datavorous/entrance-exam-dataset huggingface dataset:
Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_3$ are 60,40 and $50 \mathrm{~J} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. For the reaction
$\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at
equilibrium, the temperature should be

Options:
A $750 \mathrm{~K}$
B $1000 \mathrm{~K}$
C $1250 \mathrm{~K}$
D $500 \mathrm{~K}$

The correct answer is:
$750 \mathrm{~K}$

Solution:
For the reaction,
$$
\begin{aligned}
& \frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ} \\
& \Delta \mathrm{S}^{\circ}=\mathrm{S}_{\left(\mathrm{XY}_3\right)}^{\circ}-\left[\frac{1}{2} \mathrm{~S}_{\mathrm{X}_2}^{\circ}+\frac{3}{2} \mathrm{~S}_{\mathrm{Y}_2}^{\circ}\right] \\
& =50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \\
& =50-[30+60]=50-90 \\
& =-40 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$$
We know that,
$$
\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}
$$
At equilibrium,
$$
\begin{aligned}
\Delta \mathrm{G}^{\circ} & =0 \\
\Delta \mathrm{H} & =\mathrm{T} \Delta \mathrm{S}^{\circ} \\
\mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}^{\circ}}=\frac{-30 \times 10^3 \mathrm{~J}}{-40 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}=750 \mathrm{~K}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2010 (Screening)
Standard entropy of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} \mathrm{X}_{2}+\frac{3}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{3}, \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be

Options:
A $1250 \mathrm{~K}$
B $500 \mathrm{~K}$
C $750 \mathrm{~K}$
D $1000 \mathrm{~K}$

The correct answer is:
$750 \mathrm{~K}$

Solution:
For a reaction to be at equilibrium $\Delta \mathrm{G}=0$. Since $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ so at equilibrium $\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=0$
or $\Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S}$
For the reaction
$$
\frac{1}{2} \mathrm{X}_{2}+\frac{3}{2} \mathrm{Y}_{2} \longrightarrow \mathrm{XY}_{3} ; \quad \Delta \mathrm{H}=-30 \mathrm{~kJ}
$$
(given)
Calculating $\Delta \mathrm{S}$ for the above reaction, we get
$$
\begin{array}{l}
\Delta S=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \mathrm{JK}^{-1} \\
=50-(30+60) \mathrm{JK}^{-1}=-40 \mathrm{JK}^{-1} \\
\text { At equilibrium, } \mathrm{T} \Delta \mathrm{S}=\Delta \mathrm{H}
\end{array}
$$
$[\because \Delta \mathrm{G}=0]$
$\therefore \mathrm{T} \times(-40)=-30 \times 1000[\because 1 \mathrm{~kJ}=1000 \mathrm{~J}]$
or $\quad \mathrm{T}=\frac{-30 \times 1000}{-40}$ or $750 \mathrm{~K}$

Topic: Chemistry, Thermodynamics (C), VITEEE, VITEEE 2016
Standard free energies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) at $298 \mathrm{~K}$ are $-237.2,-394.4$ and -8.2 for $\mathrm{H}_2 \mathrm{O}(l) \mathrm{CO}_2(g)$ and pentane $(g)$, respectively. The value of $E_{\text {cell }}^0$ for the pentane-oxygen fuel cell is

Options:
A $1.968 \mathrm{~V}$
B $2.0968 \mathrm{~V}$
C $1.0968 \mathrm{~V}$
D $0.0968 \mathrm{~V}$

The correct answer is:
$1.0968 \mathrm{~V}$

Solution:
\(\begin{aligned}
& \mathrm{C}_5 \mathrm{H}_{12(g)}+8 \mathrm{O}_{2(9)} \longrightarrow 5 \mathrm{CO}_{2(g)}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{e})} \\
& \therefore \Delta \mathrm{G}_r=5\left(\Delta \mathrm{G}_r\right)_{\mathrm{CO}_2}+6\left(\Delta \mathrm{G}_{\mathrm{r}}\right)_{\mathrm{H}_2 \mathrm{O}}-\left(\Delta \mathrm{G}_{\mathrm{r}}\right)_{\text {Pentane }} \\
& =5(-5394.4)+6(-237.2)-(-8.2) \\
& =-1972-1423.2+8.2 \\
& \Delta \mathrm{G}_{\mathrm{R}}=-3387 \mathrm{~kJ} / \mathrm{mole} \\
& \mathrm{C}_5^{\frac{-12}{5}} \mathrm{H}_{12} \longrightarrow 5 \stackrel{4}{\mathrm{C}_2}
\end{aligned}\)
For 5 carbon atoms, there is a change of electrons from - 12 to 20
\(\begin{aligned}
& \therefore \text { No. of electrons involved }=32 \mathrm{e}^{-} \\
& \Delta G_r=-3387 \times 10^3=-32 \times E^{\circ} \times 96500 \\
& E^{\circ}=1.097 V
\end{aligned}\)

