BITSAT 2023 Question Paper with Solutions

To determine the best word that expresses the meaning of the underlined word "magnanimous" in the sentence, we need to understand the definition of "magnanimous."

The word "magnanimous" means generous or noble, especially in forgiving or giving. A magnanimous person shows a lofty and courageous spirit. In the given sentence, the context indicates an act of giving a large amount for flood victims, which aligns with the characteristic of being generous or noble.

Let's evaluate the options:

Option A: Vindictive

"Vindictive" means having or showing a strong or unreasoning desire for revenge. It does not align with the context of generosity.

Option B: Chimerical

"Chimerical" refers to something that is imaginary or fantastical. This option is also not relevant to the concept of generosity or nobility.

Option C: Chivalrous

"Chivalrous" means being courteous, respectful, and gallant, especially towards women. It also encompasses qualities like bravery and honor, which can be related to nobility and generosity.

Option D: Cherubic

"Cherubic" describes something that resembles a cherub, typically characterized by an innocent and angelic appearance. This term does not fit the context of generosity.

In conclusion, the best word that expresses the meaning of "magnanimous" in this context is Option C: Chivalrous.

The correct order of the sentences is Option C: CDBA. Here's why:

C. Whether or not Facebook friendships are lasting is debatable. This sentence introduces the topic of Facebook friendships and their longevity, setting the stage for the following sentences.

D. Some people believe that real friendships are collapsing in modern times. This sentence presents one side of the debate, arguing that real friendships are declining.

B. However, such social networking sites help us to keep in touch with old friends or make new ones. This sentence provides a counterargument to the previous one, highlighting the positive aspects of social networking sites.

A. There are others who claim that they have never been so well connected. This sentence further supports the counterargument by stating that people feel more connected than ever before.

Therefore, the order CDBA creates a coherent paragraph that presents both sides of the argument about Facebook friendships and their impact on real-life connections.

The correct order of the sentences is Option D: BCAD.

Here's the explanation:

B. This sentence introduces the main topic: the importance of human capital. It sets the stage for the following sentences.

C. This sentence provides evidence to support the claim in sentence B, stating that "new economic superpowers" are a testament to the power of human capital.

A. This sentence further elaborates on the superpowers mentioned in sentence C, highlighting that they aren't naturally resource-rich and have experienced political difficulties.

D. Finally, this sentence concludes by emphasizing the remarkable achievement of these superpowers in achieving economic success despite these challenges.

Therefore, the order BCAD creates a logical flow, starting with the general point about human capital, then providing examples, and finally explaining the context and achievement of those examples.

The answer to your question is Option C: Exegesis.

Exegesis refers to the critical explanation or interpretation of a text, especially of scripture. It involves an in-depth study and analysis of the text to understand its meaning and context. Scholars and theologians often perform exegesis to gain a deeper understanding of religious texts.

Here's a brief explanation of the other options for clarification:

Oblivion: This term refers to the state of being unaware or unconscious of what is happening around one. It can also mean the state of being forgotten.

Nascence: This word refers to the process of coming into existence; beginning to develop.

Naivete: This term refers to the state of being naive, which means having or showing a lack of experience, judgment, or information.

The word "culmination" refers to the highest or climactic point of something, especially as attained after a long time. It signifies the peak or final stage of a process or event. To find the opposite in meaning, we need to consider a word that signifies the start or initial stage rather than the endpoint.

Let's analyze each option:

Option A: Completion

Completion implies finishing something, reaching the end, or the finalization. This is very similar to "culmination" and therefore, not the opposite.

Option B: Climax

Climax also denotes the peak or most intense point of something, similar to "culmination." It is not an opposite.

Option C: Conclusion

Conclusion signifies the end or last part of something. This word also closely aligns with "culmination" and is not its opposite.

Option D: Beginning

Beginning means the start or initial point of something. This is opposite in meaning to culmination, which denotes the end or peak.

Therefore, the word opposite in meaning to "culmination" is Option D: Beginning.

