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$Let,\; I={\displaystyle \int si{n}^{-1}}\left(\frac{2x}{1+x^2}\right)dx$

Putting x = tanθ tan^-1x = θ dx = sec²θdθ we get,

$I={\displaystyle \int si{n}^{-1}}\left[\frac{2tan\theta }{1+ta{n}^2\theta }\right]\cdot se{c}^2\theta d\theta$

This is a short answer type question as classified in NCERT Exemplar

Volume occupied by 1 mole = 1mole of the gas at NTP= 22400mL=22400cc

So number of molecules in 1cc of hydrogen= $\frac{6.023\times {10}^{23}}{22400}=2.688\times {10}^{19}$

H₂ is a diatomic gas, having a total of 5 degrees of freedom

So total degrees possesed by all the molecules

= 5 $\times 2.688$ $\times {10}^{19}=1.344\times {10}^{20}$

Shikshapeeth College of Management and Technology provides great placements to its students. Check out the table below to know about the Shikshapeeth College of Management and Technology placement statistics 2025:

$\begin{array}{l}Let,\; I={\displaystyle \int e^{2x}}sinxdx.\\ =sinx{\displaystyle \int e^{2x}}-{\displaystyle \int \frac{d}{dx}}sinx{\displaystyle \int e^{2x}}dxdx\\ =sinx\cdot \frac{e^{2x}}{2}-{\displaystyle \int cos}x\cdot \frac{e^{2x}}{2}dx\\ =\frac{e^{2x}}{2}sinx-\frac{1}{2}{\displaystyle \int cosx}\cdot \times e^{2x}dx\\ =\frac{e^{2x}}{2}\cdot sinx-\frac{1}{2}\left\{cosx{\displaystyle \int e^{2x}}dx-{\displaystyle \int \frac{d}{dx}}cosx{\displaystyle \int e^{2x}}dxdx\right\}\\ =\frac{e^{2x}}{2}sinx-\frac{1}{2}\left\{cosx\cdot \frac{e^{2x}}{2}+{\displaystyle \int sinx\frac{e^{2x}}{2}dx}\right\}\\ =\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx-\frac{1}{4}I+\; C\\ \Rightarrow I+\frac{I}{4}=\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx+\; C\\ \Rightarrow \frac{5I}{4}=\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx+\; C\\ \Rightarrow I=\frac{4}{5}\left[\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx\right]+\; C\\ =\frac{e^{2x}}{5}[2sinx-cosx]+\; C\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $?x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_o^2}{g}\theta =45$ …………1

Horizontal component of initial velocity = V_ocos $\theta$

Vertical component of initial velocity = -V_osin $\theta$

So h = (-V_osin $\theta$ )t + $\frac{1}{2}gt$ ²………….2

R+ $?x$  = V_ocos $\theta \times t$

So t= $\frac{R+?x}{v_{o}cos\theta }$

Substituting value of t in 2 we get

So h = (-V₀sin $\theta$ ) $\left(\mathrm{}\frac{R+?x}{v_{o}cos\theta }\right)+\frac{1}{2}g({\mathrm{}\frac{R+?x}{v_{o}cos\theta })}^2$

H = -(R+ $?x$ )tan $\theta$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2{cos}^2\theta }\right)$

$\theta =45$ ,  h = -(R+ $?x$ )tan $45$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2{cos}^{2}45}\right)$

So h = -(R+ $?x$ ) $1$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2\left(\frac{1}{2}\right)}\right)$

So h = -(R+ $?x$ )+ $\frac{{(R+?x)}^2}{R}$

So h = -R- $?x$ +(R+ $\frac{{?x}^2}{R}+2?x$ )

 h= $?x+\frac{{?x}^2}{R}$

The placement rate for Shikshapeeth College of Management and Technology was above 90% in 2025. Check out the table below to know the placement rate recorded in the recent years:

To be eligible for the BBA programmes at Nazareth College of Arts and Science, applicants must have successfully completed their Class 12 education from a recognised board with a minimum aggregate score of 40%. Meeting this academic criterion is essential for submitting an application to these programmes.

