Ncert Solutions Maths class 11th

Tangents making angle $\frac{\pi }{4}$ $\frac{}{}$ with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$

So, these tangents are $\perp$ $$ . So ASB is a focal chord.

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\sqrt[]{3}}{4}{\mathcal{l}}^2=\frac{25}{2\sqrt[]{3}}$

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$ $\frac{}{}\frac{}{{}^{}}\frac{}{{}^{}}$

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$

$\frac{25S}{36}=1+\frac{3/5}{1}$

$S=\frac{288}{125}$

$\left|z\right|=3\Rightarrow$ $$ circle with radius = 3

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4},$ $\frac{}{}\frac{}{}$ part of a circle (with radius $\text{exception 25:}$ $\sqrt[]{2}$ ). no common points

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$   …… (i)

Now,   ${\left(-2\right)}^3\equiv -1mod\left(7\right)$  

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$       ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$  

$\therefore$  Remainder = 5

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

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

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

pva -> $\left(\sim rvp\right)$  

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$  

its negation as asked in question

$\left(p\vee q\right)\wedge \left(\sim p\wedge r\right)$  

=  $\left(p\wedge \sim p\wedge r\right)\vee \left(q\wedge r\wedge \sim p\right)$  

=  $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$  

Total number of possible relation = $2^{n^2}=2^4=16$  

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability =  $\frac{5}{16}$  

${\left(x-\frac{1}{\sqrt[]{2}}\right)}^2+{\left(y-\frac{1}{\sqrt[]{2}}\right)}^2=1$                                                    

here AB =  $\sqrt[]{2}$ , BC = 2, AC = $\sqrt[]{2}$

area =  $\frac{1}{2}*\sqrt[]{2}*\sqrt[]{2}=1$  

Tangents making angle $\frac{\pi }{4}$  with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$  

So, these tangents are  . So ASB is a focal chord.