Topic: Chemistry, Hydrocarbons, NEET, NEET 2008 (Screening)
Standard Gibbs energy change in $\mathrm{kJ} \mathrm{mol}^{-1}$ for a galvanic cell
$\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{A}_{(\mathrm{aq})}^{3+}+\mathrm{B}_{(\mathrm{s})}$ that has standard emf of $0.5 \mathrm{~V}$ is : $\left(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right)$

Options:
A -193.50
B -96.50
C -144.75
D -289.50

The correct answer is:
-144.75

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 2)
Standard hydrogen electrode (SHE) is a

Options:
A Primary reference electrode
B Secondary reference electrode
C Metal – sparingly soluble salt electrode
D Metal – metal ion electrode

The correct answer is:
Primary reference electrode

Solution:
Standard hydrogen electrode (SHE) is primary reference electrode. SHE is represented by PtsH2gH+aq, is assigned a zero potential at all temperatures corresponding to reaction.
H+(aq)+e-→12H2(g)

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2019 (Shift 2)
Standard molar enthalpies of formation of $\mathrm{CaCO}_{3}(s), \mathrm{CaO}(s)$ and $\mathrm{CO}_{2}(g)$ are $-1206.92 \mathrm{~kJ} \mathrm{~mol}^{-1},-635.09 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-393.51 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The $\Delta_{r} H$, for decomposition of $\mathrm{CaCO}_{3}(s)$ is

Options:
A $178.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad$
B $-178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $1448.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad$
D $-1448.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$178.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad$

Solution:
$\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$
$$
\begin{aligned}
\Delta_{r} H^{\circ} &=\Delta_{f} H^{\circ}(\mathrm{CaO})+\Delta_{f} H^{\circ}\left(\mathrm{CO}_{2}\right)-\Delta_{f} H^{\circ}\left(\mathrm{CaCO}_{3}\right) \\
&=-635.09+(-393.51)-(-1206.92) \\
&=178.3 \mathrm{~kJ} \mathrm{~mol}^{-1} .
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), COMEDK, COMEDK 2014
Standard molar enthalpy of formation, ∆fH⊖is just a special case of enthalpy of reaction, ∆rH⊖. Is the ∆rH⊖for the following reaction same as ∆fH⊖? Give reason for your answer. CaO(s) + CO2(g) → CaCO3(s); ∆fH⊖= –178.3 kJ mol–1

Options:
A Yes,since CaCO3has been formed from other compounds and not from its constituent elements.
B Yes,since CaO has been formed from other compounds and not from its constituent elements.
C No,since CaCO3has been formed from other compounds and not from its constituent elements.
D No,since CaO has been formed from other compounds and not from its constituent elements.

The correct answer is:
No,since CaCO3has been formed from other compounds and not from its constituent elements.

Solution:
Correct Option is : (C)
No,since CaCO3has been formed from other compounds and not from its constituent elements.

Topic: Chemistry, Thermodynamics (C), NEET
Standard molar enthalpy of formation of CO2 is equal to

Options:
A zero
B the standard molar enthalpy of combustion of gaseous carbon
C the sum of standard molar enthalpies of formation of CO and O2
D the standard molar enthalpy of combustion of carbon (graphite)

The correct answer is:
the standard molar enthalpy of combustion of carbon (graphite)

Solution:
Standard molar enthalpy of formation of CO2 is equal to the standard molar enthalpy of combustion of carbon (graphite).

