Maths NCERT Exemplar Solutions Class 12th Chapter One

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 = $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$

 $\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$

= cos x – 2 cos² x= $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

 $a=\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{2n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}}{\displaystyle \sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}}$

$\therefore a={\displaystyle {\int }_0^1\frac{2}{1+x^2}dx=2ta{n}^{-1}x}]{}_0^1=\frac{\pi }{2}$

$f\text{'}\left(\frac{a}{2}\right)=\sqrt[]{2}f\left(\frac{a}{2}\right)$

 $f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$

If f (x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$ ……. (i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$ ……. (ii)

From (i) and (ii) we get $b\in [-6,-2)\cup (2,6]$

 $f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x\le 0\hfill \\ \hfill \left|x-4\right|,\hfill & \hfill x>0\hfill \end{array}andg\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x-4\right)}^2+b,\hfill & \hfill x\ge 0\hfill \end{array}$

$?$ f (x) and g (x) are continuous on R $\therefore$ a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (2) = g (2) + f (-1) = -11 + 3 = -8

 $f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

$\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$

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

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

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

Since a is a odd natural number then $\left|{\displaystyle \underset{1}{\overset{3}{\int }}{y}^{a}dy}\right|=\frac{364}{3}\Rightarrow \left|{\left(\frac{y^{a+1}}{a+1}\right)}_1^3\right|=\frac{364}{3}\Rightarrow \frac{3^{a+1}}{a+1}=\frac{364}{3}$

a = 5

| (A + I) (adj A + I)| = 4 |A adj A + A + Adj A + I| = 4 | (A)I + A + adj A + I|= 4|A| = 1

|A + adj A| = 4

$A=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow adjA=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill -c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow \left|\begin{array}{cc}\hfill \left(a+d\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \left(a+d\right)\hfill \end{array}\right|=4\Rightarrow a+d=\pm 2$

 $\Delta =\left|\begin{array}{ccc}\hfill 8\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill \lambda \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|=12-3\lambda$

So for $\lambda$ = 4, it is having infinitely many solutions. ${\Delta }_x=\left|\begin{array}{ccc}\hfill -2\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \mu \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|$ = 6 $3\mu =0\Rightarrow -6-3\mu =0$

For $\mu =-2$ distance of $\left(4,-2,-\frac{1}{2}\right)$ from 8x + y + 4z + 2= 0 $\left|\frac{32-2-2+2}{\sqrt[]{64+1+16}}\right|=\frac{10}{3}$ units