Ncert Solutions Maths class 11th

19. Given, x-coordinate of R = 4

Let R divides line segment joining points P(2, –3, 4) and Q(8, 0,10) internally in the ratio k : 1. Then coordinate of R is

$\left(\frac{k(8)+1(2)}{k+1},\frac{k(0)+1(-3)}{k+1},\frac{k(10)+1(4)}{k+1}\right)$

= $\left(\frac{8k+2}{k+1},\frac{-3}{k+1},\frac{10k+4}{k+1}\right)$

Then,

$\frac{8k+2}{k + 1}$ = 4

=> $8k+2=4\left(k+1\right)$

=> $8k+2=4k+4$

=> $8k-4k=4-2$

=> $4k=2$

=> $k=\frac{2}{4}$

=> $k=\frac{1}{2}$

Hence, $y=$ $\frac{-3}{k + 1}$

= $\frac{-3}{\frac{1}{2} + 1}$

= $-3÷\frac{1+2}{2}$

= $-\mathrm{3*} \frac{2}{3}$

= $-2$

And,

z = $\frac{10k + 4}{k + 1}$

= $\frac{\left(10\frac{1}{2}\right) + 4}{\frac{1}{2} + 1}$

= $\left(5+4\right)÷\frac{1+2}{2}$

= $9\frac{2}{3}$

= 6

Therefore, coordinates of R is (4, –2, 6).

18. Let Q be the point on y-axis which are at a distance  from point P. As Q is on y-axis it has the coordinates of form (0, y, 0).

=> ${\left(0-3\right)}^2+{\left(y+2\right)}^2+{\left(0-5\right)}^2$  = 25 x 2

=> 9 +  $y^2+2^2+2.y.2+25=50$

=> $y^2+4y+34+4-50=0$

=> $y^2+4y-12=0$

=> $y^2+6y-2y-12=0$

=> $y\left(y+6\right)-2\left(y+6\right)=0$

=> $\left(y+6\right)\left(y-2\right)=0$

=> $y=-6, y=2$

So the coordinates Q are (0, 2, 0) and (0, –6, 0).

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

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