Topic: Chemistry, Thermodynamics (C), NEET
Standard reduction electrode potentials of three metals A,B and C are respectively + 0.5 V, – 3.0 V and – 1.2 V. The reducing powers of these metals are:

Options:
A A>B>C
B C>B>A
C A>C>B
D B>C>A

The correct answer is:
B>C>A

Solution:
More the reduction potential, more is the power to get itself reduced or lesser is reducing power or greater is oxidizing power

Topic: Chemistry, Electrochemistry, NEET
State as to why sodium is found to be more useful than potassium?

Options:
A Sodium is more abundant in nature compared to potassium.
B Sodium has a lower ionization energy and is less reactive than potassium.
C Sodium forms more stable compounds with common elements than potassium.
D Sodium is cheaper and easier to obtain than potassium.

The correct answer is:
Sodium is cheaper and easier to obtain than potassium.

Solution:
Correct Option is : (D)
Sodium is cheaper and easier to obtain than potassium.

Topic: Chemistry, s Block Elements, NEET
Statement A: Atherosclerosis is a type of Coronary Artery Diseases (CAD).

Statement B: Monocytes are the least abundant granulocytes.

Options:
A Only statement A is correct
B Only statement B is correct
C Both statements A and B are correct
D Both statements A and B are incorrect.

The correct answer is:
Only statement A is correct

Solution:
Coronary Artery Disease (CAD)
- It is often referred to as atherosclerosis, which affects the vessels that supply blood to the heart muscle.
- In this condition, the lumen of arteries gradually becomes narrower due to the deposits of calcium, fat, cholesterol, and fibrous tissues.
Basophils are the least common leukocytes, accounting for less than one percent of the total leukocyte count.

Topic: Biology, Body Fluids and Circulation, NEET
Statement $\mathbf{I}:$ If $-1 \leq x < 0$, then $\cos \left(\sin ^{-1} x\right)=-\sqrt{1-x^{2}}$
Statement II : If $-1 \leq x < 0$, then $\sin \left(\cos ^{-1} x\right)=\sqrt{1-x^{2}}$
Which one of the following is correct in respect of the aobve statements?

Options:
A Both statements I and II are independently correct and statement II is the correct explanation of statement I
B Both statements I and II are independently correct but statement II is not the correct explanation of statement I
C Statement I is correct but statement II is false.
D Statement I is false but statement II is correct.

The correct answer is:
Statement I is false but statement II is correct.

Solution:
$\quad(\mathrm{I}) \cos \left(\sin ^{-1} x\right)=\cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right)=\sqrt{1-x^{2}}$
$($ II $) \sin \left(\cos ^{-1} x\right)=\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)=\sqrt{1-x^{2}}$
Hence, statement I is false and II is true.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2010 (Phase 1)
Statement I: Red data book has a record of all animals.
Statement II: Ex-situ conservation is the best method to protect endangered species.

Options:
A If both statement I and statement II are true and statement II is a correct explanation of statement I
B If both statement I and statement II are true but statement II is not a correct explanation of statement I
C If statement I is true but statement II is false.
D If statement I is false but statement II is true

The correct answer is:
If statement I is false but statement II is true

Solution:
Red data book contains a record of species of an area which are known to be in danger. When there are situations where an animal or plant is endangered or threatened and needs urgent measures to save it from extinction, ex-situ (off-site) conservation is the desirable approach.

Topic: Biology, Biodiversity and its Conservation, NEET
Statement I : $y=-\tan ^{-1}\left(x^{-1}\right)+1$ is an increasing function of $x$ Statement II : $\frac{d y}{d x}$ is positive for all values of $x$. Which one of the following is correct in respect of the above statements?

Options:
A Both statements I and II are independently correct and statement II is the correct explanation of statement I
B Both statements I and II are independently correct but statement $\Pi$ is not the correct explanation of statement $\mathrm{I}$
C Statement I is correct but statement II is false.
D Statement I is false but statement II is correct.

The correct answer is:
Both statements I and II are independently correct and statement II is the correct explanation of statement I

Solution:
Let $y=-\tan ^{-1}\left(x^{-1}\right)+1$
$\therefore \quad \frac{d y}{d x}=-\frac{1}{1+x^{-2}}(-1) x^{-2}=\frac{1}{1+x^{2}}$
Since, $\frac{d y}{d x}$ is positive for all values of $x$. Therefore, $y$ is an increasing function of $x$ Hence, option (a) is correct.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
Statement-I: The mesophyll of Pinus shows no distinction as spongy mesophyll and palisade.

