Maths

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2*\frac{1}{2}\left(2+5\right)*3=21}$

Let y = mx + c is the common tangent

$soc=\frac{1}{m}=\pm \frac{3}{2}\sqrt[]{1+m^2}\Rightarrow m^2=\frac{1}{3}$

so equation of common tangents will be

$y=\pm \frac{1}{\sqrt[]{3}}x\pm \sqrt[]{3}$

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^2=1-\frac{1}{4}=\frac{3}{4}$

and length of latus rectum

$=\frac{2b^2}{a}=3$

Hence

$\frac{\mathcal{l}}{e^2}=\frac{3}{3/4}=4$

Boys (10)            Girls (5)

(3)                        (3)

B₁ & B₂ should not be selected together

Total number of ways

           

= (56 + 56) * 10 = 1120

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

$?x=\pm 1$

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