Maths

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   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.

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

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$  

Therefore square of  $ar\left(\Delta PQR\right)$ = 153.

$f\text{'}\left(x\right)=\frac{4x^2-1}{x}$ so f (x) is decreasing in $\left(0,\frac{1}{2}\right)and\left(\frac{1}{2},\infty \right)$ $\Rightarrow a=\frac{1}{2}$  

Tangent at y² = 2x is y = mx + $\frac{1}{2m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2}or\frac{1}{4}$  

So tangent may be  $y=\frac{1}{2}x+1ory=\frac{1}{4}x+2buty=\frac{1}{2}x+1$  passes through (-2, 0) so rejected.

Equation of normal  $\frac{x}{9}+\frac{y}{36}=1$  

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$  

slope of HC =  $\frac{16p-p^2-31}{16p-32}$  

slope of BC * slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$  

After solving we get  $y^2=\frac{x^4}{3}+c{x}^2$  also curve passes through (3, 3) Þ c = -2

$\therefore y^2=\frac{x^4}{3}-2x^2$ which passes through $\left(\alpha ,6\sqrt[]{10}\right)$ $\therefore \frac{{\alpha }^4-6{\alpha }^2}{3}=360\Rightarrow \alpha =6$  

${\displaystyle \underset{0}{\overset{1}{\int }}1.{\left(1-x^n\right)}^{2n+1}dx}$ using by parts we get

$\left(2n^2+n+1\right){\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx=1177{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx}}$

$\Rightarrow 2n^2+n+1=1177\Rightarrow n=24or-\frac{49}{2}\therefore n=24$

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$  

              Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$  

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$  

              Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$  

$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$  

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$  

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$  are roots of above equation and x⁵ + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.