Maths NCERT Exemplar Solutions Class 12th Chapter Three

Circle $\left|z-3\right|\le 1\Rightarrow {\left(x-3\right)}^2+y^2\le 1$

and line $z\left(4+3i\right)+\overline{z}\left(4-3i\right)\le 24$

$\Rightarrow 4x-3y\le 12\therefore slope=tan\theta =\frac{4}{3}$

$?f_{\lambda }\left(x\right)=4\lambda x^3-36\lambda x^2+36x+48$

$f_{\lambda }\left(x\right)=12\left(\lambda x^2-6\lambda x+3\right)$

For increasing $f_{\lambda }\left(x\right)\ge 0$

$f_{\lambda }^{*}\left(x\right)=\frac{4}{3}{x}^3-12x^2+36x+48\therefore f_{\lambda }^{*}\left(1\right)+f_{\lambda }^{*}\left(-1\right)=73\frac{1}{2}-1\frac{1}{2}=72$

 $?-\frac{dx}{dy}=\frac{x^2}{xy-x^2{y}^2-1}$

$\therefore \frac{dy}{dx}=\frac{xy-x^2{y}^2-1}{x^2}$

$Letxy=v\Rightarrow x\frac{dy}{dx}+y=\frac{dv}{dx}$

Put x = 1, y = 1 $\Rightarrow ta{n}^{-1}=c\Rightarrow c=\frac{\pi }{4}$

$\therefore ta{n}^{-1}\left(xy\right)=\mathcal{l}nx=\frac{\pi }{4}$

$e\left(y\left(e\right)\right)=tan\left(1+\frac{\pi }{4}\right)=\frac{tan1+1}{1-tan1}$

$?$ given statement is

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

$Let{x}^3=\theta \Rightarrow \frac{\theta }{2}\in \left(\frac{\pi }{4},\frac{3\pi }{4}\right)$

$\therefore y=ta{n}^{-1}\left(sec\theta -tan\theta \right)$

$ta{n}^{-1}\left(\frac{1-sin\theta }{cos\theta }\right)$

$\Rightarrow \frac{dy}{dx}=\frac{-3x^2}{2}\Rightarrow \frac{d^{2}y}{d{x}^2}=-3x$

$\therefore x^2\frac{d^{2}y}{d{x}^2}-6y+\frac{3\pi }{2}=0\Rightarrow x^2{y}^{11}-6y+\frac{3\pi }{2}=0$

$\left|\overrightarrow{a}+\overrightarrow{b}+2\left(\overrightarrow{a}*\overrightarrow{b}\right)\right|=2,\theta \in \left(0,\pi \right)$

squaring on both sides, we get = $\frac{2\pi }{3}$

where θ is angle between $\widehat{a}and\widehat{b}.$

$2\left|\widehat{a}*\widehat{b}\right|=\sqrt[]{3}=\left|\widehat{a}-\widehat{b}\right|,S_1$ is correct.

and projection of $\widehat{a}on\widehat{a}+\widehat{b}=\left|\frac{\widehat{a}.\left(\widehat{a}+\widehat{b}\right)}{\left|\widehat{a}+\widehat{b}\right|}\right|=\frac{1}{2}$

So (S₂) is correct.

Let P (x, y, z) be any point on plane P₁ then

${\left(x+4\right)}^2+{\left(y-2\right)}^2+{\left(z-1\right)}^2={\left(x-2\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2$

$\therefore cos\theta =\frac{\left|6-2+3\right|}{14}\Rightarrow \theta =\frac{\pi }{3}$

  $\left(1,1\right)\left(1,4\right)\left(4,1\right)\left(2,4\right)\left(4,2\right)\left(3,4\right)\left(4,3\right)\left(4,4\right)$  all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$\therefore$ total function = 3 * 3 * 2 * 2 * 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 * 2 * 1 * 1 * 1 = 4

Case II

None of the pre images have 2 as image then number of function = 2⁵ = 32

Case III

None of the pre image have either 3 or 2 as image

Total function = 1⁵ = 1

$\therefore$ Total number of onto function

= 72 – 4 – 32 + 1 = 37

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

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

$z+\overline{Z}=z^2\left(i+1\right)$ 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   ${\left|z\right|}^2={\left(\frac{1+\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+{\left(\frac{1-\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+0^2+O^2$

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