Maths

127. Let $y=x^x+x^a+a^x+a^a.$

So, $\frac{dy}{dx}=\frac{d{x}^x}{dx}+\frac{d{x}^a}{dx}+\frac{d{a}^x}{dx}+\frac{d{a}^a}{dx}.$

$\to \frac{dy}{dx}=\frac{dy}{dx}+a{x}^{a-1}+a^{x}loga+0.$ _________(1)

Where $u=x^x$

$logu=xlogx$ (Taking log)

$\to \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}logx+logx\frac{dx}{dx}$ (Differentiation w r t 'x')

$\to \frac{1}{y}\frac{dy}{dx}=\frac{x}{x}+logx$

$\to \frac{dy}{dx}=u[1+logx]$

$=x^x[1+logx]$

Hence eqn (1) becomes,

$\frac{dy}{dx}=x^x[1+logx]+a{x}^{a-1}+a^{x}loga.$

10. General term of the expansion (1 + x)²ⁿ is

T_r+1 = ²ⁿC_r (1)^2n-r(x)^r

So, co-efficient of xⁿ (i.e. r = n) is ²ⁿC_n

Similarly general term of the expansion (1 + x)^2n–1 is

T_r+1 = ^2n-1C_r (1)^2n–1–rx^r

And co-efficient of xⁿ i.e. when r = n is ^2n-1C_n

Therefore,

$\frac{co-efficient of x^n in {\left(1+x\right)}^{2n}}{co-efficient of x^n in {\left(1+x\right)}^{2n-1}}$

=

= $\frac{2n!}{n!\left(2n-n\right)!}$ ÷ $\frac{\left(2n-1\right)!}{n!\left(2n-1-n\right)!}$

= $\frac{2n!}{n!n!}$ * $\frac{n!\left(n-1\right)!}{\left(2n-1\right)!}$

= $\frac{2n}{n}$

= 2

Thus, co-efficient of $x^n$ in ${\left(1+x\right)}^{2n}$ = 2x co-efficient of $x^n$ in ${\left(1+x\right)}^{2n-1}$

126. Let $y={(sinx-cosx)}^{sinx-cosx}$

Taking log,

$logy=(sinx-cosx)log(sinx-cosx)$

Differentiating w r t 'x' we get,

$\frac{1}{y}\frac{dy}{dx}=(sinx-cosx)\frac{dlog}{dx}(sinx-cosx)+log(sinx-cosx)\frac{d}{dx}(sinx-cosx).$

$=(sinx-cosx)*\frac{1}{(sinx-cosx)}*(cosx+sinx)+log(sinx-cosx)\left(cosx+sinx\right)$

$=(cosx+sinx)[1+log(sinx-cosx)]$

$\frac{dy}{dx}=y(cosx+sinx)[1+log(sinx-cosx)]$

$\frac{dy}{dx}={(sinx-cosx)}^{sinx-cosx}*(cosx+sinx)[1+log(sinx-cosx)]$

125. Let $y=cos(acosx+bsinx)$

So,  $\frac{dy}{dx}=-sin(acosx+bsinx)\frac{d}{dx}(acosx+bsinx)$

$=-sin(acosx+bsinx)[-asinx+bcosx].$

$=sin(acosx+bsinx)(asinx-bcosx)$

124. Let $y={(logx)}^{logx}$

Taking log,

$logy=logxlog(logx)$

Differentiating w r t. x, we get,

$\frac{1}{y}\frac{dy}{dx}=logx\frac{d}{dx}log(logx)+log(logx)\frac{d}{dx}(logx)$

$=logx*\frac{1}{logx}\frac{d}{dx}logx+\frac{log(logx)}{x}$

$=\frac{1}{x}+\frac{log(logx)}{x}$

$\frac{dy}{dx}=\frac{y}{x}[1+log(logx)].$

$={(logx)}^{logx}\left[\frac{1+log(logx)}{x}\right]$

8. The general term of the expansion (1 + a)^m+n is

T_r+1 = ^m+nC_ra^r   [since, 1^m+n-r = 1]

At r = m we have,

T_m+1 = ^m+nC_ma^m

= $\frac{\left(m+n\right)!}{m!\left(m+n-m\right)!}$ (a)^m

= $\frac{\left(m+n\right)!}{m! n!}$ a^m - (1)

Similarly at r = n we have,

T_n+1 = ^m+nC_naⁿ

= $\frac{\left(m+n\right)!}{n!\left(m+n-n\right)!}$ (a)ⁿ

= $\frac{\left(m+n\right)!}{n! m!}$ aⁿ - (2)

Hence from (1) & (2),

Co-efficient of am = Co-efficient of aⁿ = $\frac{\left(m+n\right)!}{m!n!}$

123. Kindly go through the solution

122. Kindly go through the solution