The word "wither" means to dry up or shrivel, often used in the context of plants losing vitality and freshness. Therefore, we need to choose a word that conveys the opposite meaning, indicating growth, vitality, or flourishing.

Let's examine the options:

Option A: Shine - This word means to emit light or glow and is not directly related to the vibrancy of plants.

Option B: Bloom - This word means to produce flowers, to flourish, or thrive. It represents the exact opposite sense of withering.

Option C: Excel - This word means to do exceptionally well in a particular area and is, again, not related to the context of plants withering.

Option D: Wilt - This word is a synonym for wither as it also means to lose freshness and vitality.

Thus, the word opposite in meaning to "wither" is:

Option B: Bloom

The given sentence "Did everybody miss the first bus?" is in active voice. To change it into passive voice, we need to make the object of the verb (the first bus) the subject of the sentence, and use the appropriate form of the verb "to be" along with the past participle of the main verb (miss).

Let's analyze the options:

Option A: The first bus was missed by everybody?

This option is in passive voice, but the sentence structure is incorrect. It should be a question, which requires an inversion of the subject and auxiliary verb.

Option B: Was the first bus missed by everybody?

This option is correct. It correctly inverts the auxiliary verb "was" and the subject "the first bus," making it a proper question in passive voice.

Option C: Everybody missed the first bus?

This option is in active voice and is essentially restating the original sentence without the question format being clear.

Option D: Had the first bus been missed by everybody?

This option is in passive voice, but it changes the tense to past perfect, which doesn't match the original simple past tense of the sentence.

Therefore, the correct option that best expresses the same sentence in passive voice is:

Option B: Was the first bus missed by everybody?

To determine which word best expresses the meaning of the underlined word "hoax" in the sentence "The discovery of treasure under the Bermuda Triangle is probably a $\underline {hoax}$," it is important to understand what "hoax" means.

A "hoax" is a deliberate deception made to trick people into believing something that is not true. Therefore, we need a synonym that captures this sense of deception and trickery.

Let's analyze the options:

Option A: Analogy

An analogy is a comparison between two things, typically for the purpose of explanation or clarification. This does not capture the essence of deception or trickery.

Option B: Delusion

A delusion is a false belief or judgment about external reality, held despite proof or evidence to the contrary. While it is somewhat related to false belief, it does not specifically imply intentional deception.

Option C: Bonanza

A bonanza is a situation or event that creates a sudden increase in wealth, good fortune, or profits. This is the opposite of a hoax and does not fit at all.

Option D: Trick

A trick is an action that is intended to deceive someone. This is directly aligned with the meaning of "hoax," making it the best choice.

Therefore, the word which best expresses the meaning of the underlined word "hoax" is:

Option D: Trick

The sentence "Sohan was too weak to play" implies that Sohan's weakness prevented him from playing. The structure "too...to..." indicates an excessive degree of weakness that stops him from engaging in the activity of playing. Now, let's analyze the given options:

Option A: Sohan was so weak that he could not play.

This option essentially means the same thing as the original sentence. It conveys that Sohan's level of weakness prevented him from playing. The use of "so...that..." effectively translates the meaning of "too...to...". Hence, Option A is correct.

Option B: Sohan was as weak as he could not play.

This sentence is structurally incorrect and does not convey the same meaning as the original sentence. The phrase "as weak as" is not appropriately followed by the clause "he could not play".

Option C: Sohan was very weak because he could not play.

This sentence implies that Sohan's weakness is a result of his inability to play, which is the opposite of the original meaning. The original sentence means his weakness prevented him from playing, not the other way around.

Option D: Sohan was so weak as he could not play.

Similar to Option B, this sentence is grammatically incorrect and does not convey the same meaning as the original sentence. The structure "so...as" is not used correctly here.

Therefore, the correct option is Option A: Sohan was so weak that he could not play.

Holocaust is a situation in which there is great destruction and a lot of people die. So, it is the best substitute for the words.

Accounting to the given information, the arrangement is as follow,

So, it is clear 'B' is sitting between A and F.

