Maths

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

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

If n=1, we get

P(1): 1.2.3 = 6 = $\frac{1(1+1)(1+2)(1+3)}{4}$ = $\frac{1.2.3.4}{4}=6$

which is true.

considerP(k) is true for some positive integer k

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

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

Here,1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

By eqn (1), we get,

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

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

= $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$

By further Simplification,

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

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

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

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

83. Given, xy = ex-y.

Taking log,

log (x + y) = log (ex-y).

=logx + log y = (x-y) log e.

= logx +log y = x -y {Q log e = 1}

Differentiating w r t 'x' we get,

$\Rightarrow \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}+\frac{dy}{dx}=1-\frac{1}{2}$

$\Rightarrow \frac{dy}{dx}\left(\frac{1}{y}+1\right)=\left(\frac{x-1}{x}\right)$

$\Rightarrow \frac{dy}{dx}\left(\frac{1+y}{y}\right)=\left(\frac{x-1}{x}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{y(x-1)}{x(1+y)}$

The highest order derivation present in the D.E. is $\frac{d^{2}s}{d{t}^2}$ so its order is 2 .

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

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

P(n): 1+ $\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+?+\frac{1}{(1+2+3+?n)}=\frac{2n}{(n+1)}$

For n=1,

we get,P(1)=1= $\frac{2.1}{1+1}=\frac{2}{2}=1$

which is true.

Let us assume that P(k) is true for some positive integer k.

i.e., $1+\frac{1}{1+2}+\dots +\frac{1}{(1+2+3+?k)}=\frac{2k}{(k+1)}$ ------------------ (1)

Which is true.

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

$1+\frac{1}{1+2}+\frac{1}{1+2+3}$ + … $+\frac{1}{(1+2+3+?k)}+\frac{1}{1+2+3+?+k+(k+1)}$

By using eqn (1)

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

?We know that, 1+2+3+ … +n= $\frac{n(n+1)}{2}$

So, we get

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

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

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

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

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

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

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

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

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

82. Given, (cos x)y = (cos y)x

Taking log, y log (cos x) = x log (cos y)

Differentiating w r t 'x' we get,

= $y\frac{d}{dx}$ log (cos x) + log (cos x) $\frac{dy}{dx}=x\frac{d}{dx}$ log (cos y) + dog (cos y) $\frac{dx}{dx}$

= y´ $\frac{1}{cosx}\frac{d}{dx}$ cos x + log (cos x) $\frac{dy}{dx}$ = x´ $\frac{1}{cosy}\frac{d}{dx}cosy+log(cosy).$

$\Rightarrow -y\frac{sinx}{cosx}+log(cosx)\frac{dy}{dx}=-x\frac{siny}{cosy}\frac{dy}{dx}+log(cosy)$

= log (cos x) $\frac{dy}{dx}$ + x tan $\frac{dy}{dx}=yctanx+log(cosy)$

$\Rightarrow \frac{dy}{dx}[log(cosx)+xtany]$ = y tan x + log (cos y )

$\Rightarrow \frac{dy}{dx}=\frac{ytanx+log(cosy}{log(cosx)+xtany}.$

The highest order derivation present in the differential equation (D.E.) is $\frac{d^{4}y}{d{x}^4}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative, its degree is not defined.

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

P(n)=1³+2³+3³+ … +n³= ${\left(\frac{n(n+1)}{2}\right)}^2$

For, n=1, P(n)=1³=1= ${\left(\frac{1(1+1)}{2}\right)}^2={\left(\frac{1.2}{2}\right)}^2=1^2=1$

which is true.

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

1³+2³+3³+ … +k³= ${\left(\frac{k(k+1)}{2}\right)}^2$ ---------- (1)

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

Here,  1³+2³+3³+ … +k³+(k+1)³

By using eq (1)

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

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

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

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

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

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

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