2} \end{matrix}]

Expanding and simplifying, we find:

x^{T} y = x_{1} (x_{2} - α x_{3}) + x_{2} (- x_{1} + 5 x_{3}) + x_{3} (α x_{1} - 5 x_{2}) = 0

Therefore, the correct answer is

C

Quiz - Question 137

Q27. Suppose $ \sin \theta=\frac{3}{5} $, where $ \theta $ is an acute angle. Then the value of $ \left(500 \sin^4 \frac{\theta}{2} + 400 \sin^2 \frac{\theta}{2} \right) $ is:

Interactive Math Quiz - Question 28

Q28. The value of the integral $\int_{0}^{π / 2} \frac{\sin x}{\sin x + \cos x} d x$ is:

Solution:
Let

I = \int_{0}^{π / 2} \frac{\sin x}{\sin x + \cos x} d x

.

Using the property

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

, we get:

I = \int_{0}^{π / 2} \frac{\sin (π / 2 - x)}{\sin (π / 2 - x) + \cos (π / 2 - x)} d x

Simplifying further:

I = \int_{0}^{π / 2} \frac{\cos x}{\cos x + \sin x} d x

Adding the original integral and its transformed version:

2 I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sin x + \cos x} d x = \int_{0}^{π / 2} 1 d x = \frac{π}{2}

Therefore:

I = \frac{π}{4}

**Final Answer:**

A

Interactive Math Quiz - Question 29

Q29. The value of $\frac{2}{0! + 1! + 2!} + \frac{3}{1! + 2! + 3!} + \dots + \frac{n}{(n - 2)! + (n - 1)! + n!}$ is:

Solution:
General term

T_{n} = \frac{n}{(n - 2)! + (n - 1)! + n!}

n! = n \cdot (n - 1) \cdot (n - 2)!, (n - 1)! = (n - 1) \cdot (n - 2)!

Substituting, we get:

T_{n} = \frac{n}{(n - 2)! + (n - 1) (n - 2)! + n (n - 1) (n - 2)!}

Factoring

(n - 2)!

T_{n} = \frac{1}{(n - 2)!} [\frac{n}{1 + (n - 1) + n (n - 1)}]

Simplifying further:

T_{n} = \frac{1}{n!} - \frac{1}{(n - 1)!}

Expanding the series, we observe cancellation diagonally:

S_{n} = (1 - \frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + \dots + (\frac{1}{(n - 1)!} - \frac{1}{n!})

Final result:

S_{n} = 1 - \frac{1}{n!}

**Final Answer:**

C

Interactive Math Quiz - Question 31

Q31. The coefficient of $x^{2} y^{6} z^{4}$ in the expansion of $(y z + x y)^{6} {(1 - \frac{z}{x})}^{6}$ is:

Solution:
Expanding

(y z + x y)^{6} {(1 - \frac{z}{x})}^{6}

{[(y z + x y) (1 - \frac{z}{x})]}^{6}

Simplifying the inner expression:

y z - \frac{y z^{2}}{x} + x y - y z = x y - \frac{y z^{2}}{x}

The expression becomes:

{(x y - \frac{y z^{2}}{x})}^{6}

General term:

T_{r + 1} =^{6} C_{r} (x y)^{6 - r} {(\frac{y z^{2}}{x})}^{r}

Simplify:

T_{r + 1} =^{6} C_{r} (x)^{6 - r} (y)^{6} (z)^{2 r} (x)^{- r}

Combine powers of

x

T_{r + 1} =^{6} C_{r} (x)^{6 - 2 r} (y)^{6} (z)^{2 r}

Comparing with

x^{2} y^{6} z^{4}

6 - 2 r = 2 \Rightarrow r = 2

Coefficient is:

^{6} C_{2} = \frac{6 \cdot 5}{2} = 15

**Final Answer:**

B

Interactive Math Quiz - Question 32

Q32. A man standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is $60^{\circ}$ . When he goes 48 meters away from the bank along the line joining the person and the tree, he finds the angle to the tree to be $30^{\circ}$ . Then the height of the tree is:

Solution:
Let

A B = h

(Height of tree),

C B = x

(Initial distance),

D B = x + 48

(Final distance).

