datavorous/entrance-exam-dataset huggingface dataset:
The geometric mean of three numbers was computed as 6 . It was subsequently found that, in this computation, a number 8 was wrongly read as 12 . What is the correct geometric mean?

Options:
A 4
B $\sqrt[3]{5}$
C $2 \sqrt[3]{18}$
D None of these

The correct answer is:
$2 \sqrt[3]{18}$

Solution:
We know geometric mean of 3 numbers $x_{1}, x_{2}, x_{3}$ is. $\sqrt[3]{x_{1} \cdot x_{2} \cdot x_{3}}$
Given, if observations are $x_{1}, x_{2}, 12, \mathrm{G.M}$. is 6
$\Rightarrow \sqrt[3]{x_{1} \cdot x_{2} \cdot 12}=16$
$\Rightarrow x_{1} \times x_{2} \times 12=6^{3}=216$
$\Rightarrow \quad x_{1} \times x_{2}=\frac{216}{12}=18$ ...(i)
Also, given that actual number is 8 .
Actual G.M. $=\sqrt[3]{x_{1} \cdot x_{2} \cdot 8}=\sqrt[3]{18 \times 8} \quad$ (from(i))
$=\sqrt[3]{18 \times 2 \times 2 \times 2}=2 \cdot \sqrt[3]{18}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 1)
The given diagram represents

Options:
A The injection of ovum and sperms to the female uterus.
B The blastomere is transferred to the uterus for further development of the baby.
C In vitro fertilization followed by zygote transfer into a fallopian tube or uterus of the female.
D Both a and c

The correct answer is:
The blastomere is transferred to the uterus for further development of the baby.

Solution:
In the Test tube baby program an Ovum is collected from the wife/donor (female) and sperms are collected from husband/ donor (male) and are induced to form zygote under simulated conditions in the laboratory and are transferred to the uterus for further development.

Topic: Biology, Reproductive Health, NEET
The given figure shows T.S. of

Options:
A Dicot root
B Monocot root
C Secondary growth in dicot root
D Secondary growth in monocot root

The correct answer is:
Monocot root

Solution:
The anatomy of the monocot root has epidermis, cortex, endodermis, pericycle, vascular bundles, and pith. As compared to the dicot root which has fewer xylem bundles, there are usually more than six (polyarch) xylem bundles in the monocot root. Pith is large and well developed. Monocotyledonous roots do not undergo any secondary growth.

Topic: Biology, Anatomy of Flowering Plants, NEET
The growth of a quantity $N(t)$ at any instant $t$ is given by
$\frac{d N(t)}{d t}=\alpha N(t)$. Given that $N(t)=c e^{k \pi}, c$ is a constant. What
is the value of $\alpha ?$

Options:
A $c$
B $k$
C $c+k$
D $c-k$

The correct answer is:
$k$

Solution:
Given $\mathrm{N}(\mathrm{t})=c e^{k t}$
Diff. both side w.r.t. ' $t^{\prime}$
$\begin{aligned} \therefore \quad \frac{d N(t)}{d t} &=\frac{d}{d t} c e^{k t}=k\left(c e^{k t}\right) & & \\=k[N(t)] & &(\text { by Defn. of } \mathrm{N}(t)) \end{aligned}$
But $\frac{d N(t)}{d t}=\alpha N(t) \quad$ (given)
$\Rightarrow \alpha=k$

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 1)
The growth of the given system per unit time expressed on a common basis, per unit initial parameter is called

Options:
A Absolute growth rate
B Arithematicgrowth
C Relative growth rate
D Geometric growth

The correct answer is:
Relative growth rate

Solution:
The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter is called the relative growth rate. The relative growth rate is determined by the growth of the plant in its size at certain time intervals.

