Maths NCERT Exemplar Solutions Class 12th Chapter Six

$x^4?3x^3?2x^2+3x+1=0$

$?x=\pm 1$

and let, are roots of x2 – 3x – 1 = 0

$??+?=3??=?1$

$?1^3+{\left(?1\right)}^3+{?}^3+{?}^3$

= 27 + 3 (3) = 36

Case – I $\Delta \equiv \nabla \equiv \wedge$

$\left(p\wedge q\right)\to \left(\left(p\wedge q\right)\wedge r\right)$

it can be false if r is false,

so not a tautology

Case – II If $\Delta \equiv \nabla \equiv \vee$

$\left(p\vee q\right)\text{?}\to \left(\left(p\vee q\right)\vee r\right)\equiv$ tautology

then $\left(p\vee q\right)\vee r\equiv \left(p\Delta r\right)\vee q$

Case – III If $\Delta \equiv v,\nabla \equiv \wedge$

then $\left(p\wedge q\right)\to \left\{\left(p\vee q\right)\wedge r\right\}$

Not a tautology

Case – IV If $\Delta \equiv \wedge ,\nabla \equiv \vee$

$\left(p\wedge q\right)\to \left\{\left(p\wedge q\right)\vee r\right\}$

Not a tautology

 $f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 ( + 5) (5 + 9) = 11 - π

 $\overline{x}=6=\frac{a+b+8+5+10}{5}\Rightarrow a+b=7$ …… (i)

and

${\sigma }^2=\frac{a^2+b^2+8^2+5^2+10^2}{5}-6^2=6.8$

a² + b² = 25 ……. (ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

$M=\frac{1}{5}\left(3+2+2+1+4\right)=\frac{12}{5}$

25M = 60

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)

               

6x = 5 = 0     $x=\frac{5}{6}$        

P (atmost 2H)

$P\left(OH,5T\right)+P\left(1H,4T\right)+P\left(2H,3T\right)$

$=\frac{1}{6^5}\left(1+25+250\right)=\frac{276}{6^5}=\frac{4^6}{6^4}$                   

 $\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=3\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{u}$

$\overrightarrow{b}*\left(\overrightarrow{c}*\overrightarrow{a}\right)=\overrightarrow{c}-2\overrightarrow{a}=\overrightarrow{v}$

$\overrightarrow{c}*\left(\overrightarrow{b}*\overrightarrow{a}\right)=3\overrightarrow{b}-2\overrightarrow{a}=\overrightarrow{w}$

$\therefore \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{w}$

so vectors 

$\overrightarrow{u},\overrightarrow{v}and\overrightarrow{w}$

are coplanar, hence their Scalar triple product will be zero.

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$

0 $\lambda =7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

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

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$

and $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$

clearly

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$

So, a : b : c = 8 : 5 : 4

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

a = 3

Now angle between ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

Let P (at², 2 at) where

$a=\frac{3}{2}$                

T : yt = x + at² so point Q is

$\left(-a,at-\frac{a}{t}\right)$                

N : y = -tx + 2at + at³ passes through (5, -8)

$-8=-5t+3t+\frac{3}{2}{t}^3$

⇒ $3t^3-4t+16=0$                

⇒ t = -2

So ordinate of point Q is $-\frac{9}{4}$  

Equation of perpendicular bisector of AB is

$y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right)\Rightarrow x+5y=10$

Solving it with equation of given circle,

${\left(x-5\right)}^2+{\left(\frac{10-x}{5}-1\right)}^2=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}or\frac{15}{2}$

But

$x\ne \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$\left(\frac{15}{2},\frac{1}{2}\right)$

Now,

$r^2={\left(\frac{15}{2}-2\right)}^2+{\left(\frac{1}{2}+1\right)}^2=\frac{65}{2}$