Maths

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.

The highest order derivation present in the D.E. is y, so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

99. Let $y=x^{3}logx$

So,  $\frac{dy}{dx}=x^3\frac{d}{dx}logx+log2.\frac{d{x}^3}{dx}$

$=x^3.\frac{1}{x}+logx\cdot 3x^2$

$=x^2+logx\left(3x^2\right)$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(x^2+logx\cdot 3x^2\right)$

$=2x+6xlogx+3x^2*\frac{1}{x}$

$=2x+6xlogx+3x$

$=5x+6xlogx$

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.

The given equation of curve is  $y = x$ .

Differentiating with respect to x, we get:

$\frac{dy}{dx}=1 ..........(1)$

Again, differentiating with respect to x, we get:

$\frac{d^{2}y}{d{x}^2}=0..........(2)$

Now, on substituting the values of y,  $\frac{d^{2}y}{d{x}^2}$ and $\frac{dy}{dx}$   from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0-x^2.1+x.x\\ =-x^2+x^2\\ =0\end{array}$

Therefore, option (C) is correct.

98. Let $y=logx$

So,  $\frac{dy}{dx}=\frac{d}{dx}logx=\frac{1}{x}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^{-1}=-1x^{-1-1}=\frac{-1}{x^2}$

Given: $y=c_1{e}^x+c_2{e}^{-x}.........(1)$

Differentiating with respect to x, we get:

$\frac{dy}{dx}=c_1{e}^x-c_2{e}^{-x}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=c_1{e}^x+c_2{e}^{-x}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=y\\ \Rightarrow \frac{d^{2}y}{d{x}^2}-y=0\end{array}$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

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.

Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

$\begin{array}{l}{x}^2+{\left(y-b\right)}^2=3^2\\ \Rightarrow x^2+{\left(y-b\right)}^2=9..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2(y-b).y^{\text{'}}=0\\ \Rightarrow (y-b).y^{\text{'}}=-x\\ \Rightarrow y-b=\frac{-x}{y^{\text{'}}}\end{array}$

Substituting the value of $(y-b)$ in equation (1), we get:

$\begin{array}{l}{x}^2+{\left(\frac{-x}{y^{\text{'}}}\right)}^2=9\\ \Rightarrow x^2\left[1+\frac{1}{{\left(y^{\text{'}}\right)}^2}\right]=9\\ \Rightarrow x^2\left({\left(y^{\text{'}}\right)}^2+1\right)=9{\left(y^{\text{'}}\right)}^2\\ \Rightarrow (x^2-9){\left(y^{\text{'}}\right)}^2+x^2=0\end{array}$

This is the required differential equation.