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Topic: Biology, Plant - Growth and Development, NEET
The ΔHf0 for CO2(g) , CO(g) and H2O(g) are -393.5 , -110.5 and -241.8kJmol-1 respectively. The standard enthalpy changes (inkJ) for the reaction
CO2g+H2g→COg+H2O(g) is

Options:
A 524.1
B 41.2
C -262.5
D –41.2

The correct answer is:
41.2

Solution:
ΔHreaction°=ΔHf°CO+ΔHf°H2O,g-ΔHf°CO2,g=–110.5–241.8+393.5=41.2

Topic: Chemistry, Thermodynamics (C), NEET
The harmonic mean H of two numbers is 4 and the arithmetic mean A and geometric mean G satisfy the equation $2 \mathrm{~A}+\mathrm{G}^{2}=27$. The two numbers are

Options:
A 6,3
B 9,5
C 12,7
D 3,1

The correct answer is:
6,3

Solution:
Let the two numbers be $\mathrm{a}, \mathrm{b}$.
Given, $H . M .=4 \Rightarrow \frac{2 a b}{a+b}=4 \Rightarrow a b=2(a+b)$ ...(i)
Also, given $2 \mathrm{~A}+\mathrm{G}^{2}=27$
$\Rightarrow 2\left(\frac{a+b}{2}\right)+a b=27$
$\Rightarrow \quad a+b+a b=27$
$\Rightarrow a+b+2(a+b)=27 \quad$ (from(i))
$\Rightarrow \quad 3 \mathrm{a}+3 \mathrm{~b}=27$
$\Rightarrow \quad a+b=9$ ...(ii)
From (i), $a b=2(9)=18$ ...(iii)
Solving (ii), (iii) we get
$$
\mathrm{a}=3, \mathrm{~b}=6 \text { or } \mathrm{a}=6, \mathrm{~b}=3
$$

Topic: Mathematics, Sequences and Series, NDA, NDA 2013 (Phase 1)
The harmonic mean of two numbers is $21.6$.If one of the numbers is 27 , then what is the other number?

Options:
A $16.2$
B $17.3$
C 18
D 20

The correct answer is:
18

Solution:
Harmonic mean $=21.6$ and $\mathrm{a}=27$ We know that,
Harmonic mean $=\frac{2 a b}{a+b} \Rightarrow 21.6=\frac{2 \times 27 \times b}{27+b}$
$\Rightarrow 583.2=54 \mathrm{~b}-21.6 \mathrm{~b}$
$\Rightarrow \quad b=\frac{583.2}{32.4}=18$

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 1)
The helical structure of protein is stabilized by:

Options:
A dipeptide bonds
B hydrogen bonds
C ether bonds
D peptide bonds

The correct answer is:
hydrogen bonds

Solution:
The helical structure of proteins or the alpha helix is the secondary structure of proteins and it is stabilized by hydrogen bonds. In helical structure, the carbonyl oxygen atoms in $\mathrm{C}=$ $\mathrm{O}$ points in one direction, towards the amide hydrogen atoms in $\mathrm{N}-\mathrm{H}$ groups, 4 residues away.

Topic: Chemistry, Biomolecules, NEET, NEET 2004
The high growth at the time of the exponential phase of the human population growth curve is due to:

Options:
A low birth and death rates
B high birth and death rates
C high birth and low death rates
D low birth and high death rates

The correct answer is:
high birth and low death rates

Solution:
Graphical representation of the population growth is referred to as population growth curve. There are four phases of growth curve-
Lag phase - initial phase of growth where population is adapting to the environment and resources are plenty.
Exponential phase - rapid increase in population size where birth rate is higher than death rate. Natality is higher and mortality is low as there is abundance of resources and reduced resistance from environment.
Transitional phase - Due to limited resources competition between organisms of population, the natality starts reducing and increasing mortality.
Plateau phase - the natality is nearly equal to the mortality and population growth becomes static.

Topic: Biology, Organisms and Populations, NEET
The historic Convention on Biological Diversity, 'The Earth Summit' was held in Rio de Janeiro in the year:

Options:
A 2002
B 1985
C 1992
D 1986

The correct answer is:
1992

Solution:
Biodiversity knows no political boundaries and its conservation is therefore a collective responsibility of all nations. The historic Convention on Biological Diversity, known as ‘The Earth Summit’ was held in Rio de Janeiro in1992. This summit called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2023 (All India)
The hormones that activate the membrane-bound receptors are?

Options:
A Steroid hormones
B Thyroxine
C Peptide hormones
D Estrogen

The correct answer is:
Peptide hormones

Solution:
Secondary messengers are associated with response to peptide hormones they are insoluble in lipid and soluble in water. They do not enter inside the cell but bind with the extracellular receptors and generate the second messenger. Whereas steroid hormones, thyroxine, estrogen all are lipid-soluble, they directly enter the cell and bind with intracellular messengers. They do not need a second messenger.

