Maths

9. Let the given statement be P(n) l.e.,

P(n)= $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+?+\frac{1}{2^n}=1-\frac{1}{2^n}$

If n=1, we get

P(1)= $\frac{1}{2}=1-\frac{1}{2^1}=\frac{1}{2}$

which is true.

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

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+?+\frac{1}{2^k}=1-\frac{1}{2^k}$ (1)

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

Here, $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+?+\frac{1}{2^k}+\frac{1}{2^{k+1}}$

By using eqn. (1)

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

we can write as,

= $1-\frac{1}{2^k}+\frac{1}{2^k.2}$

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

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

It can be written as,= $1-\frac{1}{2^{k+1}}$

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

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

Given, y= √1 + x²

87. Given, x = 2at2 and y = at4. Differentiation w r t we get,

$\frac{dx}{dt}=4at.$ and $\frac{dy}{dt}=4a{t}^3.$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{4a{t}^3}{4at}=t^2.$

Given,  $f{x}^n$ is $y=cosx+c$

So,  $y^{|}=-sinx$

Putting the value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}+sinx=-sinx+sinx=0=R.H.S$

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

Given,  $f{x}^n$ is $y=x^2+2x+c$

So,  $y^{|}=2x+2$

Substituting value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}-2x-2=2x+2-2x-2=0=R.H.S$

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

Given $f{x}^n$ is $y=e^x+1$

Differentiating with $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of $y^{\left|\right|}$ and $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

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

86. To prove $\frac{d}{dx}(uv·w)=\frac{du}{dx}v·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}.$

By repeating application of produced rule

$=\frac{d}{dx}(uv:u)$

$=u\frac{d}{dx}(u·w)+v·w\frac{dy}{dx}$

$u\left\{v\frac{dw}{dx}+w\frac{dv}{dx}\right\}+\frac{dy}{dx}v:w.$

$=u·v·\frac{dw}{dx}+u·\frac{dv}{dx}w+\frac{dy}{dx}·v·w$

$=\frac{dy}{dx}u·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}$ = R*H*S*

By togarith differentiating,

Let y = u v w

Taking log, log y = log u + log v + log w

Differentiating w r t 'x'

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}.$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{1}{4}\frac{dy}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]$

$\Rightarrow \frac{d}{dx}(uvw)=uvw·\left[\frac{1}{u}\frac{dy}{dx}+\frac{1}{v}\frac{du}{dx}+\frac{1}{w}\frac{dur}{dx}\right]$

$=\frac{dy}{dx}v·w+u\frac{dv}{dx}·w+uv·\frac{dw}{dx}$

The highest order derivative present in the given D.E. is $\frac{d^{2}y}{d{x}^2}$ and its order is 2.

$\therefore$ Option (A) is correct.