Maths

17. We know that, the centroid of a triangle with vertices (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is

$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$

So, $\left(\frac{2a+(-4)+8}{3},\frac{2+3b+14}{3},\frac{6+(-10)+2c}{3}\right)$  = (0, 0, 0)

Equating the coordinates we get,

$\frac{2a+\left(-4\right)+8}{3}$ = 0

=> $2a-4+8=0$

=> $2a+4=0$

=> $2a=-4$

=> $a=\frac{-4}{2}$

=> $a=-2$

And,

$\frac{2+3b+14}{3}=0$

=> $2+3b+14=0$

=> $3b+16=0$

=> $3b=-16$

=> $b=\frac{-16}{3}$

And,

$\frac{6+\left(-10\right)+2c}{3}=0$

=> $6-10+2c=0$

=> $-4+2c=0$

=> $2c=4$

=> $c=\frac{4}{2}$

=> $c=2$

16. In a triangle ABC, the medians are the line segment that joins a vertex to the mid-point of the side that is opposite to that vertex. So, AE, BF and CG are the three medians.

15. Let D(x, y, z) be the fourth vertex of the parallelogram ABCD.

In a parallelogram, the diagonal AC and BD bisects each other at point say O.

=> $\left(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\right)=\left(\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2}\right)$

=> (1, 0, 2) = $\left(\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2}\right)$

Equating the coordinates we get,

$\frac{1+x}{2}$ = 1

=> $1+x=2$

=> $x=1$

And

$\frac{2 + y}{2}$ = 0

=> $y=-2$

And

$\frac{- 4 + z }{2}$ = 2

=> $-4+z=4$

=> $z=4+4$

=> $z=8$

So, coordinates of fourth vertex is (1, –2, 8)

47. Given, f (x) = (ax + b) (cx + d)²

So, f?(x) = (ax +b) $\frac{d}{dx}{(cx+d)}^2+{(cx+d)}^2\frac{d}{dx}(ax+b)$

$=(ax+b)\frac{d}{dx}\left[(c+d)(cx+d)\right]+{(cx+d)}^{2}a$

$=(ax+b)\left[(c+d)\frac{d}{dx}(x+d)+(c+d)\frac{d}{dx}(c+d)\right]+a{(cx+d)}^2$

$=(ax+b)[c(cx+d)+c(cx+d)]+a{(cx+d)}^2$

$=(ax+b)·2·c(cx+d)+a{(cx+d)}^2$

. (i) f(x)=sin x cos x

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{sin(x+h)cos(x+h)?sinxcosx}{h}$

$=\underset{h?0}{lim}\frac{1}{2h}*[2sin(x+h)cos(x+h)?2sinxcosx]$

$=\underset{h?0}{lim}\frac{1}{2h}[sin2(x+h)?sin2x]$

$=\underset{h?0}{lim}\frac{1}{2h}\left[2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)?2x}{2}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[cos\left(2x+h\right)sinh\right]$

$=\underset{h?0}{lim}cos(2x+h)*\underset{h?0}{lim}\frac{sinh}{h}$

$=cos(2x+0)$

$=cos2x$

$\mathrm{(ii)\; f}(x)=secx$

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[sec(x+h)?secx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{?2sin\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]$

$=\underset{h?0}{lim}\frac{(?1sin\left(\frac{2x+h}{2}\right)}{cos(x+h)cosx}*\underset{h?0}{lim}(?1)\frac{sinh/2}{h/2}$

$=\frac{sinx}{cosx?cosx}*1$

$=tanx?secx.$

(iii) Given f(x)=5 sec x+4 cosx.

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$\begin{array}{l}\\ =\underset{h?0}{lim}\frac{5sec(x+h)+4cos(x+h)?[5secx+4cosx]}{h}.\end{array}$

$=\underset{h?0}{lim}\frac{5}{h}[sec(x+h)?secx]+\underset{h?0}{lim}\frac{4}{h}[cos(x+h)?cosx]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{1}{cos(x+h)}?\frac{1}{cosx}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h?x}{2}\right)\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[\frac{cosx?cos(x+h)}{cos(x+h)(cosx)}\right]+\underset{h?0}{lim}\frac{4}{h}\left[?2sin\left(\frac{2x+h}{2}\right)sin\frac{h}{2}\right]$

$=\underset{h?0}{lim}\frac{5}{h}\left[?\frac{2sin\left(\frac{2x+h}{2}\right)sin(?h/2)}{cos(x+h)cosx}\right]?4\underset{h?0}{lim}sin\left(\frac{2x+h}{2}\right)\underset{h?0}{lim}\frac{sinh/2}{h/2}$

$=\frac{sin\left(\frac{2x+0}{2}\right)}{cos(x+0)cosx}*1?4sin\left(\frac{2x}{2}\right)$

$=5\frac{sinx}{cosx}?\frac{1}{cosx}?4sinx$

$=5tanx?secx?4sinx$

$\mathrm{(iv)\; Given\; f}(x)=cosecx$

$f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$

$=\underset{h?0}{lim}\frac{1}{h}[cosec(x+h)?cosecx]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{1}{sin(x+h)}?\frac{1}{sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{sinx?sin(x+h)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)]}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{1}{h}\left[\frac{2cos\left(\frac{2x+h}{2}\right)sin(?h/2)]}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{cos\left(\frac{2x+h}{2}\right)}{sin(x+h)sinx}*\frac{(?1)sin(2)}{h/2})$

$=\frac{cos\left(\frac{2x+0}{2}\right)}{sin(x+0)sinx}*(?1)$

$=?\frac{cosx}{sinx}*\frac{1}{sinx}$

$=?cotx?cosecx$

(v) Given,f(x)=3 cot x+5cosecx.

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$ $=\frac{2}{cos(x+0)cosx}*1+7sin\left(\frac{2x+0}{2}\right)*(?1)$

$=\underset{h?0}{lim}3cot(x+h)+5cosec(x+h)?[3cotx+5cosx)$

${}_{h?0}\frac{3}{h}[cot(x+h)?cotx]+\underset{h?0}{lim}$

$=\text{?}\frac{5}{h}[cosec(x+h)?cosecx]$

$=\underset{h?0}{lim}\frac{3}{h}\left[\frac{cos(x+h)}{sin(x+h)}?\frac{cosx}{sinx}\right]+\underset{h?0}{lim}\frac{5}{h}\left[\frac{1}{sin(x+h)}?\frac{1}{sinx}\right]$

$=\underset{h?0}{lim}\frac{3}{h}\left[\frac{cos(x+h)sinx?cosxsin(x+h)}{sin(x+h)sinx}\right]+\underset{h?0}{lim}\frac{5}{h}\left[\frac{sinx?sin(x+h)}{sin(x+h)sinx}\right]$

$=\underset{h?0}{lim}\frac{3}{h}[\frac{sin(x?(x+h))}{sin(x+h)sinx}+\underset{h?0}{lim}\frac{5}{h}[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x?(x+h)}{2}\right)}{sin(x+h)sinx}$

$=\underset{h?0}{lim}\frac{3}{sin(x+h)sinx}*\underset{h?0}{lim}(?1)\frac{sinh}{h}+\underset{h?0}{lim}\frac{5}{h}[\frac{2cos\left(\frac{2x+h}{2}\right)sin(?h/2)}{sin(x+h)sinx}$

$=\frac{3}{sinx?sinx}*(?1)+5\underset{}{lim}$

(i) f(x)=sin x cos x

So, $f^{?}(x)=\underset{h?0}{lim}\frac{f(x+h)?f(x)}{h}$