Ncert Solutions Maths class 12th

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) =  $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

-> b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$        

Δ ≠ 0
| 1 -λ -1 |
| λ -1 -1 | ≠ 0
| 1 -1 |

1 (1+λ) + λ (-λ+1) - 1 (λ+1) ≠ 0
1 + λ - λ² + λ - λ - 1 ≠ 0
-λ² + λ ≠ 0
λ² = 1 ⇒ λ = 1, -1

a² + b² = 2

p (10hr) = .1
p (7hr) = .2
p (4hr) = .7

p (s / 10hr) = .8
p (s / 7hr) = .6
p (s / 4hr) = .4

p (s) = .1 * .8 + .2 * .6 + .7 * .4
p (4hr / s) = (.7 * .4) / (.1 * .8 + .2 * .6 + .7 * .4) = 28 / (8 + 12 + 28) = 28 / 48 = 7 / 12

$\lambda =1+\mu$

$\begin{array}{l}\underset{\_}{2\lambda =3\lambda \mu }\\ \lambda =2\to S\left(2,4,6\right)\end{array}$

$\mu =1$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right|=\left(7,-2,-1\right)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

g (1) =?

$=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$

$f\left(1\right)={\left(2\left(1-\frac{1}{2}\right)\left(2+1\right)\right)}^{\frac{1}{50}}=3^{\frac{1}{50}}$

$3^{\frac{1}{50}}+13^{\frac{1}{50}}>1^{\frac{1}{50}}$

3 > 1

$3^{\frac{1}{50}}>2$   $2>3^{\frac{1}{50}}>1$

$3<2^{50}$   $\left[3^{\frac{1}{50}}+1\right]=1+1=2$

  $\frac{2}{-4}=\frac{-1}{2}$  

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

$5x^2-8x-21=0$  

$z\left(\frac{-7}{5},\frac{-24}{5}\right)$  

$limPQ:\frac{x-1}{1}=\frac{y-0}{1}=\frac{3-1}{1}=\lambda$  

              $M\left(1+\lambda ,\lambda ,1+\lambda \right)$

x + y + z = 5

$\Rightarrow \lambda =1$  

M (2, 1, 2)

Q (3, 2, 3)

$L:\frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}=t$  

 R (3, 1, 1), QR² = 1 + 4 = 5

$P\left(E_1/E_2\right)=\frac{1}{2},\Rightarrow \frac{P\left(E_1\cap E_2\right)}{P\left(E_2\right)}=\frac{1}{2}$

$P\left(E_2/E_1\right)=\frac{3}{4}$ $P\left(E_1\right)=\frac{1}{6}$

$P\left(E_1\cap E_2\right)=\frac{1}{8}$ $P\left(E_2\right)=\frac{1}{4}$

$P\left(E_1\cup E_2\right)=\frac{1}{6}+\frac{1}{4}-\frac{1}{8}=\frac{4+6-3}{24}=\frac{7}{24}$

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

$\Rightarrow 11x+y+17z+38=0$           

Any point on line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$