Maths

In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

92. Kindly go through the solution

The number of arbitrary constant is general solution of D.E of 4^th order is four.

$\therefore$ Option (D) is correct.

13. We can write given statement as

P(n): $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2n+1)}{n^2}\right)={(n+1)}^2$

If n=1, we get

P(1): $\left(1+\frac{3}{1}\right)$ =4=(1+ 1)²=2²=4

which is true.

Consider P(k) be true for some positive integer k.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)={(k+1)}^2$ (1)

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

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)+\left(1+\frac{(2(k+1)+1)}{{(k+1)}^2}\right)$

By using (1)

=(k+1)²$\left(1+\frac{2(k+1)+1}{{(k+1)}^2}\right)$

=(k+1)² $\left[\frac{{(k+1)}^2+2(k+1)+1}{{(k+1)}^2}\right]$

=(k+1)²+2(k+1)+1

={(k+1)+1}²

P(k+1) is true whenever P(k) is true.

Therefore, by principle of mathematical induction, the P(n) is true for all natural number n.

Kindly go through the solution

91. Given, $x=a\left(cost+logtan\frac{t}{2}\right)y=asint$

Differentiating w r t we get,

$\frac{dx}{dt}=a\frac{d}{dt}\left[cost+log\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{dt}\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.se{c}^2\frac{t}{2}\frac{d}{dt}\left(\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{co{s}^2\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{2}\right]$

$=a\left[-sint+\frac{1}{2sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sin2*\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sint}\right]=a\left[\frac{1-si{n}^{2}t}{sint}\right]$

$=a\frac{co{s}^{2}t}{sint}\left\{?1=co{s}^{2}x+si{n}^{2}x\right\}$

$\frac{\mathrm{bd}y}{dt}=\frac{d}{dt}\left(asint\right)=acost$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{acost}{\raisebox{1ex}{$aco{s}^{2}t$}\!\left/ \!\raisebox{-1ex}{$sint$}\right.}=\frac{sint}{cost}=tant$

12. Let the given statement be P(n) i.e.,

P(n)=a+ar+ar²+ … +ar^n-1== $\frac{a\left(r^n-1\right)}{r-1}$

If n = 1, we get

P(1)=a= $\frac{a\left(r^1-1\right)}{r-1}$ =a

which is true.

Consider P(k) be true for some positive integer k

a+ar+ar²+ … +ar^k-1= $\frac{a\left(r^k-1\right)}{r-1}$ (1)

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

Here, {a+ar+ar²+ … +ar^k-1}+ar^{(k+1) –1}

By using (1),

= $\frac{a\left(r^k-1\right)}{r-1}+a{r}^k$

= $\frac{a\left(r^k-1\right)+a{r}^k(r-1)}{r-1}$

= $\frac{a{r}^k-a+a{r}^{k+1}-a{r}^k}{r-1}$

= $\frac{a{r}^{k+1}-a}{r-1}$

= $\frac{a\left(r^{k+1}-1\right)}{r-1}$

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e.,

90. Given, x = 4t and y = $\frac{4}{t}$ Differentiating w r t. 't' we get,

$\frac{dx}{dt}=4\frac{dy}{dt}=\frac{4d\left(t^{-1}\right)}{dt}$

$=-4t^{-2}$

$\frac{-4}{t^2}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$t^2$}\right.}{4}=-\frac{1}{t^2}.$

Given,  $x+y=ta{n}^{-1}y$

Differentiate with 'x' we get

$\begin{array}{l}1+\frac{dy}{dx}=\frac{1}{1+y^2}\frac{dy}{dx}\\ =1+y^{|}=\frac{1}{1+y^2}{y}^{|}\\ =\left(1+y^{|}\right)\left(1+y^2\right)=y^{|}\\ =1+y^2{y}^{|}+y^{|}+y^2=y^{|}\\ =y^2{y}^{|}+y^2+1=0\end{array}$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E

11. we can write the given statement as

$P(n)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$ = $\frac{n(n+3)}{4(n+1)(n+2)}$

If n=1,

P(1)= $\frac{1}{1.2.3}$ = $\frac{1}{6}$ = $\frac{1(1+3)}{4(1+1)(1+2)}$ = $\frac{4}{4*2*3}$ = $\frac{1}{6}$

which is true.

Consider P(k) be true for some positive integer k

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}$ = $\frac{k(k+3)}{4(k+1)\left(k+2\right)}$

Let us prove that P(k+1) is true,

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$ .

By equation (1), we get

= $\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

= $\frac{1}{(k+1)(k+2)}\left[\frac{k(k+3)}{4}+\frac{1}{(k+3)}\right]$

= $\frac{1}{(k+1)(k+2)}\left[\frac{k{(k+3)}^2+4}{4(k+3)}\right]$

= $\frac{1}{(k+1)(k+2)}\left[\frac{k\left(k^2+6k+9\right)+4}{4(k+3)}\right]$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+6k^2+9k+4}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k\left(k^2+2k+1\right)+4\left(k^2+2k+1\right)}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k{(k+1)}^2+4{(k+1)}^2}{4(k+3)}\right\}$

= $\frac{{(k+1)}^2(k+4)}{4(k+1)(k+2)(k+3)}$

= $\frac{{(k+1)}^2\{(k+1)+3\}}{4(k+1)(k+2)(k+3)}$ = $\frac{(k+1)\{(k+1)+3\}}{4\{(k+1)+1\}\{(k+1)+2\}}$

P(k+1) is true whenever P(k) is true.

Hence, By the principle of mathematical induction, the P(n) is true for all natural number n.