Principle of Mathematical Induction

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 + b = 5  $\therefore a-b=8-\left(-3\right)=11$       

(b = -3) ® from (i) & (ii)

 $\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$

$\Rightarrow \frac{dy}{dx}=\left(\frac{6x}{x^2+4}\right)y+2x$

$e^{-3\mathcal{l}n\left(x^2+4\right)}=\frac{1}{{\left(x^2+4\right)}^3}$

So

$\frac{y}{{\left(x^2+4\right)}^3}={\displaystyle \int \frac{2x}{{\left(x^2+4\right)}^3}dx+c}$

$\Rightarrow y=-\frac{1}{2}\left(x^2+4\right)+c{\left(x^2+4\right)}^3$

When x = 0, y = 0 gives

$c=\frac{1}{32}$

So, for x = 2, y = 12

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

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

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

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

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

ofn=1

P (1): 2 $*1+7<{(1+3)}^2$

$9<4^2$

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  (Using 1)

= k²+ 9 + 6k +2 = k²+6k +11

Adding and subtracting (2k + k) in the R. H. S.

$=k^2+6k+11+2k+5-(2k-5)$

$=\left(k^2-8k+16\right)-(2k-5)$

$={(k+4)}^2-(2k-5)$

$<{(k+4)}^2\mathrm{,\; since\; 2}k+5>0$ for all$k\in N$

$\Rightarrow P(k+1)$ is true.

$\therefore$ By the principle of mathematical induction, P (n)is true for all n $\in$ N.

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)

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

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

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

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

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

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

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

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

22. Let P(n): $3^{2n+2}-8n-9$ is divisible by 8

put n= 1,

P(1): $3^{2+2}-8.1-9$

3⁴ – 8 – 9 = 81– 17 = 64= is divisible by 8

Which is true.

Assume that P(k) is true for some natural numbers k.

i.e, $3^{2k+2}-8k-9$ be divisible by 8

$3^{2k+2}-8k-9=8a$ where,a $\in z$

$3^{2k+2}=8a+8k+9$ (1)

We want to prove thatP(k+ 1) is true.

$P(k+1):3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8, is also true.

Now,

${3^{2k+2+}}^2-=8(k+1)-9$

= $3^{2k+4}-8k+8-9$

$3^{(2k+2)+2}-8k+8-9$

3^{(2k +2)}. 3² $-$ 8k $-$ 17

$=9(8a+8k+9)-8k-17$ (Using 1)

$=72a+72k+81-8k-17$

= 72a + 64k+ 64 = 8(9a + 8k + 8)

= 8b, where b = 9a + 8b + 8 $a\in z$

$\Rightarrow$ 3^{2k + 4}– 8(k+1) – 9 is divisible by 8.

$\therefore$ P(k+1) is true when P(k) is true. Hence, By P.M.I. P(n) is true for all positive integer n.

21. Let 

$\mathrm{Putting\; x}=1,$

$P(1)=x^2-y^2\mathrm{is\; divisible\; by\; x}+y$ or

$\left(x+y\right)(x-y)isdivisiblebyx+y,$ which is true.

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

$P(k)=x^{2k}-y^{2k}$ is divisible by x + y

i.e. $x^{2k}-y^{2k}=a(x+y)$ where $\in z$

$\Rightarrow x^{2k}=a(x+y)+y^{2k}$ (1)

Now, let us prove P(k +1) is true.

$P(k+1):x^{2(k+1)}-y^{2(k+1)}$

$=x^{2x+2}-y^{2x+2}$

$=x^2\cdot x^{2k}-y^2\cdot y^{2k}$

$=x^2\left[a(x+y)+y^{2k}\right]-y^2\cdot y^{2k}[using(1)$
$]$

$=a{x}^2(x+y)+x^2\cdot y^{2k}-y^2\cdot y^{2k}$

$=a{x}^2(x+y)+y^{2k}\left(x^2-y^2\right)$

$=a{x}^2(x+y)+y^{2k}(x+y)(x-y)$

$=(x+y)\left[a{x}^2+y^{2k}(x-y)\right]$

$=b(x+y)whereb=\left[a{x}^2+y^{2k}(x-y)\right]\in z$

$\Rightarrow x^{2k+2}-y^{2k+2}\mathrm{is\; divisible\; by\; x}+y$

$\therefore$ P(k+ 1) is true where P (k) is true.

Hence, by P.M.I. P(n) is true for all natural number i.e.

20. LetP(n): $10^{2n-1}+$ 1 is divisible by 11.

Putting n = 1

$P(1)=10^{\prime }+1=11$ is divisible by 11.

Which is true. Thus, P(1) is true.

Let us assume that P(k) is true for some natural no. k.

P(k)= $10^{2k-1}+$

$\mathrm{}\mathrm{1\; is\; divisible\; by\; 11.}$

$10^{2k-1}+1=11aa\in z$

$\to 10^{2k-1}=11a-1$ (1)

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

$P(k+1):10^{2(K+1)-1}+1=10^{2k+1}+1\mathrm{is\; divisible\; by\; 11.}$

$10^{2k+1}+1=10^{(2k-1)+2}+1$

$=10^{2k-1}\cdot 10^2+1$

$=100(11a-1)+1(using(1))$

=1100a $-$ 99= 11(100a $-$ 9)

11b where b= (100a $-$ 9) $\in z$

$\Rightarrow 1{0^2}^{k+1}+1$ is divisible by 11.

$\Rightarrow P(k+1)$

 is true when p(k) is true.

Hence by P.M.I. P(n) is true for every positive integer.