Ncert Solutions Maths class 12th

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$  

$l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$        .(A)

Put   $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$            .(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$  

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$    .(ii)

Adding (i) and (ii) we get

  $2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$         

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

Let f(x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$                   Non zero finite

So, d = e = f = 0

f(x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$          Non zero finite

f'(x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$  

f'(1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 .(i)

f'(-1) = 0

-6 + 5a – 4b + 3 = 0 .(ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$  

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$      

-> $\overrightarrow{r}*\overrightarrow{a}=\overrightarrow{c}*\overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$  

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$        

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$           

Critical point of function are x = $\frac{-1}{2},$ 1 and -2 but x = -2 is making zero .

$\therefore nondifferentiableatx=\frac{-1}{2},1$      

P {a person chosen from the group has chest disorder}

$=\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10$

$P\left(\frac{Personissmoker\&non-vegetarian}{Personhaschestdisorder}\right)$

$=\frac{\frac{160}{400}*0.35}{\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10}$

$=\frac{40*0.7}{40*0.7+25*0.4+35*0.2}=\frac{23}{23+10+7}=\frac{28}{45}$

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$  

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability =  $\frac{5}{216}$   

Let sinθ= t ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$ dθ

sinθ = t

cos θ. dθ = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

$x={\displaystyle \sum _{n=0}^{\infty }co{s}^{2n}\theta =co{s}^0\theta +co{s}^2\theta +co{s}^4\theta +.....=1+co{s}^2\theta +co{s}^4\theta +.....}$  

a = 1, r = cos² θ

$x=S_{\infty }=\frac{a}{1-r}=\frac{1}{1-co{s}^2\theta }=\frac{1}{si{n}^2\theta }$           

Similarly, y = $\frac{1}{co{s}^2\theta }\Rightarrow \frac{1}{y}=co{s}^2\theta$  

$z=\frac{xy}{xy-1}\Rightarrow xyz-z=xy...........\left(i\right)$           

Also, $\frac{1}{x}+\frac{1}{y}=1\Rightarrow x+y=xy..........\left(ii\right)$  

 (i) & (ii) ->xyz = xy + z -> (x + y) z = xy + z