To understand the relationship between "Growth" and "Expansion," let's break down the pair. "Growth" refers to the process of increasing in size, quantity, or significance, while "Expansion" refers to the process of something becoming larger or more extensive. Therefore, the relationship can be summarized as a cause and effect or state that follows from an initial state.

Now, let's evaluate each option to identify which one shares a similar relationship:

Option A: Child : Birth

The relationship here can be described as the origin or commencement. A "Child" is the result or outcome of "Birth." This relationship, however, is not exactly similar to growth leading to expansion but rather an origin-result relationship.

Option B: Urge : Request

This relationship describes a situation where an "Urge" (a strong desire or impulse) can lead to a "Request" (an act of asking for something). This is quite similar to the cause-effect relationship seen in Growth leading to Expansion.

Option C: First : Last

This relationship is one of sequential order, where "First" and "Last" represent the beginning and ending points in a series. This does not match the growth-expansion relationship, which is transformational rather than positional.

Option D: Temporary : Permanent

This relationship is one of contrast, where "Temporary" means lasting for a limited period and "Permanent" means lasting indefinitely. This does not fit the pattern of growth leading to expansion.

Given the options, the correct word-pair that shares a similar relationship to "Growth" and "Expansion" is:

Option B: Urge : Request

Here, 1 is common and it is on the same face in both the positions. So, 4 and 2 are opposite, 6 and 3 are opposite. Hence, 3 will be at the top if 6 is at the bottom.

The correct mirror image of the given combination is,

To determine which conclusions definitely follow from the given statement, let's analyze each conclusion in detail.

Statement: The old order changed yielding place to new

This statement implies that there is a process where old systems or ideas are being replaced by new ones.

Conclusion I: Change is the law of nature.

This conclusion is a broad generalization stating that change is a fundamental principle of the natural order. The statement supports this idea by illustrating an example of change (the old order making way for new). Thus, Conclusion I follows from the statement.

Conclusion II: Discard old ideas because they are old.

This conclusion implies that ideas should be discarded merely because they are old. The statement does not provide sufficient information to support the notion that "old ideas should be discarded only due to their age." It merely indicates that the old order changes to make way for something new but does not necessarily suggest discarding ideas solely because of their age.

Based on this analysis, it is clear that:

Option A: Only Conclusion I follows.

The pattern of the series is as follow,

So, 'GDWWI' will replace the question mark.

It will appear when unfolded like as,

Chandigarh is the capital of two states Haryana and Punjab.

$\therefore$ Chandigarh is different from others.

The correct answer is Option A: inert gas. Here's why:

The question is asking about the relationship between elements based on their position in the periodic table.

Flourine (F) is a member of the halogen group (Group 17). Halogens are highly reactive nonmetals.

Helium (He) is a member of the noble gas group (Group 18). Noble gases are known for being extremely unreactive, hence the name "inert gases."

So, the relationship is that both Flourine and Helium are in the same column of the periodic table, indicating they share similar chemical properties.

Let's look at why the other options are incorrect:

Option B: Gas: While both Fluorine and Helium are gases at room temperature, this is not the specific relationship the question is looking for. Many elements are gases.

Option C: Radioactive: Neither Fluorine nor Helium are inherently radioactive. Some isotopes of these elements may be radioactive, but that's not the primary characteristic defining their groups.

Option D: Metal: Both Fluorine and Helium are nonmetals.

Let's analyze the given information step by step to determine the ages of P, Q, R, S, and T relative to each other:

1. $\mathrm{T}$ is older than $\mathrm{P}$.

This means $\mathrm{T} > \mathrm{P}$.

2. $\mathrm{Q}$ is younger than $\mathrm{T}$.

This means $\mathrm{T} > \mathrm{Q}$.

3. $\mathrm{S}$ is older than $\mathrm{R}$.

This means $\mathrm{S} > \mathrm{R}$.

4. $\mathrm{P}$ is older than $\mathrm{S}$ but younger than $\mathrm{Q}$.

This means $\mathrm{Q} > \mathrm{P} > \mathrm{S}$.