Statement-II: Parenchymatous cells are present in the mesophyll of Pinus.

Options:
A Both statements are correct
B Only statement II is correct
C Only statement I is correct
D Both statements are false

The correct answer is:
Only statement I is correct

Solution:
Within the hypodermis of Pinus leaf is the parenchymatous mesophyll. It is comparted and shows no differentiation into palisade and spongy tissues. It consists of thin-walled cells which contain numerous chloroplasts and abundant starch. The mesophyll thus functions as the chlorophyll bearing tissue which manufactures food for plants.

Topic: Biology, Plant Kingdom, NEET
Statement-I: Tiger is not an endangered animal in India.

Statement-II: Project Tiger was launched to improve the tiger population.

Options:
A Both the statements are true
B Both the statements are false
C Statement-I is true and statement-II is false
D Statement-I is false and statement-II is true

The correct answer is:
Both the statements are true

Solution:
There are 21 tiger reserves in India. To save the tiger from extinction in India Project Tiger was launched on April 1, 1973. This project planned to create tiger reserves in selected areas of India. Due to this effort considerable improvement has been observed in tiger population.

Topic: Biology, Biodiversity and its Conservation, NEET
Stirred-tank bioreactors have been designed for:

Options:
A Purification of product
B Addition of preservatives to the product
C Availability of oxygen throughout the process
D Ensuring anaerobic conditions in the culture vessel

The correct answer is:
Availability of oxygen throughout the process

Solution:
Stirred-tank bioreactor is provided with a stirrer for the availability of oxygen throughout the process.

In stirred tank bioreactors, the air is bubbled through the reactor using a sparger. Air bubbles increase the surface area for transfer of oxygen and result in better mixing.

The function of the agitation system is to increase mass transfer rates through the bulk liquid and bubble boundary layers by providing proper mixing.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2016 (Phase 2)
Students of three sections of a class, having 30,30 and 40 students appeared for a test of 100 marks. The arithmetic means of the marks of the three sections are $72.2,69.0$ and $64.1$ in that order. What is the arithmetic mean of the marks of all the students of the three sections?

Options:
A $66.6$
B $67.3$
C $68.0$
D $70.6$

The correct answer is:
$68.0$

Solution:
Required Arithmetic mean $=\frac{(30 \times 72.2)+(30 \times 69.0)+(40 \times 64.1)}{100}$
$=\frac{6800}{100}=68$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
Students of two schools appeared for a common test carrying 100 marks. The arithmetic means of their marks for school I and II are 82 and 86 respectively. If the number of students of school II is $1.5$ times the number of students of school I, what is the arithmetic mean of the marks of all the students of both the schools? $\quad$

Options:
A $84.0$
B $84.2$
C $84.4$
D This cannot be calculated with the given data

The correct answer is:
$84.4$

Solution:
Let the number of students of school $I=x$ Number of students of School II $=1.5 \mathrm{x}$ As given Mean of marks for school $I=82$ and mean of marks for school II $=86$
Combined mean $=\frac{x \times 82+1.5 x \times 86}{x+1.5 x}$
$=\frac{\mathrm{x}(82+129)}{2.5 \mathrm{x}}=\frac{211}{2.5}=84.4$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
Study the following table:
\begin{array}{c|c|c}
\hline \begin{array}{c}\text { Buffer } \\
\text { Solution }\end{array} & \begin{array}{c}\text { Volume (in } \mathrm{mL} \text { ) of } \\
\text { 0.M Weak acid }\end{array} & \begin{array}{c}\text { Volume (in } \mathrm{mL} \text { ) of } \\
0.1 \mathrm{M} \text { sodium salt of } \\
\text { weak acid }\end{array} \\
\hline I. & 4.0 & 4.0 \\
II. & 4.0 & 40.0 \\
III. & 40.0 & 4.0 \\
IV. & 0.1 & 10.0 \\
\hline
\end{array}
Which of the two sets of butter solutions have least $\mathrm{pH}$ ?

Options:
A I and II
B I and III
C II and III
D II and IV

The correct answer is:
I and III

Solution:
For a buffer solution, $\mathrm{pH}$ is given by Henderson, equation,
$$
\mathrm{pH}=\mathrm{pKa}+\log \frac{[\text { salt }]}{[\text { acid }]}
$$
Hence, $\mathrm{pH}$ is least when concentration of acid is higher than that of salt.

