Maths

$f\left(x\right)=\left|x^2-2x-3\right|e^{\left|9x^2-12x+4\right|}$

$\Rightarrow f\left(x\right)=\{\begin{array}{l}\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x<-1\\ -\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},-1\le x<3\\ \left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x\ge 3\end{array}$

$f\text{'}\left(-1^{-}\right)\ne f\left(-1^{+}\right),$

$f\text{'}\left(3^{-}\right)\ne f\left(3^{+}\right)$

Number of non differential points is 2 at x = -1, 3.

$e^{4x}+2e^{3x}-e^x-6=0$

$e^x=t\in \left(0,\infty \right)$

$t^4+2t^3-t-6=0$

Let f (t) = t⁴ + 2t³ – t – 6

f' (t) = 4t³ + 6t² – 1f

Þf' (0) = -1, f' (+ $\infty )$ = + $\infty$

For t > 0

Þ f' (t) = 0 has only one root.

One solution of f (t) = 0 is possible

Equation of plane is 3x – 2y + 4z – 7 + $\lambda$ (x + 5y – 2z + 9) = 0. (i)

It passes through (1, 4, -3) and we get $\lambda =\frac{2}{3}$

$\therefore$ from (i) we get 11x + 4y + 8z – 3 = 0 Þ -11x – 4y – 8z + 3 = 0

$\therefore \alpha +\beta +\lambda =-11-4-8=-23$  

$\left(p\wedge \sim q\right)\to \left(p\vee q\right)$ is tautology

$*=\wedge ,?=\vee$

Let $l={\displaystyle \int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}}dx={\displaystyle \int \frac{2e^{2x}+3}{4e^{2x}+7}dx}$  

$={\displaystyle \int \frac{2e^{2x}}{4e^{2x}+7}dx+{\displaystyle \int \frac{3e^{-2x}}{4+7e^{-2x}}dx}}$               

Put $4e^{2x}+7=t4+7e^{-2x}=\lambda$  

$8e^{2x}dx=dt-14e^{-2x}dx=d\lambda$              

$=\frac{1}{4}{\displaystyle \int \frac{dt}{t}-\frac{3}{14}{\displaystyle \int \frac{d\lambda }{\lambda }=\frac{1}{4}lnt-\frac{3}{14}ln\lambda +c}}$

$u=\frac{13}{2},v=\frac{1}{2}$              

->so, u + v = 7

Since A (sec θ, 2 tanθ) & B $\left(sec?,2tan?\right)$  lies on 2x² – y² = 2 then

2sec² θ - 4tan^2_θ = 2 or sec² θ - 2 tan² θ = 1

-> $ta{n}^2\theta =0so\theta =0$

Similarly    $?=0but\theta +?=\frac{\pi }{2}$ (given) so not possible

Hence question is not correct

Given variance of boys ${\sigma }_b^2=2\&{\overline{x}}_b=12$ (average marks of boys)

& variance of girls  ${\sigma }_g^2=2\&\mu \to$ average marks of girls

$Now,{\overline{x}}_g=\mu =\frac{50*15-12*20}{30}=17$           

${\sigma }^2=\frac{20*2+30*2}{20+30}+\frac{20*30}{{\left(20+30\right)}^2}{\left(12-17\right)}^2=8$            

$So,\mu +{\sigma }^2=17+8=25$  

$x^2+9y^2-4x+3=0,x,y\in R.........\left(i\right)$

$x^2-4x+9y^2+3=0$        

Since $x\in R,D\ge 0i.e.16-4\left(9y^2+3\right)\ge 0$

or $9y^2-1\le 0$

$\Rightarrow y\in \left[-\frac{1}{3},\frac{1}{3}\right]$

$\left(i\right)\Rightarrow 9y^2=-x^2+4x-3$

Since L.H.S. $\ge 0soR.H.S.\ge 0i.e.-x^2+4x-3\ge 0$

or $x^2-4x+3\le 0$

$\left(x-3\right)\left(x-1\right)\le 0$

$\Rightarrow x\in \left[1,3\right]$

Let $z_1=x_1+i{y}_1,z_2=x_2+i{y}_2\&z=x+iythen$  

and $\left(z_1-z_2\right)=\frac{\pi }{4}gives\frac{y_1-y_2}{x_1-x_2}=1or{y}_1-y_2=x_1-x_2............\left(i\right)$  

y² – 6x + 9 = 0 .(ii)

as z₁ & z₂ lies on (ii) so ${y_1}^2-6x_1+9=0$    .(iii)

& $y_{2}2-6x_2+9=0.........\left(iv\right)$  

(iii) & (iv) $\left(y_1+y_2\right)\left(y_1-y_2\right)-6\left(x_1-x_2\right)=0$

$\left(x_1-x_2\right)\left(y_1+y_2-6\right)=0from\left(i\right)$

-> $y_1+y_2-6=0i.e.,y_1+y_2=6$  

Let $t=\frac{3}{2}{x}^2+\frac{5}{3}{x}^3+\frac{7}{4}{x}^4+......$

$=\left(2-\frac{1}{2}\right)x^2+\left(2-\frac{1}{3}\right)x^3+\left(2-\frac{1}{4}\right)x^4+......$   

= $2\left(x^2+x^3+x^4+.....\right)-\left(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+.....\right)$

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

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