Ncert Solutions Maths class 11th

$\left\{a\in \left(1,2,3,......,100\right):HCF\left(a,24\right)=1\right\}$

HCF of (a, 24) = 1 $\therefore$ a = 1, 5, 7, 11, 13, 17, 19, 23 sum of these numbers = 96

$\therefore$ There are four such blocks and a number 97 is there upto 100.

$\therefore$ complete sum = 96 + (24 * 8 + 96) + (48 * 8 + 96) + (72 * 8 + 96) + 97 = 1633

Sum of all given numbers = 31

Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.

$\therefore$ Hence 11 is possible.

This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.

Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.

$\therefore$ Total possible ways = (4! * 3!) * 4 = 576

 $S=\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right),a,b\in 1,2,3,.....100$

$\therefore A=\left(\begin{array}{cc}\hfill -1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right)$ then even power of A as A = $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$

If b = 1 & $a\in \left\{1,2,3.....100\right\}$ and n (n + 1) is always even

$\therefore T_1,T_2,T_3,........,T_n$ are all 1 for b = 1 and each value of a.

$\therefore {\displaystyle \underset{n=1}{\overset{1000}{\cap }}{T}_n=100}$

$?$ given statement is

$\left(A\wedge C\right)\to B$ then its negation is $\left\{\left(A\wedge C\right)\to B\right\}$

$cos\left(x+\frac{\pi }{3}\right)cos\left(\frac{\pi }{3}-x\right)=\frac{1}{4}co{s}^{2}2x$

$x=-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi$

$\therefore$ total number of solution = 7.

$\Delta =4\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill \alpha \hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill 1\hfill \end{array}\right|=4$

$\Rightarrow \alpha =\pm 8$

$\left(\alpha ,-\alpha \right),\left(-\alpha ,\alpha \right)and\left({\alpha }^2,\beta \right)$ are collinear.

$\therefore \left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill -\alpha \hfill & \hfill 1\hfill \\ \hfill -\alpha \hfill & \hfill \alpha \hfill & \hfill 1\hfill \\ \hfill {\alpha }^2\hfill & \hfill \beta \hfill & \hfill 1\hfill \end{array}\right|=0\Rightarrow \beta =-64$

 $\frac{x^2}{a^2}+\frac{y^2}{4}=1$

$\therefore \Delta =\frac{1}{2}*a\left(1+cos\theta \right).4sin\theta$

$\therefore {\Delta }_{max}^{\Delta }=6\sqrt[]{3}\Rightarrow a=4$

$\therefore e=\frac{\sqrt[]{3}}{2}$

Equation of tangent at P (x, y) is Y = $\frac{dy}{dx}\left(X-x\right)$

$A/q,2x-y\frac{dx}{dy}=0\Rightarrow 2\frac{dy}{y}=\frac{dx}{x}$

$\begin{array}{l}\Rightarrow 2\mathcal{l}ny=\mathcal{l}nx+\mathcal{l}nc\\ y^2=xc\end{array}$

It passes through (3, 3), c = 3

$\therefore y^2=3x\therefore$ Length of latus rectum = 3

 $x_{1}y>0and{x}^3{y}^2=2^{15}$

$AM\ge GM$

$\frac{3x+2y}{5}\ge {\left(x^3{y}^2\right)}^{\frac{1}{5}}$

$\Rightarrow 3x+2y\ge 40$

$\tilde{z}=i{z}^2+z^2-z$

$z+\overline{Z}=z^2\left(i+1\right)$

$z+\overline{Z}=z^2\left(i+1\right)$ Let z be equal to (x + iy)

(x + iy) + (x – iy) = (x + iy)² (i + 1)

$2x=\left(x^2-y^2+2ixy\right)\left(i+1\right)$               

Equating the real & in eg part.

$\left(x^2-y^2+2ixy\right)=0.........\left(i\right)$

$\left(x^2-y^2-2xy\right)=\left(2x\right)..............\left(ii\right)$               

(i) & (ii)

 4xy = -2x Þ x = 0 or y = $\left(\frac{-1}{2}\right)$  

(for x = 0, y = 0)

For y = $\frac{-1}{2}$  

x²   $-\frac{1}{4}+2\left(\frac{-1}{2}\right)x=0$

x =   $\frac{4\pm \sqrt[]{16+16}}{2.4}$

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

$\therefore sum$ of   ${\left|z\right|}^2={\left(\frac{1+\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+{\left(\frac{1-\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+0^2+O^2$

=   $\frac{3}{4}+\frac{\sqrt[]{2}}{2}+\frac{1}{4}+\frac{3}{4}-\frac{\sqrt[]{2}}{2}+\frac{1}{4}=\frac{3}{2}+\frac{1}{2}=2$