Class 11th

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

35. Given, f (x)= 99x, f   (100)=?

So, f (100)= ${}_{h\to 0}\frac{f(100+h)-f(100)}{h}$

$=\underset{h\to 0}{lim}\frac{99(100+h)-99(100)}{h}$

$=\underset{h\to 0}{lim}\frac{99*100+99*h-99*100}{h}$

$=\underset{h\to 0}{lim}\frac{99h}{h}$

$=\underset{h\to 0}{lim}99$

=99

39. As out of the total 9 seats 4 women are to be at even places we can have the following arrangement.

Seat places

Also from this arrangement the women and men can rearrange among themselves.

Therefore, the required number of ways = 4! * 5!

= (4 * 3 * 2 * 1) * (5 * 4 * 3 * 2 * 1)

= 24 * 120

= 2880