Now let's combine these relationships to form an order:

From (1) and (2), we have: $\mathrm{T} > \mathrm{Q}$.

From (4), we have: $\mathrm{Q} > \mathrm{P} > \mathrm{S}$.

From (3), we have: $\mathrm{S} > \mathrm{R}$.

Combining all the information, we get:

$\mathrm{T} > \mathrm{Q} > \mathrm{P} > \mathrm{S} > \mathrm{R}$.

Therefore, the oldest among all is T.

Correct Answer: Option B - T

Option figure (a) is not embedded in the given figure as its part.

The pattern of the series is as follows,

Hence, 281 will replace the question mark.

On observing the given pattern, the arrow moving $90^{\circ}$ in clockwise direction along with the pointed shape in each step. So, figure of option (a) will replace the question mark.

Given,

There are 26 triangles in the given figure, they are $\triangle \mathrm{ABK}, \triangle \mathrm{BKC}, \triangle \mathrm{ABC}, \triangle \mathrm{DCL}, \triangle \mathrm{DEL}, \triangle \mathrm{CDE}, \triangle \mathrm{AJK}, \triangle \mathrm{KJC}, \triangle \mathrm{AJC}, \triangle \mathrm{CJL}, \triangle \mathrm{LJE}, \triangle \mathrm{CJE}, \triangle \mathrm{ACE}, \triangle \mathrm{HGI}, \triangle \mathrm{GIF}, \triangle \mathrm{HGF}, \triangle \mathrm{AGJ}, \triangle \mathrm{EGJ}, \triangle \mathrm{AGE}, \triangle \mathrm{AGC}, \triangle \mathrm{ECG}, \triangle \mathrm{JCD}, \triangle \mathrm{JED}, \triangle \mathrm{ABJ}, \triangle \mathrm{BJC}$ and $\triangle \mathrm{BJD}$.

To analyze whether the given conclusions follow from the statement, let's carefully examine the statement and each conclusion individually.

Statement: In a one-day cricket match, the total runs made by a team were 200. Out of these, 160 runs were made by spinners.

Conclusion I: $80 \%$ of the team consists of spinners.

This conclusion points out a percentage related to the team composition. However, the statement only talks about the runs made by spinners and does not provide any direct information about what percentage of the team members are spinners. Therefore, we cannot conclusively determine the team composition based only on the runs scored.

Conclusion II: The opening batsmen were spinners.

This conclusion assumes that the opening batsmen were spinners based on the runs scored. The statement, however, does not provide any information about who were the opening batsmen - it only mentions the total and specific runs scored by spinners. Therefore, this conclusion cannot be definitively drawn from the given statement.

We see that neither conclusion I nor conclusion II can be conclusively derived from the given statement.

Answer: Option C: Neither I nor II follows.

The pattern is follow as,

As, $12: 36=\frac{(12)^2}{4}=36$

Similarly, $18: 81=\frac{(18)^2}{4}=81$

So, the required number is 81.

As,

Given,

# - Husband, @ - Daughter

$\&$ - Sister

We have,

P&Q@D#K@S#T

Here, in option (d), D is the daughter-in-law of 'T' is not a correct statement.

Let's denote the amounts received by Rajesh, Vivek, and Kripal as $R$, $V$, and $K$ respectively.

According to the problem, Rajesh gets double the amount that Vivek gets, so we have:

$R = 2V$

Similarly, Kripal gets double the amount that Rajesh gets, so we have:

$K = 2R$

The total amount to be divided is ₹ 5110. Therefore, we can write the equation:

$R + V + K = 5110$

Substitute $R$ and $K$ in terms of $V$ from the above equations:

$R = 2V$

$K = 2R = 2(2V) = 4V$

Now substitute these into the total amount equation:

$2V + V + 4V = 5110$

Simplify the equation:

$7V = 5110$

Solve for $V$:

$V = \frac{5110}{7} = 730$

Now, calculate the amounts received by Rajesh and Kripal:

$R = 2V = 2 \times 730 = 1460$

$K = 4V = 4 \times 730 = 2920$

So, the amount of money Kripal will get is ₹ 2920.