In

△ A B C

\tan 60^{\circ} = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} (i)

△ A B D

\tan 30^{\circ} = \frac{h}{x + 48} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 48} \Rightarrow x + 48 = \sqrt{3} h \Rightarrow x = \sqrt{3} h - 48 (ii)

Equating (i) and (ii):

\frac{h}{\sqrt{3}} = \sqrt{3} h - 48

Simplify:

h (\sqrt{3} - \frac{1}{\sqrt{3}}) = 48 \Rightarrow h = 24 \sqrt{3}

**Final Answer:**

A

Interactive Math Quiz - Question 33

Q33. A government office assigns a distinct license plate for each vehicle registered under it. Each license plate contains two letters of the English alphabet followed by four digits. In order to avoid confusion, no license plate is allowed to contain both the letter $O$ and the number $0$ . What is the maximum number of vehicles that can be registered by this office?

Solution:
Total possible license plates (no restrictions):

26^{2} \times 10^{4} = 6760000

Now, calculate unfavorable cases where both "O" and "0" are present:

$C_{1}$ : One "O" & One "0": $^{2} C_{1} \times 25 \times (^{4} C_{1} \times 9^{3})$
$C_{2}$ : One "O" & Two "0"s: $^{2} C_{1} \times 25 \times (^{4} C_{2} \times 9^{2})$
$C_{3}$ : One "O" & Three "0"s: $^{2} C_{1} \times 25 \times (^{4} C_{3} \times 9)$
$C_{4}$ : One "O" & Four "0"s: $^{2} C_{1} \times 25 \times^{4} C_{4}$
$C_{5}$ : Two "O"s & One "0": $^{2} C_{2} \times (^{4} C_{1} \times 9^{3})$
$C_{6}$ : Two "O"s & Two "0"s: $^{2} C_{2} \times (^{4} C_{2} \times 9^{2})$
$C_{7}$ : Two "O"s & Three "0"s: $^{2} C_{2} \times (^{4} C_{3} \times 9)$
$C_{8}$ : Two "O"s & Four "0"s: $^{2} C_{2} \times^{4} C_{4}$

Total unfavorable cases:

C_{1} + C_{2} + C_{3} + C_{4} + C_{5} + C_{6} + C_{7} + C_{8}

Subtract from the total possible license plates:

6760000 - Unfavorable cases = 6584611

**Final Answer:**

B

Interactive Math Quiz - Question 34

Q34. The area enclosed by the curves $y = \cos^{- 1} x$ and $y = \sin^{- 1} x$ over the range $0 \leq x \leq \frac{1}{\sqrt{2}}$ is:

Solution:
The area is given by:

Area = \int_{0}^{1 / \sqrt{2}} (\cos^{- 1} x - \sin^{- 1} x) d x

Let

I_{1} = \int_{0}^{1 / \sqrt{2}} \cos^{- 1} x d x

and

I_{2} = \int_{0}^{1 / \sqrt{2}} \sin^{- 1} x d x

. Then:

Area = I_{1} - I_{2}

Calculating

I_{1}

using integration by parts:

I_{1} = \cos^{- 1} x \cdot x |_{0}^{1 / \sqrt{2}} - \int_{0}^{1 / \sqrt{2}} \frac{- x}{\sqrt{1 - x^{2}}} d x

Substituting

1 - x^{2} = t^{2} \Rightarrow - 2 x d x = 2 d t

I_{1} = \frac{π}{4} - [1 - \frac{1}{\sqrt{2}}]

Similarly, calculate

I_{2}

I_{2} = \frac{π}{4} + [1 - \frac{1}{\sqrt{2}}]

Finally:

Area = I_{1} - I_{2} = 2 - \sqrt{2}

**Final Answer:**

A

Interactive Math Quiz - Question 35

Q35. The value of $lim_{x \to 0} (1 + 2 x)^{(x + 3) / x}$ is:

Solution:
The given limit is in the indeterminate form

1^{\infty}

. Using the standard approach for

f (x)^{g (x)}

format:

lim_{x \to a} f (x)^{g (x)} = e^{lim_{x \to a} g (x) \cdot (f (x) - 1)} .

Substituting the given expression:

lim_{x \to 0} (1 + 2 x)^{(x + 3) / x} = e^{lim_{x \to 0} (\frac{x + 3}{x}) (1 + 2 x - 1)} .

