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The types of Prepositions in English Grammar are divided based on their functions and structure. Have a look:

This is a multiple choice answer as classified in NCERT Exemplar

(d) Let radius of sphere be R . as the temperature increases radius of the sphere increases.

Original volume V_o= $\frac{4}{3}\pi R$ ³

Coefficient linear expansion = $\alpha$

Coefficient of volume expansion =3 $\alpha$

$\frac{dV}{VdT}=$ 3 $\alpha$  , dV= 4 $\pi$ R³  $\alpha ?T$ =increase in volume

This depends on the campus placement. If the company finds suitable candidates, they can offer them a good package; otherwise, getting a very high salary package just after graduation is difficult. The placements also vary from year to year. However, with relevant experience and expertise, the salary package can increase.

$\begin{array}{l}si{n}^{4}x=si{n}^{2}x.si{n}^{2}x\\ =\left(\frac{1-cos2x}{2}\right)\left(\frac{1-cos2x}{2}\right)\\ =\frac{1}{4}{(1-cos2x)}^2=\frac{1}{4}\left[1+co{s}^{2}2x-2cos2x\right]\\ =\frac{1}{4}\left[1+\left(\frac{1+cos4x}{2}\right)-2cos2x\right]\\ =\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2}cos4x-2cos2x\right]\\ =\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2}cos4x-2cos2x\right]\end{array}$

$\begin{array}{l}={\displaystyle \int si{n}^4}xdx=\frac{1}{4}{\displaystyle \int [\frac{3}{2}}+\frac{1}{2}cos4x-2cos2x]dx\\ =\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2}\left(\frac{sin4x}{4}\right)-\frac{2sin2x}{2}\right]+\; C\\ =\frac{1}{8}\left[3x+\frac{si{n}^{4}x}{4}-2sin2x\right]+\; C\\ =\frac{3x}{8}-\frac{1}{4}sin2x+\frac{1}{32}sin4x+\; C\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a) We know that temperature increases vibrations of molecules about their mean position increases. Hence kinetic energy associated with random motion of molecules increases.

Have a look at the 5 examples of on Preposition below:

• Put this painting on the wall.
• The CAT is lying on the floor.
• I will go on the 10:00 train.
• He goes to the school on foot.
• I insisted on going today.

$\begin{array}{l}\frac{cosx}{1+cosx}=\frac{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}\\ =\frac{1}{2}\left[1-ta{n}^2\frac{x}{2}\right]\\ ={\displaystyle \int \frac{cosx}{1+cosx}}dx=\frac{1}{2}{\displaystyle \int \left(1-ta{n}^2\frac{x}{2}\right)}dx\\ =\frac{1}{2}\left(1-se{c}^2\frac{x}{2}+1\right)dx\\ =\frac{1}{2}\left(2-se{c}^2\frac{x}{2}\right)dx=\frac{1}{2}\left[2x-\frac{tan\frac{x}{2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]+\; C\\ =x-tan\frac{x}{2}+\; C\end{array}$

$\begin{array}{l}\frac{1-cosx}{1+cosx}=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}\\ =ta{n}^2\frac{x}{2}=se{c}^2\frac{x}{2}-1\\ ={\displaystyle \int \frac{1-cosx}{1+cosx}}dx={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\\ =\left[\frac{tan\frac{x}{2}}{\frac{1}{2}}\right]+\; C\\ =2tan\frac{x}{2}+\; C\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

As the temperature is increased length of the pendulum increases.

T=2 $\pi \sqrt[]{\frac{L}{g}}$

So if we increase L then T also increases.

The most commonly used 10 Prepositions are: in, on, at, for, by, with, during, before, after and about. However, apart from these, there are several Preposition words such as within, without, due to, because of, among, beyond, above, below, against, along, behind, besides, despite, in spite of, regarding, during, through, towards, into, upto, onto, etc.

$\begin{array}{l}sinAsinB=\frac{-1}{2}\left\{cos(A+B)-cos(A-B)\right\}\\ ={\displaystyle \int sin}4xsin8xdx={\displaystyle \int \frac{1}{2}}\{cos(4x-8x)-cos(4x+8x)\}dx\\ =\frac{1}{2}{\displaystyle \int cos}(-4x)-cos12xdx\\ =\frac{1}{2}{\displaystyle \int (cos4x-cos12x)}dx\\ =\frac{1}{2}\left[\frac{sin4x}{4}-\frac{sin12x}{12}\right]+\; C\end{array}$

Yes, the Preposition can be used at the end of sentence if the object is interrogative or relative pronoun. The examples of the same are shared below:

• Preposition at the end in case on relative pronoun: Here is the file that you asked for.
• Preposition at the end in case on interrogative pronoun: What are you looking at?

$\begin{array}{l}Here,\\ sinAsinB=\frac{-1}{2}\{cos(A+B)-cos(A-B)\}\\ I={\displaystyle \int sinx}sin2xsin3x\\ ={\displaystyle \int \left[sinx\cdot \frac{1}{2}\{cos(2x-3x)-cos(2x+3x)\}\right]}dx\\ ={\displaystyle \int \{[sin}x\cdot \frac{1}{2}\{cos(-x)-cos5x\}\}dx\\ =\frac{1}{2}{\displaystyle \int sin}xcosx-sinxcos5xdx\\ Now,sin2x=2sinxcosx,\\ =\frac{1}{2}{\displaystyle \int \frac{sin2x}{2}}dx-\frac{1}{2}{\displaystyle \int sin}xcos5xdx\\ =\frac{1}{4}\left[\frac{-cos2x}{2}\right]-\frac{1}{2}{\displaystyle \int \frac{1}{2}}sin(x+5x)+sin(x-5x)dx\\ =\frac{-cos2x}{8}-\frac{1}{4}{\displaystyle \int sin}6x+sin(-4x)dx\\ =\frac{-cos2x}{8}-\frac{1}{4}\left[\frac{-cos6x}{6}+\frac{cos4x}{4}\right]+\; C\\ =\frac{-cos2x}{8}-\frac{1}{8}\left[\frac{-cos6x}{3}+\frac{cos4x}{2}\right]+\; C\\ =\frac{-cos2x}{8}+\frac{cos6x}{24}-\frac{cos4x}{16}+\; C\\ =\frac{1}{4}\left[\frac{1}{6}cos6x-\frac{cos4x}{4}-\frac{cos2x}{2}\right]+\; C.\end{array}$