Maths

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.

  $\left|x-2\right|>1givesx-2<-1orx-2>1$ $$

i.e. x < 1 or x > 3 . (i) represent set A

$\sqrt[]{x^2-3}>1gives{x}^2-3>1or{x}^2-4>0$

x < 2 or x > 2 . (ii) represent set B

$\left|x-4\right|\ge 2givesx-4\le -2orx-4\ge 2$

$$  $x\le 2orx\ge 6$ . (iii)respect set C

so number of subset = 28 = 256

Given $2\mathcal{l}+2m-n=0........\left(i\right)$

$mn+n\mathcal{l}+\mathcal{l}m=0...........\left(ii\right)$

$\&wehave{\mathcal{l}}^2+m^2+n^2=1..........\left(iii\right)$

$\left(i\right)\Rightarrow 2\left(\mathcal{l}+m\right)=n$  

  $\left(ii\right)\Rightarrow \mathcal{l}m+n\left(\mathcal{l}+m\right)=0$

$2{\mathcal{l}}^2+2m^2+5\mathcal{l}m=0$             

(a) $\frac{\mathcal{l}}{m}=-2$  

(i) $\Rightarrow \frac{2\mathcal{l}}{m}+2-\frac{n}{m}=0$  

$\frac{n}{m}=-2$  

$So,\left(\mathcal{l},m,n\right)=\left(-2m,m,-2m\right)$  

= (-2, 1, -2)

(b)   $\frac{\mathcal{l}}{m}=\frac{-1}{2}givesn=-2\mathcal{l}$

$\left(\mathcal{l},m,n\right)=\left(\mathcal{l},-2\mathcal{l},-2\mathcal{l}\right)=\left(1,-2,-2\right)$

$Now,cos\theta =\frac{-2-2+4}{3*3}=0$

$\Rightarrow \theta =\frac{\pi }{2}$  

f(x) = tan^-1 (sin x + cos x)

$asx\in \left[0,\frac{\pi }{2}\right]$              so  1 $\le sinx+cosx\le \sqrt[]{2}$  

so $f\left(x\right)\in \left[ta{n}^{-1}1,ta{n}^{-1}\sqrt[]{2}\right]$  

->M = tan^-1   $\sqrt[]{2}\&m=ta{n}^{-1}1=\frac{\pi }{4}$

Now, tan (M – m) = tan $\left(ta{n}^{-1}\sqrt[]{2}-\frac{\pi }{4}\right)$

$=\frac{\sqrt[]{2}-1}{1+\sqrt[]{2}.1}=\frac{\sqrt[]{2}-1}{\sqrt[]{2}+1}*\frac{\sqrt[]{2}-1}{\sqrt[]{2}-1}=3-2\sqrt[]{2}$