datavorous/entrance-exam-dataset huggingface dataset:
The lines $(\mathrm{p}+2 \mathrm{q}) \mathrm{x}+(\mathrm{p}-3 \mathrm{q}) \mathrm{y}=\mathrm{p}-\mathrm{q}$ for different values of
p and qipass through the fixed point given by which one of the following ?

Options:
A $\left(\frac{3}{2}, \frac{5}{2}\right)$
B $\left(\frac{2}{5}, \frac{2}{5}\right)$
C $\left(\frac{3}{5}, \frac{3}{5}\right)$
D $\left(\frac{2}{5}, \frac{3}{5}\right)$

The correct answer is:
$\left(\frac{2}{5}, \frac{3}{5}\right)$

Solution:
As given, $(p+2 q) x+(p-3 q) y=p-q$ $\Rightarrow \quad p x+2 q x+p y-3 q y=p-q$
$\Rightarrow \quad p(x+y)-q(3 y-2 x)=p-q$
Equation co-efficient of $\mathrm{p}$ and $\mathrm{q}$ $\Rightarrow \quad x+y=1$ and $3 y-2 x=1$
Solving these, we get $\mathrm{x}=\frac{2}{5}, \mathrm{y}=\frac{3}{5}$.
So, line passes through $\left(\frac{2}{5}, \frac{3}{5}\right)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2006 (Phase 1)
The locus of a point equidistant from three collinear points is:

Options:
A a straight line
B a pair of points
C a point
D the null set

The correct answer is:
the null set

Solution:
There can not be any point which is equidistant from three collinear points. $\therefore$ Locus $=$ null set.

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
The locus of the point of intersection of the tangents at the
endpoints of normal chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

Options:
A $\frac{a^6}{x^2}+\frac{b^6}{y^2}=\left(a^2+b^2\right)^2$
B $\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2$
C $\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2-b^2\right)^2$
D $\frac{a^6}{x^2}+\frac{b^6}{y^2}=\left(a^2-b^2\right)^2$

The correct answer is:
$\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2$

Solution:
Let $p(h, k)$ be point of intersection of tangents at end points of normal chord
Standard equation of normal chord of hyperbola
$a x \cos \theta+b y \cot \theta=a^2+b^2$ ...(i)
for $p(h, k)$, equation of chord of contact of tangents to hyperbola is
$\frac{h x}{a^2}-\frac{k y}{b^2}=1$ ...(ii)
(i) and (ii) represent some line
$\frac{a \cos \theta}{\left(\frac{h}{a^2}\right)}=\frac{b \cot \theta}{\left(\frac{-k}{b^2}\right)}=\frac{a^2+b^2}{1}$
$\frac{a^3 \cos \theta}{h}=\frac{-b^3 \cot \theta}{k}=\frac{a^2+b^2}{1}$
$\begin{aligned} \cos \theta & =\frac{h\left(a^2+b^2\right)}{a^3} \\ \sec \theta & =\frac{a^3}{h\left(a^2+b^2\right)}\end{aligned}$
$\begin{aligned} & \text { and } \cot \theta=\frac{-k\left(a^2+b^2\right)}{b^3} \\ & \tan \theta=\frac{-b^3}{k\left(a^2+b^2\right)}\end{aligned}$
$\sec ^2 \theta-\tan ^2 \theta=1$
$\begin{aligned} & \frac{a^6}{h^2\left(a^2+b^2\right)^2}-\frac{b^6}{k^2\left(a^2+b^2\right)^2}=1 \\ & \frac{a^6}{h^2}-\frac{b^6}{k^2}=\left(a^2+b^2\right)^2 \\ & \frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2\end{aligned}$

Topic: Mathematics, Hyperbola, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
The logarithm of the equilibrium constant of the cell reaction corresponding to the cell X(s)|X2+(aq)||Y+(aq)|Y(s) with standard cell potential, Eocell=1.2V is given by

Options:
A 40.2
B 47.2
C 12.5
D 21.5

The correct answer is:
40.2

Solution:
Eo=0.0591nlogKeq
So, logK eq = E o ×n 0.0591 = 1.2×2 0.0591 =40.6

Topic: Chemistry, Electrochemistry, NEET
The lower end of capillary tube is immresed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?

