datavorous/entrance-exam-dataset huggingface dataset:
Silencing of mRNA has been used in producing transgenic plants resistant to

Options:
A bollworms
B nematodes
C white rusts
D bacterial blights

The correct answer is:
nematodes

Solution:
RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary ds RNA molecule that binds to and prevents translation of the mRNA (silencing). The source of this complementary RNA could be from an infection by viruses having RNA genome or mobile genetic elements (transposons) that replicate via an RNA intermediate. Using Agrobacterium vectors, nematode-specific genes were introduced into the host plant. The introduction of DNA was such that it produced both sense and anti-sense RNA in the host cells. These two RNA's being complementary to each other formed a ds RNA that initiated RNAi and thus, silenced the specific mRNA of the nematode. The result was that the parasite could not survive in a transgenic host expressing specific interfering RNA. The transgenic plant, therefore, got itself protected from the parasite.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2011 (Mains)
Silicon dioxide is formed by the reaction of

Options:
A $\mathrm{SiCl}_4+2 \mathrm{H}_2 \mathrm{O}$
B $\mathrm{SiO}_2+4 \mathrm{HF}$
C $\mathrm{SiO}_2+\mathrm{NaOH}$
D $\mathrm{SiCl}_4+\mathrm{NaOH}$

The correct answer is:
$\mathrm{SiCl}_4+2 \mathrm{H}_2 \mathrm{O}$

Solution:
Silicon dioxide is formed by the reaction of Silicon Tetrachloride and water

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Silicon doped with arsenic is

Options:
A $p$-type semiconductor
B $n$-type semiconductor
C intrinsic semiconductor
D insulator

The correct answer is:
$n$-type semiconductor

Solution:
When, Si (having 4 valence electrons) is doped with As (having 5 valence electrons), 4 electrons form covalent bonds, while the fifth electron remains free and increase conductivity. Hence, Si doped with As is $n$-type semiconductor.

Topic: Chemistry, p Block Elements (Group 13 & 14), COMEDK, COMEDK 2015
Silicon is an important constituent of

Options:
A Rocks
B Amalgams
C Chlorophyll
D Haemoglobin

The correct answer is:
Rocks

Solution:
Silicate minerals and compounds, which incorporate silicon oxide in their lattice, are found in rocks.
The names and characteristics of rocks are influenced by the amount of silica in them. We may now state that silicon is a significant component of rocks.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Silicon is main constituent of

Options:
A Alloys
B Rocks
C Animals
D Vegetables

The correct answer is:
Rocks

Solution:
1. Silicate minerals and compounds, which incorporate silicon oxide in their lattice, are found in rocks.
2. The names and characteristics of rocks are influenced by the amount of silica in them. We may now state that silicon is a significant component of rocks.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Silver benzoate reacts with bromine to form

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Silver electrode acts as anode and $E_{\text {cell }}^{\circ}$ is $-0.34 \mathrm{~V}$

Options:
A Copper electrode acts as anode and $E_{\text {cell }}^{\circ}$ is $+0.46 \mathrm{~V}$
B Copper electrode acts as anode and $E_{\text {cell }}^{\circ}$ is $+0.46 \mathrm{~V}$
C Silver electrode acts as a cathode and $E_{\text {cell }}^{\circ}$ is $-0.34 \mathrm{~V}$.
D Copper electrode acts as cathode and $E_{\text {cell }}^{\circ}$ is $+0.46 \mathrm{~V}$

The correct answer is:
Copper electrode acts as anode and $E_{\text {cell }}^{\circ}$ is $+0.46 \mathrm{~V}$

Solution:
Electrode having higher reduction potential has to be cathode.
Hence, silver electrode act as a cathode. Because reduction potential of silver electrode $\left(E^{\circ}\right)$ is $+80 \mathrm{~V}$. Which is more than given reduction potential of $\mathrm{Cu}(+34 \mathrm{~V})$.
Hence, $E_{\text {cell }}^{\circ}$ will determine as:
$\begin{aligned}
E_{\text {cell }}^{\circ} & =E_{(\text {Reduction }) \text { cathode }}^{\circ}-E_{(\text {Reduction) anode }}^{\circ} \\
& =+80-(+0.34)=+0.46 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
Silver halides are used in photography because

Options:
A They are photosensitive
B Soluble in hypo
C Soluble in $\mathrm{NH}_4 \mathrm{OH}$
D Soluble in acids

The correct answer is:
They are photosensitive

Solution:
Silver halides are used in photography because they are photosensitive as they react with light to form the image, silver halides being reduced to silver.They are also soluble in hyposolution.

