Maths

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 $\left(ta{n}^{-1}\sqrt[]{2}-\frac{\pi }{4}\right)$

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

$\left(p\wedge q\right)\Rightarrow \left(\left(r\wedge q\right)\wedge p\right)$

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

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

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

$\Rightarrow \left(p\wedge q\right)\Rightarrow \left(r\wedge p\right)$

  $ta{n}^{-1}\frac{1}{2r^2}=ta{n}^{-1}\frac{2}{1+\left(4r^2-1\right)}$             

  $=ta{n}^{-1}\frac{\left(2r+1\right)-\left(2r-1\right)}{1+\left(2r+1\right)\left(2r-1\right)}$

${\displaystyle \sum _{r=1}^{50}\left[ta{n}^{-1}\left(2r+1\right)-ta{n}^{-1}\left(2r-1\right)\right]}$

$=ta{n}^{-1}101-ta{n}^{-1}1=ta{n}^{-1}\frac{100}{102}$

$\Rightarrow tanP=\frac{100}{102}=\frac{50}{51}$             

$\frac{dy}{dx}+\frac{e^y-2x}{2x^2}=0$  

$e^{-y}\frac{dy}{dx}-\frac{e^{-y}}{x}=-\frac{-1}{2x^2}$              

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

 I. F. $=e^{{\displaystyle \int \frac{dx}{x}}}=e^{lnx}=x$  

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$              

$P\left(-2\sqrt[]{6},\sqrt[]{3}\right)lieson\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$              

$\Rightarrow \frac{24}{a^2}-\frac{3}{b^2}=1.........\left(i\right)$              

$b^2=\frac{a^2}{4}.........\left(ii\right)$

solving (i) & (ii) a² = 12 Þ b² = 3 hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$  

Cuts conjugate axis at $R\left(0,R\sqrt[]{3}\right)$     

$\therefore QR=6\sqrt[]{3}$