Therefore, the correct option is Option A: ₹ 2920.

Given, $(1+x)^2 d y+2 x y d x=\cot x d x$

$\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{\cot x}{1+x^2}$, which is a linear differential equation of the form of $\frac{d y}{d x}+P y=Q$

Here, $P=\frac{2 x}{1+x^2}$ and $Q=\frac{\cot x}{1+x^2}$

Now, IF $=\exp \left(\int \frac{2 x}{1+x^2} d x\right)=\exp \left(\log \left(1+x^2\right)\right)$

$=1+x^2$

and the solution of this linear differential equation be

$$\begin{aligned} & y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C \\ & \Rightarrow \quad y\left(1+x^2\right)=\int\left(\frac{\cot x}{1+x^2} \times\left(1+x^2\right)\right) d x+C \\ & \Rightarrow \quad y\left(1+x^2\right)=\int \cot x d x+C \\ & \Rightarrow \quad y\left(1+x^2\right)=\log |\sin x|+C \\ & \Rightarrow \quad y=\frac{\log |\sin x|}{1+x^2}+\frac{C}{1+x^2} \end{aligned}$$

Let $A E$ be the tower $T_2$ of height $80 \mathrm{~m}$ and $B C$ be the tower $T_1$ of height $60 \mathrm{~m}$. Let $A B=C D=h \mathrm{~m}$

In $\triangle E D C$,

$$\begin{aligned} & \frac{E D}{C D}=\tan x \Rightarrow \frac{20}{h}=\tan x \\ & \Rightarrow \quad h=\frac{20}{\tan x} \quad \text{.... (i)} \end{aligned}$$

$$\begin{array}{ll} \text { In } \triangle A B C, \frac{B C}{A B}=\tan 2 x \\ \Rightarrow \frac{60}{h}=\tan 2 x \\ \Rightarrow \frac{60}{h}=\frac{2 \tan x}{1-\tan ^2 x} \\ \Rightarrow \frac{30}{h}=\frac{\tan x}{1-\tan ^2 x} \\ \Rightarrow h=\frac{30\left(1-\tan ^2 x\right)}{\tan x} \quad \text{... (ii)} \end{array}$$

From Eqs. (i) and (ii), we get

$$\begin{aligned} & \frac{30\left(1-\tan ^2 x\right)}{\tan x}=\frac{20}{\tan x} \\ & \Rightarrow \quad 3-3 \tan ^2 x=2 \\ & \Rightarrow \quad \tan ^2 x=\frac{1}{3} \Rightarrow \tan x= \pm \frac{1}{\sqrt{3}} \\ & \Rightarrow \quad x=\frac{\pi}{6} \text { or } \frac{5 \pi}{6} \\ \end{aligned}$$

So, on putting the value of $x$ in Eq. (i), we get

$h=\frac{20}{\tan \left(\frac{\pi}{6}\right)} \text { or } h=20 \sqrt{3} \mathrm{~m}$

$$\begin{aligned} & \text { Let } z=(1)^{1 / n}=(\cos 2 k \pi+i \sin 2 k \pi)^{1 / n} \\ & \Rightarrow \quad z=\left(\frac{2 k \pi}{n}\right)+i \sin \left(\frac{2 k \pi}{n}\right), k=0,1,2 \ldots, n-1 \\ & \text { Let } z_1=\cos \left(\frac{2 k_1 \pi}{n}\right)+i \sin \left(\frac{2 k_1 \pi}{n}\right) \\ & \text { and } z_2=\cos \left(\frac{2 k_2 \pi}{n}\right)+i \sin \left(\frac{2 k_2 \pi}{n}\right) \end{aligned}$$

be the two value of $z$ such that they subtend right angle at origin.

