Maths NCERT Exemplar Solutions Class 12th Chapter Five

x² = 1 – 2i

so 2 = 1 – 2i = 2

8 = 8

now $\left|{\alpha }^8+{\beta }^8\right|=2\left|{\alpha }^8\right|$

$=2{\left|\left({\alpha }^2\right)\right|}^4$

$=2{\left|{\alpha }^2\right|}^4$

$=2{\left|1-2i\right|}^4$

$=2*25$

Draw g(t) = t³ – 3t

g'(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³ – 3t) = $\{\begin{array}{cc}\hfill t^3-3t\hfill & \hfill ;-2<t<-1\hfill \\ \hfill 2\hfill & \hfill ;-1<t<2\hfill \end{array}$

$I={\displaystyle \underset{-2}{\overset{2}{\int }}f}\left(x\right)dx$

= ${\displaystyle \underset{-2}{\overset{-1}{\int }}\left(t^3-3t\right)dt}+{\displaystyle \underset{-1}{\overset{2}{\int }}2dt}$

I = $\frac{27}{4}$

again rewrite the f(x)

$f\left(x\right)=\left\{\begin{array}{ccc}\hfill \begin{array}{cc}\hfill x^3-3x\hfill & \hfill 2\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill x\le -1\hfill & \hfill -1<x\le 2\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\hfill x^2+2x-6\hfill & \hfill 9\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill 2<x<3\hfill & \hfill 3\le x<4\hfill \end{array}\hfill \\ \hfill \begin{array}{ccc}\hfill 10\hfill & \hfill 11\hfill & \hfill 2x+1\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{ccc}\hfill 4\le x<5\hfill & \hfill x=5\hfill & \hfill x>5\hfill \end{array}\hfill \end{array}\right\}$

$f\text{'}\left(x\right)=\left\{\begin{array}{cccccc}\hfill 3x^2-3\hfill & \hfill ;\hfill & \hfill x<-1\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill -1<x<2\hfill \\ \hfill 2x+2\hfill & \hfill ;\hfill & \hfill 2<x<3\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill 3<x<4\hfill \\ \hfill 0\hfill & \hfill ;\hfill & \hfill 4<x<5\hfill & \hfill 2\hfill & \hfill ;\hfill & \hfill x>5\hfill \end{array}\right\}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

 $\beta =\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)},\alpha \in R\underset{x\to 0}{lim}\frac{\frac{\alpha }{3}-\left(\frac{e^{3x}-1}{3x}\right)}{\alpha x\left(\frac{e^{3x}-1}{3x}\right)}$

$=\underset{x\to 0}{lim}\frac{1-\frac{\left(1+3x+\frac{9x^2}{2}+..........-1\right)}{3x}}{1+3x+\frac{9x^2}{2}+........-1}=-\frac{1}{2}\therefore \alpha +\beta =\frac{5}{2}$

$|z-i|=\left|z+5i\right|$ So, z lies on perpendicular bisector of (0, 1) and (0, 5) i.e., line y = 2 as |z| = 2 z = 2i x = 0 and y = 2 so, x + 2y + 4 = 0

 $\left|z^2\right|=\left|\overline{z}\right|.2^{1-\left|z\right|}\Rightarrow \left|z\right|=1$

$z^2=\overline{z}\Rightarrow z^3=1$

$\therefore z=wor{w}^2$

$w^n={\left(1+w\right)}^n={\left(-w^2\right)}^n$

Note : n should be given as a natural number:

$f\left(x\right)=\{\begin{array}{l}\begin{array}{ccc}\hfill \frac{-sin\left(x-1\right)}{x-1}\hfill & \hfill ,\hfill & \hfill x<-1\hfill \\ \hfill -\left(sin2+1\right)\hfill & \hfill ,\hfill & \hfill x=-1\hfill \\ \hfill cos2\pi x\hfill & \hfill ,\hfill & \hfill -1<x<1\hfill \end{array}\\ 1,x=1\end{array}$

$\frac{-sin\left(x-1\right)}{x-1},x>1$

f (x) is discontinuous at x = 1 and x = 1

 $a=\frac{-1}{{\alpha }^2}-\frac{1}{{\beta }^2}-2,b=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+1+\frac{1}{{\alpha }^2{\beta }^2}$

$6x^2+17x+7=0,x=\frac{-7}{3},x=\frac{-1}{2}$ are roots

Both roots, real and negative

Since,  ${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x}$ exist $\Rightarrow f\left(0\right)=0$

Now,  $f^{\mathrm{\text{'}}}\left(x\right)={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)+x{h}^2+x^{2}h}{h}($ take $y=h)$

$={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)}{h}+{\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to .0}?\left(xh\right)+x^2$

$\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1+0+{\mathrm{x}}^2$

$\Rightarrow f^{\mathrm{\text{'}}}\left(3\right)=10$

$\mathrm{u}=\frac{2\mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{k}\mathrm{i}}$

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

Since $\mathrm{R}\mathrm{e}?\left(u\right)+\mathrm{I}\mathrm{m}?\left(u\right)=1$

$\Rightarrow 2{\mathrm{x}}^2+(2\mathrm{y}+1)(\mathrm{y}-\mathrm{k})+\mathrm{x}(2\mathrm{y}+1)-2\mathrm{x}(\mathrm{y}-\mathrm{k})={\mathrm{x}}^2+(\mathrm{y}-\mathrm{k}{)}^2$

$\begin{array}{c}\mathrm{P}\left(0,{\mathrm{y}}_1\right)\\ \mathrm{Q}\left(0,{\mathrm{y}}_2\right)\end{array}?\Rightarrow {\mathrm{y}}^2+\mathrm{y}-\mathrm{k}-{\mathrm{k}}^2=0?\begin{array}{c}{\mathrm{y}}_1+{\mathrm{y}}_2=-1\\ {\mathrm{y}}_1{\mathrm{y}}_2=-\mathrm{k}-{\mathrm{k}}^2\end{array}$

$?\mathrm{P}\mathrm{Q}=5$

$\Rightarrow \left|{\mathrm{y}}_1-{\mathrm{y}}_2\right|=5\Rightarrow {\mathrm{k}}^2+\mathrm{k}-6=0$

$\Rightarrow \mathrm{k}=-\mathrm{3,2}$

So, $k=2(k>0)$

 $x^2-3x+p=0$

$\alpha ,\beta ,\gamma ,\delta$ in G.P.

$\alpha +\alpha r=3$

$x^2-6x+q=0$

$\alpha r^2+\alpha r^3=6$

$\left(2\right)÷\left(1\right)\Rightarrow r^2=2$

So,  $\frac{2q+p}{2q-p}=\frac{2r^5+r}{2r^5-r}=\frac{2r^4+1}{2r^4-1}=\frac{9}{7}$