Options:
A $5.9$
B $4.9$
C $2.9$
D $1.9$

The correct answer is:
$5.9$

Solution:
Given data:
Level of mercury in the tube is, \(\mathrm{h}^{\prime}=2 \mathrm{~cm}\).
Level of mercury in the tube is given as:
\(\mathrm{h}^{\prime}=\frac{2 \mathrm{~T}^{\prime} \cos \theta^{\prime}}{\mathrm{r} \rho^{\prime} \mathrm{g}} . \quad \ldots (1)\)
where \(\rho^{\prime}\) is the density of mercury.
Level of water in the tube is given as:
\(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \quad \ldots (2)\)
where \(\rho\) is the density of water.
Here, the same capillary is immersed in both the liquids hence, the radius of the capillary is same i.e. \(r\)
Using the standard values for variables and taking \(\theta^{\prime}=135^{\circ}\), in equation (1) gives:
\(\begin{aligned}
& 2=\frac{2(484.5) \cos 135^{\circ}}{r(13.6)(980)} \\
& r=\frac{685.083}{(26656)} \\
& r=0.025 \mathrm{~cm}
\end{aligned}\)
Using value of \(r\) and \(\theta=0^{\circ}\) in equation (2), gives:
\(\begin{aligned}
& \mathrm{h}=\frac{2(72.7) \cos ^{\circ}}{(0.025)(1)(980)} \\
& \mathrm{h}=\frac{145.4}{(24.5)} \\
& \mathrm{h}=5.9 \mathrm{~cm}
\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, JIPMER, JIPMER 2015
The magnetic field in a plane electromagnetic wave is given by: By=2×10-7sinπ×103x+3π×1011t T. Calculate the wavelength.

Options:
A π×103 m
B 2×10-3 m
C 2×103 m
D π×10-3 m

The correct answer is:
2×10-3 m

Solution:
Recall the standard form of the equation of electromagnetic wave in terms of the magnetic field, By=B0sinkx+ωt T, here B0 is the magnitude of magnetic field, k=2πλ & ω=2πf. The equation is given in the question, By=2×10-7sinπ×103x+3π×1011t T

⇒λ=2πk=2ππ×103⇒λ=2×10-3 m.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2020 (Phase 2)
The magnetic field of a beam emerging from a filter facing a floodlight is given by $\mathrm{B}_0=12 \times 10^{-8} \sin \left(1.20 \times 10^7 \mathrm{z}-3.60 \times 10^{15} \mathrm{t}\right) \mathrm{T}$. What is the average intensity of the beam?

Options:

None

Solution:
As we know that, The standard equation of magnetic field
$$
\mathrm{B}=\mathrm{B}_0 \sin (\mathrm{kx}-\omega \mathrm{t})
$$
And, the given equation is
$$
B=12 \times 10^{-6} \sin \left(1.20 \times 10^7 \mathrm{z}-3.60 \times 10^{15} \mathrm{t}\right) \mathrm{T} \text {. }
$$
On comparing this equation with standard equation (i), we have.
$$
B_0=12 \times 10^{-8}
$$
So, the average intensity of the beam
$$
\begin{aligned}
I_{\text {av }}=\frac{1}{2} \frac{B_0^2}{\mu_0} \cdot \mathrm{c} &=\frac{1}{2} \times \frac{\left(12 \times 10^{-8}\right)^2 \times 3 \times 10^8}{4 \pi \times 10^{-7}} \\
&=1.72 \mathrm{~W} / \mathrm{m}^2
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves
The magnitude of the scalar $\mathrm{p}$ for which the vector
$\mathrm{p}(-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})$ is of unit length is :