Topic: Biology, Chemical Coordination and Integration, NEET
The hydrolysis of optically active \( 2 \)-bromobutane with aqueous \( \mathrm{NaOH} \) result in the formation of

Options:
A \( (+) \) butan-2-ol
B (-) butan-2-ol
C ( \( \pm \) ) butan-1-ol
D ( \( \pm \) ) butan-2-ol

The correct answer is:
( \( \pm \) ) butan-2-ol

Solution:
2-Bromobutane is a secondary alkyl halide, therefore, hydrolysis will proceed through \( S_{N} 1 \) mechanisms, hence, racemic
mixture will be formed, that is, (±) butan-2-ol.
(+) Butan-2-ol
(-) Butan-2-ol

Topic: Chemistry, Hydrocarbons, KCET, KCET 2015
The hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ passes through the point
$(3 \sqrt{5}, 1)$ and the length of its latus rectum is $\frac{4}{3}$ units. The
length of the conjugate axis is

Options:
A 2 units
B 3 units
C 4 units
D 5 units

The correct answer is:
4 units

Solution:
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Hyperbola passes through $(3 \sqrt{5}, 1)$
$\therefore \quad \frac{(3 \sqrt{5})^{2}}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}=1$...(i)
$\frac{45}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}=1$
Now length of latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
$\Rightarrow \frac{4}{3}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
$\Rightarrow \quad \frac{2}{3}=\frac{\mathrm{b}^{2}}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{3 \mathrm{~b}^{2}}{2}$...(ii)
Putting the value of ' $a$ ' from equation (ii) in equation (i),
$\Rightarrow \frac{45 \times 4}{9 b^{4}}-\frac{1}{b^{2}}=1$
$\Rightarrow \frac{20}{b^{4}}-\frac{1}{b^{2}}=1$
$20-b^{2}=b^{4}$
$b^{4}+b^{2}-20=0$
$b^{4}+5 b^{2}-4 b^{2}-20=0$
$b^{2}\left(b^{2}+5\right)-4\left(b^{2}+5\right)=0$
$\left(b^{2}-4\right)\left(b^{2}+5\right)=0$
$b^{2}=4, b^{2}=-5$
$\therefore \quad b^{2}=4 \Rightarrow b=2$
Nowlength of conjugate axis $=2 \mathrm{~b}=2(2)=4$ $\therefore$ Option (c) is correct.

Topic: Mathematics, Hyperbola, NDA, NDA 2015 (Phase 1)
The idiom/phrase of To wash one's dirty linen in public

Options:
A To criticise one's nature in public
B To quarrel in the open
C To do some ugly work in public
D To discuss dirty and scandalous matters of personal nature in the presence of strangers

The correct answer is:
To discuss dirty and scandalous matters of personal nature in the presence of strangers

Solution:
No Solution Available

Topic: Biology, Human Health and Diseases, JIPMER, JIPMER 2008
The immature male germ cell undergo division to produce sperms by the process of spermatogenesis. Choose the correct one with reference to above.

Options:
A
Spermatogonia have 46 chromosomes and always undergo meiotic cell division

B
Primary spermatocytes divide by mitotic cell division

C
Secondary spermatocytes have 23 chromosomes and undergo second meiotic division

D
Spermatozoa are transformed into spermatids

The correct answer is:
Secondary spermatocytes have 23 chromosomes and undergo second meiotic division

Solution:
In testis, the immature male germ cells (spermatogonia) produce sperms by the process of spermatogenesis. The spermatogonia present on the inside wall of seminiferous tubules multiply by mitotic division and increase in numbers.
Each spermatogonia is diploid that contains $46(2 \mathrm{n})$ chromosomes. Some of the spermatogonia called primary spermatocytes periodically undergo meiosis. A primary spermatocyte completes the first meiotic division (reduction division) leading to formation of two equal, haploid cells called secondary spermatocytes, which contains only 23 chromosomes each (n).
The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid spermatids. The spermatids are transformed into spermatozoa (sperms) by the process called spermiogenesis.

Topic: Biology, Human Reproduction
The immature male germ cell undergoes division to produce sperms by the process of spermatogenesis. Choose the correct one with reference to above.

Options:
A Spermatogonia have 46 chromosomes and always undergo meiotic cell division.
B Primary spermatocytes divide by mitotic cell division.
C Secondary spermatocytes have 23 chromosomes and undergo second meiotic division.
D Spermatozoa are transformed into spermatids.