Simplify:

= e^{lim_{x \to 0} (\frac{x + 3}{x}) (2 x)} = e^{lim_{x \to 0} (2 x + 6)} .

Since

lim_{x \to 0} (2 x + 6) = 6

, the final result is:

e^{6} .

**Final Answer:**

B

Quiz - Question 139

Q36. The equation of the locus of point $ P $ which maintains the distance from two fixed points $ R=(0,2) $ and $ S=(0,-2) $ satisfying the equality $ |RP + SP| = 6 $ is:

Interactive Math Quiz - Question 37

Q37. Let $PQR$ be a right-angled triangle with the right angle at $P$ and the angles $θ_{1}$ and $θ_{2}$ at $Q$ and $R$ , respectively. Let the length of $PQ$ be $\sqrt{2}$ and the length of $RP$ be 3. Then the value of $\cot (θ_{1}) + \cot (θ_{2})$ is:

Solution:
Given:

\cot θ_{1} = \frac{adjacent side}{opposite side} = \frac{PQ}{PR} = \frac{\sqrt{2}}{3}

and

\cot θ_{2} = \frac{adjacent side}{opposite side} = \frac{PR}{PQ} = \frac{3}{\sqrt{2}} .

Adding these:

\cot θ_{1} + \cot θ_{2} = \frac{\sqrt{2}}{3} + \frac{3}{\sqrt{2}} .

Simplifying:

\cot θ_{1} + \cot θ_{2} = \frac{\sqrt{2} \cdot \sqrt{2} + 3 \cdot 3}{3 \sqrt{2}} = \frac{2 + 9}{3 \sqrt{2}} = \frac{11}{3 \sqrt{2}} .

**Final Answer:**

B

Interactive Math Quiz - Question 38

Q38. The number of functions $f : {1, 2, 3, 4, 5} \to {1, 2, 3, 4, 5}$ such that $f (6 - i) = f (i)$ for $i = 1, 2, 3, 4, 5$ is:

Solution:
The condition

f (6 - i) = f (i)

for

i = 1, 2, 3, 4, 5

implies the following:

$f (1) = f (5)$
$f (2) = f (4)$
$f (3) = f (3)$

This essentially reduces the number of entities in the domain of

f

to 3 groups:

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

.
Each group can independently take any of 5 possible values from the codomain

{1, 2, 3, 4, 5}

.
Therefore, the total number of such functions is:

5 \times 5 \times 5 = 125.

**Final Answer:**

B

Interactive Math Quiz - Question 39

Q39. If the line $3 x + 4 y - 7 = 0$ divides the line segment joining the points $(2, 1)$ and $(- 2, 1)$ in the ratio $λ : 1$ , then the value of $λ$ is:

Solution:
Let

A = (2, 1)

B = (- 2, 1)

, and the dividing point

C = (x, y)

.
Using the section formula for

C

, where

λ : 1

x = \frac{λ \times (- 2) + 1 \times 2}{λ + 1} = \frac{- 2 λ + 2}{λ + 1}, y = \frac{λ \times 1 + 1 \times 1}{λ + 1} = 1.

Substituting

x

and

y

into the line equation

3 x + 4 y - 7 = 0

3 (\frac{- 2 λ + 2}{λ + 1}) + 4 - 7 = 0.

Simplify:

\frac{- 6 λ + 6}{λ + 1} - 3 = 0.

Multiply through by

λ + 1

- 6 λ + 6 - 3 (λ + 1) = 0 ⟹ - 6 λ + 6 - 3 λ - 3 = 0.

Combine terms:

- 9 λ + 3 = 0 ⟹ 9 λ = 3 ⟹ λ = \frac{1}{3} .

**Correct Answer:**

A

Interactive Math Quiz - Question 40

Q40. Let

Math input error

be a continuous function. Then:

Solution:
To ensure

f (x)

is continuous at

x = 0

:

The left-hand limit (LHL) at

x = 0

is:

LHL = f (0^{-}) = lim_{h \to 0} [2 (- h)^{3} + (- h)^{2} - 2 (- h) + β] = β .

The right-hand limit (RHL) at

x = 0

is:

RHL = f (0^{+}) = lim_{h \to 0} h^{α} \sin (\frac{1}{h}) .

For continuity, LHL = RHL. Since

\sin (\frac{1}{h})

oscillates between -1 and 1,

h^{α}

must approach 0 as

h \to 0

, requiring

α \geq 1

. Additionally,

β = 0

for the left and right limits to match.