Topic: Chemistry, d and f Block Elements, JEE Main
Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident on it a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. The potential required to stop the photoelectrons, when light of wavelength 200 nm is incident on it, is

Options:
A 15.4 V
B 2.35Â V
C 3.85Â V
D 1.5 V

The correct answer is:
1.5 V

Solution:
ev0+ ϕ=hcλ
and ev0′+ ϕ=hcλ′
ev0′+ϕev0+ϕ=λλ′
=100200=12
2ev0′+2ϕ
=ev0+ ϕ
ev0′=ev0-ϕ2
=7.7-4.72
=1.5 V

Topic: Physics, Dual Nature of Matter, JEE Main
Silver ions are added to the solution with
Br −  =  Cl −  =  CO 3 2−  =  AsO 4 3−  = 0.1 M
Which compound will precipitate at the lowest [Ag+] ?

Options:
A AgBr;   (K sp  = 5 ×  10 −13 )
B AgCl;  ( K sp =1.8× 10 −10 )
C Ag 2 CO 3 ;  ( K sp =8.1× 10 −12 )
D Ag 3 AsO 4 ;  ( K sp = 10 −22 )

The correct answer is:
AgBr;   (K sp  = 5 ×  10 −13 )

Solution:
In case of AgBr, the ionic product will exceed the solubility product before AgCl, Ag2CO3 and Ag3AsO4. Thus, AgBr will start precipitation before other salts.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Silver Mirror test is given by which one of the following compounds ?

Options:
A
Acetone

B
Formaldehyde

C
Benzophenone

D
None of these

The correct answer is:
Formaldehyde

Solution:
Formaldehyde and Acetaldehyde can be oxidized by tollen's reagent to give silver mirror.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2011
Silver salt of carboxylic acid on reaction with $R-X$ gives

Options:
A alkyl halide
B ester
C aldehyde
D alcohol

The correct answer is:
ester

Solution:
$R^{\prime} \mathrm{COOAg}+R-X$
(silver salt
alkyl halide $\stackrel{-\mathrm{Ag} X}{\begin{array}{c}R^{\prime} \mathrm{COOR} \\ \text { ester } \\ \text { (fruity odour) }\end{array}}$
of acid)

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Similarity in chemical properties of the atoms of elements in a group of the Periodic table is most closely related to:

Options:
A
atomic numbers

B
atomic masses

C
number of principal energy levels

D
number of valence electrons

The correct answer is:
atomic numbers

Solution:
If elements are arranged in order of their increasing atomic numbers, element coming at intervals of $2,8,8,18,18,32$ and 32 will have similar physical and chemical properties and thus grouped in one particular group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2014 (12 Apr Online)
Simple squamous epithelia are mainly concerned with:

Options:
A Diffusion
B Secretion
C Absorption
D Both (b) and (c)

The correct answer is:
Diffusion

Solution:
Simple squamous epithelia: Single layer of flattened cells and irregular boundaries. The main function is to facilitate diffusion, e.g., wall of blood vessels, lining of alveoli.