Options:
A $1 / 8$
B $1 / 64$
C $\sqrt{182}$
D $\frac{1}{\sqrt{182}}$

The correct answer is:
$\frac{1}{\sqrt{182}}$

Solution:
Let $\overrightarrow{\mathrm{a}}=\mathrm{p}(-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})$
$=(-3 p) \hat{i}+(-2 p) \hat{j}+(13 p) \hat{k}$
It is given that a is of unit length
$\therefore|\overrightarrow{\mathrm{a}}|=1 \Rightarrow|\overrightarrow{\mathrm{a}}|^{2}=1$
$\Rightarrow \quad(-3 \mathrm{p}))(-3 \mathrm{p})+(-2 \mathrm{p})(-2 \mathrm{p})+(13 \mathrm{p})(13 \mathrm{p})=1$
$\quad 9 \mathrm{p}^{2}+4 \mathrm{p}^{2}+169 \mathrm{p}^{2}=1$
$\Rightarrow \quad \mathrm{p}^{2}=\frac{1}{182} \Rightarrow \mathrm{p}=\frac{1}{\sqrt{182}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
The magnitude of the vectors $\vec{a}$ and $\vec{b}$ are equal and the
angle between them is $60^{\circ}$. If the vectors $\lambda \vec{a}+\vec{b}$ and $\vec{a}-\lambda \vec{b}$ are perpendicular to each other, then what is the value of $\lambda ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
Since, vectors $\lambda \vec{a}+\vec{b}$ and $a-\lambda \vec{b}$ are perpendicular
to each other therefore $(\lambda \vec{a}+\vec{b}) \cdot(\vec{a}-\lambda \vec{b})=0$
$\Rightarrow \lambda \vec{a} \cdot \vec{a}-\lambda^{2} \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\lambda \vec{b} \cdot \vec{b}=0$
$\Rightarrow \lambda|\vec{a}|^{2}+\left(1-\lambda^{2}\right) \vec{a} \cdot \vec{b}-\lambda|\vec{b}|^{2}=0$
$\Rightarrow \lambda|\vec{a}|^{2}+\left(1-\lambda^{2}\right) \vec{a} \cdot \vec{b}-\lambda|\vec{a}|^{2}=0 \quad(\because|\vec{a}|=|\vec{b}|)$
$\Rightarrow\left(1-\lambda^{2}\right) \vec{a} \cdot \vec{b}$
Since, $\cos 60^{\circ}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{a}|}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}=\vec{a} \cdot \vec{b}$
$\therefore\left(1-\lambda^{2}\right) \vec{a} \cdot \vec{b}=\left(1-\lambda^{2}\right) \cos 60^{\circ}$
$\therefore\left(1-\lambda^{2}\right) \frac{1}{2}=0 \Rightarrow \lambda=\pm 1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 2)
The major product of the following reaction is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3$
B $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$
C $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$
D $\mathrm{CH}_3 \mathrm{COCH}_3$

The correct answer is:
$\mathrm{CH}_3 \mathrm{COCH}_3$

Solution:
When vapours of primary or secondary alcohols are passed over heated $\mathrm{Cu}$ at $573 \mathrm{~K}$ dehydrogenation $\left(-\mathrm{H}_2\right)$ takes place and aldehyde or ketone is formed whereas tertiary alcohols undergo dehydration $\left(-\mathrm{H}_2 \mathrm{O}\right)$.

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 1)
The major product of the following reactions is

Options:
A

B

C

D

The correct answer is:

Solution:
Alkene gives the hydroboration oxidation reaction. It formed the alkene to alcohol. After that the alcohol is oxidised into the aldehyde by the PCC. After that Grignard reagent followed by the hydrolysis gives the nucleophilic addition reaction. It converts the aldehyde into the secondary alcohol.

Reaction :-

Hence, the major product of this reaction is .

Topic: Chemistry, Hydrocarbons, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
The male sex accessory duct includes

Options:
A Vasa efferentia
B Epididymis
C Vas deferens
D All of the above

The correct answer is:
All of the above

Solution:
The vasa efferentia, epididymis and vas deference do not form gametes hence referred as secondary sex duct.

