Maths

$L.R.=\frac{\left|3*2+4x-3-5\right|}{\sqrt[]{3^2+4^2}}=\frac{11}{5}$              

Equation of family of parabolas

${\left(x-h\right)}^2=\frac{11}{5}\left(y-k\right)$              

Differentiate 2 (x – h) = $\frac{11}{5}\frac{dy}{dx}$  

Again differentiate 2 = $\frac{11}{5}\frac{d^{2}y}{d{x}^2}$  

$11\frac{d^{2}y}{d{x}^2}=10$              

Given 2x + y – z = 3         . (i)

x – y – z = α        . (ii)

3x + 3y + βz = 3                . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

(y – 2)² = (x – 1)

2 (y – 2)    $\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}\left(2,3\right)=\frac{1}{2\left(3-2\right)}=\frac{1}{2}$              

Equation of tangent at P (2, 3):

$y-3=\frac{1}{2}\left(x-2\right)$  

2y – 6 = x – 2

x – 2y + 4 = 0

Q (-4, 0)

Required area = ${\displaystyle \underset{0}{\overset{3}{\int }}\left({\left(y-2\right)}^2+1-\left(2y-4\right)\right)dy=9}$

For divisibility by 5 last digit must be 0 or 5 but 0 is not possible in palindrome

so it will be 5.

So, required no. = 10 * 10 = 100

Equation of tangent to given ellipse at

$P:\frac{x.cos\theta }{b}+\frac{y.sin\theta }{2a}=1$

A (b sec θ. 0) 7 B (0, 2a cosec θ)

area of    $\Delta OAB$

$=\frac{2ab}{2sin\theta cos\theta }=\frac{2ab}{sin2\theta }$

For minimum area sin 2θ = 1

So minimum area = 2ab

=>k = 2

  $\underset{x\to \infty }{lim}\left(\sqrt[]{x^2-x+1}-ax\right)=b$

$\underset{x\to \infty }{lim}\left|\frac{x^2-x+1-a^2{x}^2}{\sqrt[]{x^2-x+1}+ax}\right|=b$

For existence of limit 1 – a² = 0 i.e. a = 1 only

$\underset{x\to \infty }{lim}\frac{1-x}{\sqrt[]{x^2-x+1}+x}=b$

$\Rightarrow b=\frac{-1}{2}$

So, (a, b) =   $\left(1,-\frac{1}{2}\right)$

  $\overrightarrow{b}*\overrightarrow{c}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9-2\beta \right)-\widehat{j}\left(-3+\beta \right)+\widehat{k}\left(5\right)$             

$\left|\overrightarrow{b}*\overrightarrow{c}\right|=5\sqrt[]{3}gives\sqrt[]{{\left(9-2\beta \right)}^2+{\left(\beta -3\right)}^2+25}=5\sqrt[]{3}$         

$81+4{\beta }^2+36\beta +{\beta }^2+9-6\beta +25=75$

$\beta =-2,-4$    

also $\overrightarrow{a}\perp \overrightarrow{b}so\overrightarrow{a}.\overrightarrow{b}=0i.e.1+15+\alpha \beta =0$

So, ${\left|\overrightarrow{a}\right|}^2=1+25+{\alpha }^2=90$

Let f (x) = 3x⁴ + 4x³ – 12x² + 4

So, f' (x) = 12x³ + 12x² – 24x = 12x (x² + x – 2)

=12x (x + 2) (x – 1)

So, number of distinct real roots = 4

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{\sqrt[]{x}dx}{\left(1+x\right)\left(1+3x\right)\left(3+x\right)}}$

Put x = t² then dx = 2tdt

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+t^2\right)\left(3+t^2\right)}-{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+3t^2\right)\left(3+t^2\right)}}}$

$=\frac{1}{2}{\left(ta{n}^{-1}t\right)}_0^1-\frac{3}{8\sqrt[]{3}}{\left(ta{n}^{-1}\frac{t}{\sqrt[]{3}}\right)}_0^1-\frac{8}{8\sqrt[]{3}}{\left(ta{n}^{-1}\sqrt[]{3}t\right)}_0^1$

$=\frac{\pi }{8}\left(1-\frac{\sqrt[]{3}}{2}\right)$      

$r=\sqrt[]{{\left(\frac{p}{2}\right)}^2+{\left(\frac{1-p}{2}\right)}^2-5}=\sqrt[]{\frac{p^2+1+p^2-2p-20}{4}}=\sqrt[]{\frac{2p^2-2p-19}{2}}$

Since $r\in \left(0,5\right)so0<\sqrt[]{2p^2-2p-19}<10$

$\Rightarrow 2p^2-2p-19>0\&2p^2-2p-19<100$     

$P\in \left[\frac{1-\sqrt[]{239}}{2},\frac{1-\sqrt[]{39}}{2}\right)\cup \left(\frac{1+\sqrt[]{39}}{2},\frac{1+\sqrt[]{239}}{2}\right]$

So number of integral values of P² is 61.