Maths NCERT Exemplar Solutions Class 12th Chapter Six

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

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

$\therefore Areaof\Delta PQR=\frac{\text{exception 25:}}{4}{\mathcal{l}}^2=$=25/2√3

$\left|x^2-9\right|=3$

$\Rightarrow x=\pm 2\sqrt[]{3},\pm \sqrt[]{6}$

Required area = A

$\frac{A}{2}={\displaystyle \underset{0}{\overset{\sqrt[]{6}}{\int }}\left(9-x^2-3\right)dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{9+y}-\sqrt[]{9-y}\right)dy}}$

$A=16\sqrt[]{6}+32\sqrt[]{3}-72=8\left[2\sqrt[]{6}+4\sqrt[]{3}-9\right]$

Note : No option in the question paper is correct.

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

$\Rightarrow \frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}+\frac{n}{b}{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{b}{a}{\left(\frac{bx}{ay}\right)}^{n-1}$

$\frac{dy}{d{x}_{\left(a,b\right)}}=-\frac{b}{a}$

So line always touches the given curve.

 $f\left(x\right)=\left|x^2-3x-2\right|-x$

$=\left|\left(x-\frac{3-\sqrt[]{17}}{2}\right)\left(x-\frac{3+\sqrt[]{17}}{2}\right)\right|-x$

$\Rightarrow f\left(x\right)=\left[\begin{array}{cc}\hfill x^2-4x-2-1\le x\le \frac{3-\sqrt[]{17}}{2}\hfill & \hfill -x^2+2x+2\frac{3-\sqrt[]{17}}{2}<x\le 2\hfill \end{array}\right]$

absolute minimum $f\left(\frac{3-\sqrt[]{17}}{2}\right)=\frac{-3+\sqrt[]{17}}{2}$

absolute maximum = 3

$\therefore sum3+\frac{-3+\sqrt[]{17}}{2}=\frac{3+\sqrt[]{17}}{2}$

GOF is differentiable at x = 0

So R.H.D = L.H.D.

$\frac{d}{dx}\left(4e^x+k_2\right)=\frac{d}{dx}\left({\left(-\left|x+3\right|\right)}^2-k_1\left|x+3\right|\right)$                

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

= 2 (2e⁴ – 1)

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\text{exception 25:}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-$$\frac{1}{}$

$2021\equiv -2$ 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

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -8\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 3\\ -1\end{array}\right)$

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 15\hfill & \hfill -24\hfill & \hfill 27\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 3a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 9\\ -3\end{array}\right)$

$R_3-2R_1,R_2-5R_1$

for no solution

3a + 9 = 0 but $\frac{3}{2}-\frac{9b}{2}\ne 0$

$\Rightarrow a=-3\Rightarrow b\ne \frac{1}{3}$

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

Let z = x + iy

$\left|z-2\right|\le 1\Rightarrow \left|x-2+iy\right|\le 1$

(x – 2)² + y² $\le 1$

$z\left(1+i\right)+\overline{z}\left(1-i\right)\le 2$

$\left(x+iy\right)\left(1+i\right)+\left(x-iy\right)\left(1-i\right)\le 2$

$x+ix+iy-y+x-ix-iy-y\le 2$

x – y $\le 1$

PD will be least as CP – r = DP, general pts on circle with centre (2, 0) is (2 + r cos, 0 + r sin )

here r = 1, (2 + cos , sin ) now slope of CP is $\frac{4-0}{0-2}=-2$

tan = 2

so D point will be $\left(2-\frac{1}{\sqrt[]{5}},\frac{2}{\sqrt[]{5}}\right)AP$ will be the greatest. A(1, 0)

now $\left|z_1^2\right|+\left|z_2^2\right|$

$={\left|1+0i\right|}^2+{\left|2-\frac{1}{\sqrt[]{5}}+\frac{2i}{\sqrt[]{5}}\right|}^2$

$=1+{\left(2-\frac{1}{\sqrt[]{5}}\right)}^2+\frac{4}{5}$

$=6-\frac{4}{\sqrt[]{5}}$

now $5\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)=\alpha +\beta \sqrt[]{5}$

$5\left(6-\frac{4}{\sqrt[]{5}}\right)=\alpha +\beta \sqrt[]{5}$

= 30, = 4

+ = 26