$\therefore \frac{2 k_1 \pi}{n}-\frac{2 k_2 \pi}{n}= \pm \frac{\pi}{2} \Rightarrow 4\left(k_1-k_2\right)= \pm n$

As, $k_1$ and $k_2$ are integer and $k_1 \neq k_2$

$\therefore n=4 k, k \in I$

We have, $f(x)=x^2-2 x+1$ and

$$\begin{array}{ll} & f \circ g(x)=x^2+2 x+1 \\ \therefore & f(g(x))=x^2+2 x+1 \\ \Rightarrow & (g(x))^2-2\left(g(x)+1=x^2+2 x+1\right. \\ \Rightarrow & (g(x))^2-2\left(g(x)-x^2-2 x=0\right. \\ \Rightarrow & g(x)=\frac{-(-2) \pm \sqrt{(-2)^2-4\left(-x^2-2 x\right)(1)}}{2(1)} \\ \Rightarrow & g(x)=\frac{2 \pm \sqrt{4+4 x^2+8 x}}{2} \\ \Rightarrow & g(x)=\frac{2 \pm 2 \sqrt{x^2+2 x+1}}{2} \\ \Rightarrow & g(x)=1 \pm(x+1) \\ \Rightarrow & g(x)=x+2-x \end{array}$$

Let $r$ be the radius and $h$ be the height of cone.

$\therefore \quad \frac{r}{h}=\frac{7}{35}=\frac{1}{5} \Rightarrow r=\frac{h}{5} \mathrm{~cm}$

Now, volume of cone $=V=\frac{1}{3} \pi r^2 h=\frac{1}{75} \pi h^3$

$\Rightarrow \quad \frac{d \dot{V}}{d t}=\frac{1}{75} \pi\left(3 h^2\right) \frac{d h}{d t}$

$\Rightarrow \quad \frac{d h}{d t}=\frac{1 \times 75}{\pi \times 3 h^2}$

$\left[\because \text { Given, } \frac{d V}{d t}=1 \mathrm{~cm}^3 / \mathrm{sec}\right]$

$$\begin{aligned} & \text { Now, } A=\pi r l=\pi r \sqrt{r^2+h^2} \\ & \Rightarrow \quad A=\pi r^2 \sqrt{1+\frac{h^2}{r^2}}=\frac{\pi h^2}{25} \sqrt{1+25} \\ & \Rightarrow \quad A=\frac{\sqrt{26}}{25} \pi h^2 \\ & \therefore \frac{d A}{d t}=\frac{\sqrt{26}}{25} \pi \cdot 2 h \frac{d h}{d t} \\ & \left.\Rightarrow \quad \frac{d A}{d t}\right|_{h=10}=\frac{\sqrt{26}}{25} \pi \times 2 \times 10 \times \frac{1 \times 75}{\pi \times 3 \times 10^2} \\ & =\frac{\sqrt{26}}{5} \mathrm{~cm}^2 / \mathrm{s} \\ & \end{aligned}$$

The equation of tangent to the given circle at $(\cos \theta, \sin \theta)$ be, $x \cos \theta+y \sin \theta=1$

It meets the $X$-axis at $(\sec \theta, 0)$ and $Y$-axis at $(0, \operatorname{cosec} \theta)$ Let $R(h, k)$ be the mid-point of $P Q$.

$\begin{gathered} \therefore \quad \frac{\sec \theta+0}{2}=h \text { and } \frac{0+\operatorname{cosec} \theta}{2}=k \\ \Rightarrow \quad \frac{1}{2 h}=\cos \theta \quad \text{... (i)} \end{gathered}$

and $\frac{1}{2 k}=\sin \theta \quad \text{.... (ii)}$

On squaring and adding Eqs. (i) and (ii), we get

$\frac{1}{4 h^2}+\frac{1}{4 k^2}=\cos ^2 \theta+\sin ^2 \theta$

$$\begin{aligned} & \Rightarrow \quad \frac{1}{4}\left(\frac{k^2+h^2}{h^2 k^2}\right)=1 \\ & \Rightarrow \quad h^2+k^2=4 h^2 k^2 \\ & \Rightarrow \quad h^2+k^2-4 h^2 k^2=0 \\ \end{aligned}$$

