Maths

4. i. The x-axis and y-axis taken together determine a plane known as XY plane.

ii. The coordinates of points in the XY-plane are of the form (x, y, 0).

iii. Coordinate planes divide the space into 8 (eight) octants.

3. 

(1, 2, 3) lies in octant I.

(4, –2, 3) lies in octant IV.

(4, –2, –5) lies in octant VIII.

(4, 2, –5) lies in octant V.

(–4, 2, –5) lies in octant VI.

(–4, 2, 5) lies in octant II.

(–3, –1, 6) lies in octant III.

(–2, –4, –7) lies in octant VII.

2. When a point lies on XZ-plane, its y-coordinate is zero.

1. When a point lies on x-axis, its y-coordinate and z-coordinate are zero.

(c) Given, f(x)=(x-a)(x-b)

where a and b are constants.

So,

$f^{\prime }(x)=(x-a)\frac{d}{dx}(x-b)+(x-b)\frac{d}{dx}(x-a)$

=(x-a)+(x-b)

= 2x– a- b. $$

(ii) Given f(x)= ${\left(a{x}^2+b\right)}^2.$ where ab are constant

So, $f^{\prime }(x)=\frac{d}{dx}{\left(a{x}^2+b\right)}^2$

$=\frac{d}{dx}\left(a^2{x}^4+b^2+2a{x}^{2}b\right)$

$=\frac{d}{dx}{a}^2{x}^4+\frac{d}{dx}{b}^2+\frac{d}{dx}2a{x}^{2}b$

= $4a^2{x}^3+0+4axb$

= $4ax\left(a{x}^2+b\right).$

(iii) Given, f(x)= $\frac{x-9}{x-b}$ where a and bare constants

So, $f(x)=\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{{(x-b)}^2}$

$=\frac{(x-b)-(x-a)}{{(x-b)}^2}$

$=\frac{a-b}{{(x-b)}^2}$

38. Given, $f(x)=x^n+a{x}^{n-1}+a^2{x}^{n-2}+?+a^{n-1}x+a^n.$

We know that,

$\frac{d}{dx}\left(x^x\right)=n{x}^{x-1}$

So,

$f^{\prime }(x)=\frac{d}{dx}{x}^x+\frac{d}{dx}a{x}^{x-1}+\frac{d}{dx}{a}^2{x}^{x-2}+?+\frac{d}{dx}$ $a^{x-1}x+\frac{d}{dx}{a}^x=n{x}^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+?+a^{n-1}+0.$

$\begin{array}{l}(?\frac{dax}{dx}=a\frac{dx}{dx}\frac{\mathrm{and\; d}a}{dx}=\mathrm{0whereaisconstant}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$

37. 

Given $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+?+\frac{x^2}{2}+x+1$

$\underset{h\to 0}{lim}\frac{{(x+h)}^{100}-x^{100}}{100\cdot h}+\underset{h\to 0}{lim}\frac{{(x+h)}^{99}-x^{99}}{99\cdot h}+\dots +\underset{x\to 0}{lim}\frac{{(x+h)}^2-x^2}{2h}$ $+\underset{x\to 0}{lim}x+\underset{x\to 0}{lim}1$

$=\frac{100\cdot x^{99}}{100}+\frac{99x^{98}}{99}+?+\frac{2x}{2}+0+1$

$=x^{99}+x^{98}+?+x+1$

At x=0,

f(X)  =1.

and at x=1,

$f^{\prime }(1)=1^{99}+1^{98}+\dots +1^2+1+1$

=100 $*$ 1

=100 $*f^{\prime }(0)$

Hence, $f^{\prime }(1)=100f^{\prime }(0)$

41. In the 13 letter word ASSASSINATION there are 3-A, 4-S, 2-I, 2-N, 1-T and 1-O.

Since all the S are to be occurred together we treat them i.e. (SSSS) as single object. This single object together with 13 – 4 = 9 remaining object will account for 10 objects having 3-A, 2-I, 2-N, 1-T and 1-O and can be rearranged in $\frac{10!}{3!2!2!}$

= 10 * 9 * 8 * 7 * 6 * 5

= 151200

36. (i) $x^3-27$ (ii) (x -1)(x-2)

(iii) $\frac{1}{x^2}$ (iv) $\frac{x+1}{x-1}$

A.4.(i) Given, $f(x)=x^3-27.$

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

$=\underset{h\to 0}{lim}\frac{\left[{(x+h)}^3-27\right]-\left[x^3-27\right]}{h}$

$=\underset{h\to 0}{lim}\frac{x^3+h^3+3xh(x+h)-27-x^3+27}{h}$

$=\underset{h\to 0}{lim}\frac{h\left(h^2+3x(x+h)\right)}{h}$

$=\underset{h\to 0}{lim}{h}^2+3x(x+h)$

=0+3 x(x+ 0)

=3x²

(ii) Given, f(x) =(x-1)(x-2)

=x²- 3x+2

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

$\underset{h\to 0}{lim}\frac{\left[{(x+h)}^2-3(x+h)+2\right]-\left[x^2-3x+2\right]}{h}$

= $\underset{h\to 0}{lim}\frac{x^2+h^2+2xh-3x-3h+2-x^2+3x-2}{h}$

= $\underset{h\to 0}{lim}\frac{h(h+2x-3)}{h}$

$=\underset{h\to 0}{lim}h+2x-3$

= 2x – 3.

(iii) Given, f(x)= $\frac{1}{x^2}$

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

$=\underset{h\to 0}{lim}\frac{\frac{1}{{(x+h)}^2}-\frac{1}{x^2}}{h}$

$=\frac{\underset{h\to 0}{lim}\frac{x^2-{(x+h)}^2}{{(x+h)}^2-x^2}}{h}$

$=\underset{h\to 0}{lim}\frac{x^2-x^2-h^2-2xh}{h{x}^2(x+h){}^2}$ $$

$=\underset{h\to 0}{lim}\frac{h(-h-2x)}{h{x}^2(x+h){}^2}$

$=\underset{h\to 0}{lim}\frac{-h-2x}{x^2(x+h){}^2}$

$=\frac{-0-2x}{x^2(x+0){}^2}$

$=\frac{-2x}{x^4}$

$=\frac{-2}{x^3}$

(iv) Given, f(x)= $\frac{x+1}{x-1}$

$f(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{fx+h+1}{x+h-1}-\frac{x+1}{x-1}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x+h-1)(x-1)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{x^2-x+hx-h+x-1-x^2-hx+x-x-h+1}{(x-1)(x+h-1)}\right]$

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

$=\underset{h\to 0}{lim}\frac{-2}{(x-1)(x+h-1)}$

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

40. In a class of 25 students, 10 students are to be selected for excursion. As 3 students decided that either all of them will join or none of them will join we have the options:

For the 3 students to be selected along with 7 other students from the remaining 25 – 3 = 22 students. This can be done in ³C₃*²²C₇ ways.

For the 3 students to not be selected so that all 10 students will be from the remaining 25 – 3 = 22 students. This can be done in ³C₀*²²C₁₀ ways.

Therefore, the required number of ways

= ³C₃* ²²C₇ + ³C₀*²²C₁₀

= ²²C₇ + ²²C₁₀