Ncert Solutions Maths class 11th

Given series $\left\{3*1\right\},\left\{3*2,3*3,3*4\right\},\left\{3*5,3*6,3*7,3*8,3*9\right\}.........$  

$\therefore$  11^th set will have 1 + (10)2 = 21 terms

Also up to 10^th set total 3 * k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3*101,3*102,......3*121\right\}$ $\therefore$ Sum of elements = 3 * (101 + 102 + ….+121)

$=\frac{3*222*21}{2}=6993.$   

Factors of 36 = 2².3².1

Five-digit combinations can be 

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers  $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$  

$\sim \left(p\iff \sim q\right)\wedge q=\left(p\iff \sim q\right)\wedge qis:$

$\therefore \left(\sim \left(P\iff \sim Q\right)\right)\wedge$ q is equivalent to $\left(p\Rightarrow q\right)\wedge$

Any tangent to y² = 24x at (a, b) is by = 12 (x + a) therefore   Slope = $\frac{12}{\beta }$  

and perpendicular to 2x + 2y = 5 Þ 12 = b and a = 6 Hence hyperbola is $\frac{x^2}{6^2}-\frac{y^2}{12^2}$  = 1 and normal is drawn at (10, 16)

therefore equation of normal  $\frac{36x}{10}+\frac{144y}{16}=36+144\Rightarrow \frac{x}{50}+\frac{y}{20}=1$  This does not pass through (15, 13) out of given option.

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P(h, k) is $x^2+y^2+x-3y-2=0$  

Now intersection with x – axis are  $x^2+x-2=0\Rightarrow x=-2,1$  

Now intersection with y – axis are  $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$  

Therefore are of the quadrilateral ABCD is =  $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$  

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. =  ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20*\frac{20*21}{2}-\frac{20*21*41}{6}=1330$  

$\frac{z_2-0}{\left(1+2i\right)-0}=\left|\frac{OB}{OA}\right|e^{\frac{i\pi }{4}}\Rightarrow z_2=\left(1+2i\right)\left(1+i\right)=-1+3i\therefore arg{z}_2=\pi -ta{n}^{-1}3and\left|z_2\right|=\sqrt[]{10}$

$z_1-2z_2=3-4i\therefore arg\left(z_1-2z_2\right)=-ta{n}^{-1}\frac{4}{3}\Rightarrow \left|z_1-2z_2\right|=\left|2+4i+1-3i\right|=\sqrt[]{10}$

x = t² – t + 1        … (1)

y = t² + t + 1    … (2)

y – x = 2t & x + y = 2 (t² + 1)

__________on eliminating 't' we get

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

${\left(x-y\right)}^2=2\left(x+y-2\right)$

Axis : x – y = 0

Tangent at vertex : x + y – 2 = 0

Vertex : (1, 1) = (x, y)

$\frac{3z_1+i}{4}=\frac{2z_2+2z_3}{4}$

P = point of intersection of AD & BC

$AD=1\Rightarrow DP=\frac{3}{4}$

$BP=\sqrt[]{1-\frac{9}{16}}=\frac{\sqrt[]{7}}{4}\Rightarrow BC=\frac{\sqrt[]{7}}{2}$

area of Quad. ABCD = $\frac{1}{2}.AD*BC$

= $\frac{1}{2}*1*\frac{\sqrt[]{7}}{2}=\frac{\sqrt[]{7}}{4}$

EHINTZ

words starting with E is 5!

words starting with H is 5!

words starting with I is 5!

words starting with N is 5!

words starting with T is 5!