$\text { Hence, the required locus be } x^2+y^2-4 x^2 y^2=0$

Given, $z^2+|z|^2=0, z \neq 0$

Let $z=x+i y$ and $\bar{z}=x-i y$

$$\begin{array}{ll} \Rightarrow(x+i y)^2+(x+i y)(x-i y)=0 & \left\{\because z \bar{z}=|z|^2\right\} \\ \Rightarrow x^2+i^2 y^2+i 2 x y+x^2-i^2 y^2=0 \\ \Rightarrow x^2-y^2+i 2 x y+x^2+y^2=0 \quad \{\because i^2=-1\}\\ \Rightarrow 2 x(x+i y)=0 \\ \Rightarrow x=0 \text { or } x+i y=0 \text { (not possible) } \end{array}$$

Therefore, $x=0$ and $y \neq 0$

So, $y$ can take any non-zero real value.

Hence, infinitely many solutions exists.

Let $P$ be the probability of getting an even number. Then, by hypothesis the probability of getting an odd number be $3 P$.

Since, event of getting an odd number and an even number are mutually exclusive and exhaustive.

$\therefore \quad 3 P+P=1 \Rightarrow P=\frac{1}{4}$

Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$. If the die is thrown twice, then the sum of the numbers in two throws is odd if one number is odd and another is even.

$\therefore$ Required probability

$=\frac{1}{4} \times \frac{3}{4}+\frac{3}{4} \times \frac{1}{4}=\frac{6}{16}=\frac{3}{8}$

The described region is :

It is clear that

$$\begin{aligned} & A=\left(\frac{1}{2}, 0\right), B=(2,0), C=(2,3), D=(1,2) \\ & E=\left(\frac{1}{2}, \frac{3}{2}\right) \text { and } F=\left(\frac{1}{2}, \frac{5}{4}\right) \end{aligned}$$

$\therefore$ Required area $=$ Area of region A B C D E A $-$ Area of region EDFE

\begin{aligned} & =\frac{1}{2}\left(\frac{3}{2}+3\right)\left(2-\frac{1}{2}\right)-\int_\limits{1 / 2}^1\left[(x+1)-\left(x^2+1\right)\right] d x \\ & =\frac{1}{2} \times \frac{9}{2} \times \frac{3}{2}-\int_\limits{1 / 2}^1\left(x-x^2\right) d x \\ & =\frac{27}{8}-\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{1 / 2}^1 \\ & =\frac{27}{8}-\left[\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{8}-\frac{1}{24}\right)\right] \\ & =\frac{27}{8}-\left(\frac{1}{6}-\frac{1}{12}\right) \\ & =\frac{27}{8}-\frac{1}{12}=\frac{79}{24} \end{aligned}

Given, \lim _\limits{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}}

=\lim _\limits{x \rightarrow 0} \frac{3\left[\left(1+\frac{x}{27}\right)^{1 / 3}-1\right]}{9\left[1-\left(1+\frac{x}{27}\right)^{2 / 3}\right]}

=\lim _\limits{x \rightarrow 0} \frac{1}{3}\left[\frac{\left.\left(1+\frac{x}{3 \times 27}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!} x^2+\ldots\right)-1\right)}{1-\left(1+\frac{2 x}{3 \times 27}+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2!} x^2+\ldots\right)}\right]

\begin{aligned} & =\lim _\limits{x \rightarrow 0} \frac{1}{3}\left[\frac{x\left(\frac{1}{3 \times 27}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!} x+\ldots\right.}{-x\left(\frac{2}{3 \times 27}+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2!} x+\ldots\right)}\right] \\ & =\frac{1}{3}\left[\frac{\frac{1}{3 \times 27}+0+\ldots .}{-\left(\frac{2}{3 \times 27}+0+\ldots\right)}\right] \\ & =\frac{1}{3}\left(\frac{1}{-2}\right)=-\frac{1}{6} . \end{aligned}

$\text { Any point on the line } \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1} \text {, is of the form } (-3 \lambda-1,2 \lambda+3,-\lambda+2).$

Let the point $A \equiv(-3 \lambda-1,2 \lambda+3,-\lambda+2)$ and $B \equiv(-4,3,1)$

Then, $\mathbf{A B}=(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}}$

Now, as the line is parallel to the given plane, therefore $\mathrm{AB}$ will be parallel to the given plane and so $\mathrm{AB}$ will be perpendicular to the normal of plane.