The correct answer is:
Secondary spermatocytes have 23 chromosomes and undergo second meiotic division.

Solution:
Secondary spermatocytes have 23 chromosomes and undergo second meiotic division. Spermatogonia have 46 chromosomes and always undergo mitotic cell division. Primary spermatocytes divide by meiotic cell division. Spermatids are transformed into spermatozoa.

Topic: Biology, Human Reproduction, NEET
The immunity associated with memory is called

Options:
A Natural immunity
B Acquired immunity
C Innate immunity.
D Passive immunity

The correct answer is:
Acquired immunity

Solution:
Acquired immunity is pathogen-specific. It is characterized by memory. This means when our body encounters a pathogen for the first time it produces a response called primary response which is of low intensity. Subsequent encounter with the same pathogen elicits a highly intensified secondary or anamnestic response.

Topic: Biology, Human Health and Diseases, NEET
The incentre of the triangle with vertices $\mathrm{A}(1, \sqrt{3}), \mathrm{B}(0,0)$
and $\mathrm{C}(2,0)$ is

Options:
A $\left(1, \frac{\sqrt{3}}{2}\right)$
B $\left(\frac{2}{3}, \frac{1}{\sqrt{3}}\right)$
C $\left(\frac{2}{3}, \frac{\sqrt{3}}{2}\right)$
D $\left(1, \frac{1}{\sqrt{3}}\right)$

The correct answer is:
$\left(1, \frac{1}{\sqrt{3}}\right)$

Solution:
Vertices of triangle, $\mathrm{A}(1, \sqrt{3}), \mathrm{B}(0,0), \mathrm{C}(2,0)$
observe the figure, $\mathrm{BD}=1, \mathrm{DC}=1, \mathrm{AD}=\sqrt{3}$
So, $A B=2 . A C=2$ (Using Pythagoras theorem) So, given triangle is equilateral triangle In equilateral triangle, Incentre is controid only. $\therefore$ Incentre $=\left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right)$
$=\left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right)$
$=\left(1, \frac{1}{\sqrt{3}}\right)$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 1)
The Indian government has set up an organization that will make decisions regarding the validity of GM research and the safety of introducing GM organisms for public services, that organization is called

Options:
A GAEC
B GEAC
C GCAE
D GCEA

The correct answer is:
GEAC

Solution:
When modified living organisms are introduced into the ecosystem the results are unknown and unpredictable and therefore require regulation. These human activities along with moral issues have biological significance also for which certain ethical standards are required for their evaluation. Therefore, an organization like GEAC (Genetic Engineering Approval Committee) has been set up by the Government of India to take decisions regarding the validity of GM research and safety of GM-organisms for public services.

Topic: Biology, Biotechnology and Its Applications, NEET
The induction coil works on the principle of

Options:
A self-induction
B mutual induction
C Ampere's rule
D Fleming's right hand rule

The correct answer is:
mutual induction

Solution:
The induction coil works on the principle of mutual induction. In the phenomena of mutual induction, an emf is induced across the secondary coil if the magnetic flux due to primary coil linked with the secondary coil changes.

Topic: Physics, Electromagnetic Induction, COMEDK, COMEDK 2017
The inner layer of the eye is __(i)__ which contains layers of cells from inside to outside in an order (ii):

Options:
A i: Retina; (ii) Ganglion cells â Bipolar cells â Photoreceptor cells
B i: Cornea; (ii) Ganglion cells â Photoreceptor cells â Bipolar cells
C i: Retina; (ii) Photoreceptor cells â Bipolar cells â Ganglion cells
D i: Cornea; Photoreceptor cells â Bipolar cells â Ganglion cells

The correct answer is:
i: Retina; (ii) Ganglion cells â Bipolar cells â Photoreceptor cells

Solution:
Being part of the retina, bipolar cells exist between photoreceptors and ganglion cells. They act, directly or indirectly, to transmit signals from the photoreceptors to the ganglion cells. A radial section of the retina reveals the ganglion cells (the output neurons of the retina) lie innermost in the retina closest to the lens, and the photosensors (the rods and cones) lie outermost.
The cornea is the transparent part of the eye that covers the front portion of the eye.

Topic: Biology, Neural Control and Coordination, NEET
The integral $\int \frac{d x}{a \cos x+b \sin x}$ is of the form $\frac{1}{r} \operatorname{In}\left[\tan \left(\frac{x+\alpha}{2}\right)\right]$.
What is $\alpha$ equal to?