**Correct Answer:**

B

Interactive Math Quiz - Question 41

Q41. The domain of the function

Math input error

is:

Solution:
To find the domain of

f (x) = \sqrt{\log_{10} (\frac{3 x - x^{2}}{2})}

, the following conditions must be satisfied:

**1st Condition:** The expression inside the logarithm must be positive:

\frac{3 x - x^{2}}{2} > 0

Factorizing

3 x - x^{2} = x (3 - x)

, we get:

x \in (0, 3)

**2nd Condition:** The logarithmic value must be non-negative:

\log_{10} (\frac{3 x - x^{2}}{2}) \geq 0 ⟹ \frac{3 x - x^{2}}{2} \geq 1

This simplifies to:

3 x - x^{2} \geq 2 ⟹ x^{2} - 3 x + 2 \leq 0 ⟹ (x - 1) (x - 2) \leq 0

Thus:

x \in [1, 2]

**Domain:** The intersection of the two conditions is:

x \in [1, 2]

**Correct Answer:**

[1, 2]

Interactive Math Quiz - Question 42

Q42. The sums of the first $n$ terms of two arithmetic progressions are in the ratio $(7 n + 1) : (4 n + 27)$ . The ratio of their $11^{th}$ terms is:

Solution:
Let the sums of the first

n

terms of two arithmetic progressions be

S_{1 n}

and

S_{2 n}

.
The given ratio is:

\frac{S_{1 n}}{S_{2 n}} = \frac{7 n + 1}{4 n + 27} .

Substituting the formula for the sum of the first

n

terms of an AP:

S_{1 n} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}], S_{2 n} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}] .

We have:

\frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 1}{4 n + 27} .

Simplifying:

\frac{a_{1} + \frac{(n - 1)}{2} d_{1}}{a_{2} + \frac{(n - 1)}{2} d_{2}} = \frac{7 n + 1}{4 n + 27} .

To find the ratio of the

11^{th}

terms, substitute

n = 21

\frac{a_{1} + 10 d_{1}}{a_{2} + 10 d_{2}} = \frac{7 \times 21 + 1}{4 \times 21 + 27} .

Simplify the right-hand side:

\frac{7 \times 21 + 1}{4 \times 21 + 27} = \frac{148}{111} .

Hence, the ratio of the

11^{th}

terms is:

148 : 111 .

Interactive Math Quiz - Question 43

Q43. Consider the function:

Math input error

What is true about $f (x)$ ?

Solution:
Evaluate the left-hand limit (LHL) and right-hand limit (RHL) at

x = 0

f (0^{+}) = lim_{h \to 0} \frac{1 - \cos (h)}{h^{2}} = \frac{1}{2}, f (0^{-}) = lim_{h \to 0} \frac{1 - \cos (- h)}{(- h)^{2}} = \frac{1}{2} .

Since:

LHL = RHL \neq f (0),

the function

f (x)

has a removable discontinuity.
Hence, the correct answer is:

B. f has a discontinuity and it is removable.

Interactive Math Quiz - Question 44

Q44. If $w, x, y, z$ are positive real numbers, what is the least value of:

(w + 2 x + 3 y + 4 z) (\frac{1}{w} + \frac{1}{2 x} + \frac{1}{3 y} + \frac{1}{4 z}) ?

Solution:
Applying the inequality

A M \geq H M

for the given terms:

\frac{w + 2 x + 3 y + 4 z}{4} \geq \frac{4}{\frac{1}{w} + \frac{1}{2 x} + \frac{1}{3 y} + \frac{1}{4 z}}

Rearranging and simplifying, we get:

(w + 2 x + 3 y + 4 z) (\frac{1}{w} + \frac{1}{2 x} + \frac{1}{3 y} + \frac{1}{4 z}) \geq 16.

Therefore, the least value is

16

.
Correct answer: **A**

Interactive Math Quiz - Question 45

Q45. Consider the function:

Math input error

Which of the following statements is correct?

Solution:
To check differentiability at

x = 0

, calculate:

f^{'} (0^{+}) = lim_{h \to 0} \frac{h^{2} \sin \frac{1}{h}}{h} = lim_{h \to 0} h \sin \frac{1}{h} = 0.