Topic: Biology, Structural Organisation in Animals, NEET
Simplest alkyne is represented by

Options:
A $\mathrm{CH}$
B $\mathrm{CH}_2$
C $\mathrm{C}_2 \mathrm{H}_2$
D $\mathrm{C}_2 \mathrm{H}_4$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_2$

Solution:
$\mathrm{C}_2 \mathrm{H}_2$

Topic: Chemistry, Hydrocarbons, JEE Main
$\sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=$

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

The correct answer is:
$\mathrm{\pi}$

Solution:
$\sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$
$=30^{\circ}+30^{\circ}+120^{\circ}$
$=\pi$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} \sqrt{\frac{2}{3}}=$

Options:
A $\sin ^{-1} \frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}$
B $\pi-\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$
C $-\pi-\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$
D $\pi+\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$

The correct answer is:
$\pi-\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$

Solution:
We have,
$\begin{aligned}
& \sin ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} \sqrt{\frac{2}{3}} \\
& \quad=\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{1-\left(\sqrt{\frac{2}{3}}\right)^2}+\sqrt{\frac{2}{3}} \sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2}\right] \\
& \text {As } 0 < \frac{\sqrt{3}}{2}=0866, \sqrt{\frac{2}{3}}=0.816 \leq 1 \text { and }
\end{aligned}$
$\begin{aligned}
\left(\frac{\sqrt{3}}{2}\right)^2+ & \left(\sqrt{\frac{2}{3}}\right)^2=1.4167>1 \\
& x \sqrt{1-y^2}+y \sqrt{1-x^2}, \text { if } 0 < x, y \leq 1 \text { and } \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{1-\frac{2}{3}}+\sqrt{\frac{2}{3}} \sqrt{1-\frac{3}{4}}\right] \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{\frac{3-2}{3}}+\sqrt{\frac{2}{3}} \sqrt{\frac{4-3}{4}}\right] \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{1}{2}\right] \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right]
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin 12^{\circ} \sin 24^{\circ} \sin 48^{\circ} \sin 84^{\circ}=$

Options:
A $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
B $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$
C $\frac{3}{15}$
D None of these

The correct answer is:
$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$

Solution:
$\begin{aligned}
\sin 12^{\circ} \sin 24^{\circ} & \sin 48^{\circ} \sin 84^{\circ} \\
& =\frac{1}{4}\left(2 \sin 12^{\circ} \sin 48^{\circ}\right)\left(2 \sin 24^{\circ} \sin 84^{\circ}\right) \\
& =\frac{1}{2}\left(\cos 36^{\circ}-\cos 60^{\circ}\right)\left(\cos 60^{\circ}-\cos 108^{\circ}\right) \\
& =\frac{1}{4}\left(\cos 36^{\circ}-\frac{1}{2}\right)\left(\frac{1}{2}+\sin 18^{\circ}\right) \\
& =\frac{1}{4}\left\{\frac{1}{4}(\sqrt{5}+1)-\frac{1}{2}\right\}\left\{\frac{1}{2}+\frac{1}{4}(\sqrt{5}-1)\right\}=\frac{1}{16} \\
& \text { and } \cos 20^{\circ} \cos 40^{\circ} \cos 60 \cos 80^{\circ} \\
& =\frac{1}{2}\left[\cos \left(60^{\circ}-20^{\circ}\right) \cos 20^{\circ} \cos \left(60^{\circ}+20^{\circ}\right)\right] \\
& =\frac{1}{2}\left[\frac{1}{4} \cos 3\left(20^{\circ}\right)\right]=\frac{1}{8} \cos 60^{\circ}=\frac{1}{2} \times \frac{1}{8}=\frac{1}{16}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
sin-1sin2π3+cos-1cos7π6+tan-1tan3π4 is equal to

Options:
A 11π12
B 17π12
C 31π12
D -3π4

The correct answer is:
11π12

Solution:
Given, sin-1sin2π3+cos-1cos7π6+tan-1tan3π4

Now simplifying each term we get,

sin-1sin2π3=π-2π3=π3

cos-1cos7π6=2π-7π6=5π6

tan-1tan3π4=3π4-π=-π4

sin-1sin2π3+cos-1cos7π6+tan-1tan3π4

=π3+5π6-π4

=11π12

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (27 Jun Shift 1)
$\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right]=$

Options:
A $\sin ^{-1} x+\sin ^{-1} \sqrt{x}$
B $\sin ^{-1} x-\sin ^{-1} \sqrt{x}$
C $\sin ^{-1} \sqrt{x}-\sin ^{-1} x$
D None of these

