Maths

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.