Male Reproductive System	Gonads	Testes (1 pair)
Accessory ducts	Rete testis + Vasa efferentia + Epididymis + Vas deferens
External genitalia	Penis
Glands	Seminal vesicles (pair) + Prostate gland (one) + Bulbourethral glands (pair)

Topic: Biology, Human Reproduction, NEET
The male thallus of a bryophyte is given in the diagram below. Select the option giving incorrect information regarding it.

Options:
A Gametophyte phase is dominant phase.
B Gemma cups participate in asexual reproduction
C Given bryophyte is a Equisetum.
D It can also reproduce by fragmentation.

The correct answer is:
Given bryophyte is a Equisetum.

Solution:
2 stages of plant body of bryophytes
A. Protonema stage - It is a creeping, green, branching and filamentous stage. Developed from spore.
B. Leafy stage - It is upright, slender bearing spirally arranged leaves. It develops from secondary protonema. Attached to soil by multicellular and branched rhizoids.

Topic: Biology, Plant Kingdom, NEET
The marks obtained by 13 students in a test are $10,3,10,12$, $9,7,9,6,7,10,8,6,7$. The median of this data is?

Options:
A 7
B 8
C 9
D 10

The correct answer is:
8

Solution:
First we arrange the data in ascending order $3,6,6,7,7,7,8,9,9,10,10,10,12$
Since no. of observations is odd
Required median $=\left(\frac{13+1}{2}\right)^{\text {th }}$ observation $=\left(\frac{14}{2}\right)^{\text {th }}=7^{\text {th }}$ observation

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
The marks scored by two students A and B in six subjects are given below:
$\begin{array}{ccccccc}\text { A } & 71 & 56 & 45 & 89 & 54 & 44 \\ \text { B } & 55 & 74 & 83 & 54 & 38 & 52\end{array}$
Which one of the following statements is correct?

Options:
A The average scores of $\mathrm{A}$ and $\mathrm{B}$ are same but $\mathrm{A}$ is consistent
B The average scores of $A$ and $B$ are not same but $A$ is consistent
C The average scores of $\mathrm{A}$ and $\mathrm{B}$ are same but $\mathrm{B}$ is consistent
D The average scores of $\mathrm{A}$ and $\mathrm{B}$ are not same but $\mathrm{B}$ is consistent $\quad$

The correct answer is:
The average scores of $\mathrm{A}$ and $\mathrm{B}$ are not same but $\mathrm{B}$ is consistent $\quad$

Solution:
Average score of $\mathrm{A}=\frac{71+56+45+89+54+44}{6}$ $=\frac{359}{6}=59.83$
and average score of $\mathrm{B}=\frac{55+74+83+54+38+52}{6}=\frac{356}{6}=59.33$
Variation is lesser in case of B than A. So, the average scores of A and B are not same but B is consistent.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 2)
The marks scored by two students $A$ and $B$ in six subjects are given below \begin{array}{lllllll}
\hline A & 71 & 56 & 55 & 75 & 54 & 49 \\
\hline B & 55 & 74 & 83 & 54 & 38 & 52 \\
\hline
\end{array} Which one of the following statements is most appropriate'

Options:
A Theaverage scores of $A$ and $B$ are same but $A$ is consistent
B The average scores of $A$ and $B$ are not same but $A$ is consistent
C The average scores of $A$ and $B$ are same but $B$ is consistent
D The average scores of $A$ and $B$ are not same but $B$ is consistent

The correct answer is:
The average scores of $A$ and $B$ are not same but $A$ is consistent

Solution:
Average of marks of A $=\frac{71+56+55+75+54+49}{6}=\frac{360}{6}=60$
and SD
$=\sqrt{\frac{121+16+25+225+36+121}{6}}=\sqrt{\frac{544}{6}}=9.52$
Average of marks of B
$=\frac{55+74+83+54+38+52}{6}=\frac{356}{6}=59.3$
Thus, the average scores of $A$ and $B$ are not same but $A$ is consistent

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 1)
The matrix $\left[\begin{array}{cc}0 & -4+i \\ 4+i & 0\end{array}\right]$ is