Options:
A $\tan ^{-1}\left(\frac{a}{b}\right)$
B $\tan ^{-1}\left(\frac{b}{a}\right)$
C $\tan ^{-1}\left(\frac{a+b}{a-b}\right)$
D $\tan ^{-1}\left(\frac{a-b}{a+b}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{a}{b}\right)$

Solution:
$\begin{aligned} \mathrm{a} &=\mathrm{r} \sin \alpha \quad ...(i) \\ \mathrm{b} &=\mathrm{r} \cos \alpha \quad ....(ii) \end{aligned}$
Dividing (i) from (ii),
$\frac{\mathrm{a}}{\mathrm{b}}=\tan \alpha$
$\alpha=\tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right) .$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2015 (Phase 1)
The integral $\int \frac{d x}{a \cos x+b \sin x}$ is of the form $\frac{1}{r} \operatorname{In}\left[\tan \left(\frac{x+\alpha}{2}\right)\right]$.
What is $r$ equal to?

Options:
A $a^{2}+b^{2}$
B $\sqrt{a^{2}+b^{2}}$
C $a+b$
D $\sqrt{a^{2}-b^{2}}$

The correct answer is:
$\sqrt{a^{2}+b^{2}}$

Solution:
Given that, $\int \frac{\mathrm{dx}}{\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}}=\frac{1}{\mathrm{r}} \ln \left[\tan \left(\frac{\mathrm{x}+\alpha}{2}\right)\right]$
Let $a=r \sin \alpha, b=r \cos \alpha$
$\int \frac{d x}{r \sin \alpha \cos x+r \cos \alpha \sin x}=\frac{1}{r} \int \frac{1}{\sin (x+\alpha)}$
$=\frac{1}{r} \int \operatorname{cosec}(x+\alpha) d x=\frac{1}{r} \ln \left[\tan \left(\frac{x+\alpha}{2}\right)\right]$
$a=r \sin \alpha \Rightarrow a^{2}=r^{2} \sin ^{2} \alpha$ ...(i)
$b=r \cos \alpha \Rightarrow b^{2}=r^{2} \cos ^{2} \alpha$ ...(ii)
Adding (i) and (ii), we get $r^{2}=a^{2}+b^{2}$
$\Rightarrow r=\sqrt{a^{2}+b^{2}}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2015 (Phase 1)
The interior angles of a polygon of $\mathrm{n}$ sides are in AP. The smallest angle is $120^{\circ}$ and the common difference is $5^{\circ}$
How many possible values can n have?

Options:
A One
B Two
C Three
D Infinitely many

The correct answer is:
One

Solution:
Here, $a=120^{\circ}$ and $d=5$. Sum of angles of polygon $=(n-2) 180^{\circ}$
$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=(n-2) 180$
$\Rightarrow \frac{n}{2}[2 \times 120+(n-1) 5]=(n-2) 180$
$\Rightarrow n^{2}-25 n+144=0$
$\Rightarrow(n-9)(n-16)=0$
$\therefore n=9,16$
For $n=9, T_{9}=120+(9-1) 5=160$
For $n=16, T_{16}=120+(16-1) 5=195$ [not possible]

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
The interior angles of a polygon of $\mathrm{n}$ sides are in AP. The smallest angle is $120^{\circ}$ and the common difference is $5^{\circ}$
What is the largest interior angle of the polygon?

Options:
A $160^{\circ}$ only
B $195^{\circ}$ only
C Either $160^{\circ}$ or $195^{\circ}$
D Nither 160 nor $195^{\circ}$

The correct answer is:
$160^{\circ}$ only

Solution:
For $n=9$ Largest angle $=T_{9}=120+(9-1) 5=160$
For $n=16$
Largest angle $=T_{16}=120+(16-1) 5=195$
(Not possible).

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
The intermediate host of Schistosoma is

Options:
A snail
B mosquito
C housefly
D sheep

The correct answer is:
snail

Solution:
Schistosoma mansoni is the common human blood fluke . It belongs to class-Trematoda of Platyhelminthes. Blood fluke is digenetic, primary host is man and secondary host is snail.Sheep is the primary host of Fasciola hepatica (sheep liver fluke), causing 'liver rot'. Its secondary host is also the snail.
Mosquito and housefly do not found to be the intermediate host of any animal.

