Maths NCERT Exemplar Solutions Class 12th Chapter Eight

$L_1:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$  

$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$               

Now,

$\overrightarrow{p}*\overrightarrow{q}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|=10\widehat{i}-8\widehat{j}-4\widehat{k}$               

and

  ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\widehat{i}-4\widehat{j}-4\widehat{k}$

  $\therefore S.D=\left|\frac{60+32+16}{\sqrt[]{100+64+16}}\right|=\frac{108}{\sqrt[]{180}}=\frac{18}{\sqrt[]{5}}$      

$L_1:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$  

$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$               

Now,

$\overrightarrow{p}*\overrightarrow{q}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|=10\widehat{i}-8\widehat{j}-4\widehat{k}$               

and

  ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\widehat{i}-4\widehat{j}-4\widehat{k}$

  $\therefore S.D=\left|\frac{60+32+16}{\sqrt[]{100+64+16}}\right|=\frac{108}{\sqrt[]{180}}=\frac{18}{\sqrt[]{5}}$                           

C : 4x² + 4y² – 12x + 8y + k = 0

$?\left(1,-\frac{1}{3}\right)$               

Lies on or inside the C then

$4+\frac{4}{9}-12-\frac{8}{3}+k\le 0$

$\Rightarrow k\le \frac{92}{9}$               

Now, circle lies in 4^th quadrant centre

$\equiv \left(\frac{3}{2},-1\right)$               

$\therefore r<1\Rightarrow \sqrt[]{\frac{9}{4}+1-\frac{k}{4}}<1$               

$\Rightarrow \frac{13}{4}-\frac{k}{4}<1$

$\Rightarrow \frac{k}{4}>\frac{9}{4}$

$\Rightarrow k>9$

$\therefore K\in \left(9,\frac{92}{9}\right)$      

Given vertex is  (5, 4) and directrix 3x + y – 29 = 0

Let foot of perpendicular of (5, 4) on directrix be (x₁, y₁)

$\frac{x_1-5}{3}=\frac{y_1-4}{1}=\frac{-\left(-10\right)}{10}$               

$\therefore \left(x_1,y_1\right)=\left(8,5\right)$               

So, focus of parabola will be S = (2, 3)

Let P(x, y) be any point on parabola, then

$\Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2=\frac{{\left(3x+y-29\right)}^2}{10}$               

$\Rightarrow x^2+9y^2-6xy+134x-2y-711=0$               

And given parabola

$x^2+a{y}^2+bxy+cx+dy+k=0$               

$\therefore a=9,b=-6,c=134,d=-2,k=-711$               

$\therefore a+b+c+d+k=576$    

$\left(ta{n}^{-1}y\right)-x)dy=\left(1+y^2\right)dx$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$

$l.F=e^{{\displaystyle \int \frac{1}{1+y^2}dy}}=e^{ta{n}^{-1}y}$

$x.e^{ta{n}^{-1}y}{\displaystyle \int \frac{e^{ta{n}^{-1}y}ta{n}^{-1}y}{1+y^2}dy}$

Let

$e^{ta{n}^{-1}y}=t$

$=x{e}^{ta{n}^{-1}y}=e^{ta{n}^{-1}y}y-e^{ta{n}^{-1}y}+c...(i)$

$?$ It passes through (1, 0) = c = 2

Now put y = tan 1, then

ex = e – e + 2

$\Rightarrow x=\frac{2}{6}$

${\displaystyle {\int }_0^1\frac{1}{7^{\left(\frac{1}{x}\right)}}dx,let\frac{1}{x}=t}$

$\frac{-1}{x^2}dx=dt$

$={\displaystyle {\int }_{\infty }^1\frac{1}{-t^2{7}^{\left|t\right|}}dt={\displaystyle {\int }_1^{\infty }\frac{1}{t^2{7}^{\left|t\right|}}dt}}$

$=\frac{1}{7}{\left[-\frac{1}{t}\right]}_1^2+\frac{1}{7^2}{\left[\frac{-1}{t}\right]}_2^3+\frac{1}{7^3}{\left[-\frac{1}{t}\right]}_2^3+...$

$={\displaystyle {\sum }_{n=1}^{\infty }\frac{1}{7_n}\left(\frac{1}{n}-\frac{1}{n+1}\right)}$

$=1+6log\frac{6}{7}$

${\displaystyle {\int }_{cosx}^1{t}^{2}f\left(t\right)dt=si{n}^{3}x+cosx}$

$\Rightarrow sinxco{s}^{2}xf\left(cosx\right)=3si{n}^{2}xcosx-sinx$

$\Rightarrow f\text{'}\left(cosx\right)\left(-sinx\right)=3se{c}^{2}x-2se{c}^{2}tanx$

$cosx=\frac{1}{\sqrt[]{3}}$

$\therefore f\text{'}\left(\frac{1}{\sqrt[]{3}}\right)\left(-\frac{\sqrt[]{2}}{\sqrt[]{3}}\right)=9-6\sqrt[]{2}$

$\frac{1}{\text{exception 25:}}f\text{'}\left(\frac{1}{\text{exception 25:}}\right)=6-$$\frac{9}{}$

$f\left(x\right)={\displaystyle {\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt}$

$f\text{'}\left(x\right)=2x\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)=0$

$x=0,or\left(x^2-4\right)\left(x^2-1\right)=0$

$x=0,x=\pm 2,\pm 1$

Now,

$f\text{'}\left(x\right)=\frac{2x\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)}{e^{x^2}+2}$

Changes sign from positive to negative at x = 1, 1 So, number of local maximum points = 2

Changes sign from negative to positive at

x = 2, 0, 2 So, number of local minimum points = 3

$\therefore m=2,n=3$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenequationoflinesarex=2y+3,y=1andy=-1\\ \therefore Requiredarea={\displaystyle \underset{-1}{\overset{1}{\int }}\left(2y+3\right)}dy\\ =2.\frac{1}{2}{\left[y^2\right]}_{-1}^1+3{\left[y\right]}_{-1}^1=\left(1-1\right)+3\left(1+1\right)=6sq.units\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenequationoflinesarey=x+1,x=2andx=3\\ \therefore Requiredarea={\displaystyle \underset{2}{\overset{3}{\int }}\left(x+1\right)}dx\\ ={\left[\frac{x^2}{2}+x\right]}_2^3=\left(\frac{9}{2}+3\right)-\left(\frac{4}{2}+2\right)\\ =\frac{15}{2}-4=\frac{7}{4}sq.units\\ Hence,thecorrectoptionis\left(a\right).\end{array}$