Class 12th

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

Explanation – V_r= a? +b?

Velocity v_g= 5m/s

Velocity of rain w.r.t girl = V_r-V_g= a? +b? -5?

= (a-5)? +b?

a-5=0, a=5

case II 

v_g = 10m/s?

V_rg= V_r - V_g

 = a? +b? -10? = (a-10)? +b?

Rain appear to be fall at 45 degree so = b/a-10 =1

So b =-5

Velocity of rain = a? +b?

V_r = 5? -5?

Speed of rain V_r= $\sqrt[]{5^2+{(-5)}^2}=5\sqrt[]{2}m/s$

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

Explanation- y=O, u_y= V_ocos $\theta$

a_y=-gcos $\theta$ , t =T

applying equation of kinematics

y=u_yt+ $\frac{1}{2}{a}_y$ t²

0 = V_ocos $\theta T$ +T^{2 $\frac{1}{2}(-gcos\theta )$}

T= $\frac{2V_{o}cos\theta }{gcos\theta }$

T= 2V₀/g

X= L, u_x=V_osin $\theta$  , a_x= gsin $\theta$  , t=T= $\frac{2V_o}{g}$

X=u_xt+ $\frac{1}{2}{a}_x{t}^2$

L= V_osin $\theta T+\frac{1}{2}gsin\theta T^2$

L= $\frac{4v_o^2}{g}$ sin $\theta$

Kindly go through the solution

$\begin{array}{l}LetI={\displaystyle \int x^2{e^x}^{3}dx.}\\ let,t=x^3\Rightarrow dt=3x^{2}dx\cdot \Rightarrow \frac{dt}{3}=x^{2}dx\\ So,={\displaystyle \int \cdot }{e}^t\frac{dt}{3}=\frac{1}{3}{\displaystyle \int e^t}dt.\\ =\frac{1}{3}{e}^t+\; C\\ =\frac{e^{x^3}}{3}+C.\end{array}$

So, option (A) is correct.

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

Explanation – particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

for journey O to P

y=0,u_y= V_osin $\beta$ ,a_y= -gcos $\alpha ,t=T$

y=u_yt + $\frac{1}{2}{a}_y{t}^2$

0= V_osin $\beta T+\frac{1}{2}\left(-gcos\alpha \right)T^2$

T[V_osin $\beta -\frac{gcos\alpha }{2}$ T]=0

T = time of flight = $\frac{2V_{o}sin\beta }{gcos\alpha }$

 

Motion along OX

x= L ,u_x= V_ocos $\beta$ , a_x= -gsin $\alpha$

t =T = $\frac{2V_{o}sin\beta }{gcos\alpha }$

x= u_xt+ $\frac{1}{2}{a}_x{t}^2$

L= V₀cos $\beta T$ + $\frac{1}{2}(-gsin\alpha )T^2$

L= T[V₀cos $\beta -\frac{1}{2}gsin\alpha T$ ]

L=  $\frac{2V_{o}sin\beta }{gcos\alpha }$ [V_ocos $\beta -\frac{V_{o}sin\alpha sin\beta }{cos\alpha }$ ]

L= $\frac{2v_o^{2}sin\beta }{g{cos}^2\alpha }\mathrm{c}\mathrm{o}\mathrm{s}?\left(\alpha +\beta \right)$

Z= sin $\beta \mathrm{c}\mathrm{o}\mathrm{s}?\left(\alpha +\beta \right)$

 = sin $\beta [cos\alpha cos\beta -sin\alpha sin\beta ]$

= $\frac{1}{2}(cos\alpha +sin2\beta -2sin\alpha .{sin}^2\beta )$

= ½ [sin2] $\beta cos\alpha -sin\alpha (1-cos2\beta )$

= $\frac{1}{2}[sin2\beta cos\alpha +cos2\beta .sin\alpha -sin\alpha ]$

= $\frac{1}{2}$  [sin(2 $\beta +\alpha$ )-sin $\alpha$ ]

For z maximum

2 $\beta +\alpha =\frac{\pi }{2}$  , $\beta =\frac{\pi }{4}-\frac{\alpha }{2}$

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

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

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

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

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 $\theta *t$

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= $?x+\frac{{?x}^2}{R}$

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

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

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

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