Topic: Biology, Human Health and Diseases, JIPMER, JIPMER 2009
The inverse of a diagonal matrix is a:

Options:
A symmetricmatrix
B skew-symmetric matrix
C diagonal matrix
D None of the above

The correct answer is:
diagonal matrix

Solution:
The inverse of a diagonal matrix is a diagonal matrix.

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 2)
The inverse of the function $\mathrm{y}=5^{\operatorname{In} \mathrm{x}}$ is

Options:
A $\quad x=y^{\frac{1}{\operatorname{In} 5}}, y>0$
B $x=y^{\operatorname{In} 5}, y>0$
C $x=y^{\frac{1}{\operatorname{In} 5}}, y < 0$
D $x=5$ In $y, y>0$

The correct answer is:
$\quad x=y^{\frac{1}{\operatorname{In} 5}}, y>0$

Solution:
$\begin{aligned} & \mathrm{y}=\mathrm{h}^{\ln \mathrm{x}}=\mathrm{e}^{\ell \mathrm{n} 5 \times \ln \mathrm{x}} \\ & \Rightarrow \text { n } \mathrm{y}=\ln 5 \times \ln \mathrm{x} \\ & \Rightarrow \ell \mathrm{n} \mathrm{x}=\frac{1}{\ln 5} \cdot \ln \mathrm{y}=\ell \mathrm{ny}^{\frac{1}{\ell \mathrm{n} 5}} \\ & \Rightarrow \mathrm{x}=\mathrm{y}^{\frac{1}{\ln 5}}, \mathrm{y}>0 \end{aligned}$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 2)
The inverse of the matrix \( A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right] \) is

Options:
A \( \left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right] \)
B \( \frac{1}{24}\left[\begin{array}{lll}1 / a & 0 & 0 \\ 0 & 1 / b & 0 \\ 0 & 0 & 1 / c\end{array}\right] \)
C \( \frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right] \)
D \( \frac{1}{24}\left[\begin{array}{llll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \)

The correct answer is:
\( \frac{1}{24}\left[\begin{array}{lll}1 / a & 0 & 0 \\ 0 & 1 / b & 0 \\ 0 & 0 & 1 / c\end{array}\right] \)

Solution:
Given that \( A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right] \)
Now \( A^{-1}=\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right] \)
If \( A=\left[\begin{array}{ccc}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right] \), then \( A^{-1}=\left[\begin{array}{ccc}1 / a & 0 & 0 \\ 0 & 1 / b & 0 \\ 0 & 0 & 1 / c\end{array}\right] \)
When \( a \neq 0, b \neq 0, c \neq 0 \)

Topic: Mathematics, Sets and Relations, KCET, KCET 2014
The inverse of the matrix \( \left[\begin{array}{rrr}2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3\end{array}\right] \) is

Options:
A \( \left[\begin{array}{ccc}3 & -5 & 5 \\ -1 & -6 & -2 \\ 1 & -5 & 2\end{array}\right] \)
B \( \left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & -2\end{array}\right] \)
C \( \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] \)
D \( \left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right] \)

The correct answer is:
\( \left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right] \)

Solution:
(D)
\[
\begin{array}{l}
\text { Let } A=\left[\begin{array}{ccc}
2 & 5 & 0 \\
0 & 1 & 1 \\
-1 & 0 & 3
\end{array}\right] \\
|A|=2(3-0)-5(0+1) \\
=6-5=1 \\
\therefore|A|=1
\end{array}
\]
\[
\begin{array}{l}
\operatorname{adj} A=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]^{T}=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right] \\
A^{-1}=\frac{a d j A}{|A|}=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right]
\end{array}
\]

Topic: Mathematics, Matrices, KCET, KCET 2019
The iodine derivative of amino acid tyrosine brings about effects on target cells by

Options:
A Altering gene expression
B Activating secondary messengers
C Activating membrane-bound receptors
D More than one option is correct

The correct answer is:
Altering gene expression

Solution:
Thyroxine is a lipid-soluble hormone, which diffuses into the cell, where it binds to intracellular receptors, migrate to the nucleus, and activate specific target sequences of DNA. Thyroxine is a hormone produced by the thyroid gland. Thyroxine is important in controlling both development and maturation. It also controls metabolism in adult organisms.

Topic: Biology, Chemical Coordination and Integration, NEET