Maths NCERT Exemplar Solutions Class 12th Chapter Three

  $f\left(x\right)+f\left(x+k\right)=n\forall x\in R,\&k>0............\left(i\right)$

Replace x by x + k.          

$f\left(x+k\right)+f\left(x+2k\right)=n...............\left(ii\right)$

From (i) & (ii), f(x + 2k) = f(x).

$\therefore f\left(x\right)$  is periodic with period = 2k.

$I_1={\displaystyle \underset{0}{\overset{4nk}{\int }}f\left(x\right)}dx=2x{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(x\right)dx}...................\left(ii\right)$

$I_2={\displaystyle \underset{-k}{\overset{3k}{\int }}f\left(x\right)}dx$ put x = t + k

$={\displaystyle \underset{-2k}{\overset{2k}{\int }}t\left(t+k\right)dt}=2{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(t+k\right)dt}$

$=2n{\displaystyle \underset{0}{\overset{2k}{\int }}ndx}=4n^{2}k.$

$co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

Draw y = cos²x and y = 2x + $\text{exception 25:}$

 $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]$

$A^2=$ $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]$

$=3\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]$

A² = 3A

 A³ = 3A²

A³ = 3²A

A⁴ = 3³A

Aⁿ = 3^n-1A

now, A² + A³ +….+A¹⁰

= 3A + 3² A +…. + 3⁹A

= 3A (1 + 3 +….+ 3⁸

$=3A\cdot \frac{\left(3^9-1\right)}{3-1}$

$=\frac{3^{10}-3}{2}A$

$=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill \delta \hfill \end{array}\right|=0\Rightarrow \delta =-3$

and ${\Delta }_1=\left|\begin{array}{ccc}\hfill 7\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill 4\hfill & \hfill -3\hfill \end{array}\right|=0\Rightarrow k=6$

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$=\frac{10+4{*}^{10}\text{?}{C}_2}{10^4}=\frac{10+\frac{4*10.9}{2}}{10^4}=\frac{19}{1000}$

A = $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$

$A^2=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill a^2+bc\hfill & \hfill ab+bd\hfill \\ \hfill ac+dc\hfill & \hfill ac+d^2\hfill \end{array}\right)$

a2 + bc = bc + d2 = 1 ………. (i)

and b (a + d) = c (a + d) = 0 ……… (ii)

Case 1

b = c = 0

then possible ordered pair of

(a, d) $\equiv$  (1, 1) (-1, -1) (-1, 1) (1, -1) total 4 possible case

Case 2

a = -d

then (a, d) $\equiv$  (-1, 1) (1, -1)

then bc = 0

now if b = 0

then possible choice for {-1, 0, 1, 2, …….10} = 12

Similarly if c = 0 then possible choice for $b\in \left\{-1,0,1,2,......10\right\}$ is = 12

but (0, 0) counted twice

$\therefore$ bc = 0 in (12 + 12 – 1) = 23 ways

$\therefore$ total number of ways = 2 * 23 = 46

$\therefore$ total number of required matrices = 46 + 4 = 50

$A\text{'}BA=\left[111\right]\left[\begin{array}{ccc}\hfill 9^2\hfill & \hfill -10^2\hfill & \hfill 11^2\hfill \\ \hfill 12^2\hfill & \hfill 13^2\hfill & \hfill -14^2\hfill \\ \hfill -15^2\hfill & \hfill 16^2\hfill & \hfill 17^2\hfill \end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=$

$\left[9^2+12^2-15^2-10^2+13^2+16^{2}11^2-14^2+17^2\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

$=\left[9^2+12^2-15^2-10^2+13^2+16^2+11^2-14^2+17^2\right]=\left[539\right]$

Given that $A^T=A,B^T=-B$

(A) $C=A^4-B^4$

$=A^4-B^4=C$

(B)C = AB – BA

$=B^T{A}^T-A^T{B}^T=-BA+AB=C$

(C) $C=B^5-A^5$

$C^T={\left(B^5-A^5\right)}^T={\left(B^5\right)}^T-{\left(A^5\right)}^T=-B^5-A^5$

(D)C = AB + BA

= BA – AB = C

$\therefore$ option is true