The correct answer is:
$\sin ^{-1} x-\sin ^{-1} \sqrt{x}$

Solution:
Let $x=\sin \theta$ and $\sqrt{x}=\sin \phi$
$\begin{aligned}
& \text { Hence } \sin ^{-1}\left(x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right) \\
& =\sin ^{-1}\left(\sin \theta \sqrt{1-\sin ^2 \phi}-\sin \phi \sqrt{1-\sin ^2 \theta}\right) \\
& =\sin ^{-1}(\sin \theta \cos \phi-\sin \phi \cos \theta)=\sin ^{-1} (\sin (\theta-\phi))\\
& =\theta-\phi=\sin ^{-1}(x)-\sin ^{-1}(\sqrt{x})
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right]=$

Options:
A $\sin ^{-1} x+\sin ^{-1} \sqrt{x}$
B $\sin ^{-1} x-\sin ^{-1} \sqrt{x}$
C $\sin ^{-1} \sqrt{x}-\sin ^{-1} x$
D None of these

The correct answer is:
$\sin ^{-1} x-\sin ^{-1} \sqrt{x}$

Solution:
Let $x=\sin \theta$ and $\sqrt{x}=\sin \phi$
Hence $\sin ^{-1}\left(x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right)$
$\begin{aligned}
& =\sin ^{-1}\left(\sin \theta \sqrt{1-\sin ^2 \phi}-\sin \phi \sqrt{1-\sin ^2 \theta}\right) \\
& =\sin ^{-1}(\sin \theta \cos \phi-\sin \phi \cos \theta)=\sin ^{-1} (\sin (\theta-\phi)) \\
& =\theta-\phi=\sin ^{-1}(x)-\sin ^{-1}(\sqrt{x})
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin ^{-1} x=\frac{\pi}{5}$ for some $x \in(-1,1)$, then the value of $\cos ^{-1} x$ is

Options:
A $\frac{3 \pi}{10}$
B $\frac{5 \pi}{10}$
C $\frac{7 \pi}{10}$
D $\frac{9 \pi}{10}$

The correct answer is:
$\frac{3 \pi}{10}$

Solution:
$\begin{aligned} & \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ & \Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x=\frac{\pi}{2}-\frac{\pi}{5}=\frac{3 \pi}{10}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin ^{-1} \frac{\sqrt{x}}{\sqrt{x+a}}$ is equal to

Options:
A $\cos ^{-1} \sqrt{\frac{x}{a}}$
B $\operatorname{cosec}^{-1} \sqrt{\frac{x}{a}}$
C $\tan ^{-1} \sqrt{\frac{x}{a}}$
D None of these

The correct answer is:
$\tan ^{-1} \sqrt{\frac{x}{a}}$

Solution:
$\begin{aligned} & \text { Putting } x=a \tan ^2 \theta \\ & \sin ^{-1} \frac{\sqrt{x}}{\sqrt{x+a}}=\sin ^{-1} \frac{\sqrt{a} \sqrt{\tan ^2 \theta}}{\sqrt{a \tan ^2 \theta+a}}=\sin ^{-1} \frac{\sqrt{a} \tan \theta}{\sqrt{a} \sec \theta} \\ & =\sin ^{-1} \sin \theta=\theta=\tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\int\left(\sin ^{-1} x+\cos ^{-1} x\right) d x=$

Options:
A $\frac{1}{2} \pi x+c$
B $x\left(\sin ^{-1} x-\cos ^{-1} x\right)+c$
C $x\left(\cos ^{-1} x+\sin ^{-1} x\right)+c$
D $\frac{\pi}{2}+x+c$

The correct answers are:
$\frac{1}{2} \pi x+c$, $x\left(\cos ^{-1} x+\sin ^{-1} x\right)+c$

