Ncert Solutions Maths class 12th

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$              

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$  

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute =

$\therefore In\Delta AQP$

$tan30°=\frac{PQ}{AQ}$

$\frac{1}{\sqrt[]{3}}=\frac{h}{x+y}$

x + y = $\sqrt[]{3}h.....\left(i\right)$ $\text{exception 25:}$

$\therefore ln\Delta BQP$

$tan45°=\frac{h}{y}$

h = y . (ii)

(i) & (ii) x + y = $\sqrt[]{3}y$ $\text{exception 25:}$

$\Rightarrow x=\left(\sqrt[]{3}-1\right)y.......\left(iii\right)$

Let the speed be S

$\frac{x}{S}=20$

x = 20.S

from (iii)

$\frac{y}{S}=10\left(\sqrt[]{3}+1\right)$

(2 – i) z = (2 + i) $\overline{z}$ $\stackrel{}{}$ , put z = x + iy

$y=\frac{x}{2}........\left(i\right)$

(ii) $\left(2+i\right)z+\left(i-2\right)\overline{z}-4i=0$ $\stackrel{}{}$

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$ $\stackrel{}{}$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x=1.y=\frac{1}{2}\Rightarrow centre\left(1,\frac{1}{2}\right)$

Perpendicular distance of point $\left(1,\frac{1}{2}\right)$ $\frac{}{}$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$ $\frac{\frac{}{}}{\text{exception 25:}}$

$r=\frac{3}{2\sqrt[]{2}}$

Let

A : Missile hit the target

B : Missile intercepted

P (B) = $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$ $\frac{}{}\stackrel{}{}\frac{}{}$

$P\left(\overline{B}\right)=\frac{2}{3}$

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}*\frac{2}{3}=\frac{1}{2}$

Required Probability = $\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}=\frac{1}{8}$

$\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r . (i)

b = 2 + 10r. (ii)

a = 18r – 2 . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = 3

$$ $\therefore$  a = 14 + 14 (-3) = -56 and b = -2 30 = 32

$\left|a+b\right|\left|-56-32\right|=88$

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$           

So at least one root will lie in (-2, -1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$  

$=10\left[x^4+2x^3+3x^2+2x+1\right]$            

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$           

So, f(x) be purely increasing function so exactly one root of f(x) that will lie in (-2, 1). Hence |a| = 2

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$  

$P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                             

Now quadratic equation having roots a & b will be x² – (a + b)x + ab = 0

i.e.         x² – x – 1 = 0,     put x = a and put x = b

So          a² = a + 1           & b² = b + 1

(i)  $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$    

$P_n=P_{n-1}+P_{n-2}$                        

-> ${P_n}^2=234$

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

-> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$  

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$