Class 12th

$E_9=\frac{hc}{\lambda }=\frac{1242}{\lambda \left(nm\right)}=\frac{1242}{400}=3.105$

Each wile has resistance = $\frac{\rho 4\mathcal{l}}{\pi d^2}=r$ $\frac{}{{}^{}}$

Eight wire in parallel, then equivalent resistance is

$\frac{r}{8}=\frac{\rho \mathcal{l}}{2\pi d^2}$

Single copper wire of length $2\mathcal{l}$ $$ has resistance

$R=\rho \frac{2\mathcal{l}*4}{\pi d_1^2}=\frac{\rho \mathcal{l}}{2\pi d^2}$

d1 = 4d

No. of electric field line per unit area = electric field

$E=\frac{\rho r}{3{\in }_0}forr=R$

$E=45*10^{10}N/C$

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

Given : ${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2\&\overrightarrow{a}\cdot \overrightarrow{b}=3$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta =3$

$\left|\overrightarrow{a}\right|cos\theta =\frac{3}{\sqrt[]{6}}=\sqrt[]{\frac{9}{6}}=\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta =75$

$\left|\overrightarrow{a}\right|sin\theta =\sqrt[]{\frac{75}{6}}=\frac{5}{\sqrt[]{2}}$

$\left|\overrightarrow{a}\right|cos\theta =\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2=\frac{25}{2}+\frac{3}{2}=\frac{28}{2}=14$

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$               

y = 22x $-\frac{1939}{27}$  which is not tangent to the curve.

$3a^2-2a+1=22$ (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$               

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

 ⇒ y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, $a^3-a^2+a=b$  

For a = $\frac{-7}{3}$  

b = $\frac{-343}{27}-\frac{49}{9}-\frac{7}{3}$  

= $\frac{-343-147-63}{27}=\frac{-553}{27}$  

$f\left(x\right)=4\left|2x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right],-20<x<20$ doubtful points for differentiability : $x=\frac{-3}{2},$

$f\left(x\right)=4\left(2x+3\right)+9\left(-1\right)-12\left(-2\right)-240=8x-213forx=\frac{-3}{2}+h$

= $-8x-230$ for x = $\frac{-3}{2}-h$

Not diff. at x = $\frac{-3}{2}$

other doubtful points : $x+\frac{1}{2}=integer$

$-20+\frac{1}{2}<x+\frac{1}{2}<20+\frac{1}{2}$

$x+\frac{1}{2}=-19,-18,....,19,20$

$x=-19.5,-18.5,-17.5,.......,18.5,19.5\to$ total 40 numbers.

No. of number = 19.5 – $\left(-19.5\right)+1=40\left(-1.5\right)included$

$-20<x<90\Rightarrow x=-19,-18,.....,18,15\to 39points$

No. of number = 19 (19) + 1 = 39

Total : 40 + 39 = 79

$r{\cdot }^n\text{?}{C}_r=n{\cdot }^{n-1}\text{?}{C}_{r-1}$

${\displaystyle \sum _{k=1}^{10}{\left(k{\cdot }^{10}\text{?}{C}_k\right)}^2}={\displaystyle \sum _{k=1}^{10}{\left(10{\cdot }^9\text{?}{C}_{K-1}\right)}^2}$

= $100{\cdot }^{18}\text{?}{C}_9$

22000L = $100{\cdot }^{18}\text{?}{C}_9$

L =

$18!=2^{16}\cdot 3^8\cdot 5^3\cdot 7^2\cdot 11^1\cdot 13\cdot 17$

$9+4+2+1$ $9!=2^7\cdot 3^4\cdot 5^1\cdot 7^1$

6 + 24 + 2 + 1 $\frac{18!}{{\left(9!\right)}^2}=2^2\cdot 5\cdot 11\cdot 13\cdot 17$

3 +3 + 1

= 221