Solution:
$\begin{aligned}
& \int\left(\sin ^{-1} x+\cos ^{-1} x\right) d x=\int\left(\frac{\pi}{2}\right) d x=\frac{\pi x}{2}+c \\
&\left(\text{as, } \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)
\end{aligned}$
Also $\int\left(\sin ^{-1} x+\cos ^{-1} x\right) d x=x\left(\cos ^{-1} x+\sin ^{-1} x\right)+c$

Topic: Mathematics, Indefinite Integration, JEE Main
$\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2} \text {, }$ then $x$ is

Options:
A $0$
B $\frac{1}{\sqrt{5}}$
C $\frac{2}{\sqrt{5}}$
D $\frac{\sqrt{3}}{2}$

The correct answer is:
$\frac{1}{\sqrt{5}}$

Solution:
$\cot ^{-1} \frac{1}{2}=\cos ^{-1} \frac{1}{\sqrt{5}}$
Hence given equation can be written as
$\begin{aligned}
& \sin ^{-1} x+\cos ^{-1} \frac{1}{\sqrt{5}}=\frac{\pi}{2} \\
& \Rightarrow x=\frac{1}{\sqrt{5}}
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} x+\cos ^{-1} \frac{1}{x}=$

Options:
A $\pi$
B $\underline{\pi}$
C $\frac{3 \pi}{2}$
D None of these

The correct answer is:
$\pi$

Solution:
$\begin{aligned} & \sin ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} x+\cos ^{-1} \frac{1}{x} \\ & =\left\{\sin ^{-1}(x)+\cos ^{-1}(x)\right\}+\left\{\sin ^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}\left(\frac{1}{x}\right)\right\}\end{aligned}$
$=\frac{\pi}{2}+\frac{\pi}{2}=\pi$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin \left[\frac{\pi}{2}-\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=$

Options:
A $\frac{\sqrt{3}}{2}$
B $-\frac{\sqrt{3}}{2}$
C $\frac{1}{2}$
D $-\frac{1}{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
$\begin{aligned} & \sin \left[\frac{\pi}{2}-\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=\cos \left\{ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right\} \\ & =\cos \left\{ \cos^{-1} \sqrt{1-\frac{3}{4}}\right\}=\frac{1}{2}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sin 36^{\circ} \sin 72^{\circ} \sin 108^{\circ} \sin 144^{\circ}=$

Options:
A $1 / 4$
B $1 / 16$
C $3 / 4$
D $5 / 16$

The correct answer is:
$5 / 16$

Solution:
$\begin{aligned}
\sin 36^{\circ} \sin 72^{\circ} & \sin 108^{\circ} \sin 144^{\circ} \\
& =\sin ^2 36^{\circ} \sin ^2 72^{\circ}=\frac{1}{4}\left\{\left(2 \sin ^2 36^{\circ}\right)\left(2 \sin ^2 72^{\circ}\right)\right\} \\
& =\frac{1}{4}\left\{\left(1-\cos 72^{\circ}\right)\left(1-\cos 144^{\circ}\right)\right\} \\
& =\frac{1}{4}\left\{\left(1-\sin 18^{\circ}\right)\left(1+\cos 36^{\circ}\right)\right\} \\
& =\frac{1}{4}\left[\left(1-\frac{\sqrt{5}-1}{4}\right)\left(1+\frac{\sqrt{5}+1}{4}\right)\right]=\frac{20}{16} \times \frac{1}{4}=\frac{5}{16}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=$

Options:
A 3
B $\frac{3}{2}$
C 4
D 8

The correct answer is:
3

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Trigonometric Equations, JEE Main
â«sin4â¡xcos8â¡x dx is equal to (where C is an arbitrary constant)

Options:
A 1+tan5⁡x5+tan5⁡x7+C
B tan5⁡x5+tan7⁡x7+C
C tan7⁡x5+tan5⁡x7+C
D None of these

The correct answer is:
tan5⁡x5+tan7⁡x7+C

Solution:
â«sin4â¡xcos4â¡xcos8â¡xcos4â¡x.dx=â«tan4â¡x.sec4â¡x dx
=â«tan4â¡x1+tan2â¡xsec2â¡x dx
=â«tan4â¡x1+tan2â¡xdtanx=â«t41+t2dt=t55+t77+C