$\begin{array}{l}Let,\; I\; ={\displaystyle \int \frac{x-3}{{(x-1)}^3}}{e}^{x}dx.\\ ={\displaystyle \int \frac{(x-1)-2}{{(x-1)}^3}}{e}^{x}dx.\\ ={\displaystyle \int e^x}\left[\frac{x-1}{{(x-1)}^3}-\frac{2}{{(x-1)}^3}\right]dx\\ ={\displaystyle \int e^x}\left[\frac{1}{{(x-1)}^2}+\frac{(-2)}{{(x-1)}^3}\right]dx,whichisinfore\\ {\displaystyle \int e^x}\left[f(x)+f^{\prime }(x)\right]dx,Where\\ f(x)=\frac{1}{{(x-1)}^2}\\ f^{\prime }(x)=\frac{d}{dx}{(x-1)}^{-2}=\frac{-2}{{(x-1)}^3}.\\ \therefore I=e^{x}f(x)+c=e^x\frac{1}{{(x-1)}^2}+c=\frac{e^x}{{(x-1)}^2}+\; C\end{array}$

The housing rates for CSU, Los Angeles for the academic year 2025-26 are:

After 10th, you can do the following 3 years engineering Diploma courses.

1. Computer Science and engineering
2. ECE
3. Civil
4. Mechanical 
5. Electrical 

The highest package offered during Shikshapeeth College of Management and Technology placements 2025 stood at INR 19 LPA. The table below presents the same:

This is a short answer type question as classified in NCERT Exemplar

Number of helium = 5

T=7^oC=7+273=280K

(a) hence number of atoms = number of moles $*$ Avogadro's number

= $5*6.023*{10}^{23}=30.015*{10}^{23}=3*{10}^{24}$ atoms

(b) now average kinetic energy per molecule = 3/2 K_BT

= total internal energy

= $\frac{3}{2}{K}_{b}T*$ number of atoms

= $\frac{3}{2}*1.38*{10}^{-23}*280*3*{10}^{24}=1.74*{10}^{4}J$

$\begin{array}{l}Let,\; I={\displaystyle \int e^x}\left(\frac{1}{x}-\frac{1}{x^2}\right)dx{\displaystyle \int e^x}[f(x)+f(x)]dx\\ Where,f(x)=\frac{1}{x}\\ f(x)=\frac{dx}{dx}=-1x^{-2}=\frac{-1}{x^2}\end{array}$

So, I = e^xf (x) + C

$\begin{array}{l}=e^x\frac{1}{x}+\; c\\ =\frac{e^x}{x}+\; c\end{array}$

The placements at Shikshapeeth College of Management and Technology are good. The following table presents the placement statistics for Shikshapeeth College of Management and Technology.

Yes. Your PCC application should be made from the address mentioned in your Indian passport, as it helps the police department conduct background checks more efficiently.

$Let,={\displaystyle \int e^x}\left(\frac{1+sinx}{1+cosx}\right)dx.,$

$={\displaystyle \int e^x}\frac{1+2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx.$ {∴ sin 2x = 2sin x cos x. cos 2x = cos^{- 2}x- 1 1 + cos 2x = 2cos²x.

$\begin{array}{l}={\displaystyle \int e^x}\left[\frac{1}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]dx]\\ ={\displaystyle \int e^x}\left[\frac{1}{2}se{c}^2\frac{x}{2}+\frac{cosx\cdot sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]dx\end{array}$

$={\displaystyle \int e^x}\left[tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\frac{1}{2}si{n}^2\frac{x}{2}\right]dx$ is in the form

  e^x [f(x) + f(x)].dx where

$\begin{array}{l}f(x)=tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ f^{\prime }(x)=\frac{d}{dr}tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=se{c}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\frac{d}{dx}\left(\frac{x}{2}\right)\\ =\frac{1}{2}se{c}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ =e^{x}f(x)+c=e^{x}tan\frac{x}{2}+\; C\end{array}$

BBA programme at Nazareth College of Arts and Science is a three-year course structured across six semesters. This format offers students a consistent and well-organised academic path, allowing them to gradually deepen their understanding of commerce concepts while preparing for diverse career opportunities in the field.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2gh}$

$=\sqrt[]{2*10*500}=100m/s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x = $\sqrt[]{u^2-u_y^2}=\sqrt[]{{125}^2-{100}^2}=75m/s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2*500}{10}}=10s$

Time taken to reach the ground in 10 sec = 75 (10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

This is a short answer type question as classified in NCERT Exemplar

When air is pumped, more molecules are pumped and boyle's law is staed for situation where number of molecules remains constant . in this case as the number of air molecules keep increases, hence mass change. Boyle's law is only applicable in situations, where number of gas molecule of remains fixed.