Maths

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

121. Kindly go through the solution

6. Let (r + 1)th be the general term of ( $x^2- yx){}^{12}$

So, T_r-1 = ¹²C_r (x²)^12–r (–yx)^r

= (–1)^r12C_rx^24–2ry^rx^r

= (–1)^r12C_r $x^{24-2r+r}{y}^r$

= (-1)^r12C­_r $x^{24-r}{y}^r$

In (D), each of the terms has a degree 2.

Hence, (D) is homogenous

$\therefore$ Option (D) is correct.