Topic: Mathematics, Indefinite Integration, JEE Main
$\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}=$

Options:
A 1
B 0
C $1 / 2$
D 2

The correct answer is:
0

Solution:
$\begin{aligned}
& \sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ} \\
& =-2 \cos 60^{\circ} \sin 10^{\circ}+\sin 10^{\circ} \\
& =\sin 10^{\circ}\left(1-2 \cos 60^{\circ}\right)=0 .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
∫sin8x-cos8x1-2sin2xcos2xdx is equal to

Options:
A -12sin2x+c
B -sin2x+c
C -12sinx+c
D 12sin2x+c

The correct answer is:
-12sin2x+c

Solution:
sin8x-cos8x=sin2x-cos2xsin2x+cos2xsin4x+cos4x

and sin4x+cos4x=sin2x+cos2x2-2sin2xcos2x=1-2sin2xcos2x

∴I=∫sin8x-cos8x1-2sin2xcos2xdx

=∫sin2x-cos2xsin4x+cos4xsin4x+cos4xdx

=-∫cos2xdx
=-12sin2x+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main
∫sin8x-cos8x1-2sin2xcos2xdx is equal to

Options:
A -12sin2x+c
B -sin2x+c
C -12sinx+c
D 12sin2x+c

The correct answer is:
-12sin2x+c

Solution:
sin8x-cos8x=sin2x-cos2xsin2x+cos2xsin4x+cos4x

and sin4x+cos4x=sin2x+cos2x2-2sin2xcos2x=1-2sin2xcos2x

∴I=∫sin8x-cos8x1-2sin2xcos2xdx

=∫sin2x-cos2xsin4x+cos4xsin4x+cos4xdx

=-∫cos2xdx
=-12sin2x+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2014 (09 Apr Online)
∫ sin 8 x− cos 8 x 1−2 sin 2 x cos 2 x dx is equal to (where C is an arbitrary constant)

Options:
A 1 2 sin2x+C
B − 1 2 sin2x+C
C − 1 2 sinx+C
D − sin 2 x+C

The correct answer is:
− 1 2 sin2x+C

Solution:
∫sin8x-cos8x1-2sin2xcos2xdx

=∫sin4x-cos4xsin4x + cos4x1-2sin2xcos2xdx
=∫sin2x-cos2xsin4x + cos4x1-2 sin2xcos2xdx
=∫-cos2x dx=-12sin2x + C

Topic: Mathematics, Indefinite Integration, JEE Main
$\sin 163^{\circ} \cos 347^{\circ}+\sin 73^{\circ} \sin 167^{\circ}=$

Options:
A 0
B $1 / 2$
C 1
D None of these

The correct answer is:
$1 / 2$

Solution:
$\begin{aligned} \sin 163^{\circ} \cos 347^{\circ} & +\sin 73^{\circ} \sin 167^{\circ} \\ & =\sin \left(180^{\circ}-17^{\circ}\right) \cos \left(360^{\circ}-13^{\circ}\right)+\cos \left(90^{\circ}-17^{\circ}\right) \\ & =\sin 17^{\circ} \cos 13^{\circ}+\cos 17^{\circ} \sin 13^{\circ}=\sin 30^{\circ}=1 / 2\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
α=sin36° is a root of which of the following equation

Options:
A 16x4-20x2+5=0
B 16x4+20x2+5=0
C 10x4-10x2-5=0
D 16x4-10x2+5=0

The correct answer is:
16x4-20x2+5=0

Solution:
We know sin36°=10-254

Let x=10-254

On squaring both sides, we get

4x2=10-25⇒16x2=10-25

Again squaring, we get

52=5-8x22 ⇒5=25+64x4-80x2

⇒16x4-20x2+5=0

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2022 (27 Jun Shift 2)
sinax+cosax and |sinx|+|cosx| are periodic with the same fundamental period, if a equals to

