Maths NCERT Exemplar Solutions Class 12th Chapter One

 $\frac{7}{2}\left(1+cos2\theta \right)-\frac{3}{2}\left(1-cos2\theta \right)-2co{s}^{2}2\theta =2$

Put cos 2 $\theta$= t

Equation 2t²– 5t = 0, t (2t – 5) = 0

$t=0,\frac{5}{2}$

cos 20 = 0, 0 < 2 < 4

$x_1+x_2=2\left(ta{n}^2\theta +co{t}^2\theta \right)\Rightarrow ?$

$=2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)=16$

 $\frac{1}{\left(20-a\right)\left(40-a\right)}+\frac{1}{\left(40-a\right)\left(60-a\right)}+\frac{1}{\left(60-a\right)\left(80-a\right)}+.....+\frac{1}{\left(180-a\right)\left(200-a\right)}=\frac{1}{256}$

LHS = $\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{40-a}\right)+\frac{1}{20}\left(\frac{1}{40-a}-\frac{1}{60-a}\right)+...+\frac{1}{20}\left(\frac{1}{180-a}-\frac{1}{200-a}\right)$

$=\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{20}.\frac{180}{\left(20-a\right)\left(200-a\right)}$

$\Rightarrow a^2-220a+4000-2304=0\Rightarrow a^{2}220a+1696=0$

$a=\frac{220\pm 204}{8}=212^2$

$f\left(x\right)={\int }_1^3?\frac{\sqrt[]{x}dx}{(1+x{)}^2}={\int }_1^{\sqrt[]{3}}?\frac{t.2tdt}{{\left(1+t^2\right)}^2}\mathrm{}$ (put $\sqrt[]{x}=t$ )

$={\left(-\frac{\mathrm{t}}{1+{\mathrm{t}}^2}\right)}_1^{\sqrt[]{3}}+{\left({\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\mathrm{t}\right)}_1^{\sqrt[]{3}}\mathrm{}$ [Applying by parts]

$\begin{array}{rr}& =-\left(\frac{\sqrt[]{3}}{4}-\frac{1}{2}\right)+\frac{\pi }{3}-\frac{\pi }{4}\\ & =\frac{1}{2}-\frac{\sqrt[]{3}}{4}+\frac{\pi }{12}\end{array}$

. $?f\left(n\right)=\{\begin{array}{c}\hfill 2n,n=1,2,3,4,5\hfill \end{array}$

$\begin{array}{c}\hfill \mathrm{ \; \; \; \; \; \; \; \; \; \; 2}n-11,n=6,7,8,9,10\hfill \end{array}$

$\therefore f\left(1\right)=2,f\left(2\right)=4,....,f\left(5\right)=10$

And f(6) = 1, f(7) = 3, f(8) = 5, …., f(10) = 9

Now,

$f\left(g\left(n\right)\right)=\{\begin{array}{cc}\hfill n+1\hfill & \hfill ifnisodd\hfill \\ \hfill n-1,\hfill & \hfill ifniseven\hfill \end{array}$

$\therefore f\left(g\left(10\right)\right)=9\Rightarrow g\left(1\right)=1$

$\therefore g\left(10\right)\left(g\left(1\right)+g\left(2\right)+g\left(3\right)+g\left(4\right)+g\left(5\right)\right)=190$

$S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+........(i)$  

$\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+........(ii)$               

(i) - (ii)

$\frac{6}{7}S=2+\frac{4}{7}=\frac{6}{7^2}+\frac{8}{7^3}+...........(iii)$               

$\frac{6}{7^2}S=\frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+........(iv)$              

(iii) – (iv)

$=2+\left[\frac{1}{1-\frac{1}{7}}\right]=2\left(\frac{7}{6}\right)$

$\therefore 4S=8{\left(\frac{7}{6}\right)}^3={\left(\frac{7}{3}\right)}^3$      

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\text{'}+\text{'}isabinaryoperationonthesetNbutithasnoidentityelement.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Onlybijectivefunctionsareinvertible.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Letf(x)=2x,g(x)=x-1andh(x)=2x+3\\ fo\{goh(x)\}=fo\{g(2x+3)\}\\ =f(2x+3-1)=f(2x+2)=2(2x+2)=4x+4\\ and(fog)oh(x)=(fog)\{h(x)\}\\ =fog(2x+3)\\ =f(2x+3-1)=f(2x+2)=2(2x+2)=4x+4\\ \therefore fo\{goh(x)\}=(fog)oh(x)=4x+4\\ Hence,thestatementisTrue.\end{array}$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

This is a True or False Type Question as classified in NCERT Exemplar

Sol: