Maths

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$  

Coefficient of x^-1 ⇒ r = 10 ⇒ m = ¹⁵C₁₀ $\cdot 2^5$  

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$               

now mn² = ¹⁵C_r   $\cdot 2^r$

⇒ r = 5

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (2, 2)

Given a > b

Area common to x2 + y2 $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$

is $\pi a^2-\pi ab=30\pi ..............\left(i\right)$

Similarly $\pi ab-\pi b^2=18\pi .................\left(ii\right)$

Equation (i) and equation (ii) $\frac{a}{b}=\frac{5}{3}$

Equation (i) + equation (ii) $a^2-b^2=48$

a² = 75, b² = 27

sin x = 1 – sin² x

sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

First common term to both AP's is 9

t₇₈ of $\left(3,6,9,......\right)=78*3=234$

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$

n^th common term $\le 234$

9 + (n – 1) 12 $\le$ 234

n < $\frac{237}{12}\Rightarrow n=19$

Now sum of 19 terms with a = 9, d = 12

$=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$

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

IF $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$

= $e^{2x}\cdot {\left(x-1\right)}^2$

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y = $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$

y (2) = $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$

y (3) = $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$

Data contradiction.

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

pva ⇒$\left(\sim rvp\right)$

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$

its negation as asked in question

$\left(p\vee q\right)\wedge \left(\sim p\wedge r\right)$

= $\left(p\wedge \sim p\wedge r\right)\vee \left(q\wedge r\wedge \sim p\right)$

= $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$