Maths NCERT Exemplar Solutions Class 12th Chapter Five

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l}Let{z}_1=x_1+y_{1}iand{z}_2=x_2+y_{2}i\\ \therefore \left|x_1+y_{1}i\right|=\left|x_2+y_{2}i\right|\\ \Rightarrow \sqrt[]{x_1^2+y_1^2}=\sqrt[]{x_2^2+y_2^2}\\ \Rightarrow x_1^2+y_1^2=x_2^2+y_2^2\Rightarrow x_1^2=x_2^{2}and{y}_1^2=y_2^2\\ \Rightarrow x_1=\pm x_{2}and{y}_1=\pm y_2\\ So,z_1=x_1+y_{1}iand{z}_2=\pm x_2\pm y_{2}i\\ \therefore z_1\ne z_2\\ Hence,itisnotnecessarythat{z}_1=z_2.\end{array}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab)\frac{2b^2}{a}=10\Rightarrow b^2=5a$

Now,  $?\left(t\right)=\frac{5}{12}+t-t^2=\frac{8}{12}-{\left(t-\frac{1}{2}\right)}^2$
$?(\mathrm{t}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{4}{9}$

$\Rightarrow {\mathrm{a}}^2=81\mathrm{}$  (From (i) and (ii)

So,  $a^2+b^2=81+45=126$

$x^2-3x+p=0$

$\alpha ,\beta ,\gamma ,\delta$ 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,  $\frac{2q+p}{2q-p}=\frac{2r^5+r}{2r^5-r}=\frac{2r^4+1}{2r^4-1}=\frac{9}{7}$

$\alpha ,\beta$ are roots of x² - $4\lambda x+5=0$  

$\therefore \alpha +\beta =4\lambda and\alpha \beta =5$               

Also $\alpha ,\gamma$  are roots of

$x^2-\left(3\sqrt[]{2}+2\sqrt[]{3}\right)x+7+3\sqrt[]{3}\lambda =0,\lambda >0$

$\therefore \alpha +\gamma =3\sqrt[]{2}+2\sqrt[]{3},\alpha \gamma =7+3\sqrt[]{2\lambda }$               

$?\alpha$ is common root

$\therefore {\alpha }^2-4\lambda \alpha +5=0$     …….(i)

And

${\alpha }^2-\left(3\sqrt[]{2}+2\sqrt[]{3}\right)\alpha +7+3\sqrt[]{3}\lambda =0$      ….(ii)

From (i) – (ii) ; we get

$\alpha =\frac{2+3\sqrt[]{3}\lambda }{3\sqrt[]{2}+2\sqrt[]{3}-4\lambda }$               

$?\beta +\gamma =3\sqrt[]{2}$               

$\therefore {\left(\alpha +2\beta +\gamma \right)}^2={\left(\alpha +\beta +\beta +\gamma \right)}^2$               

$=98$             

$f\left(x\right)={\displaystyle {\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt}$

$f\text{'}\left(x\right)=2x\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)=0$

$x=0,or\left(x^2-4\right)\left(x^2-1\right)=0$

$x=0,x=\pm 2,\pm 1$

Now,

$f\text{'}\left(x\right)=\frac{2x\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)}{e^{x^2}+2}$

Changes sign from positive to negative at x = 1, 1 So, number of local maximum points = 2

Changes sign from negative to positive at

x = 2, 0, 2 So, number of local minimum points = 3

$\therefore m=2,n=3$

This is a True or False Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}False.\\ Letustakeanexample:f\left(x\right)=sinxandg\left(x\right)=cotx\\ \therefore f\left(x\right).g\left(x\right)=sinx.cotx=sinx.\frac{cosx}{sinx}=cosxwhichiscontinuousatx=0butcotxisnot\\ continuousatx=0.\end{array}$

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $True.$

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $True.Weknowthatthesumanddifferenceoftwoormorefunctionsisalwayscontinuous.$

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $True.Weknowthatthefunctioniscontinuousfunctiononitsdomain.$

This is a True or False Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}False.Giventhatf\left(x\right)=\left|x-1\right|in\left[0,2\right]\\ Weknowthatthefunctionisnotdifferentiable.So.itisfalse.\end{array}$