Maths NCERT Exemplar Solutions Class 12th Chapter One

${\overrightarrow{b}}_1*{\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -a\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$ = $-a\widehat{i}-\widehat{j}+\left(a-1\right)\widehat{k}$

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

Shortest distance = $\left|\frac{\left({\overrightarrow{a}}_1-{\overrightarrow{a}}_2\right).\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}\right|$

$=\frac{2\left(a-1\right)}{\sqrt[]{a^2+1+{\left(a+1\right)}^2}}=\sqrt[]{\frac{2}{3}}=\frac{4{\left(a-1\right)}^2}{a^2+1+{\left(a-1\right)}^2}=\frac{2}{3}$

$12{\left(a-1\right)}^2=2\left(a^2+1\right)+2{\left(a-1\right)}^2$

$10{\left(a-1\right)}^2=2\left(a^2+1\right)$

$5a^2-10a+5=a^2+1\Rightarrow 4a^2-10a+4=0$

$2a^2-5a+2=0$

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

$\therefore a=\frac{1}{2},2$

$R_2={\displaystyle \underset{0}{\overset{1}{\int }}\left(x-x^3\right)dx={\left(\frac{x^2}{2}-\frac{x^4}{4}\right)}_0^1}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$R_1={\displaystyle \underset{0}{\overset{\frac{1}{4}}{\int }}\left(\sqrt[]{x}-2x\right)dx={\left(\frac{2x^{3/2}}{3}-x^2\right)}_0^{\frac{1}{4}}}=\frac{1}{12}-\frac{1}{16}$

$=\frac{4-3}{48}=\frac{1}{48}$

$R_1+R_2={{\displaystyle \underset{0}{\overset{1}{\int }}\left(\sqrt[]{x}-x^3\right)dx=\left(\frac{2x^{3/2}}{3}-\frac{x^4}{4}\right)}}_0^1=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

$\frac{R_1+R_2}{R_1}=1+\frac{R_2}{R_1}=\frac{\frac{5}{12}}{\frac{1}{48}}=20$ $\therefore \frac{R_2}{R_1}=19$

 

$\frac{x^2}{\frac{1}{25}}+\frac{y^2}{\frac{1}{4}}=1$      

$25{\alpha }^2+4{\beta }^2=1..........(i)$

Equation of tangent to parabola y = mx + $\frac{1}{m}$  passes

though (a, b)

$\alpha m^2-\beta m+1=0$

$m_1+m_2=\frac{\beta }{\alpha },4m_1^2=\frac{1}{\alpha }$

$\therefore from(i)\&(ii)$

25(a² +  a) = 1 …………(iii)

$\therefore {\left(10\alpha +5\right)}^2+{\left(16{\beta }^2+50\right)}^2=2929$

$\therefore a=1+0+b{\left(sin-\frac{\pi }{2}\right)}^{\frac{\pi }{2}}+2b+1$           

Let f(x) = $\frac{x^2-9}{x-5}$  

  $f\text{'}\left(x\right)=\frac{2x\left(x-5\right)-\left(x^2-9\right)}{{\left(x-5\right)}^2}$

$=\frac{x^2-10x+9}{{\left(x-5\right)}^2}=\frac{\left(x-1\right)\left(x-9\right)}{{\left(x-5\right)}^2}$

$\alpha =f\left(1\right)=2,\beta =\left\{f\left(0\right),f\left(2\right)\right\}=\frac{5}{3}$

$\therefore {\displaystyle \underset{-1}{\overset{3}{\int }}max}\left\{\frac{x^2-9}{x-5},x\right\}dx+{\left[\frac{x^2}{2}\right]}_{9/5}^3$           

a₁ = 18, a₂ = 16

a₁ + a₂ = 34

  $f\left(\theta \right)=sin\theta +{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(sin\theta +tcos\theta \right)f\left(t\right)dt}$

$=sin\theta +sin\theta {\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt+cos\theta {\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}}$

  $=\left(1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt}\right)sin\theta +\left({\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}\right)cos\theta$              

f(q) = a sin q + b cos q

  $\therefore a=1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt=1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(asint+bcost\right)dt}}$

$\therefore a=2b+1..........(i)$

$b={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(asint+bcost\right)dt}$

 Solving (i) & (ii) we get a =   $-\frac{1}{3},b=-\frac{2}{3}$

    $\therefore \left|{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\theta \right)d\theta }\right|=1$           

$y=2\left|x^2-\frac{3}{2}x-\frac{7}{2}\right|$

$=2\left|{\left(x-\frac{3}{4}\right)}^2-\frac{65}{16}\right|$

  $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=r_1$

$A\left(3r_1+7,1-r_1,-2+r_1\right)$

and   $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=r_2$  

  $B\left(2r_2,7+3r_2,r_2\right)$

$A/q,\frac{3r_1-2r_2+7}{1}=\frac{3r_2+r_1+6}{4}=\frac{r_1-r_2-2}{2}$             

  $\Rightarrow r_1=-5,r_2=-3$

$\therefore A\left(-8,6,-7\right)andB\left(-6,-2,-3\right)$

AB² = 84

  $f\left(x\right)=\{\begin{array}{l}\left|2x^2-3x-7\right|,x\le -1\\ \left[4x^2-1\right],-1<x<1\\ \left|x+1\right|+\left|x-2\right|,x\ge 1\end{array}$              

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x=\pm \frac{1}{2},\pm \frac{1}{\sqrt[]{2}},\pm \frac{\sqrt[]{3}}{2}$          

$B\left(-\frac{3}{\sqrt[]{a}},\sqrt[]{a}\right)andc\left(-\frac{3}{\sqrt[]{a}},-\sqrt[]{a}\right)$

$\therefore Areaof\Delta ACD=\frac{1}{2}\left|\begin{array}{ccc}\hfill \frac{3}{\sqrt[]{a}}\hfill & \hfill \sqrt[]{a}\hfill & \hfill 1\hfill \\ \hfill \frac{-3}{\sqrt[]{a}}\hfill & \hfill -\sqrt[]{a}\hfill & \hfill 1\hfill \\ \hfill 3cos\theta \hfill & \hfill asin\theta \hfill & \hfill 1\hfill \end{array}\right|$  = 12 

$\Rightarrow \Delta =3\sqrt[]{a}\left|cos\theta +sin\theta \right|=12$

$\therefore \Delta max=3\sqrt[]{a}.\sqrt[]{2}=12\Rightarrow a=8$

Let x = correct answer, y = incorrect answer

  $\therefore 3x-2y=5,x+y\le 5,x,y\in w$              

 only possible (x, y) is (3, 2)

$\therefore$ Required number of ways =