Options:
A 0
B 1
C 2
D 4

The correct answer is:
4

Solution:
sinax+cosax=2sinax+π4
So, fundamental period =2πa
Let, T be the fundamental period of sinx+cosx
So sin(x+T)+cos(x+T)= sinx+cosx.
Squaring both sides, we get,
1+sin(2x+2T)=1+sin2x
Therefore, the fundamental period of sinx+cosx is π2
Hence, 2πa=π2
⇒a=4

Topic: Mathematics, Functions, JEE Main
$\sin \left[\cot ^{-1}\left(\operatorname{costan}^{-1} x\right)\right]=$

Options:
A $\frac{x}{\sqrt{x^2+2}}$
B $\frac{x}{\sqrt{x^2+1}}$
C $\frac{1}{\sqrt{x^2+2}}$
D $\sqrt{\frac{x^2+1}{x^2+2}}$

The correct answer is:
$\sqrt{\frac{x^2+1}{x^2+2}}$

Solution:
\(\begin{aligned}
& \sin \left[\cot ^{-1}\left(\cos \left(\tan ^{-1} x\right)\right)\right] \\
& \operatorname{let} \tan ^{-1} x=\theta \\
& \tan \theta=x \\
& \Rightarrow \cos \theta=\frac{1}{\sqrt{1+x^2}}=a
\end{aligned}\)
the function is reduced to \(\sin \left[\cot ^{-1} \mathrm{a}\right]\)
\(\begin{aligned}
& \operatorname{let} \cot ^{-1} a=\theta_2 \\
& \cot \theta_2=a=\frac{1}{\sqrt{1+x^2}} \\
& \tan \theta_2=\sqrt{1+x^2} \\
& \sin \theta_2=\sqrt{\frac{1+x^2}{2+x^2}} \\
& \sin \left[\cot ^{-1}\left(\cos \left(\tan ^{-1} x\right)\right)\right]=\sqrt{\frac{1+x^2}{2+x^2}}
\end{aligned}\)

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\int \sin (\log x) d x=$

Options:
A $\frac{1}{2} x[\cos (\log x)-\sin (\log x)]+c$
B $\cos (\log x)-x+c$
C $\frac{1}{2} x[\sin (\log x)-\cos (\log x)]+c$
D $-\cos \log x+c$

The correct answer is:
$\frac{1}{2} x[\sin (\log x)-\cos (\log x)]+c$

Solution:
$\begin{aligned} & \text { Let } I=\int \sin (\log x) d x \\ & \text { Put } \log x=t \Rightarrow x=e^t \Rightarrow d x=e^t d t \text {, then } \\ & I=\int \sin t \cdot e^t d t=\sin t \cdot e^t-\int e^t \cdot \cos t d t \\ & =\sin t \cdot e^t-\left[\cos t \cdot e^t+\int e^t \cdot \sin t d t\right]\end{aligned}$
$\begin{aligned} & \Rightarrow 2 I=\sin t . e^t-\cos t . e^t+c \\ & \Rightarrow I=\int \sin (\log x) d x=\frac{1}{2} x[\sin (\log x)-\cos (\log x)]+c\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\sin (\beta+\gamma-\alpha)+\sin (\gamma+\alpha-\beta)+\sin (\alpha+\beta-\gamma)-\sin (\alpha+\beta+\gamma)=$

Options:
A $2 \sin \alpha \sin \beta \sin \gamma$
B $4 \sin \alpha \sin \beta \sin \gamma$
C $\sin \alpha \sin \beta \sin \gamma$
D None of these

The correct answer is:
$4 \sin \alpha \sin \beta \sin \gamma$

Solution:
Combine first two terms and last two terms
$\begin{aligned}
\text { L.H.S. } & =2 \sin \gamma \cos (\beta-\alpha)+2 \sin (-\gamma) \cos (\alpha+\beta) \\
& =2 \sin \gamma[\cos (\beta-\alpha)-\cos (\alpha+\beta)] \\
& =2 \sin \gamma .2 \sin \alpha \sin \beta=4 \sin \alpha \sin \beta \sin \gamma .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main