$\Rightarrow \mathrm{AB} \cdot \mathrm{n}=0$, where $\mathrm{n}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathbf{k}}$ is normal to the plane.

$$\begin{aligned} & \Rightarrow(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=0 \\ & \Rightarrow 3 \lambda-3-4 \lambda-\lambda+1=0 \\ & \Rightarrow-2 \lambda=2 \Rightarrow \lambda=-1 \end{aligned}$$

Now, the required equation is the equation of line joining $A(2,1,3)$ and $B(-4,3,1)$, which is

$$\begin{array}{ll} & \frac{x-(-4)}{2-(-4)}=\frac{y-3}{1-3}=\frac{z-1}{3-1} \\ \Rightarrow & \frac{x+4}{6}=\frac{y-3}{-2}=\frac{z-1}{2} \\ \text { or } & \frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1} \end{array}$$

Given, $y=m_1 x+c_1$ is common tangent of circle and parabola.

For circle, $\left|\frac{c_1}{\sqrt{1+m_1^2}}\right|=\sqrt{2} \Rightarrow c_1^2=2\left(1+m_1^2\right)$ .... (i)

For parabola, $c_1=\frac{1}{4 m_1} \Rightarrow c_1^2=\frac{1}{16 m_1^2}$ .... (ii)

Also, $y=m_2 x+c_2$ is common tangent of circle and parabola.

For circle, $\left|\frac{c_2}{\sqrt{1+m_2^2}}\right|=\sqrt{2} \Rightarrow c_2^2=2\left(1+m_2^2\right)$

For parabola, $c_2=\frac{1}{4 m_2} \Rightarrow c_2^2=\frac{1}{16 m_2^2}$

Now, $2\left(1+m_1^2\right)=\frac{1}{16 m_1^2} \quad$ [From Eqs. (i) and (ii)]

$$\begin{aligned} & \Rightarrow \quad 32 m_1^4+32 m_1^2-1=0 \\ & \Rightarrow \quad m_1^2=\frac{-32 \pm \sqrt{(32)^2-4(32)(-1)}}{2 \times 32} \end{aligned}$$

$\Rightarrow \quad m_1^2=\frac{-32+24 \sqrt{2}}{64}$

[Ignoring negative value because $m^2 \geq 0$]

$\Rightarrow m_1= \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}$

Similarly, $m_2= \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}$

$$\begin{aligned} \therefore 8\left|m_1 m_2\right| =8\left|\left( \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}\right)\left( \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}\right)\right| \\ =\frac{8}{64}(-32+24 \sqrt{2}) \\ =-4+3 \sqrt{2} \end{aligned}$$

$$\begin{aligned} & \text { Given, } \frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)} \\ & \Rightarrow \quad \frac{\sin (B+C)}{\sin (A+C)}=\frac{\sin (A-C)}{\sin (C-B)} \\ & {[\because A+B+C=\pi \text { and } \sin (\pi-\theta)=\sin \theta]} \\ & \Rightarrow \quad \sin (B+C) \sin (C-B)=\sin (A+C) \sin (A-C) \\ & \Rightarrow \quad \sin ^2 C-\sin ^2 B=\sin ^2 A-\sin ^2 C \\ & \Rightarrow \quad 2 \sin ^2 C=\sin ^2 A+\sin ^2 B \\ & \Rightarrow \quad 2(2 R \sin C)^2=(2 R \sin A)^2+(2 R \sin B)^2 \\ & \Rightarrow \quad 2 c^2=a^2+b^2 \quad\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}+\frac{c}{\sin C}=2 R\right] \\ & \Rightarrow \quad a^2, c^2, b^2 \text { are in AP. } \\ & \Rightarrow \quad b^2, c^2, a^2 \text { are in AP. } \end{aligned}$$