Topic: Chemistry, Ionic Equilibrium, AP EAMCET, AP EAMCET 2004
Study the following table:
\begin{array}{c|c|c}
\hline \begin{array}{c}\text { Buffer } \\
\text { Solution }\end{array} & \begin{array}{c}\text { Volume (in } \mathrm{mL} \text { ) of } \\
\text { 0.M Weak acid }\end{array} & \begin{array}{c}\text { Volume (in } \mathrm{mL} \text { ) of } \\
0.1 \mathrm{M} \text { sodium salt of } \\
\text { weak acid }\end{array} \\
\hline I. & 4.0 & 4.0 \\
II. & 4.0 & 40.0 \\
III. & 40.0 & 4.0 \\
IV. & 0.1 & 10.0 \\
\hline
\end{array}
Which of the two sets of butter solutions have least $\mathrm{pH}$ ?

Options:
A I and II
B I and III
C II and III
D II and IV

The correct answer is:
I and III

Solution:
For a buffer solution, $\mathrm{pH}$ is given by Henderson, equation,
$$
\mathrm{pH}=\mathrm{pKa}+\log \frac{[\text { salt }]}{[\text { acid }]}
$$
Hence, $\mathrm{pH}$ is least when concentration of acid is higher than that of salt.

Topic: Chemistry, Ionic Equilibrium, TS EAMCET, TS EAMCET 2004
Sugar and amino acids are:

Options:
A Primary metabolites
B Secondary metabolites
C Tertiary metabolites
D None of the above

The correct answer is:
Primary metabolites

Solution:
Sugar and amino acids are primary metabolites. Sugars are building blocks of starch, glycogen, etc., while amino acids are the building blocks of proteins.
primary metabolites also referred to as a central metabolite. while secondary metabolite usually has an important ecological function or related function like defence for plants.

Topic: Chemistry, Biomolecules, NEET
Suggest a reagent for the following conversion.

Options:

None

Solution:
The given reactant is
which is a secondary
alcohol. Secondary alcohol $>\mathrm{CHOH})$ on oxidation by $\mathrm{CrO}_3$ or pyridinium chlorochromate gives ketones without carrying out oxidation at the doule bond.

Topic: Chemistry, Alcohols Phenols and Ethers
Suggest some important steps that you would recommend to be taken to improve the reproductive health standards in India.

Options:

None

Solution:
Following are the measures needed to improve the reproductive health standards in India.
(i) By providing professional expertise and infrastructural facilities to attain reproductive health.
(ii) By educating people about safe and hygienic sexual practices, birth control methods, care of pregnant mothers, importance of breast feeding and safeguard against STDs.
(iii) By introduction of sex education in schools in order to provide proper information to the young minds about sex-related aspects.
(iv) By the help of audio-visual and print media, one should create awareness among people about reproduction related aspects.
(v) Bythe awareness of problems due to population explosion, social evils like sex-abuse and sex-related crimes.
(vi) By statutory ban on amniocentesis to legally check female foeticides.

Topic: Biology, Reproductive Health
Sulphur containing amino acids of the following are
\begin{array}{cccc}
Serine & Cysteine & Lysine & methionine \\
A & B & C & D
\end{array}

Options:
A A, D
B A, C
C B, C
D B, D

The correct answer is:
B, D

Solution:
S-Containing amino acids $=$ Cysteine, methionine.
$\mathrm{OH}$ - Containing amino acid $=$ Serine
$\mathrm{NH}_2$ - Containing amino acid = Lysine .

Topic: Chemistry, Biomolecules, AP EAMCET, AP EAMCET 2023 (16 May Shift 1)
Sum of first $n$ natural numbers is given by $\frac{n(n+1)}{2} .$ What is the geometric mean of the series $1,2,4,8, \ldots ., 2^{n}$ ?

