Ncert Solutions Maths class 12th

Surface area, S = 4pr²

$?\frac{ds}{dt}=4?.2r\frac{dr}{dt}=8?\frac{dr}{dt}=constant=k\left(say\right)$                

$?\frac{ds}{dt}=k?s=kt+c$

$?4?r^2=kt+c$        

Initially t = 0, r = 3

c = 36 p

When t = 5, r = 7, k = 32p

When t = 9, r = r, r = 9

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

$\therefore$ Required possibilities =

${=}^4{C}_4{*}^6{C}_4{\left(\frac{3}{4}\right)}^4*{\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^2{+}^4{C}_3{*}^6{C}_5{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^5*\frac{1}{4}*\frac{3}{4}$ ${+}^4{C}_2{*}^6{C}_6*{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^2*{\left(\frac{1}{4}\right)}^6$

$=\frac{27}{4^{10}}\left[15*27+24*3+2\right]=\frac{27*479}{4^{10}}$

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 $f_{\lambda }\left(x\right)\ge 0$

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

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

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

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

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

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

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

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

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

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

S. D. = $\frac{1}{\sqrt[]{3}}$

$b_1*b_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill \lambda \hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|=\widehat{i}\left(15-4\lambda \right)+\widehat{j}\left(\lambda -10\right)+\widehat{k}\left(5\right)$

$\frac{\left|\left(-\widehat{i}-2\widehat{j}-2\widehat{k}\right).\left\{\left(15-4\lambda \right)\widehat{i}+\left(\lambda -10\right)\widehat{j}+5\widehat{k}\right\}\right|}{\sqrt[]{{\left(15-4\lambda \right)}^2+{\left(\lambda -10\right)}^2+25}}=\frac{1}{\sqrt[]{3}}$

$\Rightarrow \lambda =16$

 $?$ x is a random variable.

$\begin{array}{l}\therefore k+2k+4k+6k+8k=1\\ \therefore k=\frac{1}{21}\end{array}$

$\therefore P\left(\left(1<x<4\right)|x\le 2\right)=\frac{4k}{7k}=\frac{4}{7}$

$f\left(x\right)=x^7-7x-2$

$f\text{'}\left(x\right)=7x^6-7\Rightarrow f\text{'}\left(x\right)=0,x=\pm 1$

$\therefore f\left(1\right)=<0andf\left(-1\right)>0$

Hence number of real roots of f (x) = 0 are 3.