Options:
A symmetric
B skew-symmetric
C Hermitian
D skew-Hermitian

The correct answer is:
skew-Hermitian

Solution:
The given matrix $\mathrm{A}=\left[\begin{array}{cc}0 & -4+\mathrm{i} \\ 4+\mathrm{i} & 0\end{array}\right]$
Now, from options:
From option (a): For Symmetric matrix
$\mathrm{A}^{\mathrm{T}}=\mathrm{A}$
Now, $\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{cc}0 & 4+\mathrm{i} \\ -4+\mathrm{i} & 0\end{array}\right] \neq \mathrm{A}$
$\therefore$ The given matrix is not symmetric
$\therefore \quad$ option (a) is wrong. From option (b): For Skew-symmetric matrix $\mathrm{A}^{\mathrm{T}}=-\mathrm{A}$
$=\left[\begin{array}{cc}0 & 4+i \\ -4+i & 0\end{array}\right] \neq-\mathrm{A}$
$\therefore \quad$ Given matrix is not skew-symmetric
$\therefore \quad$ option (b) is wrong. From option (c): For Hermitian matrix
$\mathrm{A}^{\mathrm{T}}=\overline{\mathrm{A}}$, where $\overline{\mathrm{A}}$ is conjugate of matrix $\mathrm{A}$
$\overline{\mathrm{A}}=\left[\begin{array}{cc}0 & -4-\mathrm{i} \\ 4-\mathrm{i} & 0\end{array}\right] \neq \mathrm{A}^{\mathrm{T}}$
$\therefore \quad$ option (c) is wrong. From option (d): For Skew-Hermitian matrix The diagonal element of a skew-hermitian matrix are pure imaginary or zero.
$\mathrm{A}=\left[\begin{array}{cc}0 & -4+\mathrm{i} \\ 4+\mathrm{i} & 0\end{array}\right]$
Here, diagonal element indicates that the given matrix is skew-hermitian matrix.
$\therefore$ option (d) is correct.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 1)
The matrix $\mathrm{A}=\left[\begin{array}{ccc}1 & 3 & 2 \\ 1 & x-1 & 1 \\ 2 & 7 & x-3\end{array}\right]$ will have inverse for every real number $x$ except for

Options:
A $x=\frac{11 \pm \sqrt{5}}{2}$
B $x=\frac{9 \pm \sqrt{5}}{2}$
C $x=\frac{11 \pm \sqrt{3}}{2}$
D $x=\frac{9 \pm \sqrt{3}}{2}$

The correct answer is:
$x=\frac{11 \pm \sqrt{5}}{2}$

Solution:
$A=\left[\begin{array}{ccc}1 & 3 & 2 \\ 1 & x-1 & 1 \\ 2 & 7 & x-3\end{array}\right]$
$|\mathrm{A}|=1[(\mathrm{x}-1)(\mathrm{x}-3)-7]-3[(\mathrm{x}-3)-2]+2[7-2(\mathrm{x}-1)]$
$=x^{2}-11 x+29$
If inverse will not exist then $|\mathrm{A}|=0$ $x^{2}-11 x+29=0$
$\mathrm{x}=\frac{11 \pm \sqrt{5}}{2}$

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 2)
The matrix $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]$ satisfies which one of the following polynomial equations?

Options:
A $A^{2}+3 A+2 I=0$
B $A^{2}+3 A-2 I=0$
C $A^{2}-3 A-2 I=0$
D $A^{2}-3 A+2 I=0$