Options:
A $2^{n}$
B $2^{\frac{n}{2}}$
C $2^{1 / 2}$
D $2^{n-1}$

The correct answer is:
$2^{\frac{n}{2}}$

Solution:
Geometric mean $=\sqrt[n+1]{1.2 .4 .8 \ldots 2^{n}}$
$\left(\because\right.$ there are $(n+1)$ multiples from $2^{\circ}$ to $\left.2^{n}\right)$
$=\sqrt[n+1]{2^{0} \cdot 2^{1} \cdot 2^{2} \cdot 2^{3} \ldots 2^{n}}$
$=\sqrt[n+1]{2^{1+2+3+\ldots+n}}=\sqrt[n+1]{2 \frac{n(n+1)}{2}}$
$=\left[2^{ \left.\frac{n(n+1)}{2}\right]^{\frac{1}{n+1}}=2^{\frac{n}{2}}}\right.$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
Suppose $\omega_{1}$ and $\omega_{2}$ are two distinct cube roots of unity
different from 1 . Then what is $\left(\omega_{1}-\omega_{2}\right)^{2}$ equal to?

Options:
A 3
B 1
C -1
D -3

The correct answer is:
-3

Solution:
Cube root of unity are 1, $-\frac{1}{2} \pm \frac{\sqrt{3}}{2}$ i $\omega_{1}$ and $\omega_{2}$ are two distinct cube roots of unity different from 1 .
$\omega_{1}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i, \omega_{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i$
$\left(\omega_{1}-\omega_{2}\right)^{2}=\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i+\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{2}$
$=(\sqrt{3} i)^{2}=3 i^{2}$
$\left(\omega_{1}-\omega_{2}\right)^{2}=-3$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
Suppose $\cos \mathrm{A}$ is given. If only one value of $\cos \left(\frac{\mathrm{A}}{2}\right)$ is possible, then A must be

Options:
A An odd multiple of $90^{\circ}$
B A multiple of $90^{\circ}$
C An odd multiple of $180^{\circ}$
D A multiple of $180^{\circ}$

The correct answer is:
An odd multiple of $180^{\circ}$

Solution:
Given, $\cos \frac{\mathrm{A}}{2}$ has only one value.
We know, $\cos A=2 \cos ^{2} \frac{\mathrm{A}}{2}-1$
$\Rightarrow 2 \cos ^{2} \frac{\mathrm{A}}{2}=\cos \mathrm{A}+1 \Rightarrow \cos \frac{\mathrm{A}}{2}=\sqrt{\frac{\cos \mathrm{A}+1}{2}}$
Since, $\cos \frac{\mathrm{A}}{2}$ is single value, $\frac{\cos \mathrm{A}+1}{2}=0$
$\Rightarrow \cos \mathrm{A}=-1$
So, $\mathrm{A}$ is an odd multiple of $180^{\circ}$.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 1)
Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=\left\{\begin{array}{cc}1, \quad \text { if } x=1 \\ e^{\left(x^{10}-1\right)}+(x-1)^2 \sin \frac{1}{x-1}\end{array}\right., \text { if } x \neq 1 \text { then }\)

Options:
A \(f^{\prime}(1)\) does not exist
B \(f^{\prime}(1)\) exists and is zero
C \(f^{\prime}(1)\) exist and is 9
D \(f^{\prime}(1)\) exists and is 10

The correct answer is:
\(f^{\prime}(1)\) does not exist

Solution:
Hint: \(f: R \rightarrow R f(x)=\left\{\begin{array}{cc} 1 & \text { if } x=1 & \\ e^{\left(x^{10}-1\right)} &+(x-1)^2 &\cdot \sin \left(\frac{1}{x-1}\right) \text { if } x \neq 1\end{array}\right.\)
\(\begin{aligned}
& f\left(1^{-}\right)=1+0=1 \\
& f\left(1^{+}\right)=1+0=1 \\
& f(1)=1 \\
& \therefore \quad f(x) \text { is continuous at } x=1 \\
& f^{\prime}(x)=\left\{e^{\left(x^{10}-1\right)} \cdot\left(10 x^5\right)+2(x-1) \cdot \sin \left(\frac{1}{x-1}\right)-(x-1)^2 \cdot \cos \left(\frac{1}{x-1}\right) \cdot\left(\frac{1}{x-1}\right)^2\right\}
\end{aligned}\)
which does not exist.

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2023
Suppose $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}} .$ What is
the range of the function?

Options:
A $[0,1)$
B $[0,1]$
C $(0,1]$
D $(0,1)$

The correct answer is:
$[0,1)$

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$
for $\mathrm{x}=0, \mathrm{f}(0)=0$
for $\mathrm{x}=1, \mathrm{f}(1)=\frac{1}{2}=0.5$
So, range is $[0,1)$

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 1)