Ncert Solutions Maths class 12th

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{x^{2}cosx}{1+{\pi }^2}+\frac{1+si{n}^{2}x}{1+e^{{\left(sinx\right)}^{2023}}}\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{x^{2}cosx}{1+{\pi }^x}+\frac{1+si{n}^{2}x}{1+e^{{\left(sinx\right)}^{2023}}}\right)+\left(\frac{x^{2}cosx}{1+{\pi }^{-x}}+\frac{1+si{n}^{2}x}{1+e^{-{\left(sinx\right)}^{2023}}}\right)\right\}dx}$

$=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(x^{2}cosx+1+si{n}^{2}x\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}cosxdx}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$ .(1)

Let $I_1={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx}$

$I_1={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}1\cdot dx}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{1-cos2x}{2}\right)dx}$

$I_1=\frac{\pi }{2}+\frac{1}{2}\left[\frac{\pi }{2}+0\right]$

$I_1=\frac{3\pi }{4}$

Let $I_2={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}cosxdx}$

$I_2={\left[x^2\left(sinx\right)-{\displaystyle \int 2x}{\displaystyle \int cosxdx}\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^2\left(sinx\right)-2{\displaystyle \int xsinx}\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^{2}sinx-2\left(x\left(-cosx\right)+{\displaystyle \int cosx}\right)\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^{2}sinx-2\left(-xcosx+sinx\right)\right]}_0^{\frac{\pi }{2}}$

$I_2=\left(\frac{{\pi }^2}{4}-2\right)$

Put l₁ and l₂ in (1)

$\frac{{\pi }^2}{4}-2+\frac{3\pi }{4}$

$\frac{{\pi }^2}{4}+\frac{3\pi }{4}-2$

$\frac{\pi }{4}\left(\pi +3\right)-2$

α = 3

P (2 obtained on even numbered toss) = k (let)

P (2) = $\frac{1}{6}$  

P (  $\overline{2}$ )= $\frac{5}{6}$  

$k=\frac{5}{6}*\frac{1}{6}+{\left(\frac{5}{6}\right)}^3*\frac{1}{6}+{\left(\frac{5}{6}\right)}^5*\frac{1}{6}+...$

$=\frac{\frac{5}{6}*\frac{1}{6}}{1-{\left(\frac{5}{6}\right)}^2}$

$=\frac{5}{11}$

$I=9{\displaystyle \underset{0}{\overset{9}{\int }}\left[\sqrt[]{\frac{10x}{x+1}}\right]dx}$

$=9\left[{\displaystyle \underset{0}{\overset{1/9}{\int }}0dx}+{\displaystyle \underset{1/9}{\overset{2/3}{\int }}dx}+{\displaystyle \underset{2/3}{\overset{9}{\int }}2dx}\right]$

$=9\left[\frac{2}{3}-\frac{1}{9}+2\left[9-\frac{2}{3}\right]\right]$

$=9\left[\frac{5}{9}+2*\frac{25}{3}\right]$

= 5 + 6 * 25

= 5 + 150

= 155

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  $=\frac{30}{11*11}=\frac{30}{121}$

Given $\left|\overrightarrow{a}\right|=1$ , $\left|\overrightarrow{b}\right|=4$ , $\overrightarrow{a}\cdot \overrightarrow{b}=2$

$\overrightarrow{c}=2\left(\overrightarrow{a}*\overrightarrow{b}\right)-3\overrightarrow{b}$  

Dot product with  $\overrightarrow{a}$ on both sides

$\overrightarrow{c}\cdot \overrightarrow{a}=-6$ . (1)

Dot product with  $\overrightarrow{b}$  on both sides

$\overrightarrow{b}\cdot \overrightarrow{c}=-48$ . (2)

$\overrightarrow{c}\cdot \overrightarrow{c}=4{\left|\overrightarrow{a}*\overrightarrow{b}\right|}^2+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[{\left|\overrightarrow{a}\right|}^2\cdot {\left|\overrightarrow{b}\right|}^2-{\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)}^2\right]+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[\left(1\right){\left(4\right)}^2-\left(4\right)\right]+9\left(16\right)$

${\left|\overrightarrow{c}\right|}^2=4\left[12\right]+144$

${\left|\overrightarrow{c}\right|}^2=48+144$

${\left|\overrightarrow{c}\right|}^2=192$

$cos\theta =\frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}$

$cos\theta =\frac{-48}{\sqrt[]{192}\cdot 4}$

$cos\theta =\frac{-48}{8\sqrt[]{3}\cdot 4}$

$cos\theta =\frac{-3}{2\sqrt[]{3}}$

$cos\theta =\frac{-\sqrt[]{3}}{2}$ $\theta =co{s}^{-1}\left(\frac{-\sqrt[]{3}}{2}\right)$

-> $-1\le \left|\begin{array}{cc}\hfill \overline{2-\left|x\right|}\hfill & \hfill 4\hfill \end{array}\right|\le 1$

-> $\left|\frac{2-\left|x\right|}{4}\right|\le 1$

$-1\le \frac{2-\left|x\right|}{4}\le 1$

–4 £ 2 – |x| £ 4

–6 £ – |x| £ 2

–2 £ |x| £ 6

|x| £ 6

->x Î [–6, 6]              …(1)

Now, 3 – x ¹ 1

And x ¹ 2                    …(2)

and 3 – x > 0

x < 3                            (3)

From (1), (2) and (3)

->x Î [–6, 3] – {2}

a = 6

b = 3

g = 2

a + b + g = 11

$f\text{'}\left(x\right)=\frac{g\text{'}\left(x\right)-g\text{'}\left(2-x\right)}{2},f\text{'}\left(\frac{3}{2}\right)=\frac{g\text{'}\left(\frac{3}{2}\right)-g\text{'}\left(\frac{1}{2}\right)}{2}=0$  

Also  $f\text{'}\left(\frac{1}{2}\right)=\frac{g\text{'}\left(\frac{1}{2}\right)-g\text{'}\left(\frac{3}{2}\right)}{2}=0$ , f' (1) = 0

-> $f\text{'}\left(\frac{3}{2}\right)=f\text{'}\left(\frac{1}{2}\right)=0$  

->roots in $\left(\frac{1}{2},1\right)$ and $\left(1,\frac{3}{2}\right)$  

->f" (x) is zero at least twice in $\left(\frac{1}{2},\frac{3}{2}\right)$

(a – 1) * 2 + (b – 2) * 5 + (g – 3) * 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$\lambda =\frac{1}{10}$  

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$=8\lambda +5=\frac{8}{10}+5=5.8$

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

$gof\left(x\right)=\{\begin{array}{cc}\hfill f\left(x\right),\hfill & \hfill f\left(x\right)<0\hfill \\ \hfill f\left(x\right),\hfill & \hfill f\left(x\right)>0\hfill \end{array}$

$=\{\begin{array}{lll}{e}^{lnx}=x\hfill & \left(0,1\right)\hfill & e^{-x}\hfill \\ \left(-\infty ,0\right)\hfill & lnx\hfill & \left(1,\infty \right)\hfill \end{array}$

Therefore, gof (x) is many one and into