Maths

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.

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

P(n): 1+3+3²+ …+3^n-1= $\frac{\left(3^n-1\right)}{2}$

For n=1, P(1)=1= $\frac{\left(3^1-1\right)}{2}=\frac{3-1}{2}=\frac{2}{2}=1$

which is true.

Assume that P(k) is true for some positive integer k i.e.,

1+3+3²+ … +3^k–1= $\frac{\left(3^k-1\right)}{2}$

--------(1)

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

Here, 1+3+3²+ … +3^k–1+3^(k+1)–1

$\frac{3^k-1}{2}+3^{k+1-1}$

[By using eq (1)]

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

= $\frac{3^k+2.3^k-1}{2}$

= $\frac{3^k(1+2)-1}{2}$

= $\frac{3^k·3-1}{2}=\frac{3^{k+1}-1}{2}$

? 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.

42. We have,

${\left(\frac{1+i}{1-i}\right)}^m$ = 1

=> ${\left(\frac{1+i}{1-i} *\frac{1+}{1+}\right)}^m$ = 1 [multiply denominator and numerator of LHS by (1 + i)]

=> ${\left(\frac{1+i+i+i^2}{1^2- i^2}\right)}^m$ = 1 [since, (a – b) (a + b) = a² – b²]

=> ${\left(\frac{1+2i-1}{1+1}\right)}^m$ = 1 [since, i² = –1]

=> ${\left(\frac{2i}{2}\right)}^m$ = 1

=>i^m = 1

=>i^m = i^4k              [since, i^4k = 1]

So, m = 4k where k = integer

Therefore, least positive integral value of m is,

m = 4 * 1

m = 4

41. Given,

(a + ib) (c + id) (e + if) (g + ih) = A + iB

We know that,

Hence proved.