Similarly:

f^{'} (0^{-}) = lim_{h \to 0} \frac{h^{2} \sin \frac{1}{- h}}{- h} = lim_{h \to 0} h \sin \frac{1}{h} = 0.

Hence,

f (x)

is differentiable at

x = 0

. For

x \neq 0

f^{'} (x) = - \cos \frac{1}{x} + 2 x \sin \frac{1}{x} .

x = 0

, the oscillating term

\cos \frac{1}{x}

makes

f^{'} (x)

discontinuous, as

lim_{x \to 0^{+}} f^{'} (x)

does not exist. Thus,

f

is differentiable at 0, but

f^{'} (x)

is not continuous at 0.
Correct answer: **A**

Quiz - Question 140

Q46. In a $ (8 \times 8) $ chessboard, numbers are placed on each of the 64 squares such that the number on each square is the average of its neighboring squares (that is, the squares with which it shares a side). Also, it is known that the sum of all the numbers is 640. Which of the following is true?

Interactive Math Quiz - Question 47

Q47. The equation of the straight line with slope $\frac{5}{2}$ that touches the parabola $y^{2} = 5 x$ is:

Solution:
Given the parabola

y^{2} = 5 x

, differentiate both sides:

2 y \frac{d y}{d x} = 5 \Rightarrow \frac{d y}{d x} = \frac{5}{2 y} .

At the point of tangency, the slope of the tangent line

\frac{d y}{d x} = \frac{5}{2}

. Thus:

\frac{5}{2 y} = \frac{5}{2} \Rightarrow y = 1.

Using the equation of the parabola,

y^{2} = 5 x

, substitute

y = 1

1^{2} = 5 x \Rightarrow x = \frac{1}{5} .

The point of tangency is

(\frac{1}{5}, 1)

. The equation of the tangent line is:

y - 1 = \frac{5}{2} (x - \frac{1}{5}) .

Simplify:

2 y - 2 = 5 x - 1 \Rightarrow 5 x - 2 y + 1 = 0.

Correct answer: **B**

Interactive Math Quiz - Question 48

Q48. Let $P (x)$ and $Q (x)$ be two distinct polynomials with degree at most 2. Let $a_{0}, \dots, a_{n - 1}$ be distinct elements of $R$ . Consider the following set:

Math input error

Which of the following is always correct?

Solution:

Case 1: Both $P (x)$ and $Q (x)$ are parabolas.

Two parabolas can intersect at most at two points. Therefore, out of the $n$ values of $i$ , $P (a_{i}) = Q (a_{i})$ for at most two values of $i$ . Thus, the set $X$ has at least $(n - 2)$ distinct elements.

Case 2: One of $P (x)$ and $Q (x)$ is a parabola, and the other is a straight line.

A straight line and a parabola can intersect at most at two points. By similar reasoning, the set $X$ has at least $(n - 2)$ distinct elements.

Case 3: Both $P (x)$ and $Q (x)$ are straight lines.

Two distinct straight lines intersect at exactly one point. Therefore, the set $X$ has at least $(n - 2)$ distinct elements.

Hence, in all cases, the number of distinct elements in the set $X$ is at least $(n - 2)$ .

Correct Answer: A

Interactive Math Quiz - Question 49

Q49. If there are 10 red balls and 12 blue balls, and these are arranged by drawing one ball at a time at random, then what is the probability that the last ball in the order is of color red?

Solution:

Total cases: The total number of arrangements of 22 balls, including 10 red and 12 blue, is: $Total cases = \frac{22!}{10! \cdot 12!}$

Favorable cases: The number of arrangements where the last ball is red can be calculated by treating the last red ball as fixed and arranging the remaining 21 balls (9 red and 12 blue): $Favorable cases = \frac{21!}{9! \cdot 12!}$

Probability: $P = \frac{Favorable cases}{Total cases} = \frac{\frac{21!}{9! \cdot 12!}}{\frac{22!}{10! \cdot 12!}} = \frac{10}{22}$

Correct Answer: A

Quiz - Question 141

Q50. If $ p, q, r $ are strictly positive real numbers, and $ p x+q y+r z=0, q x+r y+p z=0 $ and $ r x+p y+q z=0 $, then there is a real number $ \lambda \neq 1 $ such that $ x: y: z $ is the same as:

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PGDBA 2023 Solutions

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