Ncert Solutions Maths class 12th

$\Delta =\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right|=2\left(-a-1\right)+\left(1-a\right)+2=-3a+1$

${\Delta }_3=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill b\hfill \end{array}\right|=2\left(-b-1\right)+\left(3-b\right)+5*2=-3b+7$

For a = $\frac{1}{3},b\ne \frac{7}{3},$ $\frac{}{}\frac{}{}$ system has no solution

Applying Leibniz theorem,  

$\sqrt[]{1-{\left(f\text{'}\left(x\right)\right)}^2}=f\left(x\right)\Rightarrow \frac{f\text{'}\left(x\right)}{\sqrt[]{1-{\left(f\left(x\right)\right)}^2}}=1$ on integrating both sides, we get

$f\left(x\right)=sinx+C$ put x = 0 and f (0) = 0 we get C = 0

$Now\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}}{x^2},\left(\frac{0}{0}\right)$ by L' Hospital rule $\underset{x\to 0}{lim}\frac{f\left(x\right)}{2x}=\underset{x\to 0}{lim}\frac{sinx}{2x}=\frac{1}{2}$

${\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^{3/4}{\left(x+2\right)}^{5/4}}={\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^2}put\frac{x-1}{x+2}}=t\Rightarrow \frac{3}{{\left(x+2\right)}^2}dx=dt}$

$\therefore \frac{1}{3}{\displaystyle \int \frac{dt}{t^{3/4}}=\frac{1}{3}.\frac{t^{1/4}}{\frac{1}{4}}+C=\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{1/4}+C}$

$LHS=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to \infty }{lim}\frac{-1}{h}ln\left(\frac{1-\frac{h}{a}}{1+\frac{h}{b}}\right)=\underset{h\to 0}{lim}\frac{ln\left(1-\frac{h}{a}\right)}{a\left(\frac{-h}{a}\right)}+\underset{h\to 0}{lim}\frac{ln\left(1+\frac{h}{b}\right)}{b\left(\frac{h}{b}\right)}=\frac{1}{a}+\frac{1}{b}.............\left(i\right)$

$RHS=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{cos2x-1}{x^2+1-1}\left(\sqrt[]{x^2+1}+1\right)=\underset{x\to 0}{lim}\frac{-2si{n}^{x}x}{x^2}*2=-4...............\left(ii\right)$

$and\underset{x\to 0}{lim}f\left(x\right)=k.............\left(iii\right)$

$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=-4=k$

From (i), (ii) and (iii) we get $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$

$\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi -\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{x^4}$

$=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}*\frac{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}{x^4}=4{\pi }^2\underset{x\to 0}{lim}\frac{si{n}^{4}x}{x^4}=4{\pi }^2$

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

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

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

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

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

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

$*=\wedge ,?=\vee$

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

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

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

$So,\mu +{\sigma }^2=17+8=25$  

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

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

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

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

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