The correct answer is:
$A^{2}-3 A-2 I=0$

Solution:
Given that, $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]$
$\therefore \quad A^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{cc}1+4 & 2+4 \\ 2+4 & 4+4\end{array}\right]$
$=\left[\begin{array}{ll}5 & 6 \\ 6 & 8\end{array}\right]$
Let $\mathrm{A}^{2}+\mathrm{xA}+\mathrm{yI}=0$ where $\mathrm{x}$ and $\mathrm{y}$ are constant.
$\Rightarrow\left[\begin{array}{ll}5 & 6 \\ 6 & 8\end{array}\right]+\left[\begin{array}{cc}x & 2 x \\ 2 x & 2 x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}5+x+y & 6+2 x \\ 6+2 x & 8+2 x+y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
So, $\begin{aligned} 6+2 \mathrm{x}=0 & \Rightarrow \mathrm{x}=-3 \\ 5+\mathrm{x}+\mathrm{y}=0 & \Rightarrow \mathrm{y}=-5-\mathrm{x}=-2 \\ & \Rightarrow \mathrm{A}^{2}-3 \mathrm{~A}-2 \mathrm{I}=0 \end{aligned}$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 2)
The matrix A has $x$ rows and $x+5$ columns. The matrix $B$ has y rows and $11-y$ columns. Both $\mathrm{AB}$ and BA exist. What are the values of $\mathrm{x}$ and $\mathrm{y}$ respectively?

Options:
A 8 and 3
B 3 and 4
C 3 and 8
D 8 and 8

The correct answer is:
3 and 8

Solution:
Given, Matrix A has $x$ rows and $x+5$ columns Matrix $\mathrm{B}$ has $\mathrm{y}$ rows and $11-\mathrm{y}$ columns. Also given $\mathrm{AB}$ and $\mathrm{BA}$ exists. If AB exists, then the number of rows in A most be equal to number of columns in $\mathrm{B}$.
i.e., $x=11-y$ $\ldots .(1)$
If BA exists, then the number of rows in $\mathrm{B}$ must be equal to number of rows in $\mathrm{A}$.
i.e., $x+5=y$ $\Rightarrow 11-y+5=y($ from $(1))$
$\Rightarrow 2 y=16$
$\Rightarrow y=8$
$(1) \Rightarrow x=11-8=3$
So, $x=3, y=8$.

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 2)
The matrix \(A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{2}}\end{array}\right]\) is

Options:
A Unitary
B Orthogonal
C Nilpotent
D Involutory

The correct answer is:
Nilpotent

Solution:
\(\begin{aligned}
A & =\left[\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{array}\right] \\
A^2 & =\left[\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{array}\right]\left[\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{array}\right] \\
\Rightarrow \quad A^2 & =\left[\begin{array}{cc}
\frac{1}{2}-\frac{1}{2} & \frac{1}{2}-\frac{1}{2} \\
-\frac{1}{2}+\frac{1}{2} & -\frac{1}{2}+\frac{1}{2}
\end{array}\right] \\
A^2 & =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \text { null matrix }
\end{aligned}\)
So, \(A\) is nilpotent matrix

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
The maximum value of $\quad$ $\sin \left(x+\frac{\pi}{5}\right)+\cos \left(x+\frac{\pi}{5}\right)$, where $x \in\left(0, \frac{\pi}{2}\right)$, is attained at

Options:
A $\frac{\pi}{20}$
B $\frac{\pi}{15}$
C $\frac{\pi}{10}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{20}$

Solution:
Let $\mathrm{y}=\sin \left(\mathrm{x}+\frac{\pi}{5}\right)+\cos \left(\mathrm{x}+\frac{\pi}{5}\right)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \left(\mathrm{x}+\frac{\pi}{5}\right)-\sin \left(\mathrm{x}+\frac{\pi}{5}\right)$
for maximum value, $\frac{\mathrm{dy}}{\mathrm{dx}}=0$. $\therefore \cos \left(x+\frac{\pi}{5}\right)-\sin \left(x+\frac{\pi}{5}\right)=0$
$\Rightarrow \frac{\cos \left(x+\frac{\pi}{5}\right)}{\sin \left(x+\frac{\pi}{5}\right)}=1$
$\Rightarrow \cot \left(x+\frac{\pi}{5}\right)=\cot \frac{\pi}{4}$
$\Rightarrow x+\frac{\pi}{5}=\frac{\pi}{4}$
$\Rightarrow x=\frac{\pi}{4}-\frac{\pi}{5}=\frac{5 \pi-4 \pi}{20}=\frac{\pi}{20}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 1)
The maximum value of $\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$ in the
interval $\left(0, \frac{\pi}{2}\right)$ is attained at

