Let the solution curve y = y(x) of the differential equation (4 + x²)dy – 2x(x² + 3y + 4)dx = 0 pass through the origin. They y(2) is equal to

Let base = b

$tan60°=\frac{h}{b}$

$tan30°=\frac{h-20}{b}$

$\underset{x\to 1}{lim}\frac{sin\left(3x^2-4x+1\right)-x^2+1}{2x^3-7x^2+ax+b}=-2$

For this limit to be defined 2x³ – 7x² + ax + b should also trend to 0 or x ® 1.

$\therefore 2.{\left(1\right)}^3-7{\left(1\right)}^2+a\cdot 1+b=0$

 2 – 7 + (a + b) = 0

(a + b) = 5 …………….(i)

Now this becomes % form  $\therefore$ we apply L’lopital rule

$\underset{x\to 1}{lim}\frac{\left(3x^2-4x+1\right)-x^2+1}{2x^3-7x^2+ax+b}=\underset{x\to 1}{lim}\frac{cos\left(3x^2-4x+1\right)\left(6x-4\right)-2x}{6x^2-14x+a}$

Now the numerator again ® 0 as x = 1

$\therefore$  6x² – 14x + a ® 0 as x = 1

6 . (1)² – 14 + a = 0

a = 8 …………….(ii)

a +

$?X=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore X^2=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore Y=\alpha l+\beta X+\gamma X^2=\left[\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill \gamma \hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right]$

$\begin{array}{l}\therefore \alpha =5,\beta =10,y=15\\ \therefore {\left(\alpha -\beta +y\right)}^2=100\end{array}$

24. Let P (n) be the statement “ 2n+7< (n+3)2”

ofn=1

P (1): 2 $\times 1+7<{(1+3)}^2$

$9<4^2$

9<16 which is true. This P (1) is true.

Suppose P (k) is true.

P (k)= 2k+7< (k+3)^{2   . (1)}              

Lets prove that P (k +1) is also true.

“ 2 (k + 1) + 7 < (k + 4)²=k²+ 8k + 16”

P (k +1) = 2 (k +1) +7 = (2k +7) +2

  < (k +3)²+ 2  (

23. Let $P(n):41^n-14^n\mathrm{is\; a\; multiple\; of\; 27.}$

Put n= 1,

$P(1)=41-14=27$ is a multiple of 27.

Which is true.

Assume that P(k) is true for some natural no. k.

P(k)= $41^k-14^k$ be a multiple of 27

i.e, $41^k-14^k=27a,a\in z$

$\Rightarrow 41^k=27a+14^k$ (1)

We want to prove that P(k+1) is also true.

Now,

$P(k+1)=41^{k+1}-14^{k+1}$

$=41^k\cdot 41-14\cdot 14^k$

$=41\left(27a+14^k\right)-14\cdot 14^k$ (Using 1)

$=41\times 27a+41\cdot 14^k-14\cdot 14^k$

$=-27\left(24+a+14^k\right)=41\times 27a+14^k(41-14)$

$=41\times 27a+27\cdot 14^k$

$=27\left(41a+14^k\right)$

$=27b,whereb=\left(41a+14^k\right)\in z$

$\Rightarrow 41^{k+1}-14^{k+1}\mathrm{is\; multiple\; of\; 2}7$

$\Rightarrow P(k+1)$ is true when P(k) is true.

Hence, by P.M.I. P(n) is true for every po