Ncert Solutions Maths class 12th

 ${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx}}}}$

${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x-\left[x\right]}dx+{\displaystyle \underset{1}{\overset{2}{\int }}{e}^{x-\left[x\right]}dx}}+{\displaystyle \underset{2}{\overset{3}{\int }}{e}^{x-\left[x\right]}dx+.....+{\displaystyle \underset{99}{\overset{100}{\int }}{e}^{x-\left[x\right]}dx}}$

$=e-1+\frac{1}{e}\left(e^2-e\right)+\frac{1}{e^2}\left(e^3-e^2\right)+......+\frac{1}{e^{99}}\left(e^{100}-e^{99}\right)$

$=e-1+\left(e-1\right)+\left(e-1\right)+......+\left(e-1\right),100times$

$=100\left(e-1\right)$

2nd method

${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{\left\{x\right\}}dx=}}{\displaystyle \sum _{n=1}^{100}\left({\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}\right)}}}={\displaystyle \sum _{n=1}^{100}\left(e-1\right)}=100\left(e-1\right)$

The points on the curve are (0, 0), (2, 2) and $\left(3,\frac{21}{2}\right)$

$\therefore \frac{dy}{dx}=2x^3-15x^2+36x-19$

$at\left(0,0\right),\frac{dy}{dx}=-19,at\left(2,2\right),\frac{dy}{dx}=9at\left(3,\frac{21}{2}\right),\frac{dy}{dx}=8$

Hence maximum slope at (2, 2) is 9.

 $l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{co{s}^{2}x}{1+3^x}dx..........\left(i\right)}$

Using properties, ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{co{s}^{2}x}{1+3^{-x}}dx}={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{3^{x}co{s}^{2}x}{1+3^x}dx.......\left(ii\right)}$

Adding (i) and (ii), we get

$2l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{\left(1+3^x\right)co{s}^{2}x}{1+3^x}}dx={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}co{s}^{2}xdx=2}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}co{s}^{2}xdx}$

$\therefore l=\frac{\pi }{4}$

 $\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a+2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a+3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a+4\hfill & \hfill 1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|,$ [using properties]

$=0+\left(a+1\right)\left(a+2\right)\left(-1\right)+2\left(a+3\right)\left(a+4\right)0\left(a+2\right)\left(a+3\right)+4\left(\left(a+2\right)\left(a+3\right)-3\left(a+3\right)\left(a+4\right)\right)$

$=2\left(a+2\right)-2\left(a+3\right)=2\left(a+2-a-3\right)=-2$

$\frac{si{n}^{-1}x}{a}=\frac{co{s}^{-1}x}{b}=\frac{ta{n}^{-1}y}{c}=\lambda$

$\therefore \frac{si{n}^{-1}x}{a}=\lambda$

$\therefore si{n}^{-1}x=a\lambda$

$x=sina\lambda .........\left(i\right)$

$x=cosb\lambda ..........\left(ii\right)$

$y=tanc\lambda ..........\left(iii\right)$

From (i) & (ii), we get, $sina\lambda =cosb\lambda =sin\left(\frac{\pi }{2}-b\lambda \right)$

$\therefore a\lambda =\frac{\pi }{2}-b\lambda \Rightarrow \left(a+b\right)\lambda =\frac{\pi }{2}\Rightarrow \lambda =\frac{\pi }{2\left(a+b\right)}...........\left(iv\right)$

From (iii) y = tan c $\lambda$

$\Rightarrow cos\left(\frac{2c\pi }{2\left(a+b\right)}\right)=\frac{1-y^2}{1+y^2}\Rightarrow cos\left(\frac{\pi c}{a+b}\right)=\frac{1-y^2}{1+y^2}$

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$           

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$        

$\therefore f\left(x\right)$ is continuous $\forall x\in R$  

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$     $\left(\frac{0}{0}\right)$  applying L’ Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

$\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$              

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$        

Equation of tangent y – t³ = 3t² (x – t)  

Let again meet the curve at $Q\left(t_1,t_1^3\right)$  

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$           

  $t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$          

$t_1^2+t{t}_1-2t^2=0$            

->t₁ = -2t

Required ordinate =  &nbs