Options:
A $\frac{\pi}{12}$
B $\frac{\pi}{6}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{12}$

Solution:
$\quad \sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$
$=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \left(x+\frac{\pi}{6}\right)+\frac{1}{\sqrt{2}} \cos \left(x+\frac{\pi}{6}\right)\right]$
$=\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}\right) \cos \frac{\pi}{4}+\cos \left(x+\frac{\pi}{6}\right) \sin \frac{\pi}{4}\right]$
$=\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}+\frac{\pi}{4}\right)\right]$
$[\because \sin (A+B)=\sin A \cos B+\cos A \operatorname{Sin} B]$
$=\sqrt{2}\left[\sin \left(x+\frac{5 \pi}{12}\right)\right]$
Given interval is $\left(0, \frac{\pi}{2}\right)$ For, maximum value $x+\frac{5 \pi}{12}=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{2}-\frac{5 \pi}{12}=\frac{6 \pi-5 \pi}{12}=\frac{\pi}{12}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 1)
The maximum value of the function $f(x)=x^{3}+2 x^{2}-4 x+6$ exists at

Options:
A $\quad x=-2$
B $x=1$
C $x=2$
D $x=-1$

The correct answer is:
$\quad x=-2$

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+2 \mathrm{x}^{2}-4 \mathrm{x}+6$
$\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+4 \mathrm{x}-4$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}+4$
Put $\mathrm{f}^{\prime}(\mathrm{x})=0$
$3 \mathrm{x}^{2}+4 \mathrm{x}-4=0$
$\mathrm{x}=-2, \frac{2}{3}$
$\mathrm{f}^{\prime \prime}(-2)=6 \mathrm{x}-2+4=-8 < 0$
$\mathrm{f}^{\prime \prime}\left(\frac{2}{3}\right)=6 \times \frac{2}{3}+4=8>0$
Value is maximumat $\mathrm{x}=-2$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 2)
The maximum value of $\frac{\operatorname{In} x}{x}$ is

Options:
A e
B $\frac{1}{\mathrm{e}}$
C $\frac{2}{\mathrm{e}}$
D 1

The correct answer is:
$\frac{1}{\mathrm{e}}$

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\ln \mathrm{x}}{\mathrm{x}}$
$f^{\prime}(x)=\frac{\left(\frac{1}{x}\right) x-\ell n x}{x^{2}}=\frac{1-\ell n x}{x^{2}}$
$\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \frac{1-\ln \mathrm{x}}{\mathrm{x}^{2}}=0 \Rightarrow 1-\ell \mathrm{nx}=0$
$\Rightarrow \ell \mathrm{nx}=1 \quad \Rightarrow \mathrm{x}=\mathrm{e}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
The maximum work (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) that can be derived from complete combustion of $1 \mathrm{~mol}$ of $\mathrm{CO}$ at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ is
[Standard enthalpy of combustion of $\mathrm{CO}=-283.0 \mathrm{~kJ} \mathrm{~mol}^{-1} ;$ standard molar entropies at $298 \mathrm{~K} ; \mathrm{S}_{\mathrm{O} 2}=205.1 \mathrm{~J} \mathrm{~mol}^{-1}$ $\left.\mathrm{S}_{\mathrm{CO}}=197.7 \mathrm{~J} \mathrm{~mol}^{-1}, \mathrm{~S}_{\mathrm{CO} 2}=213.7 \mathrm{~J} \mathrm{~mol}^{-1}\right]$

Options:
A 257
B 227
C 57
D 127

The correct answer is:
257

Solution:
$\mathrm{Co}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) \Delta \mathrm{N}=-283.5$
$\Delta \mathrm{S}=213.7-197.7-\frac{20501}{2}=-86.5$
$\Delta \mathrm{G}=-283-\frac{298 \times(-86.5)}{1000}=-2570 \mathrm{~kJ}$
$\mathrm{~W}_{\max }=-\Delta \mathrm{G}=-(-257 \mathrm{~kJ})=257 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), KVPY, KVPY 2017 (19 Nov SB/SX)